In 2 Physics

S C I S PHY @ HSC

Stephen Bosi John O’Byrne Peter Fletcher Joe Khachan Jeff Stanger Sydney, Melbourne, Brisbane, Perth, Adelaide and associated companies around the world

Sandra Woodward

Contents Series features How to use this book Stage 6 Physics syllabus grid

vi viii x

Module 1 Space Module 1 Introduction Chapter 1 Cannonballs, apples, planets and gravity 1.1 Projectile motion 1.2 Gravity 1.3 Gravitational potential energy Practical experiences Chapter summary Review questions

2 4 4 10 16 20 22 22

Chapter 2 Explaining and exploring the solar system 26 2.1 Launching spacecraft 26 2.2 Orbits and gravity 35 2.3 Beyond Kepler’s orbits 41 2.4 Momentum bandits: the slingshot effect 44 2.5 I’m back! Re-entry 46 Practical experiences 52 Chapter summary 53 Review questions 54 Chapter 3 Seeing in a weird light: relativity 3.1 Frames of reference and classical relativity 3.2 Light in the Victorian era 3.3 Special relativity, light and time 3.4 Length, mass and energy Practical experiences Chapter summary Review questions

58 58 61 64 69 75 76 76

Module 1 Review

80

Module 2 Motors and Generators Module 2 Introduction

82

Chapter 4 Electrodynamics: moving charges and magnetic fields 84 4.1 Review of essential concepts 84 4.2 Forces on charged particles in magnetic fields 89 4.3 The motor effect 90 4.4 Forces between parallel wires 93 Practical experiences 97 Chapter summary 98 Review questions 98

Chapter 5 Induction: the influence of changing magnetism 5.1 Michael Faraday discovers electromagnetic induction 5.2 Lenz’s law 5.3 Eddy currents Practical experiences Chapter summary Review questions

100 100 104 106 109 110 110

Chapter 6 Motors: magnetic fields make the world go around 6.1 Direct current electric motors 6.2 Back emf and DC electric motors 6.3 Alternating current electric motors Practical experiences Chapter summary Review questions

114 114 120 121 126 127 127

Chapter 7 Generators and electricity supply: power for the people 7.1 AC and DC generators 7.2 Transformers 7.3 Electricity generation and transmission Practical experiences Chapter summary Review questions

130 130 136 141 148 149 149

Module 2 Review

152

Module 3 From Ideas to Implementation Module 3 Introduction

154

Chapter 8 From cathode rays to television 8.1 Cathode ray tubes 8.2 Charges in electric fields 8.3 Charges moving in a magnetic field 8.4 Thomson’s experiment 8.5 Applications of cathode rays Practical experiences Chapter summary Review questions

156 156 160 164 165 167 170 171 171

Chapter 9 Electromagnetic radiation: particles or waves? 9.1 Hertz’s experiments on radio waves 9.2 Black body radiation and Planck’s hypothesis 9.3 The photoelectric effect 9.4 Applications of the photoelectric effect Practical experiences Chapter summary Review questions

174 174 178 182 184 185 186 187 iii

Contents

Chapter 10 Semiconductors and the electronic revolution 10.1 Conduction and energy bands 10.2 Semiconductors 10.3 Semiconductor devices 10.4 The control of electrical current Practical experiences Chapter summary Review questions

188 189 190 193 197 201 202 202

Chapter 11 Superconductivity 11.1 The crystal structure of matter 11.2 Wave interference 11.3 X-ray diffraction 11.4 Crystal structure 11.5 Electrical conductivity and the crystal structure of metals 11.6 The discovery of superconductors 11.7 The Meissner effect 11.8 Type-I and type-II superconductors 11.9 Why is a levitated magnet stable? 11.10 BCS theory and Cooper pairs 11.11 Applications of superconductors Practical experiences Chapter summary Review questions

204 204 205 207 208

Module 3 Review

224

209 211 212 212 213 215 217 220 221 221

Module 4 Quanta to Quarks Module 4 Introduction

226

Chapter 12 From Rutherford to Bohr 228 12.1 Atomic timeline 228 12.2 Rutherford’s model of the atom 229 12.3 Planck’s quantised energy 231 12.4 Spectral analysis 232 12.5 Bohr’s model of the atom 235 12.6 Bohr’s explanation of the Balmer series 236 12.7 Limitations of the Rutherford–Bohr model 239 Practical experiences 241 Chapter summary 242 Review questions 243 Chapter 13 Birth of quantum mechanics 13.1 The birth 13.2 Louis de Broglie’s proposal 13.3 Diffraction 13.4 Confirming de Broglie’s hypothesis 13.5 Electron orbits revisited 13.6 Further developments of atomic theory 1924–1930 Practical experiences Chapter summary Review questions iv

247 247 248 250 251 252 253 256 256 257

Chapter 14 20th century alchemists 14.1 Discovery of the neutron 14.2 The need for the strong force 14.3 Atoms and isotopes 14.4 Transmutation 14.5 The neutrino 14.6 Was Einstein right? 14.7 Binding energy 14.8 Nuclear fission 14.9 Chain reactions 14.10 Neutron scattering Practical experiences Chapter summary Review questions

260 260 261 262 263 265 266 268 269 270 272 273 274 275

Chapter 15 The particle zoo 15.1 The Manhattan Project 15.2 Nuclear fission reactors 15.3 Radioisotopes 15.4 Particle accelerators 15.5 The Standard Model Practical experiences Chapter summary Review questions

279 279 280 282 286 292 295 296 297

Module 4 Review

300

Module 5 Medical Physics Module 5 Introduction

302

Chapter 16 Imaging with ultrasound 16.1 What is ultrasound? 16.2 Principles of ultrasound imaging 16.3 Piezoelectric transducers 16.4 Acoustic impedance 16.5 Types of scans 16.6 Ultrasound at work Practical experiences Chapter summary Review questions

304 304 305 308 310 312 315 317 318 318

Chapter 17 Imaging with X-rays 17.1 Overview and history: types of X-ray images 17.2 The X-ray tube 17.3 Types of X-rays 17.4 Production of X-ray images 17.5 X-ray detector technology 17.6 Production of CAT X-ray images 17.7 Benefits of CAT scans over conventional radiographs and ultrasound Practical experiences Chapter summary Review questions

320 320 321 322 324 326 326 329 330 331 331

Contents

Chapter 18 Imaging with light 18.1 Endoscopy 18.2 Medical uses of endoscopes Practical experiences Chapter summary Review questions

333 333 336 338 339 339

Chapter 19 Imaging with gamma rays 19.1 Isotopes and radioactive decay 19.2 Half-life 19.3 Radiopharmaceuticals: targeting tissues and organs 19.4 The gamma camera 19.5 Positron emission tomography Practical experiences Chapter summary Review questions

340 340 343

Chapter 20 Imaging with radio waves 20.1 Spin and magnetism 20.2 Hydrogen in a magnetic field 20.3 Tuning in to hydrogen 20.4 It depends on how and where you look 20.5 The MRI scanner 20.6 Applications of MRI Practical experiences Chapter summary Review questions

354 354 355 357 359 360 362 363 364 364

Module 5 Review

366

344 346 347 350 351 351

Module 6 Astrophysics Module 6 Introduction

368

Chapter 21 Eyes on the sky 21.1 The first telescopes 21.2 Looking up 21.3 The telescopic view 21.4 Sharpening the image 21.5 Interferometry 21.6 Future telescopes Practical experiences Chapter summary Review questions

370 370 373 374 377 380 382 383 384 384

Chapter 22 Measuring the stars 22.1 How far? 22.2 Light is the key 22.3 The stellar alphabet 22.4 Measuring magnitudes 22.5 Colour matters Practical experiences Chapter summary Review questions

388 388 389 394 397 400 403 405 405

Chapter 23 Stellar companions and variables 23.1 Binary stars 23.2 Doubly different 23.3 Variable stars 23.4 Cepheid variables Practical experiences Chapter summary Review questions

407 407 411 413 415 418 418 419

Chapter 24 Birth, life and death 24.1 The ISM 24.2 Star birth 24.3 Stars in the prime of life 24.4 Where to for the Sun? 24.5 The fate of massive stars 24.6 How do we know? Practical experiences Chapter summary Review questions

422 422 423 425 428 430 433 435 436 436

Module 6 Review

438

Module 7 Skills Module 7 Introduction

440

Chapter 25 Skills stage 2 25.1 Metric prefixes 25.2 Numerical calculations 25.3 Sourcing experimental errors 25.4 Presenting research for an exam 25.5 Australian scientist 25.6 Linearising a formula

442 442 443 445 446 447 447

Chapter 26 Revisiting the BOS key terms 26.1 Steps to answering questions

448 449

Numerical answers Glossary Index Acknowledgements Formulae and data sheets Periodic table

452 454 465 471 473 474

v

S C I S Y PH @ HSC AGE FOR NSW STUDENTS CK PA S IC YS PH E ET PL M CO THE in2 Physics is the most up-to-date physics package written for the NSW Stage 6 Physics syllabus. The materials comprehensively address the syllabus outcomes and thoroughly prepare students for the HSC exam. Physics is presented as an exciting, relevant and fascinating discipline. The student materials provide clear and easy access to the content and theory, regular review questions, a full range of exam-style questions and features to develop an interest in the subject.

in2 Physics @ HSC student book • The student book closely follows the NSW Stage 6 Physics syllabus and its modular structure. • It clearly addresses both the contexts and the prescribed focus areas (PFAs). • Modules consist of chapters that are broken up into manageable sections. • Checkpoint questions review key content at 8 regular intervals throughout each chapter. • Physics Philes present short, interesting snippets of relevant information about physics or physics applications. • Physics Features highlight important real-life examples of physics. • Physics For Fun—Try This! provide hands-on activities that are easy to do. • Physics Focus brings together physics concepts in the context of one or more PFAs and provides students with a graded set of questions to develop their skills in this vital area.

From ideas to on implementati

From cathode rays to television

glass

anode (positive)

electrons 'boil' off the heated cathode

collimator

electron beam

heater

cathode (negative)

Figure 8.5.3

electrons attracted to the positive anode

The components of an electron gun used in oscilloscopes and CRT televisions

both cathode ray

Television

electron gun

magnetic coils

fluorescent screen

Figure 8.5.5

A television picture tube showing the electron gun, deflection coils and fluorescent screen

only one beam.

Each student book includes an interactive student CD containing: • an electronic version of the student book. • all of the student materials on the companion website with live links to the website. vi

mask blue beam

electron guns

red beam

R G B

green beam

focusing coils

V

Time

sawtooth voltage for timebase

Figure 8.5.4

V

Time

mask

fluorescent screen

fluorescent screen

electron beams

holes in mask

Figure 8.5.6

guns A colour CRT television set has three electron that will only strike their respective coloured phosphor dots with the aid of a shadow mask.

sinusoidal vertical voltage

A sawtooth voltage waveform on the horizontal beam deflection plates of a CRO sweeps the electron waveform across the screen to display the sinusoidal on the vertical deflection plates.

used the principles of the cathode ray Cathode ray tube (CRT) television sets are now being superseded by plasma tube for most of the 20th century. These which use different operating principles and liquid crystal display television sets, sharper image. However, the CRT and allow a larger display area with a place in this form of television holds quite a significant historical communication. television set is shown in Figure 8.5.5. A schematic diagram of a colour CRT of the CRO. The main difference is the Its basic elements are similar to those field coils placed outside the tube method of deflecting the electrons. Magnetic fields inside it. The magnitude and produce horizontal and vertical magnetic degree and direction of electron beam direction of the current determine the rule for the force on charged particles deflection. Recall your right-hand palm field will deflect the electrons in a magnetic field. The vertical magnetic deflect them vertically. horizontally; the horizontal field will scanning the beam from left to right The picture on the screen is formed by the television switches the beam on and and top to bottom. The electronics in in order to reproduce the transmitted off at the appropriate spots on the screen images, colour television sets need to picture. However, to reproduce colour green phosphors on the screen. Three and blue red, of intensity the control one aimed at one particular colour. The separate electron guns are used, each in groups of red, blue and green dots coloured dots on the screen are clustered cannot be distinguished by eye that are very close to each other and generally For this reason a method of guiding the without the aid of a magnifying glass. coloured dots was devised. A metal different electron beams to their respective 8.5.6) and consisting of an array of sheet, known as a shadow mask (Figure hole guides the three beams to Each screen. holes, is placed behind the phosphor the beams move horizontally and vertically. their respective coloured phosphor as need the shadow mask since they had Black and white television sets did not

168

phosphor dots on screen

vacuum

beam to move up or down in The vertical deflection plates cause the For example, a sinusoidal voltage will synchronisation with an input voltage. as a trace) on the screen. display a sinusoidal waveform (known

electron beam

deflecting coils

try this! Do not aDjust your horizontal! If you have access to an old black and white TV set or an old style monochrome computer monitor, try holding a bar magnet near the front of the screen and watch how the image distorts. This occurs because the magnetic field deflects the electrons that strike the screen. DO NOT do this with

a colour TV set. This can magnetise the shadow mask and cause permanent distortion of the image and its colour. You can move a bar magnet near the back of a colour TV set to deflect the electrons from the electron gun and therefore distort or shift the image without causing permanent damage to the TV set.

used Can an osCillosCope be as a television set?

ray oscilloscope (CRO) and CRT he similarity between the cathode be used as a television set. In television suggests that a CRO can that have made use of the CRO as fact, there have been some devices in principle, it can be used as a you would a computer monitor. So, ‘why did they need to deflect the television. One is then forced to ask fields rather than with electric beam in a television set with magnetic

T

fields as in the CRO?’ be made in the same design as In principle all television sets could with and cheaper to deflect the beam a CRO; however, it is much easier the tube rather than embed electrodes a magnetic field on the outside of is a little trickier. So now in the glass and inside the vacuum—this deflect the beam of the CRO using another question arises: ‘why not CROs?’. in cheaper magnetic fields, wouldn’t it result instruments. The horizontal Cathode ray oscilloscopes are precision to very high frequencies in order sweep rate must be able to be increased to quickly. Electric fields can be made to detect signals that change very extra power requirements. change very quickly without significant system requires higher and higher However, a magnetically deflected in and vertical deflection frequencies voltages with increasing horizontal same the in the coils, and therefore, order to maintain the same current a significantly greater power angle of beam deflection – thus having sets, however, only operate at requirement. Cathode ray tube television horizontal and vertical frequencies. fixed and relatively low scanning the mass market to deflect with a Thus it is simpler and cheaper for magnetic field.

CheCkpoint 8.5 1 2 3 4

Outline the purpose of a CRO. List the main parts of a CRO. in the CRO. Describe the role of each of these parts the cathode ray tube CRO and CRT TV. State the similarities and differences between 169

in2 Physics @ HSC Activity Manual Chapter 8 from cathode rays to television

MODULE

• A write-in workbook that provides a structured approach to the mandatory practical experiences, both first-hand and secondary-source investigations. • Dot point and skills focused.

3

from ideas to tion implementa

Chapter 8

from Cathode rays to television aCtivity 8.1 first-hand investigation

Changing pressure of discharge tubes

the occurrence of different striation first-hand information to observe Perorm an investigation and gather discharge tubes. patterns for different pressures in

Physics skills

in this activity include: The skills outcomes to be practised 12.1 perform first-hand investigations 12.2 gather first-hand information 14.1 analyse information syllabus grid on pages vi–viii. skills outcomes can be found in the The complete statement of these

Aim

To observe the striation patterns for different

different patterns that Because of this development, could be seen depended on the pressure. the air molecules. To do this, but it can be made to conduct by ionising Normally air is considered to be an insulator, an electric field). At high pressures these that are always in air are accelerated (with to ionise the air energy the very small fraction of free electrons sufficient gain not losing their energy and, as a result, do electrons collide frequently with the air, molecules, thereby acquiring enough travel further before colliding with air they are atoms. As pressure is reduced, these electrons in turn, can ionise other atoms. When will produce more free electrons that, energy to ionise the air molecules. This show (known as a discharge). The lower to be absorbed by atoms, we see a light discharge. a able to travel far enough to gain the energy producing and can travel before colliding with gas molecules the pressure, the further the electrons excited (increasing in energy) and electrons around the gas atom becoming will also The light that is emitted is a result of the energy they can have in an atom). Light lowest (the state ground the to return re-emitting the photon of light as they ground state, emitting photons. As every with ions and the electrons return to the be produced when free electrons recombine the element with which the electron collides. the colour of light seen will vary with element has a distinct set of energy levels, be passed through air. The it was found that electric current could

Equipment

The patterns are can carry out the experiment first hand. If you have the apparatus at school, you is very dark. • discharge tubes at different pressures • induction coil • DC power supply • connecting wires Alternatively, you can use the simulations

them.

Risk assessment

Method 1 2 3 4

8.1.1. Set up the equipment as shown in Figure in the tube. Observe the patterns and note the pressure series. Replace the tube with the next in the and Repeat the process of observing the patterns your set. in tubes the noting the pressure for each of

DC power supply

tube Figure 8.1.1 Induction coil and discharge

pressures in discharge tubes.

Hypothesis

in Part B and make observations from

hard to see unless the room

HAZARD

The voltages necessary to operate the coils and may produce unwanted X-rays. used, High voltages are produced by induction the tube. Generally, the higher the voltage the tube and the pressure of the gas in tubes depend upon the dimensions of of unwanted X-rays. the greater the danger of the production a minimum of 1 m away from the equipment. Use the lowest possible voltage and stand

Theory

which the pressure could be reduced Plücker collaborated to create a tube in Ever since Heinrich Geissler and Julius ray tube have advanced tremendously. atom and developing uses for the cathode substantially, our understanding of the

69

68

in2 Physics @ HSC Teacher Resource • Editable teaching materials, including teaching programs, so that teachers can tailor lessons to suit their classroom. • Answers to student book and activity manual questions, with fully worked solutions and extended answers and support notes. • Risk assessments for all first-hand investigations.

in2 Physics @ HSC companion website Visit the companion website in the student lounge and teacher lounge of Pearson Places • Review questions— auto-correcting multiple-choice questions for each chapter. • Web destinations—a list of reviewed websites that support further investigation.

For more information on the in2 Physics series, visit www.pearsonplaces.com.au vii

How to use this book in2 Physics @ HSC is structured to enhance student learning and their enjoyment of learning. It contains many outstanding and unique features that will assist students succeed in Stage 6 Physics. These include: • Module opening pages introduce a range of contexts for study, as well as an inquiry activity that provides immediate activities for exploration and discussion.

2 Context

• Key ideas are clearly highlighted with a and indicate where domain dot points Syllabus flags appear in the student book. The flags are placed as closely as possible to where the relevant content is covered. Flags may be repeated if the dot point has multiple parts, is complex or where students are required to solve problems. 3

Motors and Generators

l v = l0 1 − tv =

mv =

Figure 4.0.2

of interference of electromagnetic radiation, and examine how this was applied to crystals using X-rays. Then we will see how the BCS theory of superconductivity made use of the crystal structure of matter.

try thiS!

CheCkpoInT 11.1

Crystals in the kitChen

Explain what is meant by the crystal structure of matter.

Look at salt grains through a magnifying lens. Each grain is a single crystal that is made from the basic arrangement of sodium and chlorine atoms shown in Figure 11.1.1. Although the grains mostly look irregular due to breaking and chipping during the manufacturing process, occasionally you will see an untouched cubic or rectangular prism that reflects the underlying crystal lattice structure.

11.2 Wave interference The wave nature of light can be used to measure the size of very small spaces. Recall that two identical waves combine to produce a wave of greater amplitude when their crests overlap, as shown in Figure 11.2.1a (see in2 Physics @ Preliminary sections 6.4 and 7.4). The overlapping waves will cancel to produce t=0s a resulting wave of zero amplitude when the crest of one wave coincides with the trough of the other (Figure 11.2.1b). This addition and subtraction is called constructive and destructive interference respectively and is a property of all wave phenomena. t=1s As an example, two identical circular water waves in a ripple tank overlap (see Figure 11.2.2). The regions of constructive and destructive interference radiate outwards along the lines as shown. Increasing the spacing between the sources t = 3 s (Figure 11.2.2b). causes the radiating lines to come closer together a

Figure 11.2.1 Two identical waves (red, green) travelling in opposite directions can add (blue) t=1s

(a) constructively or (b) destructively.t = 5 s

The interference of identical waves from two sources can also be represented by outwardly radiating transverse waves (see Figure 11.2.3). The distance that a twave = 3 s travels is known as its path length. t = 6 s Constructive interference occurs when the difference in the path length of the two waves is equal to 0, λ, 2λ, 3λ, 4λ or any other integer multiple of the wavelength λ. Destructive interference occurs when the two waves are half a wavelength out of step. This corresponds to t=4s t=7s a path length difference of λ/2, 3λ/2, 5λ/2 etc.

t=5s

lines of destructive interference

lines of constructive interference

t=4s

t=0s

11.1 The crystal structure of matter

a

b

waves in phase

destructive interference constructive interference

204

t=7s

1 2

evil tWinS

T

he most extreme mass–energy conversion involves antimatter. For every kind of matter particle there is an equivalent antimatter particle, an ‘evil twin’, bearing properties (such as charge) of opposite sign. Particles and their antiparticles have the same rest mass. When a particle meets its antiparticle, they mutually annihilate—all their opposing properties cancel, leaving only their mass‑energy, which is usually released in the form of two gamma‑ray photons. Matter– antimatter annihilation has been suggested (speculatively) as a possible propellant for powering future interstellar spacecraft.

 1 v2  1 ≈ m0c  1 + × 2  = m0c 2 + m0v 2 2 2 c   2

E = mc 2

Figure 3.4.6

One of the four ultra-precise superconducting spherical gyroscopes on NASA’s Gravity Probe B, which orbited Earth in 2004/05 to measure two predictions of general relativity: the bending of spacetime by the Earth’s mass and the slight twisting of spacetime by the Earth’s rotation (frame-dragging)

In general relativity, Einstein showed that gravity occurs because objects with mass or energy cause this 4D spacetime to become distorted. The paths of objects through this distorted 4D spacetime appear to our 3D eyes to follow the sort of astronomical trajectories you learned about in Chapter 2 ‘Explaining and exploring the solar system’. However, unlike Newton’s gravitation, general relativity is able to handle situations of high gravitational fields, such as Mercury’s precessing orbit around the Sun and black holes. General relativity also predicts another wave that doesn’t require a medium: the ripples in spacetime called ‘gravity waves’.

where m is any kind of mass. In relativity, mass and energy are regarded as the same thing, apart from the change of units. Sometimes the term mass-energy is used for both. m0 c 2 is called the rest energy, so even a stationary object contains energy due to its rest mass. Relativistic kinetic energy therefore: m0c 2 mv c 2 − m0c 2 = − m0c 2 v2 1− 2 c Whenever energy increases, so does mass. Any release of energy is accompanied by a decrease in mass. A book sitting on the top shelf has a slightly higher mass than one on the bottom shelf because of the difference in gravitational potential energy. An object’s mass increases slightly when it is hot because the kinetic energy of the vibrating atoms is higher. Because c 2 is such a large number, a very tiny mass is equivalent to a large amount of energy. In the early days of nuclear physics, E = mc 2 revealed the enormous energy locked up inside an atom’s nucleus by the strong nuclear force that holds the protons and neutrons together. It was this that alerted nuclear physicists just before World War II to the possibility of a nuclear bomb. The energy released by the nuclear bomb dropped on Hiroshima at the end of that war (smallish by modern standards) resulted from a reduction in relativistic mass of about 0.7 g (slightly less than the mass of a standard wire paperclip).

Discuss the implications of mass increase, time dilation and length contraction for space travel.

Worked example qUESTIon When free protons and neutrons become bound together to form a nucleus, the reduction in nuclear potential energy (binding energy) is released, normally in the form of gamma rays. Relativity says this loss in energy is reflected in a decrease in mass of the resulting atom.

• Each chapter concludes with: – a chapter summary – review questions, including literacy-based questions (Physically Speaking), chapter review questions (Reviewing) and physics problems (Solving Problems). Syllabus verbs are clearly highlighted as and where appropriate – Physics Focus—a unique feature that places key chapter concepts in the context of one or more prescribed focus areas.

b

Figure 11.2.2 Interference of water waves for two sources that are (a) close together and (b) further apart

19

Imaging with gamma rays

PRACTICAL EXPERIENCES

ChAPTER 19

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Figure 11.2.3 Constructive and destructive interference between

Activity 19.1: Bone scAns

identical transverse waves from two sources

Perform an investigation to compare a bone scan with an X-ray image.

205

• Chapters are divided into short, accessible sections— the text itself is presented in short, easy-to-understand chunks of information. Each section concludes with a Checkpoint—a set of review questions to check understanding of key content and concepts.

A bone scan is performed to obtain a functional image of the bones and so can be used to detect abnormal metabolism in the bones, which may be an indication of cancer or other abnormality. Because cancer mostly involves a higher than normal rate of cell division (thus producing a tumour), chemicals involved in metabolic processes in bone tend to accumulate in higher concentrations in cancerous tissue. This produces areas of concentration of gamma emission, indicating a tumour. Compare the data obtained from the image of a bone scan with that provided by an X-ray image. Discussion questions 1 Identify the best part of the body for each of these diagnostic tools to image. 2 Compare and contrast the two images in terms of the information they provide.

Figure 19.6.1

a

Chapter summary •

• • • •

•

The number of protons in a nucleus is given by the atomic number, while the total number of nucleons is given by the mass number. Atoms of the same element with different numbers of neutrons are called isotopes of that element. Many elements have naturally occurring unstable radioisotopes. In alpha decay an unstable nucleus decays by emitting an alpha particle (α-particle). In beta decay, a neutron changes into a proton and a high-energy electron that is emitted as a beta particle (β-particle). In positron decay, a positron—the antiparticle of the electron—is emitted.

b

Activity 19.2: HeAltHy or diseAsed? Typical images of healthy bone and cancerous bone are shown. The tumours show up as hot-spots. Use the template in the activity manual to research and compare images of healthy and diseased parts of the body. Discussion questions 1 Examine Figure 19.4.2. There is a hot-spot that is not cancerous near the left elbow. Explain. 2 In the normal scan (Figure 19.6.2a), the lower pelvis has a region of high intensity. Why is this? (Hint: It may be soft tissue, not bone. Looking at Figure 19.6.2b might help you with this question.) 3 State the differences that can be observed by comparing an image of a healthy part of the body with that of a diseased part of the body.

PHysicAlly sPeAking Below is a list of topics that have been discussed throughout this chapter. Create a visual summary of the concepts in this chapter by constructing a mind map linking the terms. Add diagrams where useful.

Radioactive decay

350

Radiation

Radioisotope

Neutron

Proton

Beta decay

Gamma decay

Antimatter

Bone scan

Positron decay

Half-life

Bones scans of (a) a healthy person and (b) a patient with a tumour in the skeleton

•

mEdICAL PhySICS

When a positron and an electron collide, their total mass is converted into energy in the form of two gamma-ray photons. In gamma decay a gamma ray (g) is emitted from a radioactive isotope. The time it takes for half the mass of a radioactive parent isotope to decay into its daughter nuclei is the half-life of the isotope. Artificial radioisotopes are produced in two main ways: in a nuclear reactor or in a cyclotron. A gamma camera detects gamma rays emitted by a radiopharmaceutical in the patient’s body. PET imaging uses positron-emitting radiopharmaceuticals to obtain images using gamma rays emitted from electron–positron annihilation.

• •

• • •

Review questions

Comparison of an X-ray and bone scan of a hand

Gather and process secondary information to compare a scanned image of at least one healthy body part or organ with a scanned image of its diseased counterpart.

Figure 19.6.2

viii

−

73

constructive interference

t=6s

1 2

72

Figure 11.1.1 Crystal structure of sodium chloride. The red spheres represent positive sodium ions, and the green spheres represent negative chlorine ions.

−

1 Rearrange: mvc 2 – m0c 2 = (mv – m0)c 2 ≈ m0v 2 2 In other words, at low speeds, the gain in relativistic mass (mv – m0) multiplied by c 2 equals the kinetic energy—a tantalising hint that at low speed mass and energy are equivalent. It can also be shown to be true at all speeds, using more sophisticated mathematics. In general, mass and energy are equivalent in relativity and c 2 is the conversion factor between the energy unit (joules) and the mass unit (kg). In other words:

c2

here are two more invariants in special relativity. Maxwell’s equations (and hence relativity) requires that electrical charge is invariant in all frames. Another quantity invariant in all inertial frames is called the spacetime interval. You may have heard of spacetime but not know what it is. One of Einstein’s mathematics lecturers Hermann Minkowski (1864–1909) showed that the equations of relativity and Maxwell’s equations become simplified if you assume that the three dimensions of space (x, y, z) and time t taken together form a four‑dimensional coordinate system called spacetime. Each location in spacetime is not a position, but rather an event—a position and a time. Using a 4D version of Pythagoras’ theorem, Minkowski then defined a kind of 4D ‘distance’ between events called the spacetime interval s given by: s 2 = (c × time period)2 – path length2 = c 2t 2 – ((∆x)2 + (∆y)2 + (∆z)2) Observers in different frames don’t agree on the 3D path length between events, or the time period between events, but all observers in inertial frames agree on the spacetime interval s between events.

from ideaS to implementation

A crystal is a three-dimensional regular arrangement of atoms. Figure 11.1.1 shows a sodium chloride crystal (ordinary salt also called rock salt when it comes as a large crystal). The crystal is made from simple cubes repeated many times, with sodium and chlorine atoms at the corners of the cubes. Crystals of other materials may have different regular arrangements of their atoms. There are 14 types of crystal arrangements that solids can have. The regular arrangement of atoms in crystals was a hypothesis before Max Von Laue and his colleagues confirmed it by X-ray diffraction experiments. William and Lawrence Bragg took this method one step further by measuring the spacing between the atoms in the crystal. Let us first look at the phenomenon

 v2  m0c 2  1 − 2  c  

2

T

Surprising discovery crystal, constructive interference, destructive interference, path length, diffraction grating, Bragg law, phonons, critical temperature, type-I superconductors, type-II superconductors, critical field strength, vortices, flux pinning, BCS theory, Cooper pair, coherence length, energy gap, spin

v

 v2  = m0c 2  1 − 2  c  

m0c 2

v2 1− 2 c Using a well-known approximation formula that you might learn at university, (1 – x )n ≈ 1 – nx for small x:

1. The history of physics

InQUIRY ACtIVItY

• Chapter openings list the key words of each chapter and introduce the chapter topic in a concise and engaging way.

Just as an improved understanding of the conducting properties of semiconductors led to the wide variety of electronic devices, research into the conductivity of metals produced quite a surprising discovery called superconductivity. This is the total disappearance of electrical resistance below a certain temperature, which has great potential applications ranging from energy transmission and storage to public transport. An understanding of this phenomenon required a detailed understanding of the crystal structure of conductors and the motion of electrons through them.

mv c 2 =

m0

TwISTIng SPACETImE ... And YoUR mInd

A simple homopolar motor

83

Superconductivity

c2

c2

1−

82

11

v2

v2

PHYSICS FEATURE

Many of the devices you use every day have electric motors. They spin your DVDs, wash your clothes and even help cook your food. Could you live without them, and how much do you know about how they work? The essential ingredients for a motor are a power source, a magnetic field and things to connect these together in the right way. It’s not as hard as you think. All you need is a battery, a wood screw, a piece of wire and a cylindrical or spherical magnet. Put these things together as shown in Figure 4.0.2 and see if you can get your motor to spin. Be patient and keep trying. Then try the following activities. 1 Test the effects of changing the voltage you use. You could add another battery in series or try a battery with a higher voltage. 2 Try changing the strength of the magnet by using a different magnet or adding another. What does this affect? 3 Try changing the length of the screw, how sharp its point is or the material it is made from. Does it have to be made of iron?

A generator produces electricity in each of these wind turbines.

The kinetic energy formula K = 1 mv 2 doesn’t apply at relativistic speeds, 2 even if you substitute relativistic mass mv into the formula. Classically, if you apply a net force to accelerate an object, the work done equals the increase in kinetic energy. An increase in speed means an increase in kinetic energy. But in relativity it also means an increase in relativistic mass, so relativistic mass and energy seem to be associated. Superficially, if you multiply relativistic mass by c 2 you get mv c 2, which has the same dimensions and units as energy. But let’s look more closely at it.

t0 1−

The first recorded observations of the relationship between electricity and magnetism date back more than 400 years. Many unimagined discoveries followed, but progress never waits. Before we understood their nature, inventions utilising electricity and magnetism had changed our world forever. Today our lives revolve around these forms of energy. The lights you use to read this book rely on them and the CD inside it would be nothing but a shiny coaster for your cup. We use magnetism to generate the electricity that drives industry, discovery and invention. Electricity and magnetism are a foundation for modern technology, deeply seated in the global economy, and our use impacts heavily on the environment. The greatest challenge that faces future generations is the supply of energy. As fossil fuels dry up, electricity and magnetism will become even more important. New and improved technologies will be needed. Whether it’s a hybrid car, a wind turbine or a nuclear fusion power plant, they all rely on applications of electricity and magnetism.

Space How does this formula behave at low speeds (when v 2/c 2 is small)?

Mass, energy and the world’s most famous equation

Solve problems and analyse information using: E = mc2

BUIld YoUR own eleCtRIC motoR

Figure 4.0.1

Seeing in a weird light: relativity

Isotope

reviewing 1

Recall how the bone scan produced by a radioisotope compares with that from a conventional X-ray.

2

Analyse the relationship between the half-life of a radiopharmaceutical and its potential use in the human body.

3

Explain how it is possible to emit an electron from the nucleus when the electron is not a nucleon.

4

Assess the statement that ‘Positrons are radioactive particles produced when a proton decays’.

5

Discuss the impact that the production and use of radioisotopes has on society.

6

Describe how isotopes such as Tc-99m and F-18 can be used to target specific organs to be imaged.

7

Use the data in Table 19.6.1 to answer the questions: a Which radioactive isotope would most likely be used in a bone scan? Justify your choice. b Propose two reasons why cesium-137 would not be a suitable isotope to use in medical imaging.

Nucleon

Alpha decay

PET

Table 19.6.1 Scintillator

Properties of some radioisotopes

Radioactive souRce

Radiation emitted

Half-life

C-11 Tc-99m TI-201 I-131 Cs-137 U-238

β+, g g g β, g α α

20.30 minutes 6.02 hours 3.05 days 8.04 days 30.17 years 4.47 × 109 years

351

How to use this book

Other features

• Module reviews provide a full range of exam-style questions, including multiple-choice, short-response and extended-response questions.

3

from ideas to implementation 4

The review contains questions in a similar style and proportion to the HSC Physics examination. Marks are allocated to each question up to a total of 25 marks. It should take you approximately 45 minutes to complete this review. 5

Experimental data from black body radiation during Planck’s time showed that predicted radiation levels were not achieved in reality. Planck best described this anomaly by saying that: A classical physics was wrong. B radiation that is emitted and absorbed is quantised. C he had no explanation for it. D quantum mechanics needed to be developed.

extended response

Figure 11.13.4 shows a cathode ray tube that has been evacuated. Which answer correctly names each of the labelled features?

III

6

Explain, with reference to atomic models, why cathode rays can travel through metals. (2 marks)

7

Outline how the cathode ray tube in a TV works in order to produce the viewing picture. (2 marks)

8

Give reasons why CRT TVs use magnetic coils and CROs use electric plates in order to deflect the beams, given that both methods work. (2 marks).

9

In your studies you were required to gather information to describe how the photoelectric effect is used in photocells. a Explain how you determined which material was relevant and reliable. b Outline how the photoelectric effect is used in photocells. (3 marks)

II

I

multiple choice (1 mark each) 1 Predict the direction of the electron in Figure 11.13.1 as it enters the magnetic field. A Straight up B Left C Right D Down

A B

Figure 11.13.1

An electron in a magnetic field C

The diagrams in Figure 11.13.2 represent semiconductors, conductors and insulators. The diagrams show the conduction and valence bands, and the energy gaps. Which answer correctly labels each of the diagrams?

A B C D

3

Figure 11.13.4 An evacuated cathode ray tube

–

I

II

III

Conductor Insulator Insulator Semiconductor

Insulator Conductor Semiconductor Conductor

Semiconductor Semiconductor Conductor Insulator

D

I

A B C D

I

II

III

Critical temperature Superconductor material Critical temperature Normal material

Superconductor material Critical temperature Normal material

Normal material

Superconductor material

II

Figure 11.13.2

The graph in Figure 11.13.3 shows how the resistance of a material varies with temperature. Identify each of the parts labelled on the graph.

Superconductor material Critical temperature

II

III

Cathode Striations

Anode Cathode

Anode

Faraday’s dark space Striations

Faraday’s dark space

10

Justify the introduction of semiconductors to replace thermionic devices. (4 marks)

11

Magnetic levitation trains are used in Germany and Japan. The trains in Germany use conventional electromagnets, whereas the one in Japan uses superconductors. Compare and contrast the two systems. (3 marks)

12

a

b

Determine the frequency of red light, which has a wavelength λ = 660 nm. (Speed of light c = 3.00 × 108 m s–1) Calculate the energy of a photon that is emitted with this wavelength. (Planck’s constant h = 6.63 × 10–34 J s) (4 marks)

I

Figure 11.13.3

• Physics for Fun—Try This! activities are short, handson activities to be done quickly, designed to provoke discussion. • Physics Features are a key feature as they highlight contextual material, case studies or prescribed focus areas of the syllabus.

III

II

Normal material

I Striations Faraday’s dark space Crooke’s dark space Cathode

• A complete glossary of all the key words is included at the end of the student book.

Energy bands

Resistance (Ω)

2

• Physics Philes present short, interesting items to support or extend the text.

III

Temperature (K)

Resistance varies with temperature

224

225

Practical experiences The accompanying activity manual covers all of the mandatory practical experiences outlined in the syllabus. in2 Physics @ HSC Activity Manual is a write-in workbook that outlines a clear, foolproof approach to success in all the required practical experiences. Within the student book, there are clear cross-references to the activity manual: Practical Experiences icons refer to the activity number and page in the activity manual. In each chapter, a summary of possible investigations is provided as a starting point to get students thinking. These include PRACTICAL the aim, a list of equipment and EXPERIENCES Activity 10.2 discussion questions. Activity Man

• The final two chapters provide essential reference material: ‘Skills stage 2’ and ‘Revisiting the BOS key terms’. • In all questions and activities, except module review questions, the BOS key terms are highlighted.

in2 Physics @ HSC Student CD This is included with the student book and contains: • an electronic version of the student book • interactive modules demonstrating key concepts

MODULE

ual, Page 94

2

Chapter 6 motors: magnetic fields make the world go around

motors and generators

aCtIVItY 6.2 First-hand investigation

• the companion website on CD

Risk assessment

Motors and torque

Solve problems and analyse information about simple motors using: τ = nBIA cos θ

Method

Physics skills

1

Cut a length of cotton-covered wire so that the wire is long enough to wrap around the exterior of a matchbox three times (as shown in Figure 6.2.2).

2

Leave a straight piece (approx. 10 cm long) hanging out and then wind the remainder of the wire around the box 2½ times. Leave another straight piece the same length as at the start, on the opposite side.

3

Wrap the straight pieces around the loops so that they tie both ends.

4

Fan out the loops so that you get equally spaced loops and that it looks like a bird cage (see Figure 6.2.3).

5

Push out the middle of the paper clip as shown and Blu-Tack to the bench.

6

Slip the straight pieces of wire through the paper clip supports. Unwrap the cotton from these parts.

7

Connect an AC power supply to the paper clips.

8

Place two magnets so that a north pole and a south pole face on opposing sides of the cage.

9

Turn on. You may need to give the cage a tap to get it spinning.

The skills outcomes to be practised in this activity include: 12.4 process information 14.1 analyse information The complete statement of these skills outcomes can be found in the syllabus grid on pages vii–viii.

Aim

Hypothesis

Theory The motor effect means that a current-carrying wire experiences a force when placed in a magnetic field. This is the basis for the workings of a motor. For a motor to work as needed, the motion resulting from the motor effect needs to be circular and the force needs to be adjusted so the direction of rotation does not change.

Question Figure 6.2.1 shows the simplified workings of a motor that you will be making. Label all the parts of the motor.

48

insulated wire from which insulation can be removed easily magnets magnetic field sensor and data logger (if available) paperclips

matchbox wire b

loop wire through

Figure 6.2.2 Equipment set-up 1

cage fanned out

paper clip

alligator clip wires

power source

Figure 6.2.3 Equipment set-up 2

Record your observations of the motor.

The complete in2 Physics @ HSC package

2

How did adding more magnets affect how the motor ran?

Remember the other components of the complete package:

3

When the current is increased, what changes occurred?

Results N

S

C:

D:

Figure 6.2.1 Simplified motor

• Blu-Tack • connecting wires with alligator clips • power supply

• a link to the live companion website (Internet access required) to provide access to the latest information and web links related to the student book.

1

A: B:

Equipment • • • •

a

• in2 Physics @ HSC companion website at Pearson Places 49

• in2 Physics @ HSC Teacher Resource.

ix

Stage 6 Physics syllabus grid Prescribed focus areas 1. The history of physics

H1. evaluates how major advances in scientific understanding and technology have changed the direction or nature of scientific thinking

Feature: pp. 12, 29, 72

2. The nature and practice of physics

H2. analyses the ways in which models, theories and laws in physics have been tested and validated

Focus: p. 79

3. Applications and uses of physics

H3. assesses the impact of particular advances in physics on the development of technologies

Feature: pp. 12, 29, 307, 334, 346

Focus: pp. 25, 246, 299

Focus: pp. 57, 79, 129, 173, 223, 246, 259, 278 4. Implications for society and the Environment

H4. assesses the impacts of applications of physics on society and the environment

Feature: pp. 29, 307, 344

5. Current issues, research and developments in physics

H5. identifies possible future directions of physics research

Feature: pp. 391, 410

Focus: pp. 113, 173, 353 Focus: pp. 79, 113, 173, 223, 353, 386

Module 1 Space 1. The Earth has a gravitational field that exerts a force on objects both on it and around it Students learn to:

Page Students:

define weight as the force on an object due to a gravitational field

13

perform an investigation and gather information to determine a value for Act. 1.2 acceleration due to gravity using pendulum motion or computer-assisted technology and identify reason(s) for possible variations from the value 9.8 m s–2

Page

explain that a change in gravitational potential energy is related to work done

16

gather secondary information to predict the value of acceleration due to gravity on other planets

Act. 1.3

define gravitational potential energy as the work done to move an object from a very large distance away to a point in a gravitational field: mm EP = G 1 2 r

16

analyse information using the expression:

Act. 1.3

F = mg to determine the weight force for a body on Earth and for the same body on other planets

2. Many factors have to be taken into account to achieve a successful rocket launch, maintain a stable orbit and return to Earth Students learn to:

Page Students:

Page

describe the trajectory of an object undergoing projectile motion within the Earth’s gravitational field in terms of horizontal and vertical components

5

7, 9, 23, 24

solve problems and analyse information to calculate the actual velocity of a projectile from its horizontal and vertical components using: vx2 = ux2

v = u + at vy2 = uy2 + 2ay ∆y ∆x = ux t ∆y = uyt + 12 ay t 2

describe Galileo’s analysis of projectile motion

5

perform a first-hand investigation, gather information and analyse data to calculate initial and final velocity, maximum height reached, range and time of flight of a projectile for a range of situations by using simulations, data loggers and computer analysis

explain the concept of escape velocity in terms of the: – gravitational constant – mass and radius of the planet

18

identify data sources, gather, analyse and present information on the contribution 29 of one of the following to the development of space exploration: Tsiolkovsky, Act. 2.1 Oberth, Goddard, Esnault-Pelterie, O’Neill or von Braun

x

Act. 1.1

Stage 6 Physics syllabus grid outline Newton’s concept of escape velocity

18

identify why the term ‘g forces’ is used to explain the forces acting on an astronaut during launch

31

discuss the effect of the Earth‘s orbital motion and its rotational motion on the launch of a rocket

34

analyse the changing acceleration of a rocket during launch in terms of the: – Law of Conservation of Momentum – forces experienced by astronauts

30, 33

analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth

25, 32, solve problems and analyse information to calculate the centripetal force acting 34, 37, on a satellite undergoing uniform circular motion about the Earth using: 54, 55 mv 2 F = r

compare qualitatively low Earth and geo-stationary orbits

43

define the term orbital velocity and the 36, 40, solve problems and analyse information using: quantitative and qualitative relationship 56 r3 GM = between orbital velocity, the 2 4 π2 T gravitational constant, mass of the central body, mass of the satellite and the radius of the orbit using Kepler’s Law of Periods account for the orbital decay of satellites in low Earth orbit

46

discuss issues associated with safe re-entry into the Earth’s atmosphere and landing on the Earth’s surface

47

identify that there is an optimum angle for safe re-entry for a manned spacecraft into the Earth’s atmosphere and the consequences of failing to achieve this angle

47

37, 54, 55 Act. 2.2

39, 43, 56

3. The solar system is held together by gravity Students learn to:

Page Students:

Page

describe a gravitational field in the region surrounding a massive object in terms of its effects on other masses in it

13

present information and use available evidence to discuss the factors affecting the strength of the gravitational force

Act. 1.3

define Newton’s Law of Universal Gravitation: mm F = G 1 22 d

11

solve problems and analyse information using: mm F = G 1 22 d

23, 24, 25, 37, 54, 55

discuss the importance of Newton’s Law of Universal Gravitation in understanding and calculating the motion of satellites

35, 38

identify that a slingshot effect can be provided by planets for space probes

44

xi

Stage 6 Physics syllabus grid 4. Current and emerging understanding about time and space has been dependent upon earlier models of the transmission of light Students learn to:

Page Students:

Page

outline the features of the aether model 61 for the transmission of light describe and evaluate the MichelsonMorley attempt to measure the relative velocity of the Earth through the aether

62

gather and process information to interpret the results of the Michelson-Morley experiment

62 Act. 3.2

discuss the role of the MichelsonMorley experiments in making determinations about competing theories

62

outline the nature of inertial frames of reference

58

perform an investigation to help distinguish between non-inertial and inertial frames of reference

60 Act. 3.1

discuss the principle of relativity

58

analyse and interpret some of Einstein’s thought experiments involving mirrors and trains and discuss the relationship between thought and reality

66

describe the significance of Einstein’s assumption of the constancy of the speed of light

65

analyse information to discuss the relationship between theory and the evidence supporting it, using Einstein’s predictions based on relativity that were made many years before evidence was available to support it

78

identify that if c is constant then space and time become relative

65

discuss the concept that length standards are defined in terms of time in contrast to the original metre standard

79

explain qualitatively and quantitatively the consequence of special relativity in relation to: – the relativity of simultaneity – the equivalence between mass and energy – length contraction – time dilation – mass dilation

64, 69 solve problems and analyse information using: E = mc 2 lv = l0 1 −

v2

66, 69, 72, 77, 78

c2

t0 tv = 1 −

v2 c2

m0 mv = 1 − discuss the implications of mass increase, time dilation and length contraction for space travel

v2 c2

70, 73

Module 2 Motors and Generators 1. Motors use the effect of forces on current-carrying conductors in magnetic fields Students learn to:

Page Students:

Page

discuss the effect on the magnitude of the force on a current-carrying conductor of variations in: – the strength of the magnetic field in which it is located – the magnitude of the current in the conductor – the length of the conductor in the external magnetic field – the angle between the direction of the external magnetic field and the direction of the length of the conductor

92

Act. 4.1

xii

perform a first-hand investigation to demonstrate the motor effect

Stage 6 Physics syllabus grid describe qualitatively and quantitatively 94 the force between long parallel currentcarrying conductors: II F =k 1 2 l d

solve problems using: II F =k 1 2 l d

94

define torque as the turning moment of a force using: t = Fd

115

solve problems and analyse information about the force on current-carrying conductors in magnetic fields using: F = BIl sin θ

92 Act. 4.1

identify that the motor effect is due to the force acting on a current-carrying conductor in a magnetic field

90, 116

solve problems and analyse information about simple motors using: t = nBIA cos θ

117 Act. 6.2

describe the forces experienced by a current-carrying loop in a magnetic field and describe the net result of the forces

117

identify data sources, gather and process information to qualitatively describe the 91, 119 application of the motor effect in: Act. 6.1 – the galvanometer – the loudspeaker

describe the main features of a DC electric motor and the role of each feature

115

identify that the required magnetic fields in DC motors can be produced either by current-carrying coils or permanent magnets

115

2. The relative motion between a conductor and magnetic field is used to generate an electrical voltage Students learn to:

Page Students:

outline Michael Faraday’s discovery of the generation of an electric current by a moving magnet

100

perform an investigation to model the generation of an electric current by moving 101 a magnet in a coil or a coil near a magnet Act. 5.1

Page

define magnetic field strength B as magnetic flux density

101

plan, choose equipment or resources for, and perform a first-hand investigation to predict and verify the effect on a generated electric current when: – the distance between the coil and magnet is varied – the strength of the magnet is varied – the relative motion between the coil and the magnet is varied

Act. 5.1

describe the concept of magnetic flux in terms of magnetic flux density and surface area

101

gather, analyse and present information to explain how induction is used in cooktops in electric ranges

108 Act. 5.2

describe generated potential difference as the rate of change of magnetic flux through a circuit

103

gather secondary information to identify how eddy currents have been utilised in electromagnetic braking

Act. 5.2 113

account for Lenz’s Law in terms of conservation of energy and relate it to the production of back emf in motors

105, 120

explain that, in electric motors, back emf opposes the supply emf

120

explain the production of eddy currents in terms of Lenz’s Law

106

3. Generators are used to provide large scale power production Students learn to:

Page Students:

Page

describe the main components of a generator

131

plan, choose equipment or resources for, and perform a first-hand investigation to demonstrate the production of an alternating current

Act. 5.1

compare the structure and function of a generator to an electric motor

135

gather secondary information to discuss advantages/disadvantages of AC and DC generators and relate these to their use

135 Act. 7.1

describe the differences between AC and DC generators

135

analyse secondary information on the competition between Westinghouse and Edison to supply electricity to cities

141 Act. 7.2

gather and analyse information to identify how transmission lines are: – insulated from supporting structures – protected from lightning strikes

146 Act. 7.3

discuss the energy losses that occur as 144 energy is fed through transmission lines from the generator to the consumer assess the effects of the development of AC generators on society and the environment

147

xiii

Stage 6 Physics syllabus grid 4. Transformers allow generated voltage to be either increased or decreased before it is used Students learn to:

Page Students:

Page

describe the purpose of transformers in electrical circuits

136

perform an investigation to model the structure of a transformer to demonstrate how secondary voltage is produced

Act. 7.3

compare step-up and step-down transformers

137

solve problems and analyse information about transformers using: Vp np = Vs ns

137 Act. 7.3

identify the relationship between the ratio of the number of turns in the primary and secondary coils and the ratio of primary to secondary voltage

137

gather, analyse and use available evidence to discuss how difficulties of heating caused by eddy currents in transformers may be overcome

139 Act. 7.3

explain why voltage transformations are related to conservation of energy

139

gather and analyse secondary information to discuss the need for transformers in the transfer of electrical energy from a power station to its point of use

145 Act. 7.3

explain the role of transformers in electricity substations

142

discuss why some electrical appliances in the home that are connected to the mains domestic power supply use a transformer

136, 144

discuss the impact of the development of transformers on society

147

5. Motors are used in industries and the home usually to convert electrical energy into more useful forms of energy Students learn to:

Page

Students:

Page

describe the main features of an AC electric motor

124

perform an investigation to demonstrate the principle of an AC induction motor

Act. 6.3

gather, process and analyse information to identify some of the energy transfers 124, and transformations involving the conversion of electrical energy into more useful 153 forms in the home and industry Act. 7.3

Module 3 From Ideas to Implementation 1. Increased understandings of cathode rays led to the development of television Students learn to:

Students:

Page

explain why the apparent inconsistent 157 behaviour of cathode rays caused debate as to whether they were charged particles or electromagnetic waves

Page

perform an investigation and gather first-hand information to observe the occurrence of different striation patterns for different pressures in discharge tubes

Act. 8.1

explain that cathode ray tubes allowed the manipulation of a stream of charged particles

perform an investigation to demonstrate and identify properties of cathode rays using discharge tubes: – containing a Maltese cross – containing electric plates – with a fluorescent display screen – containing a glass wheel

Act. 8.2

analyse the information gathered to determine the sign of the charge on cathode rays

Act. 8.2

solve problem and analyse information using: F = qvB sin θ F = qE and V E= d

162, 164

157

identify that moving charged particles in a magnetic field experience a force

164

identify that charged plates produce an electric field

161

xiv

Stage 6 Physics syllabus grid describe quantitatively the force acting 164 on a charge moving through a magnetic field: F = qvB sin θ discuss qualitatively the electric field strength due to a point charge, positive and negative charges and oppositely charged parallel plates

160

describe quantitatively the electric field 161 due to oppositely charged parallel plates outline Thomson’s experiment to measure the charge/mass ratio of an electron

165

outline the role of: – electrodes in the electron gun – the deflection plates or coils – the fluorescent screen – in the cathode ray tube of conventional TV displays and oscilloscopes

167

2. The reconceptualisation of the model of light led to an understanding of the photoelectric effect and black body radiation Students learn to:

Page

Students:

Page

describe Hertz’s observation of the effect of a radio wave on a receiver and the photoelectric effect he produced but failed to investigate

182

perform an investigation to demonstrate the production and reception of radio waves

Act. 9.1

outline qualitatively Hertz’s experiments 175 in measuring the speed of radio waves and how they relate to light waves

identify data sources, gather, process and analyse information and use available evidence to assess Einstein’s contribution to quantum theory and its relation to black body radiation

Act. 9.2

identify Planck’s hypothesis that radiation emitted and absorbed by the walls of a black body cavity is quantised

179

identify data sources, gather, process and present information to summarise the use of the photoelectric effect in photocells

184 Act. 9.3

identify Einstein’s contribution to quantum theory and its relation to black body radiation

179

solve problems and analyse information using: E = hf and c = f λ

181 Act. 9.3

explain the particle model of light in terms of photons with particular energy and frequency

179

process information to discuss Einstein and Planck’s differing views about whether science research is removed from social and political forces

Act. 9.4

identify the relationships between 179 photon energy, frequency, speed of light and wavelength: E = hf and c = f λ

xv

Stage 6 Physics syllabus grid 3. Limitations of past technologies and increased research into the structure of the atom resulted in the invention of transistors Students learn to:

Page

Students:

Page

identify that some electrons in solids are shared between atoms and move freely

189

perform an investigation to model the behaviour of semiconductors, including the creation of a hole or positive charge on the atom that has lost the electron and the movement of electrons and holes in opposite directions when an electric field is applied across the semiconductor

Act. 10.1

describe the difference between conductors, insulators and semiconductors in terms of band structures and relative electrical resistance

189

gather, process and present secondary information to discuss how shortcomings in available communication technology lead to an increased knowledge of the properties of materials with particular reference to the invention of the transistor

Act. 10.2

identify absences of electrons in a 191 nearly full band as holes, and recognise that both electrons and holes help to carry current

identify data sources, gather, process, analyse information and use available evidence to assess the impact of the invention of transistors on society with particular reference to their use in microchips and microprocessors

Act. 10.2

compare qualitatively the relative number of free electrons that can drift from atom to atom in conductors, semiconductors and insulators

190

identify data sources, gather, process and present information to summarise the effect of light on semiconductors in solar cells

Act. 10.3

identify that the use of germanium in early transistors is related to lack of ability to produce other materials of suitable purity

199

describe how ‘doping’ a semiconductor can change its electrical properties

193

identify differences in p and n-type semiconductors in terms of the relative number of negative charge carriers and positive holes

193

describe differences between solid state and thermionic devices and discuss why solid state devices replaced thermionic devices

199

4. Investigations into the electrical properties of particular metals at different temperatures led to the identification of superconductivity and the exploration of possible applications Students learn to:

Page

Students:

Page

outline the methods used by the Braggs 208 to determine crystal structure

process information to identify some of the metals, metal alloys and compounds 211 that have been identified as exhibiting the property of superconductivity and their critical temperatures

identify that metals possess a crystal lattice structure

209

perform an investigation to demonstrate magnetic levitation

Act. 11.1

describe conduction in metals as a free movement of electrons unimpeded by the lattice

209

analyse information to explain why a magnet is able to hover above a superconducting material that has reached the temperature at which it is superconducting

Act. 11.1

209 identify that resistance in metals is increased by the presence of impurities and scattering of electrons by lattice vibrations

gather and process information to describe how superconductors and the effects of magnetic fields have been applied to develop a maglev train

Act. 11.1

describe the occurrence in 215 superconductors below their critical temperature of a population of electron pairs unaffected by electrical resistance

process information to discuss possible applications of superconductivity and the 219 effects of those applications on computers, generators and motors and Act. transmission of electricity through power grids 11.1

discuss the BCS theory

215

discuss the advantages of using superconductors and identify limitations to their use

217

xvi

Stage 6 Physics syllabus grid

Module 4 From Quanta to Quarks 1. Problems with the Rutherford model of the atom led to the search for a model that would better explain the observed phenomena Students learn to:

Page

Students:

Page

discuss the structure of the Rutherford model of the atom, the existence of the nucleus and electron orbits

230, 244

perform a first-hand investigation to observe the visible components of the hydrogen spectrum

Act. 12.1

analyse the significance of the hydrogen spectrum in the development of Bohr’s model of the atom

236

process and present diagrammatic information to illustrate Bohr’s explanation of the Balmer series

236 Act. 12.1

define Bohr’s postulates

236

solve problems and analyse information using: 1 1  1 = R 2 − 2  λ ni   nf

233, 245 Act. 12.1

discuss Planck’s contribution to the concept of quantised energy

231

analyse secondary information to identify the difficulties with the RutherfordBohr model, including its inability to completely explain: – the spectra of larger atoms – the relative intensity of spectral lines – the existence of hyperfine spectral lines – the Zeeman effect

Act. 12.2

describe how Bohr’s postulates led to the development of a mathematical model to account for the existence of the hydrogen spectrum: 1 1  1 = R 2 − 2  λ n n  f i 

237, 244

discuss the limitations of the Bohr model of the hydrogen atom

239

2. The limitations of classical physics gave birth to quantum physics Students learn to:

Page

Students:

Page

describe the impact of de Broglie’s proposal that any kind of particle has both wave and particle properties

250, 259

solve problems and analyse information using: h λ= mv

249, 258

define diffraction and identify that interference occurs between waves that have been diffracted

250, 257

gather, process, analyse and present information and use available evidence to assess the contributions made by Heisenberg and Pauli to the development of atomic theory

255 Act. 13.1

describe the confirmation of de Broglie’s 251, proposal by Davisson and Germer 257 explain the stability of the electron orbits in the Bohr atom using de Broglie’s hypothesis

253, 257

xvii

Stage 6 Physics syllabus grid 3. The work of Chadwick and Fermi in producing artificial transmutations led to practical applications of nuclear physics Students learn to:

Students:

Page

define the components of the nucleus 261, (protons and neutrons) as nucleons and 278 contrast their properties

Page

perform a first-hand investigation or gather secondary information to observe radiation emitted from a nucleus using Wilson Cloud Chamber or similar detection device

Act. 14.1

discuss the importance of conservation laws to Chadwick’s discovery of the neutron

261, 275

solve problems and analyse information to calculate the mass defect and energy released in natural transmutation and fission reactions

267, 277

define the term ‘transmutation’

263

describe nuclear transmutations due to natural radioactivity

263

describe Fermi’s initial experimental observation of nuclear fission

269

discuss Pauli’s suggestion of the existence of neutrino and relate it to the need to account for the energy distribution of electrons emitted in β-decay

266, 276

evaluate the relative contributions of electrostatic and gravitational forces between nucleons

261

account for the need for the strong nuclear force and describe its properties

262

explain the concept of a mass defect using Einstein’s equivalence between mass and energy

267

describe Fermi’s demonstration of a controlled nuclear chain reaction in 1942

270, 275

compare requirements for controlled and uncontrolled nuclear chain reactions

271, 275

4. An understanding of the nucleus has led to large science projects and many applications Students learn to:

Page

Students:

Page

explain the basic principles of a fission reactor

280, 298

gather, process and analyse information to assess the significance of the Manhattan Project to society

280 Act. 15.1

describe some medical and industrial applications of radioisotopes

283, 298

identify data sources, and gather, process, and analyse information to describe the use of: – a named isotope in medicine – a named isotope in agriculture – a named isotope in engineering

284, Act. 15.2

describe how neutron scattering is used 272, as a probe by referring to the properties 298 of neutrons identify ways by which physicists 286, continue to develop their understanding 299 of matter, using accelerators as a probe to investigate the structure of matter discuss the key features and components of the standard model of matter, including quarks and leptons

xviii

292, 298

Stage 6 Physics syllabus grid

Module 5 Medical Physics 1. The properties of ultrasound waves can be used as diagnostic tools Students learn to:

Page

Students:

Page

identify the differences between ultrasound and sound in normal hearing range

305

solve problems and analyse information to calculate the acoustic impedance of a range of materials, including bone, muscle, soft tissue, fat, blood and air and explain the types of tissues that ultrasound can be used to examine

312

describe the piezoelectric effect and 308 the effect of using an alternating potential difference with a piezoelectric crystal

gather secondary information to observe at least two ultrasound images of body organs

Act. 16.1

define acoustic impedance: Z = ρυ and identify that different materials have different acoustic impedances

310, 311

identify data sources and gather information to observe the flow of blood through the heart from a Doppler ultrasound video image

Act. 16.2

describe how the principles of acoustic impedance and reflection and refraction are applied to ultrasound

311

identify data sources, gather, process and analyse information to describe how ultrasound is used to measure bone density

315 Act. 16.3

define the ratio of reflected to initial intensity as:

310

solve problems and analyse information using: Z = ρυ and

310, 311

2

I r Z2 − Z 1  = Io Z + Z  2  2 1

2

I r Z2 − Z 1  = Io Z + Z  2  2 1

identify that the greater the difference in acoustic impedance between two materials, the greater is the reflected proportion of the incident pulse

310

describe situations in which A scans, B scans and sector scans would be used and the reasons for the use of each

312

describe the Doppler effect in sound waves and how it is used in ultrasonics to obtain flow characteristics of blood moving through the heart

315

outline some cardiac problems that can 316 be detected through the use of the Doppler effect

2. The physical properties of electromagnetic radiation can be used as diagnostic tools Students learn to:

Page

Students:

Page

describe how X-rays are currently produced

321

gather information to observe at least one image of a fracture on an X-ray film and X-ray images of other body parts

Act. 17.1

compare the differences between ‘soft’ and ‘hard’ X-rays

322

gather secondary information to observe a CAT scan image and compare the information provided by CAT scans to that provided by an X-ray image for the same body part

Act. 17.1

explain how a computed axial tomography (CAT) scan is produced

326

perform a first-hand investigation to demonstrate the transfer of light by optical fibres

Act. 18.1

describe circumstances where a CAT scan would be a superior diagnostic tool compared to either X-rays or ultrasound

329

gather secondary information to observe internal organs from images produced by an endoscope

Act. 18.1

explain how an endoscope works in relation to total internal reflection

334

discuss differences between the role of coherent and incoherent bundles of fibres in an endoscope

336

explain how an endoscope is used in: – observing internal organs – obtaining tissue samples of internal organs for further testing

337

xix

Stage 6 Physics syllabus grid 3. Radioactivity can be used as a diagnostic tool Students learn to:

Page

Students:

Page

outline properties of radioactive isotopes and their half-lives that are used to obtain scans of organs

340, 343, 344

perform an investigation to compare an image of bone scan with an X-ray image

Act. 19.1

describe how radioactive isotopes may be metabolised by the body to bind or accumulate in the target organ

344

gather and process secondary information to compare a scanned image of at least Act. one healthy body part or organ with a scanned image of its diseased counterpart 19.2

identify that during decay of specific radioactive nuclei positrons are given off

342

discuss the interaction of electrons and 342 positrons resulting in the production of gamma rays describe how the positron emission 349 tomography (PET) technique is used for diagnosis

4. The magnetic field produced by nuclear particles can be used as a diagnostic tool Students learn to:

Students:

Page

identify that the nuclei of certain atoms 355 and molecules behave as small magnets

Page

perform an investigation to observe images from magnetic resonance image (MRI) scans, including a comparison of healthy and damaged tissue

Act. 20.1

identify that protons and neutrons in 354 the nucleus have properties of spin and describe how net spin is obtained

identify data sources, gather, process and present information using available evidence to explain why MRI scans can be used to: – detect cancerous tissues – identify areas of high blood flow – distinguish between grey and white matter in the brain

Act. 20.1

explain that the behaviour of nuclei with a net spin, particularly hydrogen, is related to the magnetic field they produce

355

gather and process secondary information to identify the function of the Act. electromagnet, radio frequency oscillator, radio receiver and computer in the MRI 20.1 equipment

describe the changes that occur in the orientation of the magnetic axis of nuclei before and after the application of a strong magnetic field

355

identify data sources, gather and process information to compare the advantages and disadvantages of X-rays, CAT scans, PET scans and MRI scans

Act. 20.2

define precessing and relate the frequency of the precessing to the composition of the nuclei and the strength of the applied external magnetic field

356

gather, analyse information and use available evidence to assess the impact of medical applications of physics on society

Act. 20.3

discuss the effect of subjecting precessing nuclei to pulses of radio waves

357

explain that the amplitude of the signal 359 given out when precessing nuclei relax is related to the number of nuclei present explain that large differences would 360 occur in the relaxation time between tissue containing hydrogen bound water molecules and tissues containing other molecules

xx

Stage 6 Physics syllabus grid

Module 6 Astrophysics 1. Our understanding of celestial objects depends upon observations made from Earth or from space near the Earth Students learn to:

Page

discuss Galileo’s use of the telescope to 371 identify features of the Moon Act. 21.1 discuss why some wavebands can be more easily detected from space

373

define the terms ‘resolution’ and ‘sensitivity’ of telescopes

375

discuss the problems associated with ground-based astronomy in terms of resolution and absorption of radiation and atmospheric distortion

373, 378

Students:

Page

identify data sources, plan, choose equipment or resources for, and perform an investigation to demonstrate why it is desirable for telescopes to have a large diameter objective lens or mirror in terms of both sensitivity and resolution

377 Act. 21.2

outline methods by which the resolution 378, and/or sensitivity of ground-based 380 systems can be improved, including: – adaptive optics – interferometry – active optics

2. Careful measurement of a celestial object’s position in the sky (astrometry) may be used to determine its distance Students learn to:

Page

Students:

Page

define the terms parallax, parsec, light-year

388

solve problems and analyse information to calculate the distance to a star given its trigonometric parallax using: 1 d = p

Act. 22.1

explain how trigonometric parallax can be used to determine the distance to stars

388

gather and process information to determine the relative limits to trigonometric parallax distance determinations using recent ground-based and space-based telescopes

Act. 22.2

discuss the limitations of trigonometric parallax measurements

389

3. Spectroscopy is a vital tool for astronomers and provides a wealth of information Students learn to:

Page

Students:

Page

account for the production of emission and absorption spectra and compare these with a continuous black body spectrum

390

perform a first-hand investigation to examine a variety of spectra produced by discharge tubes, reflected sunlight, or incandescent filaments

Act. 22.3

describe the technology needed to measure astronomical spectra

390

analyse information to predict the surface temperature of a star from its intensity/ Act. wavelength graph 22.4

identify the general types of spectra produced by stars, emission nebulae, galaxies and quasars

393

describe the key features of stellar spectra and describe how these are used to classify stars

395

describe how spectra can provide information on surface temperature, rotational and translational velocity, density and chemical composition of stars

393

xxi

Stage 6 Physics syllabus grid 4. Photometric measurements can be used for determining distance and comparing objects Students learn to:

Page

Students:

Page

define absolute and apparent magnitude

398

solve problems and analyse information using:

400

M = m − 5 log and IA = 100 (m IB

B

d  10

– mA)/5

to calculate the absolute or apparent magnitude of stars using data and a reference star explain how the concept of magnitude can be used to determine the distance to a celestial object

399

perform an investigation to demonstrate the use of filters for photometric measurements

Act. 22.5

outline spectroscopic parallax

401

identify data sources, gather, process and present information to assess the impact of improvements in measurement technologies on our understanding of celestial objects

Act. 22.6

explain how two-colour values (i.e. colour index, B – V) are obtained and why they are useful

401

describe the advantages of photoelectric technologies over photographic methods for photometry

397

5. The study of binary and variable stars reveals vital information about stars Students learn to:

Page

Students:

Page

describe binary stars in terms of the means of their detection: visual, eclipsing, spectroscopic and astrometric

411

perform an investigation to model the light curves of eclipsing binaries using computer simulation

Act. 23.1

explain the importance of binary stars in determining stellar masses

408

solve problems and analyse information by applying: 4 π 2r 3 m 1+ m 2 = GT 2

420

classify variable stars as either intrinsic 413 or extrinsic and periodic or non-periodic explain the importance of the period– luminosity relationship for determining the distance of cepheids

xxii

416

Stage 6 Physics syllabus grid 6. Stars evolve and eventually ‘die’ Students learn to:

Page

Students:

Page

describe the processes involved in stellar formation

423

present information by plotting Hertzsprung–Russell diagrams for: – nearby or brightest stars – stars in a young open cluster – stars in a globular cluster

Act. 24.1

outline the key stages in a star’s life in terms of the physical processes involved

428

analyse information from an HR diagram and use available evidence to determine 437 the characteristics of a star and its evolutionary stage

describe the types of nuclear reactions involved in Main-Sequence and postMain Sequence stars

425, 430

present information by plotting on a HR diagram the pathways of stars of 1, 5 and 10 solar masses during their life cycle

discuss the synthesis of elements in stars by fusion

425, 430

explain how the age of a globular cluster can be determined from its zero-age main sequence plot for a HR diagram

433

explain the concept of star death in relation to: – planetary nebula – supernovae – white dwarfs – neutron stars/pulsars – black holes

429, 431

437

xxiii

1 Context

Figure 1.0.1 The knowledge of how things move through space, influenced by gravity, has transformed the way we work, play and think.

2

Space

Modern physics was born twice. The first time (arguably) was in the 17th century when Newton used his three laws of motion and his law of universal gravitation to connect Galileo’s equations of motion with Kepler’s laws of planetary motion. Then early in the 20th century, when many thought physics had almost finished the job of explaining the universe, it was unexpectedly born again. Einstein, in trying to understand the nature of light, proposed the special and general theories of relativity (and simultaneously helped launch quantum mechanics). Space was the common thread—Kepler, Galileo, Newton and Einstein were all trying to understand the motion of objects (or light) through space. Newton’s laws of mechanics and his theory of gravitation led to space exploration and artificial satellites for communication, navigation and monitoring of the Earth’s land, oceans and atmosphere. Einstein’s theory of relativity showed that mass and energy are connected, and that length, mass and even space and time are rubbery. Relativity has come to underlie most new areas of physics developed since then, including cosmology, astrophysics, radioactivity, particle physics, quantum electrodynamics, anything involving very precise measurements of time and the brain-bending ‘string theory’. So, whenever you use the global positioning system (GPS), consult Google maps, check the weather report or make an international call on your mobile phone, remember that the technology involved can be traced directly back to physics that started 400 years ago.

Figure 1.0.2 The revolution in our

Inquiry activity

understanding of the universe started with the humble question of how projectiles move.

Go ballistic! The path through the air of an object subject only to gravity and air resistance, is called a ballistic trajectory. If the object is compact and its speed is low, then air resistance is negligible and its trajectory is a parabola. Investigate parabolic trajectories using a tennis ball, an A4 piece of paper, a whiteboard or a blackboard and a digital camera. 1 On a board about 2 m wide, draw an accurate grid of horizontal and vertical lines 10 cm apart. 2 With a firmly mounted camera, take a movie of a tennis ball thrown slowly in front of the board. Try different angles and speeds to get eight or more frames with the ball on screen, and get as much of a clear parabolic shape (including the point of maximum height) as you can. 3 Using video-editing software, view the best movie, frame by frame, on a computer. If your software allows it, create a single composite image with all the ball’s positions shown on one image, to show the parabolic trajectory. 4 If you can’t do that, then for each frame, on the board, and using the grid, estimate the x- and y-coordinates of the ball’s centre to the nearest 5 cm or better. Some video software allows you to read the x- and y-coordinates (in pixels) by clicking on the image. 5 Plot a graph of x versus y to produce a graph of the parabolic trajectory. The graph might be a bit irregular because of random error in reading the blackboard scale. 6 Video the trajectory of a loosely crumpled-up piece of A4 paper. Now air resistance is NOT negligible. Does the trajectory still look like an ideal parabola?

3

1

Cannonballs, apples, planets and gravity What goes up must come down

projectile, trajectory, parabola, ballistics, vertical and horizontal components, Galilean transformation, range, launch angle, time of flight, inverse square law, law of universal gravitation, universal gravitation constant G, gravitational field g, test mass, central body, density, gravimeter, low Earth orbit, gravitational potential energy, escape velocity, gravitationally bound

One of the powers of physics is that it enables us to find connections between seemingly unconnected things and then use those connections to predict new and unexpected phenomena. What started as separate questions about the shape of the path of cannonballs through the air and the speed of the Moon’s orbit around the Earth eventually led to the law of gravitation. This explained how the solar system works, but also led to the development of artificial satellites and spacecraft for the exploration of the solar system.

1.1 Projectile motion Up and down, round and round Before Galileo Galilei (1564–1642), it was a common belief that an object such as a cannonball projected through open space (a projectile) would follow a path (trajectory) through the air in a nearly straight line until it ran out of ‘impetus’ and then drop nearly straight down in agreement with the ideas of Aristotle. However, through experiments (Figure 1.1.1) in which he rolled balls off the edge of a table at different speeds and then marked the position of collisions with the ground, Galileo demonstrated that the trajectory of a falling ball is actually part of a parabola (see Figure 1.1.2). Remember that a parabola is the shape of the graph of a quadratic equation. The immediate result of Galileo’s discovery was that the art of firing cannonballs at your enemies became a science (ballistics). However, there were also more far-reaching, constructive consequences.

Figure 1.1.1 Galileo’s laboratory notes on his experiments showing that projectiles follow parabolic paths 4

space

Vertical displacement

Opponents of Copernicus’ heliocentric universe claimed that if the Earth was rotating and orbiting the Sun, then a person jumping vertically into the air would have the ground move under their feet, so that they would land very far away from where they started. Galileo argued that a person jumping from a moving Earth is like a projectile dropped by a rider on a horse (representing the Earth) moving with a constant velocity (Figure 1.1.3). From the rider’s point of view, the projectile would appear to drop vertically, straight to the ground, accelerating downwards the whole time. A bystander who is stationary relative to the ground would see the rider, horse and projectile whoosh past and, like any other projectile, the dropped object would appear to follow a parabolic trajectory. Figure 1.1.2 Galileo argued that the parabolic motion of the projectile was made up of two separable parts: its accelerating vertical motion as seen by the rider, and its constant horizontal velocity (which is the same as that of the horse). Recall from your Preliminary physics course that these two contributions to velocity are called vertical and horizontal components (see in2 Physics @ Preliminary section 2.2, p 26). Galileo then argued that the Earth doesn’t zoom away under your feet because at the moment you jump upwards you already have the same horizontal component of velocity as the Earth’s surface. Relative to the Earth’s surface, your horizontal velocity is zero and so you land on the same spot. In connecting the two problems of projectile motion and a moving Earth, Galileo developed two important new concepts. The first is the idea that the parabolic trajectory of a projectile can be divided into vertical and horizontal components. The second is the idea of measuring motion relative to another moving observer (or ‘frame of reference’). The formula vB (relative to A) = vB – vA (see in2 Physics @ Preliminary, p 8) is used to transform velocities relative to different frames of reference. This formula is sometimes called the Galilean transformation.

Components of a trajectory The ideal parabolic trajectory is an approximation that works under two conditions: 1 Air resistance is negligible (gravity is the only external force). 2 The height and range (horizontal displacement) of the motion are both small enough that we can ignore the curvature of the Earth. a

Horizontal displacement

This graph of a parabolic trajectory shows the vertical and horizontal components of displacement separately. The projectile positions are plotted at equal time intervals.

Describe Galileo’s analysis of projectile motion.

Describe the trajectory of an object undergoing projectile motion within the Earth’s gravitational field in terms of horizontal and vertical components.

b

Figure 1.1.3 Trajectory of the rider’s projectile as seen by (a) the rider and (b) an observer on the ground 5

1

Cannonballs, apples, planets and gravity The first condition is true for compact and low-speed projectiles. The second is true in almost all human-scale situations, typically at or near the Earth’s surface. Let’s analyse an example of ideal projectile motion. Recall that the acceleration due to gravity is g = 9.8 m s–2 (see in2 Physics @ Preliminary section 1.3). Here we are going to write it as a vector g. Clearly its direction is downwards. Consider the trajectory of a ball. We start by separating the horizontal and While the ball is in the air, the only vertical components of its motion. external force on it is gravity acting downwards, so there is a constant vertical acceleration ay = g, illustrated by the changing vertical spacing of projectile positions plotted at equal time intervals in Figure 1.1.2. The net horizontal force is zero, so, consistent with Newton’s first law, horizontal velocity is constant (ax = 0), which is clear from the equal horizontal spacing of the projectile positions plotted at equal time intervals in Figure 1.1.2. We can recycle the kinematics (SUVAT) equations from the Preliminary course. (See in2 Physics @ Preliminary section 1.3.) (4) s = vt (1) s = ut + 1  at  2 2 u+v s= t (2) (5) v 2 = u2 + 2as 2 v = u + at (3)

PRACTICAL EXPERIENCES Activity 1.1

Activity Manual, Page 1

Table 1.1.1 

Equations of projectile motion

Horizontal components  

Vertical components

 

ux = u cos θi

uy = u sin θi

vx = ux

vy = uy + gt

∆x = uxt

1 2 gt  ∆y = uyt + —

vx 2 = ux 2

vy 2 = uy 2 + 2g∆y

2

75°

90°

60° 45°

Here we need to apply them separately to the vertical (y) and horizontal (x) components of motion. Instead of displacement s, we’ll use ∆x = xf – xi for horizontal displacement and ∆y = yf – yi for vertical displacement. We’ll put subscripts on the initial and final vertical velocities (uy and vy for example). We only need to use SUVAT equations 3, 4 and 5. θi is the launch angle (between Remember to adjust the sign the initial velocity u and the horizontal axis). of g to be consistent with your sign convention. In problems involving gravity, up is normally taken as positive, making the vector g negative (i.e. g = –9.8 m s–2). In the syllabus, vx2 = ux2 is included for completeness; but is unnecessary, as it can be derived from vx = ux. Some properties of ideal parabolic trajectories are: • At the maximum height of the parabola, vertical velocity vy = 0. • The trajectory is horizontally symmetrical about the maximum height position. • The projectile takes the same time to rise to the maximum height as it takes to fall back down to its original height. • For horizontal ground, initial speed = final speed. • Maximum possible height occurs for a 90° launch angle. The maximum possible range (for horizontal ground) occurs for a 45° launch angle (Figure 1.1.4). • Independent of their initial velocity, all objects projected horizontally from the same height have the same time of flight as one dropped from rest from the same height, because they all have a zero initial vertical velocity (Figure 1.1.5).

30° 15°

Figure 1.1.4 For a fixed initial speed, maximum range occurs for a 45° angle of launch and maximum height occurs for a 90° angle of launch. 6

space

100 mm

Ballistics is a drag

A

ir resistance or ‘drag’ introduces deceleration in both the vertical and horizontal directions, distorting the ballistic trajectory from an ideal parabola. As a projectile becomes less compact, air resistance increases relative to weight. The range decreases, the trajectory becomes less symmetrical, and the final angle becomes steeper. The launch angle for maximum range decreases. In extreme cases (for example, a loosely crushed piece of paper), the trajectory seems to approach Aristotle’s prediction: it moves briefly in a nearly straight line and then drops nearly vertically. no air resistance

Figure 1.1.5 Multiflash photo of two falling objects. All horizontally projected objects have the same time of flight as an object dropped from rest from the same height.

Target practice You now have all the equations you need to ‘do some damage’, so let’s launch some projectiles. Safety warning!  The following worked example may seem dangerously long because it illustrates several alternative methods of solving projectile problems rolled into one.

increasing air resistance

Figure 1.1.6 The effect of increasing air resistance

Worked example Question You throw a ball into the air (Figure 1.1.7). You release the ball 1.50 m above the ground, with a speed of 15.0 m s–1, 30.0° above horizontal. The ball eventually hits the ground. Answer the following questions, assuming air resistance is negligible. a For how long is the ball in the air before it hits the ground (time of flight)? b What is the ball’s maximum height? c What is the ball’s horizontal range? d With what velocity does the ball hit the ground?

Solve problems and analyse information to calculate the actual velocity of a projectile from its horizontal and vertical components using: v x2 = u x2 v = u + at vy2 = uy2 + 2ay ∆y ∆x = uxt 1 ∆y = uyt + ayt2 2

1.50 m

Figure 1.1.7 Throwing a ball into the air 7

1

Cannonballs, apples, planets and gravity

Solution Always draw a diagram! Divide the motion into vertical (y) and horizontal (x) components. Choose the origin to be the point of release, so xi = yi = 0. This is not always the most convenient choice of origin. ↑ Use the sign convention + → & + . Components of initial velocity u (Figure 1.1.8): ux = +u cos θi = +15.0 cos 30.0° = +13.0 m s–1 uy = +u sin θi = +15.0 sin 30.0° = +7.50 m s–1 The only external force is gravity so vertical acceleration is g = –9.80 m s–2. There is no horizontal force, therefore ax = 0 m s–2 (constant horizontal velocity). fin

al

yu

ial

t

ini

θi = 30.0°

it loc

ve

uy

ve

vy

loc

ity

v

θf

ux

vx

Figure 1.1.8 Components of initial and final velocities a The ball hits the ground when vertical displacement ∆y = –1.50 m.

Find final vertical velocity: vy 2 = uy  2 + 2g ∆y = 7.502 + 2 × –9.80 × –1.50 = 85.65

vy =

85.65 = 9.255 m s–1 (must be downwards), so vy = –9.255 m s–1

vy = uy + gt = –9.255 = +7.50 + (–9.80) × t  −9. 255 − 7. 50 Rearrange, solve: t= = 1.71 s −9 .80 The ball hits the ground 1.71 s after being thrown. Find t :

1 Alternative method using the quadratic formula  ∆y = uyt +  gt 2 = –1.50 m 2 1 2 Substitute, rearrange: 1.50 + 7.50 × t +  × –9.80 × t   = 0 2 −7.5 ± 7.502 + 4 × 4.90 × 1.50 = –0.179 s or  +1.71 s −2 × 4.90



Quadratic, solve for t : t =



Two solutions: Reject the physically irrelevant negative solution, so t = 1.71 s.

b At maximum height, vertical velocity vy = 0, so use vy 2 = uy 2 + 2g ∆y. 0 = uy2 + 2g∆ymax = 7.502 + 2 × (–9.80) × ∆ymax 7.502 Rearrange, solve: ∆ymax = = +2.87 m above the point of release, 2 × 9.80 so height above ground = 2.87 m + 1.50 m = 4.37 m above the ground.

Alternative method

1 Use vy = uy + gt  to find the time t  when vy = 0, then use ∆y = uyt + gt 2 to find 2 vertical displacement.

c From part a, we know the time of flight t = 1.71 s. 8

Horizontal displacement in this time is:  ∆x = uxt = +13.0 m s–1 × 1.71 s = +22.2 m = 22.2 m (to the right)

space d x-component of final velocity:  vx = +13.0 m s–1

y-component of final velocity:  vy = –9.255 m s–1 (down) (from part a)



To find magnitude, use Pythagoras’ theorem (see Figure 1.1.8): v = v x2 +v y2 = 132 + 9.2552 = 15.96 ≈ 16.0 m s–1 v y 9.25 Direction: tan θf = , so θf = 35.4° down from horizontal = v x 13.0 Alternative magnitude calculation





Negligible air resistance, ∴ mechanical energy = kinetic energy + gravitational potential energy and is conserved (see in2 Physics @ Preliminary section 4.2). Near the Earth’s surface, gravitational potential energy U = mgh. Using the ground as h = 0: Ki + Ui = Kf + Uf 1 2 1 Cancel m: mv + mghi = mvf2 + mghf 2 i 2 1 1 Substitute: 15.02 + 9.80 × 1.50 = vf2 + 0 2 2



Rearrange, solve:  v f = 15.02 + 2 × 9.80 × 1.50 = 15.94 ≈ 15.9 m s–1



This is the same as for the previous method within the three-figure precision of the calculation, but doesn’t tell us the direction.

In the previous example, time of flight was determined by the vertical However, if the component—the flight ended when the ball hit the ground. projectile hits a vertical barrier such as a wall, then the time of flight is determined by the horizontal component.

Worked example Question Suppose you kick a ball at 22.0 m s–1, 20.0° above the horizontal, towards a wall 21.0 m away (Figure 1.1.9). Ignore air resistance and the ball’s radius. a What is the ball’s time of flight (before hitting the wall)? b At what height does the ball hit the wall? c Is that the greatest height reached by the ball?

Solution

Solve problems and analyse information to calculate the actual velocity of a projectile from its horizontal and vertical components using: v x2 = u x2 v = u + at vy2 = uy2 + 2ay ∆y ∆x = uxt 1 ∆y = uyt + ayt 2 2

Figure 1.1.9 The ball hits the wall. ↑ Choose the origin to be the initial position, so xi = yi = 0. Use the sign convention + and + →. ux = 22.0 cos 20.0° (right) = +20.7 m s–1 uy = 22.0 sin 20.0° (up) = +7.52 m s–1 9

1

Cannonballs, apples, planets and gravity a The ball hits the wall when the horizontal displacement ∆x = +21.0 m.

Substitute:  ∆x = uxt = +21.0 m = +20.7 m s–1 × t +21.0 m

= 1.014 s ≈ 1.01 s +20.7 m s−1 1 b The ball hits the wall at a height (vertical displacement) of ∆y = uyt + gt 2. 2



Rearrange, solve:  t =



Substitute, solve:  ∆y = +7.52 × 1.014 + 0.5 × –9.80 × 1.0142 = +2.587



The ball hits the wall ≈ 2.59 m above ground.

c Check if the ball reaches maximum height of the parabola before hitting the wall.

Time of flight = 1.01 s. vy = 0 at maximum height of parabola.



Find the time taken to reach maximum height.



Substitute:  vy = 0 = uy + gt = +7.52 + –9.80 × t



Rearrange, solve:  t =



7.52 = 0.767 s  which is less than time of flight 9.80 The ball would reach the maximum height of the parabola before hitting the wall, therefore the final height is NOT the maximum height for the trajectory.

Checkpoint 1.1 1 2 3 4 5 6

Determine the horizontal acceleration of a projectile in flight. Determine its vertical acceleration. (Assume negligible air resistance.) What angle of launch gives maximum horizontal range? What angle of launch gives the maximum possible height? (Assume negligible air resistance.) What is another name for air resistance? If you throw a ball horizontally from the roof, and drop another at the same time, which one will hit the ground first? Describe the two conditions that must apply so that a trajectory is a parabola. List the 8 equations used in calculations of projectile motion. Explain why at least one of them is unnecessary.

1.2 Gravity In Ptolemy’s universe, the Sun, Moon and planets each had a separate clockworklike mechanism to keep it in motion. Copernicus and Kepler greatly improved the picture, but Isaac Newton finally showed there was a single mechanism for them all—the force of gravity. The calculations of parabolic trajectories in section 1.1 work well close to the Earth’s surface where g is constant. However, if we’re going to venture out into space, we can’t use these simple equations. We need to look at the force of gravity on a larger scale.

Newton’s law of universal gravitation Newton assumed several properties of gravity (see in2 Physics @ Preliminary section 13.5): • All ‘massive’ objects (that is, objects with mass) attract each other. The larger the masses, the larger the force. 10

space • Like light intensity, the magnitude of the force decreases with distance according to the inverse square law (see in2 Physics @ Preliminary sections 6.1 and 15.1). However, astronomer Ismael Boulliau had suggested this before him. • The law of gravitation is universal—it applies throughout the universe and is responsible for the orbits of all the planets and moons. All this is expressed mathematically as the law of universal gravitation: FG = G

m1m2 d2

where FG is the magnitude of the force of gravitational attraction between Define Newton’s Law of Universal Gravitation: two masses m1 and m2 and d is the distance between their centres of mass (see in2 Physics @ Preliminary section 3.6). The universal gravitational constant G mm F =G 1 2 (‘big G ’) is 6.67 × 10–11 N m2 kg–2 in SI units. It should not be confused with d2 –2 ‘little g’, 9.8 m s . More properties of gravitation: • The direction of the force acts along the line joining the centres of the two masses and is always attractive. • The formula is strictly correct for point masses and spheres, but works well for non-spheres. • The formula must be modified if one mass penetrates the surface of the other—gravity would not approach infinity if you were to burrow towards the centre of the Earth. You can see the feeble force of gravity acting • The resultant force on a mass due to the presence of other between objects in your garage. John Walker’s masses is the vector sum of the individual forces on the first Fourmilab website describes step by step how mass due to each of the other individual masses. you can perform a crude version of the

Try this!

Slightly attractive

Worked example Question Calculate the gravitational force between the Earth and the Moon. Data: Earth’s mass mE = 5.97 × 1024 kg

Moon’s mass mM = 7.35 × 1022 kg



Average Earth–Moon distance dEM = 3.84 × 108 m



Universal gravitational constant G = 6.67 × 10–11 N m2 kg–2

Solution

FG = G

=

Cavendish experiment in your own garage (see Physics Focus ‘How to weigh the Earth’ at the end of this chapter), using commonly found household items and a video camera. If you’re feeling too lazy to do it yourself, you can just download sped-up videos of the experiment in progress.

mEmM dEM 2

6.67 × 10–11 × 5.97 × 1024 × 7.35 × 1022 (3.84 × 108 )2

= 1.98 × 1020 N

Figure 1.2.1 Cavendish apparatus at home

11

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Cannonballs, apples, planets and gravity Now let’s try an example with more than two masses.

Worked example Moon

Question Earth

spacecraft

Figure 1.2.2 A spacecraft in the Earth–Moon system FSE θ

A 1000 kg spacecraft is in the vicinity of the Earth–Moon system. The spacecraft is at the origin, the Moon is on the positive y‑axis and the Earth is on the positive x‑axis (Figure 1.2.2). Given that the Earth–spacecraft and Moon–spacecraft distances are 3.82 × 108 m and 3.91 × 107 m respectively, calculate the resultant gravitational force on the spacecraft. Data: Earth’s mass mE = 5.97 × 1024 kg

Moon’s mass mM = 7.35 × 1022 kg



Universal gravitational constant G = 6.67 × 10–11 N m2 kg–2

Solution Force due to Moon: FSM = G

FSM

Fres



=



spacecraft

Figure 1.2.3 Gravitational force vector diagram. Note; This does not resemble the position vector diagram in Figure 1.2.2.

=

d SM 2

6.67 × 10–11 × 1000 × 7.35 × 1022

Force due to Earth: FSE = G θ

mSmM

mSmE d SE 2

(3.91 × 107 )2

6.67 × 10–11 × 1000 × 5.97 × 1024 (3.82 × 108 )2

Don’t underestimate the power of boredom Boredom part 1 ored? Don’t just write graffiti—try revolutionising physics! In 1665, an outbreak of bubonic plague around London closed Cambridge University, so Isaac Newton (aged 23) escaped for 2 years to his mother’s farm. He was not a very good farmer, so he fended off his city-boy boredom by inventing calculus and using prisms to show that white light is actually a mixture of colours (the spectrum). To top this off, when he saw an apple fall off his mother’s tree, he wondered if the force accelerating the apple downwards was also responsible for keeping the Moon orbiting the Earth. So he began formulating his theory of gravitation. His mathematics professor was so impressed that a couple of years after Newton returned to Cambridge, he resigned and handed his professorship to Newton. After this initial investigation, it took Newton another 20 years to fully develop and finally publish his law of universal gravitation. 12

= 2.729 N (+x direction)

Magnitude of resultant:  Fres = 3.2072 + 2.7292 = 4.21 N 3.207 , so θ = +49.6° from the x‑axis Direction:  tan θ = 2.729

PHYSICS FEATURE

B

= 3.207 N (+y direction)

1. The history of physics

3. Applications and uses of physics

Figure 1.2.4 Graffiti carved on a stone at the King’s School in Grantham, England, by Isaac Newton, then about 10 years old

space

Weight and gravitational fields As far as we know, the universal gravitational constant G is a fundamental constant, unchanging with position or time. But the acceleration due to gravity g is different on other astronomical bodies, at different heights and even at different positions on the Earth’s surface. Recall that weight w = mg is defined as the force on an object due to gravity (see in2 Physics @ Preliminary section 3.2); in other words, FG = w = mg. ‘Little g’, the acceleration due to gravity, can also be thought of as the strength of the gravitational field. However, the word weight is usually reserved for the case in which the gravitational field is due to a body of astronomical size, such as a planet. Any massive object can be described as being surrounded by a gravitational field, a region within which other objects experience an attractive force. Just as for electrical and magnetic fields (see in2 Physics @ Preliminary sections 10.6, 12.3 and 12.4), we can draw diagrams of gravitational field lines (Figure 1.2.6). The arrows on the field lines around a mass, point in the direction of the force acting on another (normally much smaller) test mass. Gravitational field is a vector (g). The density of the field lines at any particular point in space represents g, the magnitude of the field at that point, and the direction of the field lines represents the direction of this vector. Field lines run in radial directions from point masses or spherical masses. Using a small test mass m, let’s derive g, the magnitude of the gravitational field due to a planet of mass M. The weight w of the test mass is defined as the force on m due to the planet’s gravity; that is: mM w = mg = FG = G d2

PRACTICAL EXPERIENCES Activity 1.2

Activity Manual, Page 5

Describe a gravitational field in the region surrounding a massive object in terms of its effects on other masses in it. Define weight as the force on an object due to a gravitational field.

Boredom part 2

I

t is said that, at age 17, Galileo was attending church and, bored, was watching a lantern swing from the ceiling. Using his pulse as a stopwatch, he observed that the oscillation period of a pendulum barely changed as its amplitude gradually decreased. Back at home he started experiments confirming that the oscillation period depends on pendulum length L, but not at all on mass and only slightly on amplitude. He proposed (correctly) that pendulums could be used to create the first accurate mechanical clocks. We now know that, consistent with Galileo’s observations, for a simple mass-on-string pendulum the formula for oscillation period T is: T = 2π

L g

The formula is an approximation, but if the maximum swing angle is less than 15° from vertical, the formula is correct within 0.5%. With this formula and a pendulum, you can measure the value of ‘little g’, which varies slightly between locations around the world.

Figure 1.2.5 Young Galileo watches a swinging lantern in Pisa cathedral. 13

1

Cannonballs, apples, planets and gravity Divide both sides by test mass m:

a

g=

FG M =G 2 m d

Newton’s equation for gravitational force is symmetrical—you can choose either mass as the test mass and calculate the field around the other and still get the same magnitude of force when you multiply them together because of Newton’s third law (see in2 Physics @ Preliminary section 3.5)—the two masses are an action–reaction pair. However, if one of the masses is much larger (such as a planet), it is more convenient to calculate the field around it and use the smaller mass as the test mass. In astronomical situations where one of the bodies (such as a planet or star) is very much larger, the larger body is sometimes called the central body. Because of its large mass, the central body experiences negligible gravitational accelerations compared with a small test mass. Strictly speaking, the acceleration g is the acceleration of the test mass towards the common centre of mass of the whole system of two masses. However, if the central body is much larger than the test mass, we can ignore its acceleration, so g effectively becomes the acceleration of the test mass towards the central body. Gravitational field is a vector, so when calculating the resultant field due to several bodies, the approach is identical to calculating the resultant gravitational force due to several bodies—calculate the field due to each individual mass and then find the vector sum of the fields.

b

Worked example Question Figure 1.2.6 Gravitational field lines around the Earth (a) on an astronomical scale and (b) near the surface

Calculate gE the magnitude of the gravitational field at the Earth’s surface. Data: Earth’s mass mE = 5.97 × 1024 kg

Earth’s radius rE = 6.37 × 106 m



Universal gravitational constant G = 6.67 × 10–11 N m2 kg–2

Solution

gE =

GM E d2

The test mass is at the Earth’s surface, ∴ d = rE Substitute:

gE =

6.67 × 10–11 × 5.97 × 1024 (6.37 × 106)2

= 9.81 m s–2

This should be a very familiar result.

PRACTICAL EXPERIENCES Activity 1.3

Activity Manual, Page 11

14

Variations in gravitational field Newton’s gravitation equation says that the magnitude of a planet’s gravitational field depends on the mass of the planet and decreases with distance from the planet’s centre. For example, on Earth, the value of g is 0.28% lower at the top of Mt Everest than at sea level. Also, because the Earth has a slightly larger radius near the equator than at the poles (the ‘equatorial bulge’), g is slightly lower at the equator. Except at the poles, there is an additional (fictitious) decrease in g

space

I

nte

ractiv

e

measurements that gets more severe as one approaches the equator. Because of the Earth’s rotation, the (downward) centripetal acceleration (see in2 Physics @ Preliminary section 2.3) of the ground appears to be subtracted from the true value of g. In fact this centripetal effect is responsible for the formation of the equatorial bulge, which was predicted by Newton before it was measured. The Sun and Moon also exert a weak gravitational force on objects at the Earth’s surface, so the magnitude and direction of g vary slightly, depending on the positions of the Sun and Moon. Variation in g caused by the positions of the Sun and Moon relative to the oceans is responsible for the pattern of tides. Strictly speaking, Newton’s gravitation equation written in the form above assumes that the planet is a perfectly uniform sphere. Close to the surface of a planet, local deviations from uniform density can result in small local changes in the magnitude and direction of g. The magnitude will be slightly larger than average when measured on the ground above rock (such as iron ore) of high density ρ (mass per unit volume) and lower above rock containing low-density minerals (such as salt or oil), an effect exploited by geologists in mineral exploration. The Earth’s crust is less dense than the mantle, so variations in thickness of the crust also affect g. Variation in g is measured using a gravimeter, the simplest kind being an accurately known mass suspended from a sensitive spring balance. Variations in g on larger distance scales around the Earth can be measured using satellites orbiting in low Earth orbit. Deviations in the orbital speed of satellites indicate that, in addition to the equatorial bulge, Earth is also slightly pear-shaped—pointier at the North Pole than the South Pole.

M o d u le

Hooke’s law

I

saac Newton had enemies, and Robert Hooke (1635–1703) was probably his greatest. They argued bitterly over (among other things) who first suggested the inverse square law for gravity. Hooke was an accomplished experimental physicist, astronomer, microscopist, biologist, linguist, architect and inventor. He is best remembered for the discovery of (biological) cells and the invention of the spring balance (see in2 Physics @ Preliminary section 3.2), which exploits Hooke’s law F = ‑k x. The force F exerted by a spring is proportional to x, the change in spring length. The ‘spring constant’ k is a measure of the spring’s stiffness. A calibrated spring balance can measure weight, and, if used with an accurately calibrated mass, it can be used as a gravimeter to measure g.

haviour of springs

notes on the be Figure 1.2.7 Hooke’s

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Cannonballs, apples, planets and gravity

Checkpoint 1.2 1 2 3 4 5 6

Write down Newton’s law of universal gravitation. Define weight. What part of Newton’s formula for gravitational force is responsible for the inverse square law behaviour? What are two names for the quantity g? List three factors responsible for (real) variations in g around the Earth. Outline the differences between G and g.

1.3 Gravitational potential energy We’ve already mentioned gravitational potential energy (GPE) U = mgh (see in2 Physics @ Preliminary section 4.1) in part d of the worked example accompanying Figure 1.1.7. This formula for GPE is an approximation that only works close to the Earth’s surface, where g is very nearly constant. It’s good enough for projectile motion but, as you now know, g decreases with distance, so we need a more accurate formula to understand energy on an astronomical scale.

Work and GPE Explain that a change in gravitational potential energy is related to work done.



Define gravitational potential energy as the work done to move an object from a very large distance away to a point in a gravitational field: mm EP = −G 1 2 r

For clarity we’ll use the symbol EP instead of U to denote gravitational potential energy calculated using the more accurate formula, even though the two symbols are really interchangeable. Potential energy is energy stored by doing work against any force (such as gravity) that depends only on position; therefore, gravitational potential energy EP is energy stored by doing work against the force of gravity. It can be shown (using calculus to derive the work done against gravity by changing the separation of two masses) that: E P = −G

m1m2 r

where m1 and m2 are two masses separated by a displacement (or separation) r and G is the universal gravitational constant. Note that EP is always negative and EP approaches zero as displacement r approaches infinity (Figure 1.3.1). for a separation r is the work that would need to be done by a force opposed to gravity in moving the masses together, starting at ‘infinite’ separation where EP = 0 and bringing them to a separation of r (with no net change in speed). Equivalently, EP is the work done by gravity while the masses are moved apart, starting at a separation of r to a position of ‘infinite’ separation (with no The gravitational potential energy does not depend on net change in speed). the path taken by the masses to get to their final positions; it depends only on the final separation r. The formula isn’t affected by the choice of which mass to move, although normally we treat a large mass such as the Sun or a planet as an immoveable central body and the smaller mass as a moveable test mass. The formula seems to imply that EP approaches negative infinity as the test mass approaches the centre of a planet. However, this formula no longer applies in this form once one mass penetrates the surface of the other. 16

space

+GmtmE r E2

FG

0

rE

2rE

3rE

4rE

5rE separation

–GmtmE rE

EP

Figure 1.3.1 Plots of gravitational force (FG) and gravitational potential energy (EP) versus separation

between a test mass mt and the Earth mE, starting at one Earth radius rE. The vertical FG and EP axes are not drawn to the same scale.

Worked example Question A piece of space junk of mass mJ drops from rest from a position of 30 000 km from the Earth’s centre. Calculate the final speed vf it attains when it reaches a height of 1000 km above the Earth’s surface. Assume that above 1000 km, air resistance is negligible. Data: Earth’s mass mE = 5.97 × 1024 kg

Earth’s radius rE = 6.37 × 106 m



Universal gravitational constant G = 6.67 × 10–11 N m2 kg–2

Solution Air resistance is negligible, so total mechanical energy (kinetic + potential energy) is conserved. Assume that because of the enormous mass of the Earth, its change in velocity is negligible. Use the Earth as the frame of reference. Don’t forget to convert to SI units.

Cancel mJ: Substitute:

K i + EPi = Kf + EP m m mm 1 1  mJvi2 – G J E  =  mJvf2 – G J E   ri rf 2 2 0 – 6.67 × 10–11 ×

5.97 × 1024

1 2 5.97 × 1024 –11 = − 6.67 × 10 × v f 30.0 × 106 2 (6.37 + 1.00) × 106

 10−6 10−6  − Rearrange, solve: v f = 2 × 6.67 × 10–11 × 5.97 × 1024 ×    7.37 30.0  = 9030 m s−1 = 9.03 km s−1 Note that this result doesn’t depend on m J.

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Cannonballs, apples, planets and gravity

Escape velocity: what goes up …? Outline Newton’s concept of escape velocity.

Isaac Newton showed that what goes up doesn’t necessarily come down. Normally, if one fires a projectile straight up, the object will decelerate until its velocity changes sign and it falls back down. However, if a projectile’s initial velocity is high enough, the 1/d 2 term in the gravity equation will cause the acceleration g to decrease with height too rapidly to bring the projectile to a stop so it will never turn back—it can ‘escape’ the planet’s gravitational field. The minimum velocity that allows this is called the escape velocity. Strictly speaking, it’s really a speed, because the initial direction of the projectile isn’t critical. Newton treated the projectile as a cannonball (with no thrust) so that, other than the initial impulse from the cannon, the only force acting on it is gravity. He conceived escape velocity using his force equation, and the escape velocity formula can be derived from it. However, a more modern derivation using energy is easier and similar to the previous worked example. Let m be the mass of a projectile, M the mass of a planet, ve the initial speed and r the initial position (the planet’s radius if you are on the surface). Assume air resistance is negligible, so total mechanical energy (KE + GPE) is conserved (see in2 Physics @ Preliminary section 4.2). Ki + EPi = Kf + EPf The escape velocity represents the minimum limiting case where the projectile ‘just reaches infinite displacement’ with zero speed; in other words, Kf = EPf = 0. 1 GmM mve 2 − =0+0 2 r

Rearrange, cancel m:

ve =

Explain the concept of escape velocity in terms of the: – gravitational constant – mass and radius of the planet.

2GM r

If the initial speed is greater than this, the projectile will maintain a non-zero speed even as it approaches infinite displacement. Note that the escape velocity depends only on the planet’s mass and the projectile’s starting position r but not on the projectile’s mass. You may be puzzled that in the above derivation, the total mechanical energy (sum of KE and GPE) was exactly zero. This means that the escaping projectile has just enough (positive) KE to overcome its negative potential energy. When the mechanical energy is less than zero, there is not enough KE to overcome the GPE and the two masses are said to be gravitationally bound. When the total mechanical energy ME > 0, the KE can overcome the GPE and the two bodies are no longer bound together. This concept of binding also applies to the other three fundamental forces (including electromagnetism, which binds electrons to the nucleus of an atom). The escape velocity from the Earth’s surface is: 2 × 6.67 × 10 –11 × 5.97 × 1024 6.37 × 10

18

6

= 11 200 m s −1 = 11.2 km s −1

space This idealised escape velocity needs to be modified when applied to real spacecraft. First, the derivation ignores air resistance in the atmosphere (hundreds of kilometres thick), which would increase the escape velocity. Second, in a real rocket, engines produce an extra force—thrust—that can accelerate a craft to a higher altitude where the escape velocity is lower. It also ignores other sources of gravitational fields such as the Sun, Moon and planets. The escape velocity for a projectile under the gravitational influence of more than one body is given by: ve total = ve12 + ve22 + where ve total is the escape velocity for the total system and ve1, ve2 … are the escape velocities from the individual bodies within the system, calculated for the projectile using the same starting position in space.

Ultimate frisbee

W

as the first artificial object to leave the solar system a giant steel frisbee? In the 1950s, the US started testing nuclear bombs underground, to minimise atmospheric nuclear fallout. In 1957, during Operation Plumbbob in the Pascal-B test, a nuclear bomb was detonated at the bottom of a 150 m shaft sealed with concrete and a 900 kg, 10 cm thick steel cap. The steel cap fired upwards at enormous speed and was never seen again. Before the test, it was estimated that an extreme upper limit for the speed of the steel cap would be 67 km s–1. This is well above the escape velocity for the whole solar system (43.6 km s–1 from Earth), starting an urban myth that it beat the Voyager probes (launched in 1977) out of the solar system. A later, more realistic, estimate suggested that, at most, the cap had a speed of 1.4 km s–1, reaching an altitude of less than 95 km.

Checkpoint 1.3 1 2 3 4 5

Define under what circumstances it is suitable to use the simplified formula U = mgh for gravitational potential energy (GPE). Write down the more accurate formula for GPE. What limit does GPE approach as the separation of the two masses approaches infinity? On what factors does Newton’s idealised escape velocity depend? What other factors affect escape velocity in realistic situations?

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Cannonballs, apples, planets and gravity

PRACTICAL EXPERIENCES CHAPTER 1

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 1.1: Projectiles Perform a first-hand investigation, gather information and analyse data to calculate initial and final velocity, maximum height reached, range and time of flight of a projectile for a range of situations by using simulations, data loggers and computer analysis.

A ball is rolled down a ramp, whose dimensions will be known to you. Predict where the ball will land. Equipment: aluminium track, ball bearing, metre ruler, measuring tape, shoe. ball bearing

track

ruler

Figure 1.4.1 Equipment set-up for this activity

Discussion questions 1 List assumptions you have made in order to make an estimate of the range. 2 Assess how reliable is your method. 3 Explain how changing the original angle of the ramp will affect the range of the ball.

Activity 1.2: Determining the value of acceleration due to gravity Perform an investigation and gather information to determine a value for acceleration due to gravity using pendulum motion or computer-assisted technology and identify reason(s) for possible variations from the value 9.8 m s–2.

Use the motion of a pendulum to gather data to determine the acceleration due to gravity. Equipment: pendulum (string and mass), retort stand and clamp, stopwatch, metre ruler, data logger.

retort stand

string

mass

Figure 1.4.2 Pendulum apparatus set-up

20

space Discussion questions 1 Explain what you did in order to make the experiment reliable. 2 Galileo originally thought that the period of the pendulum did not depend at all on the amplitude of the swing. Is this true? Explain how you can take this into account in your experiment. 3 How does your value compare with the accepted value? 4 Outline another method that would allow you to achieve the same aim.

Activity 1.3: Gravity— Out of this world Use the spreadsheet template to gather appropriate information to help you predict the acceleration due to gravity at the surface of other planets.

Figure 1.4.3   Spreadsheet template Process the information you have gathered using the spreadsheet template. Complete the template to calculate the values of acceleration due to gravity on other planets. Discussion questions 1 Determine which planet has the largest value for acceleration due to gravity at its surface. (Note that the gas giants Jupiter, Saturn, Uranus and Neptune don’t have a well-defined boundary between the atmosphere and a solid planet surface. The visible ‘surface’ is fluid, i.e. gas and/or liquid.) 2 Identify the factors that affect the acceleration due to gravity.

Gather secondary information to predict the value of acceleration due to gravity on other planets. Present information and use available evidence to discuss the factors affecting the strength of the gravitational force.



Analyse information using the expression: F = mg to determine the weight force for a body on Earth and for the same body on other planets.

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•

•

•

• •

•

•

•

Chapter summary

If air resistance (drag) is negligible and g is very nearly constant (for example near a planet’s surface), then the trajectory of a projectile is a parabola. The formula for transforming velocity within one frame of reference into one relative to another frame of reference is called the Galilean transformation: vB (relative to A) = vB – vA Parabolic projectile motion can divided into vertical and horizontal components. The vertical component has a downward acceleration of g and the horizontal component has a constant velocity. In parabolic projectile motion, the equations of motion are: Horizontal components: ux = u cos θi, vx = ux, ∆x = uxt, vx2 = ux2 Vertical components: uy = u sin θi, vy = uy + gt, ∆y = uyt + 1 gt2, 2 vy2 = uy2 + 2g∆y For horizontal ground, the maximum possible range occurs for a 45° launch angle. The maximum possible height occurs for a 90° launch angle. All objects projected horizontally from a particular height have the same time of flight as one dropped from rest from the same height. If a trajectory ends when the projectile hits the ground, time of flight is determined by the vertical component. If the projectile hits a vertical barrier, then time of flight is determined by the horizontal component. Newton’s law of universal gravitation: mm FG = G 1 2 2 d

•

•

•

•

• •

Gravitational acceleration (g) towards a central body such as a planet is also called its gravitational field. It depends on the central body mass M and the distance d from its centre: M g =G 2 d The force of gravity on an object in that field is called its weight: w = mg. Gravitational field g measured near the Earth’s surface varies slightly with distance from the Earth’s centre and density of the surrounding material. The centripetal acceleration of the Earth’s surface also decreases measured values of g (only an apparent effect). Gravitational potential energy (GPE) is the work done by a force opposing gravity in moving masses together starting at ‘infinite’ separation and bringing them to a separation of r (with no net change in speed). The simple formula for GPE (U = mgh) is an approximation that only works at or near the surface of a planet. The more accurate expression is: mm E P = −G 1 2 r EP approaches zero as separation of the two masses approaches infinity. The minimum initial velocity that a projectile needs to have in order to escape a planet’s gravitational field is called escape velocity:



ve =

2GM r

Review questions Physically speaking Complete each definition by using a keyword taken from the list at the beginning of the chapter. To approach infinite distance from a massive central body, a projectile must start with _________________ . The path of a projectile is known as a _________________ . The formula for converting velocities between frames of reference is the _________________ . A projectile’s maximum horizontal displacement is its _________________ . Universal gravitation and the intensity of light both follow the _________________ . Close to the Earth’s surface and subject only to gravity, a projectile’s path is a _________________ . The acceleration of a _________________  near the central body equals the gravitational field. Close to the Earth’s surface, all objects projected horizontally from the same height have the same_________________ . A _________________  is apparatus used to assist in mineral exploration. If drag is negligible, then a projectile’s range is determined only by initial velocity and _________________ .

22

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Reviewing 1 Given that the Earth rotates, account for why when you jump straight up, you land on the same spot.

2 The high jump and the long jump both involve a run-up and then a jump. Using ideas from projectile motion, briefly compare and contrast the ideal characteristics of the run-up and jump for the two sports.

3 A projectile takes 1.25 s to reach its maximum height. What is its time of

Solve problems and analyse information to calculate the actual velocity of a projectile from its horizontal and vertical components using: vx 2 = ux 2 v = u + at vy 2 = uy 2 + 2ay∆y 1 ∆x = uxt; ∆y = uyt + ayt 2 2

flight, assuming the ground is horizontal and drag is negligible?

4 Explain why (assuming negligible air resistance) all objects projected horizontally from the same height have the same time of flight as an object dropped from that height, regardless of their initial speed.

5 Predict what would happen to the magnitude of the gravitational force between two masses: a if one of the masses were doubled b if both masses were doubled c if the distance between the masses were doubled.

6 Describe how (and explain why) g would differ slightly from average at a point on the Earth’s surface above an oil deposit.

7 You’ve seen diagrams of electrical field lines around positive charges in which



Solve problems and analyse information using: mm F =G 1 2 d2

the arrows point outwards (see in2 Physics @ Preliminary section 10.6). Briefly discuss the possibility of a planet with gravitational field lines that point outwards. Propose how you would expect a test mass to behave there.

8 Without doing a calculation, deduce the speed at which a meteorite would hit the Earth’s surface if it started from rest at a very large distance from the Earth. Justify your answer. Ignore air resistance and gravity of other astronomical bodies. (Hint: The value is one already calculated elsewhere in this chapter.)

9 Read the definition of gravitational potential energy EP in section 1.3 page 16. Explain why it is necessary to specify in the definition that the work is done with no net change in speed. (Hint: What other form of energy is involved?)

Solving problems 10 Repeat the calculation in the worked example accompanying Figure 1.1.7, assuming that the ball lands on the flat roof of a 2.5 m high garage, instead of the ground.

11 Consider the worked example accompanying Figure 1.1.9. Keeping everything unchanged except initial speed: a What would the initial speed of the ball need to be if the ball hit the wall when it was just at its maximum possible height? What would be its time of flight? b What would the initial speed of the ball need to be if the ball hit the ground just in front of the wall? What be would its time of flight?

Solve problems and analyse information to calculate the actual velocity of a projectile from its horizontal and vertical components using: vx 2 = ux 2 v = u + at vy 2 = uy 2 + 2ay∆y 1 ∆x = uxt; ∆y = uyt + ayt 2 2

12 By considering the vertical component of velocity and ignoring air resistance,



derive an expression (containing initial speed u and launch angle θ) for the time taken for a projectile near the Earth’s surface to reach its maximum height. Then show that the time of flight for a projectile fired over horizontal ground is given by: 2u t= sin θ g

23

1

Cannonballs, apples, planets and gravity

13 A marble rolls horizontally off the edge of a 1.00 m high table with a speed of 3.00 m s–1. Calculate the speed with which it hits the ground, by: a using the equations of projectile motion b assuming the conservation of mechanical energy (using the simple version of the equation for GPE).

Solve problems and analyse information using: mm F =G 1 2 d2



14 Repeat the calculation in the worked example accompanying Figure 1.2.2, with the positions of the Moon and Earth swapped.

15 Using your own mass, calculate the maximum force of gravity exerted by

the planet Mars (m = 6.42 × 1023 kg) on you, given that the closest approach of Mars to Earth is approximately 5.6 × 1010 m. How close would you need to stand to the centre of mass of a 10 tonne truck for the magnitude of the gravitational force it exerts on you to be the same? (1 tonne = 1000 kg)

16 Show that g is 0.28% lower on top of Mt Everest (8848 m) than at sea level. Data: Mean Earth radius r E = 6.367 × 106 m.

17 Calculate the change in GPE in moving a 10 kg object from an initial position 1000 km above the surface of the Earth to a final position at a distance from the Earth equivalent to the mean orbital radius of the Moon (r = 3.84 × 108 m). Assume the Moon is on the opposite side of its orbit at the time and you can ignore its gravitational effect.

18 Using the data and answer from Question 17, calculate the speed at which you would need to project the 10 kg object radially outwards from the initial position so that it would just reach the final position, stop and fall back to Earth. (You can ignore air resistance above an altitude of ~1000 km.)

19

a Calculate the velocity required for a projectile to escape the Sun’s gravitational field (m Sun = 1.99 × 1030 kg) if launched from the orbital radius of the Earth (1.50 × 1011 m), if the Earth and other planets weren’t there. b Using part a and Earth’s escape velocity (11.2 km s–1), show that the total escape velocity from the solar system for a projectile launched from Earth is 43.6 km s–1. Assume the projectile doesn’t pass near other planets.

Extension Solve problems and analyse information to calculate the actual velocity of a projectile from its horizontal and vertical components using: v x2 = u x2 v = u + at vy2 = uy2 + 2ay∆y 1 ∆x = uxt; ∆y = uyt + ayt2 2

20 By considering the horizontal component of displacement for a projectile and your answer for Question 12, derive an expression (containing initial speed u and launch angle θ) for the horizontal range. Either by using calculus or by considering the properties of trigonometric functions, show that the maximum range is attained for a launch angle of 45°.

21 A wildlife reserve ranger needs to hit a monkey in a tree with a tranquiliser dart in order to capture and examine it. The barrel of the dart gun is pointing exactly at the monkey. The angle between the barrel of the dart gun and the horizontal is not 90°. At the instant the ranger fires, the monkey is startled and drops from rest to the ground below.

Re

24

iew

Q uesti o

n

s

v



Show that the dart will hit the monkey. (Hint: Show that by the time the dart reaches the horizontal position of the monkey, both the dart and the monkey have the same vertical position. Assume that air resistance is negligible.)

space

PHYSICS FOCUS How to weigh the Earth

a

b

torsion wire

1. The history of physics Solve problems and analyse information using: mm F =G 1 2 d2 Analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth.

Newton first tested his law of universal gravitation by showing that gravity was responsible for both the acceleration of a falling apple (9.8 m s–2) and the centripetal acceleration (see in2 Physics @ Preliminary section 2.3) of the orbiting Moon. However, he didn’t know the Earth’s mass ME or the value of G, but by using ratios of acceleration and distance squared, he showed that GME = ad 2 is the same for an apple and the Moon, confirming that the same law of gravity applied to both. In 1798, the Earth’s mass was finally measured. Henry Cavendish (1731–1810) (who discovered hydrogen) measured the average density of the Earth ρE to be 5.448 times denser than water (1000 kg m–3). The experiment was designed by John Michell (1724 –1793) (who first predicted the existence of black holes). Since the Earth’s radius was accurately known, this was equivalent to both ‘weighing the Earth’ and measuring the value of G. Cavendish used an extremely sensitive ‘torsional balance’ (Figure 1.4.4) to measure the tiny gravitational attraction between two small lead spheres m (attached to a thin 1.86 m rod) and two nearby large lead spheres M. From the angle of twist θ in the calibrated torsion wire, he determined the gravitational force between the spheres. From this, and using Newton’s equation for gravitational force, he calculated the Earth’s average density. Vibrations, temperature variations and slight air movements would disturb the apparatus, so it was built into a small sealed building, with Cavendish outside, operating the apparatus via cords and pulleys, and making observations through telescopes in the walls.

F m

M

M

θ m

F

r

Figure 1.4.4 (a) Schematic and (b) cutaway view of the apparatus used by Cavendish to ‘weigh the Earth’

The shift in position of the smaller masses was about 4 mm. 1 Because of Earth’s gravitational field, the Moon must accelerate towards Earth. Why doesn’t the Moon crash into Earth? 2 Using Cavendish’s value for Earth’s density ρ,

the definition  ρ = 

mass and the mean radius volume

rE = 6.37 × 106 m, calculate the Earth’s mass and compare it with the modern value. 3 Using Newton’s universal gravitation equation, Cavendish’s value for mE, the modern values for rE and g = 9.8 m s–2, calculate G and compare it with the modern value. 4 Using the modern value for G, calculate the total gravitational force between the spheres measured by Cavendish. (Hint: Calculate the force between a small and a large sphere in a single pair and double it. Ignore the thin rod etc.) Distance between sphere centres r = 0.225 m, large sphere mass M = 158 kg, small sphere mass m = 0.73 kg 5 Typical laser printer paper weighs 0.080 kg m–2. Calculate the size (in mm) of a square piece of printer paper that would have a weight on Earth equivalent to the force in Question 4.



25

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Explaining and exploring the solar system Getting up there

propellant, impulse, exhaust velocity, reaction device, thrust, payload, g-force, effectively weightless, lift-off, Kepler’s laws, satellite, ellipse, orbital velocity, eccentric, semimajor axis, periapsis, apoapsis, perihelion, aphelion, perigee, apogee, hyperbola, closed or stable orbit, geosynchronous, geostationary, medium Earth orbit, semi-synchronous, gravity assist, slingshot effect, re-entry, orbital decay, drag, lift, supersonic, hypersonic, shock wave, heat shield, ablation

How many times have you been told to ‘stop dreaming and be practical’? For scientists and engineers, both dreams and practical know-how were potent tools to turn the understanding of the physics of gravity and motion into the technology of space travel. Most of the important pioneers of rocketry were inspired to pursue dreams of space travel by reading Jules Verne’s (1828–1905) story From the Earth to the Moon, or the stories of HG Wells (1866–1946). But they also had a solid grounding in physics and engineering.

2.1 Launching spacecraft In his book Philosophiae Naturalis Principia Mathematica (Principia for short), Newton used his law of gravity and laws of motion to explain Kepler’s laws of planetary motion, but also predicted the launching of artificial satellites and projectiles capable of escaping Earth’s gravity. Once you understand the physics behind something, it becomes possible to create new technology. In the case of space flight, it took 300 years to release the potential buried within Newton’s equations, via the 2000-year-old Chinese technology of fireworks.

A bite-size history of rocketry

Figure 2.1.1 The Apollo 11 mission: the launch of a Saturn 5 booster—the largest rocket in history—on its way to deliver the first humans to the Moon 26

For most of the history of rocketry, starting with the invention by the Chinese of gunpowder (the first rocket fuel or propellant) sometime between 300 bc e and 850 c e , the technology was driven mainly by military applications. The Chinese invented the first rockets or ‘fire arrows’ (fireworks tied to arrows). Some of the early milestones of this history are summarised in Table 2.1.1 in the Physics Feature ‘Fire Arrows’ on page 29. Only in the 20th century were civilian and scientific applications of rocketry (space exploration, Earth monitoring and communications) finally considered to be potentially as important as the military ones.

space Here we’ll concentrate on the important rocket researchers of the 20th century, the period in which the most rapid scientific advances took place. Below is a list their most important contributions.

PRACTICAL EXPERIENCES Activity 2.1

Activity Manual, Page 14

Konstantin Tsiolkovsky (1857–1935) Tsiolkovsky (also Tsiolkovskii), a Russian mathematics teacher, derived the basic rocketry equations including the ‘Tsiolkovsky rocket equation’ (see Physics Phile ‘This is rocket science’, p 30), used Newton’s definition of escape velocity to calculate it for Earth, and proposed multi-stage rockets and steerable thrusters. He advocated the use of liquid propellants (including liquid hydrogen) because they could be controlled using valves and would give a larger impulse than solids (see in2 Physics @ Preliminary section 4.5). He also wrote science fiction, predicting space stations, and space colonies using biological recycling of food and oxygen and airlocks for moving between a spacecraft and vacuum. Robert Goddard (1882–1945) Goddard, a US physicist, invented and tested many practical aspects of rockets, launching the first liquid-propellant rockets (liquid oxygen–gasoline) in 1926. He confirmed experimentally that rockets work in vacuum and showed that an hourglass-shaped de Laval steam nozzle greatly increased rocket efficiency. He launched the first scientific payload (camera, thermometer and barometer) that parachuted back to Earth, and steered rockets using vanes to direct exhaust gas and a gimballed (pivoted) nozzle under the automatic control of a gyroscope. He even experimented with very futuristic ion thrusters. Goddard attracted public ridicule by predicting travel to the Moon (see in2 Physics @ Preliminary Physics Phile p 43). He was mostly ignored by the US government, but he strongly influenced Oberth, von Braun and Korolyov (see below). Robert Esnault-Pelterie or REP (1881–1957) REP, a French aircraft designer, wrote on interplanetary travel, calculated the energies and flight times for trips to the Moon, Venus and Mars and proposed atomic energy to power interplanetary craft. With André Hirsch, he established the REP–Hirsch Prize for aeronautics, the first winner being Oberth (below). In 1931, Esnault-Pelterie conducted early experiments with liquid propellants (petrol–liquid oxygen, benzene–nitrogen peroxide and tetranitromethane) and developed a gimballed nozzle. Herman Oberth (1894–1989) The German physicist Oberth’s PhD thesis describing space travel was initially rejected as ‘utopian’ (though it was later accepted), so he published it as an influential book By Rocket into Planetary Space. In it he developed equations for space flight, proposed a design for a two-stage rocket using hydrogen–oxygen propellant and described craft for human space exploration. A follow-up book won him the REP–Hirsch Prize, which he used to purchase rocket engines for research assisted by his student Wernher von Braun. He worked (with von Braun) on both the Nazi V‑2 rocket program and later the American rocket program. In 1953 he published Man in Space, proposing space stations, space-based telescopes and space suits.

Figure 2.1.2 Konstantin Tsiolkovsky

Figure 2.1.3 Robert Goddard

Figure 2.1.4 Robert Esnault-Pelterie

Figure 2.1.5 Herman Oberth 27

2

Explaining and exploring the solar system

Figure 2.1.6 Wernher von Braun

Figure 2.1.7 Sergey Korolyov

Figure 2.1.8 Gerard O’Neill

28

Wernher von Braun (1912–1977) As a student, von Braun (German physicist and aeronautical engineer) tested Oberth’s rocket engines. He was an early amateur researcher in the Spaceflight Society, which was taken over by the Nazis. Under the Nazis von Braun led the team that developed the alcohol–oxygen-fuelled A4 (or V‑2) rocket used on Allied cities including London, killing and wounding thousands. After the war, he joined the US army’s nuclear missile program. He dreamed of a civilian space program. In magazines and television, he publicly promoted exploration to the Moon and Mars with permanent colonies and orbiting space stations serviced by re-usable shuttle-type craft. In 1957 the USSR launched Sputnik, the first artificial satellite, shocking the US and leading to the ‘space race’ of the 60s between the USSR and the US. In response, a civilian space agency, the National Aeronautics and Space Administration (NASA), was formed, and in 1960 von Braun became director of its Marshall Space Flight Center. He became a major figure in the race to the Moon (the Apollo missions) announced in 1961 by President Kennedy. He led the project to construct the largest rocket ever built—the Saturn 5 (Figure 2.1.1). As is well known, the US won the race to the Moon in 1969, although they spent much of the 60s catching up to many USSR space ‘firsts’. The race also led to rapid development of civilian satellites for communications, Earth surface–atmospheric monitoring and scientific space exploration. Sergey Korolyov (also Sergei Korolev) (1907–1966) Korolyov, a Ukrainian-born Russian aircraft designer, was known only as the ‘Chief Designer’ of the USSR space program—his name was kept secret until his death. He helped set up the Jet Propulsion Research Group, which launched liquid-fuelled rockets in 1933, and led to the USSR government forming the Jet Propulsion Research Institute, with Korolyov as Deputy Chief. During Stalin’s Great Purge of 1938, Korolyov was imprisoned for 6 years, then released to become a rocket designer in the nuclear missile program, where he quickly improved on the design of captured Nazi V‑2 missiles. Like his US rival von Braun, he dreamed of space travel and tried to convince his government to allow civilian projects. In 1957, he was allowed to launch the first artificial satellite Sputnik into orbit, starting the space race. He oversaw a string of space firsts (and failures): first animal (dog) in orbit, first unmanned Moon landing, first image of the unseen side of the Moon, first man and first woman in orbit, first extra-vehicular activity (space walk), first fly-pasts of Venus and Mars and more. Launch failures of four N1 boosters (rival to von Braun’s Saturn 5) and Korolyov’s death in 1966 helped to lose the race to the Moon for the USSR. Gerard O’Neill (1927–1992) O’Neill, a US physicist, invented the particle storage ring used in particle accelerators, and an early wireless computer network. He led development of a satellite positioning system—a precursor to the US global positioning system (GPS). Through conferences, papers and books, he was an energetic advocate of space travel. He proposed colonies in cylindrical spacecraft positioned at

space ‘Lagrange points’. (These are five stable locations around pairs of orbiting bodies such as Earth and Moon at which a test mass can remain indefinitely, requiring little or no thrust.) He suggested that colonists would live on the inner surface of these cylinders 3 km in radius and 20 km long. The cylinders would spin, using centripetal force, to simulate gravity, and the inside would be covered with Earth-like geography.

Identify data sources, gather, analyse and present information on the contribution of one of the following to the development of space exploration: Tsiolkovsky, Oberth, Goddard, EsnaultPelterie, O’Neill or von Braun.

PHYSICS FEATURE Fire arrows

T

he following table is a very incomplete summary of some of the highlights of the 24-century long history of rocketry.

Figure 2.1.9 The Chinese character for ‘rocket’ translates literally as ‘fire-arrow’.

1. The history of physics

3. Applications and uses of physics

4. Implications for society and the environment

Table 2.1.1  Some milestones in the pre-20th century history of rocketry 300 BCE to 850

At some time between these dates, the Chinese invent gunpowder and fireworks.

1150–1200

The Chinese develop the first rockets, ‘fire arrows’ (fireworks tied to arrows), and projectile weapons including grenades and cannons are used against invading Mongols.

1200–1300

Invading Mongols bring Chinese rocket technology to Europe and the Arabian Peninsula.

1529–1556

Conrad Haas (Austria) proposes the first designs for multi-staged rockets.

1687

Isaac Newton publishes Philosophiae Naturalis Principia Mathematica containing his three laws of motion and the law of universal gravitation. He defines escape velocity and predicts artificial satellites.

~1730

German Colonel von Geissler manufactures rockets (up to 54 kg) for warfare.

1792, 1799

Sultan Tipu (India) uses iron-cased 1 km range rockets against British troops.

1803–1806

Impressed by Tipu, Sir William Congreve (Britain) develops more accurate 3 km range rockets up to 136 kg, which were used successfully against Napoleon’s ships and against the Americans in the war of 1812.

19th century

Engineers, scientists, inventors and crackpots experiment with non-military applications of rockets.

1821 1861–1865

Rocket-propelled harpoons are used to hunt whales. Rockets are used in the American Civil War.

1865

Science fiction writer Jules Verne (France) publishes From the Earth to the Moon.

1903

Konstantin Tsiolkovsky (Russia) publishes reports in which he applies rigorous physics to rocketry and discusses the possibility of space travel.

29

2

Explaining and exploring the solar system

Forces and rockets

This IS  rocket science!

U

sing calculus and Newton’s second law, Tsiolkovsky derived his famous ‘delta v’ rocket equation: m ∆v = v e ln i mf where ∆v is the magnitude of velocity change during a rocket ‘burn’, ve is the exhaust velocity and mi and mf are the initial and final masses of the rocket (plus remaining propellant). However, a recently discovered pamphlet by William Moore showed he had derived a similar equation in 1813.

Analyse the changing acceleration of a rocket during launch in terms of the: • Law of Conservation of Momentum • forces experienced by astronauts.

To understand the forces exerted on rockets and astronauts during take-off, we first need to define some terms.

Thrust Tsiolkovsky called a rocket a reaction device. This is because the burning propellant forms hot, high-pressure exhaust gas that is forced through the nozzle at high exhaust speed ve. By Newton’s third law (see in2 Physics @ Preliminary section 3.5), the large force exerted on the exhaust gas results in a reaction (called thrust) back on the rocket, pushing it forward. You can use momentum to calculate the thrust. (See in2 Physics @ Preliminary Worked example, p 70.) Let’s analyse the [rocket + propellant + exhaust] as our system. The forces ejecting exhaust out of the nozzle are internal forces, so they can’t change the net momentum of the system; therefore, momentum gained by the ejected exhaust ∆(meve) must be cancelled by the [rocket + propellant] gaining ‘equal and opposite’ momentum. An increase in momentum (impulse) of the [rocket + propellant] implies acceleration and, hence, a force (called thrust FT). Suppose the speed of the exhaust gas ve is constant over a time period ∆t. The impulse J = FT∆t exerted on the [rocket + propellant] and the impulse ∆(meve) exerted on the exhaust gas are equal in magnitude: J = FT∆t = ∆(meve) = ve∆me Rearrange:

FT = v e

∆me ∆t

∆me is the mass of exhaust gas lost per unit time. Increasing exhaust ∆t speed ve is important in rocket design because it increases thrust FT . Even if thrust FT = mRa is constant, because the mass mR of the [rocket + propellant] is rapidly reducing, the acceleration a rapidly increases during launch.

FTwhere = ve

Worked example Question The thrust equation doesn’t only apply to rockets. A fireman holding a hose was not prepared when the water was turned on and was knocked over by the unexpected thrust. Water exited the spout with a speed 39.0 m s–1 and with a flow rate 470 L min–1.   Calculate the force that knocked him over. (Water density is 1000 kg m–3 = 1.00 kg L–1.)

Solution Mass flow rate of water = 470 L min–1 × 1.00 kg L–1/60 s = 7.833 kg s–1 ∆me Equation for thrust: FT = ve = 39.0 m s–1 × 7.833 kg s–1 = 305 N ∆t

Rocket engines and stages There are two basic kinds of rocket engine, those using solid and those using liquid propellant (Figure 2.1.10). Solid propellant engines are simpler and can achieve maximum thrust faster, but cannot be controlled once they start. Liquid propellant engines are more complicated and slower to start, but can be controlled and produce greater thrust. A multi-stage rocket can deliver a heavier payload (space cargo) because, when the propellant in each stage is finished, that stage can be jettisoned (falling into the ocean) reducing the rocket’s mass and so allowing for greater payload mass. Rockets from the ‘space-race’ era were typically liquid-propelled stages stacked 30

space vertically. More recent rockets such as the US Space Shuttle and the European Ariane 5 use a combination of liquid- and solid-propelled stages stacked side-byside. Russian rockets, such as the Proton, use only liquid propellant. Of course, multi-stage rockets are also more complicated and so have more ways to fail. In the US, smaller satellites and military missiles are launched using simpler, solid-fuel rocket engines.

a

b

fuel solid fuel– oxidiser mixture

g-force oxidiser Maybe you’ve heard that astronauts are ‘squashed by g-force’ when a rocket accelerates on take-off. Often the term ‘g-force’ is used to quantify the effects on pumps your body of accelerations experienced in a roller-coaster, car or aeroplane. It’s not combustion an accurate name, because its value is more closely related to acceleration than chamber force, but it is used extensively in aeronautics and astronautics, so we’ll give you combustion a commonly used definition. The ‘g’ refers to acceleration expressed in units of chamber g = 9.8 m s–2. The ‘force’ refers to the fact that a net external force is responsible for nozzle nozzle that acceleration and it is this force and resulting reaction forces within a body that are responsible for the effects of g-force. Sometimes the term ‘g-load’ is used instead. exhaust exhaust Let’s start with the vertical component of motion and g-force. Consider three Figure 2.1.10 Typical (a) solid-propellant situations: and (b) liquid-propellant 1 While you are sitting or standing still (not accelerating), there is no net force rocket engines on you. Your body is compressed by a pair of balanced forces—weight mg downwards, and the upward normal force N from the seat or floor. (See in2 Physics @ Preliminary p 45.) This compression causes the internal effects of weight. Your body is experiencing the compressive effect of 1 unit of Earth gravity (i.e. g-force = 1). 2 If you are accelerating upwards, such as riding the bottom of a curve on a roller-coaster or in a rocket during take-off, the net force is upwards— the normal force from the seat is larger than your weight. Your body compresses more than usual, as though you are heavier. If your net upward acceleration is 9.8 m s–2 (1g), then your body is compressed as though you are in a gravitational field of 1 + 1 = 2 units of Earth gravity (i.e. g-force = 2). 3 In free-fall (or in orbit), the normal force from the chair disappears and you Identify why the term ‘g forces’ are no longer compressed. You feel effectively weightless (see in2 Physics @ is used to explain the forces Preliminary pp 37–38), even though at typical Space Shuttle altitudes you acting on an astronaut during actually have ~90% of your weight on Earth. In this case you are experiencing launch. g-force = 0, the same effect as 0 units of Earth gravity. The term g-force usually means apparent weight divided by true weight on Earth. Apparent weight is what appears on a set of bathroom scales. Bathroom scales actually measure the magnitude of the normal force, not true weight, so, to calculate g-force, first calculate normal force. Consider vertical motion only. Weight mg is down and normal force N is up. Let av be vertical acceleration and let up be positive: Fnet = mav = N + (–mg) Apparent weight N:

N = mav + (+mg)

The apparent weight increase is caused by the increase in normal force N due to the term mav, which is simply the net force accelerating you. Vertical g-force = N mav + ( + mg) = mg mg a Vertical g-force = gv + 1 31

2

Explaining and exploring the solar system

The First Astronauts?

A

ccording to Chinese legend, in 1500 a senior bureaucrat (a Mandarin) called Wan Hu tried to launch himself into space by tying 47 gunpowder rockets to a chair. He failed to become the first astronaut by dying in the explosion at launch. In about 1806 in France, Claude Ruggieri launched a rocket containing a sheep ~300 m into the air, parachuting it back to Earth alive. The police prevented him from turning a small boy into the first astronaut by the same method.

The agv   term in the g-force formula represents the effects of your acceleration due to the net force and the +1 represents the background effects of your real weight. Notice that in free-fall your acceleration is –g, so the g-force is –1 + 1 = 0. If you are holding a rope or are strapped into a harness being pulled upwards instead of sitting or standing, then the above discussion and formulae still apply except that now your body is being stretched and the normal force is replaced by the tension force. The horizontal component of g-force, such as when you accelerate or decelerate in a car, is easier to calculate. (See in2 Physics @ Preliminary Physics Phile ‘g‑Whiz’ p 11.) As there is no horizontal component of weight mg, the g-force equation simplifies: Horizontal g-force = agh To calculate the resultant g-force, combine the vertical and horizontal components of g-force using vector addition.

Worked example Question You round the bottom of an upturning curve on a roller-coaster at a speed of 36.0 km h–1. The curve is circular with radius 5.00 m. Calculate the g-force you would experience.

Solution Analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth.

At that moment, you experience (upward) centripetal acceleration (see in2 Physics @ Preliminary section 2.3). Calculate the vertical g-force. 36 v = 36 km h−1 = m s−1 = 10 m s−1 3.6 ac = Vertical g-force =

v 2 102 = = 20.0 m s−2 R 5.00 av +20.0 +1 = + 1 = 3.04 g 9.80

One effect of g-force is that the apparently increased weight of the blood drains it from the head, affecting vision and consciousness. On average, 4 –5 g causes dimming of vision, 5–6 g visual blackout and above 6 g you experience loss of consciousness (‘g-LOC’). Much larger g-forces can be tolerated for periods of less than about 4 seconds. To increase g-force tolerance during launch, astronauts face the direction of acceleration. This orientation is called ‘eyeballs in’, because the eyeballs are effectively pushed into their sockets. Also, the seats are oriented with the head and body lying horizontally (Figure 2.1.11). In this way, g-force doesn’t easily force blood into or out of the head. Fighter pilots and astronauts also wear ‘g‑suits’ containing inflatable bladders in the trousers which squeeze blood out of the legs and back into the head. Using a powerful cannon to launch a satellite (see Newton’s thought experiment in section 2.2) would not work because of the enormous g-force from the initial explosion. 32

space

Figure 2.1.11 Gemini 3 astronauts Gus Grissom and John Young strapped into their horizontally oriented seats are being prepared for launch (1965).

Warning! The terms g-force and g-load are not SI quantities. They are informal terms and are sometimes used carelessly. Sometimes g-force and g-load are used to mean the same thing. Sometimes g-load is used to mean only the net acceleration in units of g, not including the effect of gravity. When reading g-force or g-load data, be careful to check which definition is being used.

Force during take-off Here we’ll account for the forces, accelerations and g-force experienced during a typical launch. This example is for a Space Shuttle, but the principles apply to other craft. In the Shuttle, the solid rockets boosters and the main liquidpropelled engines fire-up at the same time, while in a more traditional multistage rocket each stage fires sequentially. Figure 2.1.12 is a graph of g-force experienced by everything within the Shuttle during launch. Just before lift-off (or take-off ), the vector sum of thrust plus the force exerted by the gantry (the crane-like structure holding the rocket) plus the rocket’s weight is zero, so there is no acceleration. Because acceleration is zero, the net force on the astronaut is also zero, so the astronaut’s weight is balanced by the normal force exerted by the seat: g-force = 1 (point A). After lift-off, thrust is larger in magnitude than rocket weight plus air resistance, so the rocket (and astronaut) accelerate upwards. Now the seat exerts a normal force greater than the astronaut’s weight. The g-force is greater than 1, increasing steadily, along with acceleration as the mass of remaining propellant decreases. Note that throughout the launch process, the craft is ‘pitching over’ from vertical to horizontal motion, so the gravitational contribution to g-force is becoming progressively smaller in the direction of motion. The ~0.2 drop in g-force between points A and B is due to ‘throttle down’; air resistance-induced pressure on the Shuttle surface reaches a dangerous maximum (‘max Q’), so thrust is deliberately reduced until the atmosphere thins out. Thrust is increased again and g-force increases to between 2 and 3 (point B). As the mass of remaining propellant decreases, acceleration (and g-force) increases until fuel in the boosters (or lower stage in a traditional rocket) runs out. Acceleration decreases dramatically, but the boosters (or empty stages) are discarded (point C), and the remaining engines provide the thrust. Acceleration increases again as propellant mass decreases. To avoid the astronauts and payload

Analyse the changing acceleration of a rocket during launch in terms of the: • Law of Conservation of Momentum • forces experienced by astronauts.

33

2

Explaining and exploring the solar system external tank separation

C solid rocket

E in orbit

D accelerating up

booster separation

to orbital speed

3.5

D

3.0

B reduced

air resistance

B

A

g-force

2.5

lift-off

2.0 1.5 1.0

A

C

0.5 0

E 100

200

300

400

500

600

Time (s)

Figure 2.1.12 g-force during a typical Shuttle launch

Analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth.

Discuss the effect of the Earth’s orbital motion and its rotational motion on the launch of a rocket.

Earth’s rotation

being subjected to a dangerously high g-force, the thrust must be reduced to limit g-force to 3 g (point D). Once the rocket is in orbit, the rockets stop firing (point E). The only force acting now is weight (providing the centripetal force of orbit), so the rocket and the astronaut are both in free-fall and effectively weightless; the astronaut experiences a zero g-force.

Running start It takes a lot of fuel to get a spacecraft to a high enough altitude and high enough speed to achieve orbit. You can get higher if (like a pole vaulter) you get a ‘run-up’ before lift-off. Given that Earth rotates rapidly, a rocket already has a large easterly tangential velocity at launch. So, if you launch towards the east, you can use less propellant, carry a larger payload or go into a higher orbit. The closer you are to the equator, the faster your initial speed u (Figure 2.1.13). At the equator u = 465 m s–1.

Worked example Question Compare Earth’s rotational tangential speed vT at the rocket launch facilities at Woomera, Australia (used during the 60s and 70s), and Kourou, French Guiana.

v1 v2 tangential velocity

Figure 2.1.13 Rockets are usually launched towards the east, to take advantage of Earth’s rotation. The effect is greatest at the equator.

Data:

Earth’s radius rE = 6.37 × 106 m



Earth’s rotational period T = 86 164 s



Woomera: Latitude 31.1°S, longitude 136.8°E.



Kourou:

Solution Tangential speed: v T =

Latitude 5.2° N, longitude 52.8°W 2πR T

(See in2 Physics @ Preliminary p 29.)

Radius of rotation R depends on θlat, the latitude angle: R = rE cos θlat 34

space

vT =

2πrE 2π × 6.37 × 106 cos θlat = cos θlat = 465 cos θlat T 86164

Not enough hours in a day

Woomera: 465 cos 31.1° = 398 m s–1 Kourou:

 

 

–1

465 cos 5.2° = 463 m s

To explore the solar system, you need to reach Earth’s escape velocity. As the Earth also orbits the Sun, you can get an extra boost from Earth’s orbital speed (about 3.0 × 104 m s–1) if you launch at a time of the year when the Earth’s orbital motion points in the desired direction (Figure 2.1.14).

direction of Sun

orbital motion

T

he absolute period of Earth’s rotation as determined from the orientation of distant stars is only 23 h, 56 min and 4 s, or 86 164 s (a sidereal day). The normal 24 hour mean solar day is longer because it includes the extra time needed for the Earth’s rotation to catch up with the extra component of the Sun’s apparent motion in the sky due to the Earth’s orbit.

Earth

Figure 2.1.14 Interplanetary missions require higher launch rotational motion

velocities. These launches need to take advantage of the Earth’s orbital velocity.

Checkpoint 2.1 1 2 3 4 5 6 7 8

Which two people led the space race between the USSR and the US? What event triggered the US government to fund a large civilian rocket program? What are Lagrange points? Name the two basic kinds of rocket engine. Explain why a multi-stage rocket allows a heavier payload. What is the vertical g-force on someone: a standing stationary? b in free-fall? c in orbit? Explain why the acceleration of a firing rocket increases with time. Explain one reason why most rockets are launched towards an easterly direction.

2.2 Orbits and gravity How and why do spacecraft stay up there? Newton, using his three laws of motion and his law of universal gravitation, showed not only that gravity provided the centripetal force required to keep the Moon in orbit around the Earth, but also that he could use his laws to explain all of Kepler’s laws of planetary motion. Newton developed a thought experiment to understand orbit. He imagined standing on a mountain (Figure 2.2.1) and firing a projectile horizontally from a powerful cannon. Gravity would accelerate the projectile towards the ground,

Discuss the importance of Newton’s Law of Universal Gravitation in understanding and calculating the motion of satellites.

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Explaining and exploring the solar system curving it downwards until impact. If you increased the projectile’s initial velocity, it would travel further around the Earth and its trajectory would be less steeply curved. Eventually you would reach an initial velocity at which the curvature of the trajectory exactly matched the curvature of the Earth itself, so that the projectile would never catch up to the ground—the projectile was now in a circular orbit. The projectile would become an artificial satellite (an object in orbit around a much larger one) in the same way that the Moon is a natural satellite of Earth. Newton also showed that if you increased the velocity further, the orbit would become an ellipse. At high enough velocity (escape velocity), the projectile would never return.

Figure 2.2.1 This illustration from Newton’s Principia illustrates the principles behind orbits, and launching an artificial satellite.

PRACTICAL EXPERIENCES Activity 2.2

Activity Manual, Page 17

Vi B

A

Vf R

θ

R Vf

ΔV

θ –Vi

Figure 2.2.2 This diagram shows the derivation of the centripetal acceleration formula. Define the term orbital velocity and the quantitative and qualitative relationship between orbital velocity, the gravitational constant, mass of the central body, mass of the satellite and the radius of the orbit using Kepler’s Law of Periods.

Deriving centripetal force You have seen the formula for centripetal acceleration ac, but where does it come from? Look at Figure 2.2.2. Suppose an object moves uniformly in a circle. The magnitudes of the initial and final velocities (vi and vf ) are the same, so call them both v (v = vi = vf ). The distance travelled d = v∆t in going from A to B is R θ (using radians), giving R θ = v∆t, which can be rearranged to: v∆t θ= R To find the instantaneous acceleration rather than the average over a long time, use a ∆t (and therefore θ) that approaches zero. Consider the ∆v vector diagram on the right in Figure 2.2.2. It is an isosceles triangle with two equal sides of length v. For θ approaching zero, the length of ∆v approaches the length of the arc v θ between vf and ‑vi so: ∆v → v θ If you combine this with the previous equation, eliminating θ, then in the limit: v 2∆t ∆v = R ∆v v2 Divide both sides by ∆t : = ac = ∆t R Then of course, to get centripetal force, multiply centripetal acceleration by mass (see in2 Physics @ Preliminary p 46). Don’t forget that the magnitude of tangential velocity is v  = 2pr ⁄ T. For objects in orbit, the tangential velocity is also called the orbital velocity.

Heaven and Earth Newton showed that the force acting in the ‘heavens’ to keep the Moon in orbit was the same one acting on small projectiles on Earth. He assumed that in both cases the force is given by his law of universal gravitation. At that time, neither the Earth’s mass ME nor the value of G were known, but if you rearrange the formula: M m F = ma = G E d2 their product GME = ad 2 should evidently be the same constant for the orbiting Moon and a falling apple on Earth. Newton showed that this was true.

36

space Worked example Question Show that GME = ad 2 is the same for a falling apple and the orbiting Moon. Assume that the Moon’s orbit is circular. Data:



Earth’s radius rE = 6.37 × 106 m



Average Earth–Moon distance dEM = 3.84 × 108 m



Moon’s orbital period T = 27.3 days

Solution Apple on Earth: GME = ad 2 = 9.80 m s–2 × (6.37 × 106 m)2 = 3.98 × 1014 m3 s–2 Moon in orbit: Centripetal acceleration ac =

v r

2

2πr 4 π 2r ∴a = 2 T T Orbit radius r = dEM; T = 27.3 × 24 × 3600 = 2.359 × 106 s

Solve problems and analyse information using: mm F =G 1 2 d2 Solve problems and analyse information to calculate the centripetal force acting on a satellite undergoing uniform circular motion about the Earth using: mv 2 F = r Analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth.

Orbital velocity v =



GME = ad 2 =

4 π2dEM 3 T

2

=

4 π2 × (3.84 × 108 m)3 6

2

(2.359 × 10 s)

= 4.02 ×1014 m3 s –2

The two values agree within ~1%.

Extension: (Hard) Can you think of a reason for the ~1% discrepancy? (Hint: See Physics Phile ‘Finding new planets’ page 40.)

Kepler’s laws and satellites Johannes Kepler (1571–1630), using Tycho Brahe’s data, showed that the known planets and Earth orbit the Sun in ellipses that obey his three laws of planetary motion, but he had no idea why. An ellipse is a circle, stretched along one dimension. The more stretched the ellipse, the more eccentric it is. Most of the orbits of planets of our solar system are very nearly circular, not very eccentric. Comets have highly eccentric orbits. Figure 2.2.3 defines some properties of elliptical orbits. The semimajor axis is half the length of the ellipse’s longest axis. It can also be thought of as a kind of average radius for the orbit. For a circular orbit, the semimajor axis is the radius. The point of closest approach of the orbit to the central body is called the periapsis. The furthest point is the apoapsis. If the central body is the Sun, then these points can also be called perihelion and aphelion (helios is ‘Sun’ in Greek). If the central body is the Earth, then they are the perigee and apogee (geos is ‘Earth’ in Greek). Let a be the semimajor axis and let dA and dP be the distances from the central body to the aphelion and perihelion respectively, then look at Figure 2.2.3a and confirm that a is the average of these: a = (dA + dP)/2. Kepler’s laws of planetary motion in the You had a sneak preview of Preliminary text (see in2 Physics @ Preliminary p 250). Here they are again: 1 The orbits of the planets are ellipses, with the Sun at one focus (Figure 2.2.3a).

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2

Explaining and exploring the solar system 2 A line joining the planet to the Sun sweeps out equal areas in equal times. This means the planet travels faster when it is closer to the Sun (Figure 2.2.3b). 3 For all planets orbiting the Sun, the square of the orbital period T is proportional to the cube of the semimajor axis a. This is the law of periods. T2 a3

constant = constant

These laws hold for any orbiting system of two bodies if the mass of the central body is very much larger than the mass of the other. They also apply to circular orbits, since a circle is a special case of an ellipse. If c is the half the distance between the foci of an ellipse, then eccentricity e is defined as e = ac . A circle is an ellipse with zero eccentricity (e = 0); both foci are For a circle, the semimajor axis becomes the together at the circle centre. radius, a = r. Orbits of planets in our solar system are very nearly circular, the two most eccentric being Mercury with e = 0.2056 and Mars with e = 0.0934. Pluto has e = 0.2482 but, sadly for Pluto-fans, it was demoted to a dwarf planet in 2006. a

b

Sun

semimajor axis

focus

Discuss the importance of Newton’s Law of Universal Gravitation in understanding and calculating the motion of satellites.

perihelion (closest to the Sun)

focus

aphelion (farthest from Sun)

focus

Area

=

Area

Figure 2.2.3 (a) A highly eccentric elliptical orbit. (b) Kepler’s law of areas: a line joining the planet to the Sun sweeps out equal areas in equal times.

Newton derived Kepler’s laws from his laws of motion and gravitation. He showed that Kepler’s first law (law of elliptical orbits) follows from the inverse square law. By including his three laws of motion, he also proved Kepler’s second law (law of equal areas). These derivations are beyond the syllabus; however, showing Kepler’s third law, the law of periods, for a circular orbit is very easy. Careful! Don’t confuse the semimajor axis a with centripetal acceleration ac. Suppose a satellite of mass m orbits a central body of mass M and m  Vin; the trip around Jupiter has increased the speed of the probe (Figure 2.4.1b). Momentum is conserved, so the probe’s gain in momentum is exactly balanced by the planet’s loss of momentum. However, because of the planet’s enormous mass, the decrease in its speed is immeasurably small.

space We can also treat gravity assist as a perfectly elastic collision (see in2 Physics @ Preliminary p 66). The planet pulls the probe rather than pushing and the ‘collision’ is gradual, but the conservation of momentum still applies. It is elastic because gravity conserves mechanical energy—there is no ‘friction’. The planet is like an enormously heavy cricket bat and the probe is a small, highly elastic ‘superball’. In the bat’s frame of reference, the bat is a stationary ‘immovable’ object (see in2 Physics @ Preliminary p 66), so in a highly elastic collision the ball’s speed before and after the collision is practically unchanged. In the frame of reference of the batsman, however, the ball has been given an increase in speed by the moving bat. In 1973, Mariner 10 first used gravity assist (from Venus) to achieve a flyby of Mercury to gather images and other measurements. The Galileo probe was launched from a Space Shuttle in 1989, reached Jupiter in 1995 and studied Jupiter’s moons until 2003. To reduce the explosion hazard to the Shuttle’s astronauts during launch, Galileo’s fuel requirement was decreased by using gravity assist, once from Venus and twice from Earth, to slingshot Galileo to Jupiter. The reduction in Earth’s orbital velocity was of the order of only 10–18 m s–1.

spacecraft’s velocity outbound vout = vin

spacecraft’s trajectory

Jupiter

spacecraft’s velocity inbound

resultant Vout

Vout > Vin

Worked example

Jupiter

Question For gravity assist, the maximum possible speed increase occurs for the (unrealistic) extreme limit at which the spacecraft executes a nearly 180° turn, parallel to the planet’s orbital motion (Figure 2.4.2).

Jupiter’s velocity relative to the Sun resultant Vin

Figure 2.4.1 Gravitational slingshot in

Show that for this special case, the change in speed of the craft in the Sun’s frame of reference is twice the orbital speed of the planet Vp. You can assume that in the Sun’s frame, the probe’s speed is always larger thanVp.

(a) Jupiter’s frame of reference and (b) the Sun’s frame of reference

Solution Be careful! Change in speed Vout – Vin is the change in the magnitude of the velocity, but it is not the same as the magnitude of change in velocity |Vout – Vin|. All labelled velocities in Figure 2.4.2 lie in one dimension. Upper case variables denote quantities in the Sun’s reference frame, and lower case variables the planet’s frame. As usual, bold = vectors, italic = magnitudes and sign = direction. Vin, Vout and Vp are respectively the spacecraft’s incoming and outgoing speeds and the planet’s orbital speed in the Sun’s frame of reference. Because of orbital symmetry, in the planet’s frame the spacecraft’s incoming and outgoing speeds vin and vout are equal. Let them both equal v.

a

vout

vin b

Vout

Using the Galilean transformation formula vB (rel. to A) = vB ‑ vA to convert velocities in Figure 2.4.2a from the planet’s frame and using the sign convention + → :

vin = (–v) = Vin – Vp = (–Vin) – (+Vp)

(1)

vout = (+v) = Vout – Vp = (+Vout) – (+Vp) ∴ v = Vout – Vp (2)

Equate (1) and (2): Rearrange:

∴ v = Vin + Vp

v = Vin + Vp = Vout – Vp

∴ Vout – Vin = ∆V = 2Vp

Vp

Vin

Figure 2.4.2 Hypothetical 180° hyperbolic orbit in (a) the planet’s frame of reference (b) the Sun’s frame

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Explaining and exploring the solar system

Hitchhiker’s guide to the solar system

I

nstead of selling burgers, two university students took summer holiday jobs at the NASA Jet Propulsion Laboratory and made the next half century of solar system exploration possible. In 1961, Mike Minovitch proved that gravity assist was possible, and in 1965 Gary Flandro showed (just in the nick of time) that in 1976, 1977 and 1978 the planets would, by luck, be suitably aligned for a space probe to hitchhike across the solar system using gravity assist from Jupiter and Saturn to explore the outer planets Jupiter, Saturn, Uranus and Neptune. This ‘grand tour’ would not be possible again until about 2157. Two of the resulting missions, Voyager 1 and Voyager 2, provided enormous advances in planetary science, including the nowfamiliar spectacular images of the outer planets and their moons. Voyager 1 is now at the boundary between the solar system and interstellar space.

Voyager 1 Saturn 12 Nov 80

Jupiter 9 July 79

Saturn 26 Aug 81

Jupiter 5 Mar 79

Earth 5 Sept 77 20 Aug 77

Pluto Aug 89

Uranus 27 Jan 86

Voyager 2 Neptune 01 Sept 89

Figure 2.4.3 Voyager 1 and Voyager 2 take the grand tour of the solar system using gravity assist.

Checkpoint 2.4 1 2 3 4

In a gravity-assist manoeuvre, a probe increases its momentum. Explain how momentum is conserved and why this might not be obvious to an observer. During a slingshot manoeuvre, what is the shape of the probe’s orbit in the planet’s reference frame? What can we say about the probe’s speed (far from the planet), before and after the slingshot manoeuvre, when viewed from the planet’s frame? If VP is the orbital speed of the planet, what is the maximum possible speed increase for a probe executing gravity assist?

2.5 I’m back! Re-entry It is time to come home. You made it through the launch safely so you’re feeling lucky. However, you still have another dangerous hurdle to jump—coming back to Earth or re-entry.

Orbital decay Account for the orbital decay of satellites in low Earth orbit.

46

Craft in low Earth orbits do experience some drag or air resistance, because the atmosphere gradually thins out with altitude in a very roughly exponential trend (Figure 2.5.1). At altitudes above ~1000 km, drag is considered negligible. Most Earth-sensing satellites orbit below this altitude, and so have a limited lifetime. Drag converts orbital kinetic energy into thermal energy, causing the orbital radius to decrease (orbital decay). The lower the orbit, the greater the air density (and drag) and so the faster the orbital decay. Sustained orbits are not possible at

space 100

Density (kg m–3)

altitudes below ~160 km (sub-orbital). Satellites in low orbits can counteract decay by firing rockets from time to time, but only until the propellant runs out. Eventually, a craft in a decaying orbit spirals down towards the Earth’s surface, to burn up catastrophically like a meteor due to the thermal energy produced by the enormous drag (see in2 Physics @ Preliminary p 57). Drag can be enhanced by solar activity such as temporary increases in the Sun’s output of ultraviolet radiation, which inflates the upper atmosphere, making the orbital decay rate unpredictable. Normally a satellite will burn up completely in the atmosphere, although parts of larger craft (such as the Russian Mir Space Station or the US Skylab) occasionally make it to Earth’s surface.

10–5

10–10

10–15

0

100

200

300

400 500 600 Altitude (km)

700

800

900 1000

Figure 2.5.1 Graph of typical air density in the Earth’s atmosphere versus altitude

Rest In Pieces

I

n the late 70s, NASA’s first space station, Skylab, was crippled with low propellant and damaged gyroscopes. Enhanced solar activity increased its orbital decay and the first launch of the newly developed Space Shuttle was delayed, preventing repair missions. So, in July 1979, it fell on Australia. Pieces too large to burn up on re-entry landed in a line between Esperance and Rawlinna, Western Australia. Many pieces are in the Esperance Museum, but the largest (Figure 2.5.2) is now in the United States Space & Rocket Center. The Shire of Esperance sent the US government a $400 fine for littering. The fine was finally paid in April 2009 in a Californian radio publicity stunt, using listeners’ donations.

of Skylab Figure 2.5.2 The largest surviving fragment

Safe re-entry corridor If a spacecraft is carrying a human crew, or if the craft needs to be retrieved, then plunging into a fiery re-entry like a meteorite is not an option. A safer return from low Earth orbit normally starts by retro-firing rockets to slow the craft down so it begins to fall into a lower energy, lower altitude orbit. There the higher air density starts to slow the craft further. At the bottom of LEO, orbital speed is nearly 7.8 km s–1. Orbiting spacecraft could not carry enough propellant to ease down to the surface. The crew has no choice but to head bravely towards the Earth at the correct angle, using only high technology and clever physics to protect them. Safe re-entry is a balance between two forces: drag and lift. Drag is the deceleration force; lift is the force that keeps an aeroplane in the air. Air moving relative to the craft creates pressure differences. If pressure underneath is greater than that above, lift results. The shape and orientation of the craft and the re-entry angle all affect the ratio of these two forces.

Discuss issues associated with safe re-entry into the Earth’s atmosphere and landing on the Earth’s surface. Identify that there is an optimum angle for safe re-entry for a manned spacecraft into the Earth’s atmosphere and the consequences of failing to achieve this angle.

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Explaining and exploring the solar system

s in

u

cie ffi

nt d

rag

re-entry corridor sive drag c ex es

Figure 2.5.3 The safe re-entry corridor

The three main issues behind safe re-entry are: 1 minimising the effects of deceleration (g-force) 2 managing the effects of heating 3 landing the craft safely in the right place. The first two issues lead to the existence of a narrow range of safe re-entry angles and speeds (Figure 2.5.3). Drag is both good and bad. Drag provides the spacecraft with brakes, but it also produces the copious amounts of thermal energy that could destroy the craft. If the approach into the atmosphere is at too shallow an angle, drag will be too small, the air flow will provide too much lift and the craft will skip over the atmosphere instead of entering. If the angle is too steep, drag will be too large, producing excessive heat and deceleration g-force, which would destroy the craft and crew.

Deceleration As at launch, astronauts’ seats during re-entry are oriented perpendicular to, and facing (‘eyeballs in’) the direction of acceleration, but this time acceleration is opposite to velocity, so they look backwards. Traditional re-entry vehicles (such as were used in the 1960s and 1970s) were teardrop-shaped capsules with the blunt end pointing forward (Figure 2.5.4). They allowed very little or no control once re-entry had begun and provided very little lift. Such a re-entry is called ballistic re-entry and requires larger re-entry angles. This kind of capsule subjected the astronauts to a maximum re-entry g-force of anywhere between 6 and 12. The Apollo re-entry angle was between 5.2° and 7.2° The Space Shuttle introduced in 1981 has wings that provide lift and flightcontrol structures (such as elevons, a rudder/speed brake and a body flap) that allow considerable control over the descent, adjusting the vehicle’s aerodynamics to the changing density of the air, and making re-entry more gentle, with a maximum g-force of about 2–3. This degree of control also widens the safe re-entry corridor, allowing a gentle, low-g 1–2° re-entry. This is called glide re-entry. To further decrease descent speed without excessive g-force, the Shuttle performs a series of S‑shaped turns by rolling and banking, gently enhancing drag. The Russian Soyuz capsule, in use continuously (in modified form) since the 1960s, is a more spherical variation of the traditional capsule shape but with attitude control thrusters, which provide some glide control during re-entry. It usually yields a g-force of 4–5, but sometimes up to about 8 for a completely ballistic re-entry. Ballistic re-entries are high acceleration but quick—between 10 and 15 minutes. A full glide re-entry is low acceleration but slow—Shuttle re-entry, for example, takes about 45 minutes. Soyuz is intermediate and takes about 30 minutes. Heating On re-entry, vehicles travel at well above the speed of sound. The speed of sound is sometimes called Mach 1 (after Ernst Mach (1838–1916) a physicist and philosopher who studied gas dynamics). Twice the speed of sound is called Mach 2 and so on. Supersonic means travelling faster than Mach 1. Hypersonic usually means faster than Mach 5 (oversimplifying somewhat). Pressure builds up in front of projectiles. Sudden pressure changes normally propagate away as sound (at the speed of sound). In supersonic flight, however, 48

space this pressure wave is too slow to move out of the projectile’s way, so the pressure builds up to very high levels, forming a shock wave—the air equivalent of the bow wave in front of a speed boat. The enormous mechanical energy of orbit must go somewhere. Drag converts it to thermal energy. Contrary to common sense, in hypersonic flight, a blunt projectile with more drag actually gets less hot than a more streamlined one. In the 1950s, Harvey Julian Allen proved this theoretically and explained why the sharp nose cones of intercontinental ballistic missiles were vaporising on re-entry. Hypersonic wind tunnel tests (see Figure 2.5.5) confirmed his theory. In hypersonic flight, much of the heat generation takes place in the shock wave (the dark line wrapping around the front of both projectiles in Figure 2.5.5). The shock wave does not touch the blunt projectile (Figure 2.5.5a) and so doesn’t transfer the heat efficiently to it, but it does touch the tip of the sharp projectile (Figure 2.5.5b), which gets much hotter. For this reason, re-entry vehicles (including the Space Shuttle) are blunt at the front, and hence the traditional teardrop shape of capsules. The blunt front of the vehicle is also coated with a suitable heat shield with very high melting and vaporisation temperatures. It is also highly insulating to slow the rate of thermal conduction. Thermal insulating materials in most applications are almost always very porous because tiny pockets of gas are very poor thermal conductors. Some insulator materials are also designed to be highly light-absorbing (black) in the visible and near infra-red parts of the spectrum because such surfaces, when hot, also radiate thermal energy away more efficiently (radiative cooling). Tiles on the Shuttle surface are made of 90% porous silica fibre, which is an excellent high melting point insulator, but it is brittle. The tiles on the hottest parts (the underside and leading edges) are also coated with a tough black glass to enhance radiation of thermal energy, but also to provide mechanical strength. Broken tiles were believed to be responsible for the destruction during re-entry of the Shuttle Discovery in 2003. In more traditional space capsule ballistic re-entry, drag is higher, so heat is generated more rapidly, and insulation and radiation alone are not enough. In these cases, the insulating heat shield is also designed to vaporise and erode (ablation). The hot, vaporised and ablated material carries thermal energy away rather than conduct it to the capsule, similar to the way in which evaporating sweat carries away excess heat from your skin. The pressure from this ablation also helps to push away the hot gas convecting from the shock wave. The shield must be thick enough to last the journey and provide sufficient insulation. Two modern examples of ablating materials are phenolic impregnated carbon ablator (PICA) and silicone impregnated reusable ceramic ablator (SIRCA). In the Chinese space program, one of the ablation materials used is blocks of oak wood. It’s cheap and easy to work. As it chars, it forms charcoal, which is porous and almost pure carbon, making it an extraordinarily good thermal insulator with a very high melting point. Another advantage is that porous carbon is very black and radiates thermal energy efficiently. However, it is mechanically weaker than more ‘high-tech’ ablation materials. During re-entry, superheated air surrounding the vehicle is ionised. The air becomes a plasma—a conductive soup of free positive and negative charges that, like the Earth’s ionosphere (see in2 Physics @ Preliminary pp 153–4), reflects

a

b

c

d

Figure 2.5.4 Re-entry vehicles: (a) Gemini 1964–1966 (b) Apollo 1966–1975 (c) Soyuz 1960–present (d) Space Shuttle 1981–present 49

2

Explaining and exploring the solar system a

b

Figure 2.5.5 Hypersonic wind tunnel tests. (a) The crescent-shaped shock wave is detached from the blunt projectile, but (b) touches the tip of the sharp projectile.

radio waves, so the astronauts cannot communicate with the Earth for several minutes during re-entry. This problem has been solved for the Shuttle by communicating via a satellite above it, since only the bottom of the Shuttle has significant ionisation.

Landing Drag depends on the projectile’s cross-sectional area and speed. Drag cannot stop a projectile completely because, during deceleration, drag decreases until it exactly cancels the weight of the projectile and deceleration stops—the projectile has reached terminal speed (see in2 Physics @ Preliminary p 45). The terminal speed of a capsule is too high for it to land safely. To slow the capsule further for the landing, drag is enhanced (and terminal speed decreased) by using parachutes to increase the effective area of the capsule. The final ‘touchdown’ could be on land (typical of Russian missions) or a ‘splashdown’ in the water (typical of US missions pre-Shuttle). Russian Soyuz also has soft-landing engines that fire just before it touches the ground. The Space Shuttle lands on a runway, much like an aeroplane (Figure 2.5.6) but it uses parachutes to help it brake. During landing, the Shuttle (which has been described as being ‘like flying a brick with wings’) is controlled entirely by computer. Another issue is accurate targeting of the landing site. The steeper the re-entry angle, the smaller the horizontal component of motion (range) and so the more accurate the prediction of the final landing site. However, the Shuttle makes up for its shallow re-entry, because its aeroplane-like flight-control structures allow adjustment of the landing path. The shape of the landing path is also designed to be more forgiving. The Shuttle approaches the runway roughly opposite to the landing direction. Four minutes from touchdown, it does a ‘heading-alignment’ loop, to adjust precisely to the direction of the runway (Figure 2.5.6).

50

space

Altitude 25 000 m

Mojave Runw ay 23 E d Airfor wards ce Ba se

Figure 2.5.6 Scale drawing of the relatively gentle descent of the Space Shuttle. The Shuttle is drawn at 1 minute intervals to touchdown. The squares on the ground are 10 nautical miles (18.5 km) wide.

Checkpoint 2.5 1 2 3 4 5 6 7 8 9 10 11

Define orbital decay and explain what causes it. Because of drag, satellites at altitudes below ~1000 km can do nothing to combat orbital decay. True or False? Explain. What other astronomical body can affect the rate of orbital decay? Explain. Discuss how drag is ‘good and bad’ for re-entry. Outline what can happen if a spacecraft attempts re-entry with too shallow or too steep an angle. Explain why astronauts face backwards during re-entry, unlike at launch. Outline why occupants of the Space Shuttle experience lower g-force during re-entry than in the more traditional re-entry vehicles. Define the terms supersonic and hypersonic. What is a shock wave? Outline why a pointy hypersonic projectile is more likely to melt than a blunt one. Explain why a capsule with a parachute slows down more than without one.

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Explaining and exploring the solar system

PRACTICAL EXPERIENCES CHAPTER 2

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 2.1: Development of space exploration Identify data sources, gather, analyse and present information on the contribution of one of the following to the development of space exploration: Tsiolkovsky, Oberth, Goddard, EsnaultPelterie, O’Neill or von Braun.

Use the template provided in the activity manual to extract information about your chosen scientist. Process this information to make a short oral presentation to the class. Discussion questions 1 For the scientist that you have researched, list their main contributions to space exploration. 2 Explain how later scientists have benefited from this research. Extension 3 Werner von Braun’s great Russian rival, the ‘Chief Designer’ for the USSR space program Sergey Korolyov, is not as familiar as some of the names mentioned in the syllabus, despite leading the launch of the first artificial satellite, Sputnik, in 1957. This is probably because his name was kept secret by the communist government of the USSR until after his death in 1966. You may also want to research his contribution to space exploration.

Activity 2.2: Uniform circular motion Solve problems and analyse information to calculate the centripetal force acting on a satellite undergoing uniform circular motion about the Earth using:

F =m

v2 r

Perform an experiment that will allow you to determine the relationship between the radius of a satellite’s orbit around the Earth and its gravitational force. Equipment: string, rubber stopper, mass carrier and masses, electronic scales, glass or plastic tube, paperclip, sticky tape, metre ruler, stopwatch. Discussion questions 1 From your experimental data, determine the mathematical relationship between the orbital radius of a satellite and its tangential velocity for a given centripetal force. 2 Describe the method you would use to determine the centripetal force on a small model satellite.

tension glass or plastic tube paperclip

string

mass carrier

Figure 2.6.1 Force on mass moving in a horizontal circle

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mg

Chapter summary • • • • • • • •

•

• • •

• • •

•

•

Rocket thrust is the reaction to the force exerted on the exhaust exiting the nozzle. For a fixed mass of exhaust per unit time, thrust increases as exhaust speed increases. As the mass of remaining propellant decreases, a rocket’s acceleration (and g-force) increases. There are two kinds of rocket engines: liquid and solid propelled. By jettisoning used stages, a rocket’s mass is decreased, allowing more payload to be carried. g-force is the ‘apparent weight’ experienced during acceleration, divided by true weight on Earth. a a Vertical g-force = v + 1. Horizontal g-force = h . g g To increase g-force tolerance, astronauts are seated with bodies horizontal, and looking in the direction of the acceleration (‘eyeballs in’) for both lift-off and re-entry. A spacecraft launched eastwards has the extra initial velocity of Earth’s rotation. This is greatest at the equator (465 m s–1). Launches in the direction of Earth’s orbital velocity obtain an initial velocity boost of 3.0 × 104 m s–1. Gravity provides the centripetal force for satellite orbits:

mv 2 (for circular orbits) R Kepler’s laws (apply to any two bodies if the central body has a very much larger mass): 1 Orbits of planets are ellipses, with the Sun at one focus. 2 Planets sweep out equal areas in equal times. T2 3 Law of periods: = a constant a3 a 3 GM Explicit form of Kepler’s third law:  = T 2 4π2 Eccentricity is a measure of the elongation of an ellipse. A circle is an ellipse of zero eccentricity. Periapsis is the position of closest approach to the central mass and fastest orbital speed (perihelion for the Sun, perigee for Earth). Apoapsis is the position of furthest distance from the central mass and slowest orbital speed (aphelion for the Sun, apogee for Earth). If the mass of a satellite is not negligible when compared to that of the central body, then both masses orbit with the same period around the system’s centre of mass.

• •

• • • •

• • •

Fc =

•

•

• •

•

• • •

space

GM r Two-body orbital mechanical energy 1 GmM : ME = mv 2 − 2 r Magnitude of orbital speed: v =

– ME  0 (unbound), orbit is hyperbolic; velocity > escape velocity GmM . For stable orbits, ME = − 2a Orbits are symmetrical in shape and speed. Low Earth orbit (LEO): altitude between ~160 and ~2000 km. At altitudes below ~1000 km, drag causes orbital decay. Upper atmosphere can be inflated by increased solar UV radiation, increasing drag. Orbits below ~1000 km are protected from the Van Allen radiation belts by the atmosphere and distance. Geosynchronous orbit: T = 1 sidereal day (23 h 56 m 4 s). Circular geosynchronous orbits over the equator are called geostationary. These orbits are used extensively for communication satellites (r = 42 164 km). A space probe entering a temporary hyperbolic orbit behind an orbiting planet can gain momentum via gravity assist or the slingshot effect. For gravity assist, the maximum possible change in speed of the probe in the Sun’s frame of reference is twice the planet’s orbital speed Vp. Drag converts orbital KE into thermal energy. Safe re-entry angle: if angle is too low, the craft will skip off atmosphere; if angle is too high, g-force and heating rate are too large. Much heating takes place in the hypersonic shock wave. Blunt-fronted re-entry vehicles are used because the shock wave is detached from the craft. Heating of a spacecraft on re-entry is reduced by an insulating and radiating heat shield. Traditional capsules also use ablation of the heat shield to dissipate heat. Parachutes decrease terminal velocity by increasing the effective cross-sectional area.

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2

Review questions

Explaining and exploring the solar system

Physically speaking For each type of orbit, fill in the missing information. One has been done already.

Name of orbit

Sign of two-body ME

Open or closed

v >, =, < vescape

Bound or unbound

Negative

Closed

<

Bound

Geostationary Slingshot (in planet’s frame) Elliptical Hohmann (see Physics Focus) Parabolic Hyperbolic Halley’s Comet Circular Molniya

Reviewing Solve problems and analyse information to calculate the centripetal force acting on a satellite undergoing uniform circular motion about the Earth using: 2

mv r Solve problems and analyse information using: mm F =G 1 2 d2 Analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth. F =

1 What was the first solid rocket propellant and who invented it? 2 Assuming that propellant is burned at a constant mass per unit time, use the equation for thrust to explain why forcing exhaust gas through a narrow nozzle increases thrust.

3 List the advantages of both solid and liquid propellants. 4 Discuss why the vertical g-force formula has a ‘+ 1’ term, but the horizontal formula doesn’t.

5 Describe a situation during launch in which astronauts would experience a g-force greater than zero but less than 1.

6 At the bottom of a bungee jump with the cord attached to the ankles, one can easily experience a g-force of 3, the maximum normally allowed for Shuttle launches. Describe three important differences in the way the g-force is experienced in these two situations.

7 Explain why launch facilities are usually built as close to the equator as is practical.

8 Describe the circumstances under which a star would not sit at the focus of a planet’s elliptical orbit.

9 Discuss why we only briefly see Halley’s Comet with an unaided eye every 76 years, even though it is in orbit continuously around the Sun.

10 Outline what happens to the period of a satellite if its semimajor axis is reduced by a factor of 4.

11 A space probe approaches a planet in a hyperbolic orbit. Discuss the condition that must be fulfilled to move it into a stable orbit of the planet and describe how it might be achieved.

12 By re-examining the gravity assist worked example on page 45, show that the magnitude of change in the probe’s velocity in the Sun’s reference frame is twice the probe’s initial speed in the planet’s reference frame (that is, |Vout – Vin| = 2v).

13 A satellite is in a highly elliptical orbit around Earth such that, at perigee, it is briefly at an altitude of less than 1000 km. Over many orbits, the altitude at apogee decreases (the orbit becoming more circular). Explain why this occurs.

14 List two reasons why (human-crewed) space stations are always in low Earth orbit. 54

space

Solving Problems 15 Calculate the two-body gravitational potential energy for a system consisting

Analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth.

of a 1.00 kg test mass sitting on the surface of the Earth. How far would the test mass need to be from the Sun so that the two-body GPE of the test mass–Sun system is the same value? Estimate roughly where that position would be in relation to the orbital radii of the planets.

16 Typically, at launch, the Shuttle’s main engines, with an effective exhaust velocity of 4460 m s–1, produce a thrust of 5.45 × 106 N. The two solidfuel rocket boosters, with an effective exhaust velocity of 2640 m s–1, produce 1.250 × 107 N each. a Calculate the combined rate (in kg s–1) at which propellant is used at launch. b Assuming a mass at launch of 2.03 × 106 kg, calculate the Shuttle’s acceleration at launch and 1 minute later, assuming the above specifications remain constant. (Hint: Don’t forget gravity.)

17 On a roller-coaster, you round the top of a circular hump in the track of



5.00 m radius. You have a g-force meter with you and at the moment you’re at the top it reads a vertical g-force of 0.00. a What is your weight at that moment? b What is the magnitude of the normal force exerted on you by the seat at that moment? c What is your centripetal acceleration? d Calculate your speed at the top. e Assuming friction and air resistance are negligible, calculate your horizontal g-force at that moment.

18 Prunella spins a weight (mass m) on a string (length L) in a horizontal

Solve problems and analyse information to calculate the centripetal force acting on a satellite undergoing uniform circular motion about the Earth using:

circle (Figure 2.6.2) to illustrate the relationship between orbital speed and centripetal force for an orbiting satellite. Renfrew says: ‘Because of the weight, the string isn’t horizontal so the orbital radius is R = L sin θ, and the centripetal force is Fc = T sin θ.’

Prunella then says: ‘Yeah, but as long as the orbital speed v is high



enough, θ will be very close to 90° so you can use the approximation that string tension T is the centripetal force and the string length L represents orbital radius R.’ v2 Show that as long as orbital speed v fulfils the condition > 7g, R then L is no more than 1% larger than the true orbital radius R and T is no more than 1% larger than the true centripetal force Fc.

Solve problems and analyse information using: mm F =G 1 2 d2



mv 2 r Analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth. F =

T sin θ θ

L

mg

T θ

R

Figure 2.6.2 Spinning weight model of a satellite

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2

Explaining and exploring the solar system

19 In Gerard O’Neill’s proposed colonies in space, people would live on the inner surfaces of rotating cylinders 3 km in radius and 20 km long. Gravity would be simulated via the reaction force to the centripetal force exerted by the cylinder on the occupants inside. Calculate the rotation rate (in revolutions per hour) that would yield artificial gravity, mimicking the magnitude of gravity at the Earth’s surface. Solve problems and analyse information using: r 3 GM = T 2 4 π2



Define the term orbital velocity and the quantitative and qualitative relationship between orbital velocity, the gravitational constant, mass of the central body, mass of the satellite and the radius of the orbit using Kepler’s Law of Periods.

20 Using the mass of Jupiter calculated in the worked example on page 39, 21

predict the period of Jupiter’s moon Callisto, given that its semimajor axis is 1.89 × 106 km. 2πr and the explicit form Using the equation for tangential velocity v = T of Kepler’s law of periods, re-derive the expression from section 2.2, for the magnitude of orbital velocity of a satellite in circular orbit: v=

GM

r 22 Using the explicit form of Kepler’s third law, calculate the radii of: a a geostationary orbit b a circular semi-synchronous orbit.

23 Draw a table in which you compare the kinetic energy, the two-body



potential energy and the two-body mechanical energy of a geosynchronous and a circular semi-synchronous satellite of mass 2000 kg. Use the data from Question 22 to confirm that the two-body mechanical energy for both orbits is: GmM ME = − 2a

24 A comet passes the Sun (MS = 1.99 × 1030 kg). At perihelion, the

centre-to-centre distance is 8.55 × 1010 m and the speed of the comet is 6.69 × 106 m s–1. What kind of orbit is it? Do you expect the comet to return?

25 Gravity assist can also be used as a brake. Show that if the diagram in the planet’s frame in Figure 2.4.1a is unchanged, but the planet’s orbital velocity in Figure 2.4.1b is reversed, then the probe’s speed in the Sun’s frame would decrease.

26 A Soyuz capsule with a crew of three (7460 kg) is in a circular orbit at

Re

56

iew

Q uesti o

n

s

v

an altitude of 336 km, having completed a mission to the International Space Station. a Calculate its kinetic energy. b Ignoring the effect of its soft-landing engines, calculate how much thermal energy is generated by drag during its re-entry. Earth’s mass ME = 5.97 × 1024 kg Earth’s radius at the landing site rE = 6366 km

space

PHYSICS FOCUS

final orbit

 L  Driving lesson in orbit initial orbit

3. Applications and uses of physics satellite

Once in a stable orbit, a spacecraft is coasting. One only needs to do work (burn fuel) to swap between orbits. The following are two important examples of manoeuvres between orbits. Changing lanes: Circular orbits around Earth (or other body) are like lanes on a highway, the smaller radius orbits being the fast lanes. In 1925, Walter Hohmann devised a fuel-efficient method for ‘changing lanes’. To move from a small orbit to a larger one, an accelerating engine burn of just the right impulse will convert the circular orbit to an elliptical Hohmann transfer orbit (Figure 2.6.3), which joins the initial and final orbits. Once in the elliptical orbit, the engine is shut off again. Once the craft reaches the apoapsis of its elliptical orbit, a second accelerating burn of the right impulse will convert the elliptical orbit into a (now larger radius) circular orbit. This method can also be used in reverse to drop down to a smaller orbit, in which case the two engine burns must be decelerations. Hohmann transfer orbits around the Sun can also be used for travel between planets. The launch must be timed so the planet is at the periapsis of the Hohmann orbit just as the probe arrives. Overtaking: Sometimes spacecraft need to rendezvous (meet) in orbit, such as when the Space Shuttle needs to dock with the International Space Station. If you’re way behind a craft in the same orbit and you need to catch up, you should paradoxically ‘slam on the brakes’—briefly retrofire (fire your engines in reverse) to reduce speed (Figure 2.6.4). This temporarily reduces KE (and total mechanical energy) so you drop down into a lower energy, smaller radius ‘overtaking’ orbit, more than regaining your speed. Smaller radius orbits are faster, so you eventually catch up with the other craft. When you’re about to overtake it in the ‘fast lane’, fire your engines forwards to increase mechanical energy, so you pop back up into the original orbit rendezvous.

elliptical transfer orbit

Figure 2.6.3 A Hohmann elliptical transfer orbit

Figure 2.6.4 To overtake, first slow down. 1 When you fire a rocket to change your speed, does this violate momentum conservation? Explain. 2 Why doesn’t one just fly from the initial orbit to the final orbit in a straight line instead of following a Hohmann orbit? Does this violate Newton’s first law? Explain. 3 Show that to transfer from an initial circular orbit of radius Ri to a larger final one of Rf, you must use a Hohmann transfer orbit with a semimajor axis a, where Ri + Rf = 2a. (Hint: Draw some diagrams.) 4 Why can’t the Hohmann transfer orbit be a circular orbit? (Hint: Draw another diagram.) 5 Why is the Shuttle in Figure 2.6.4 pointing the ‘wrong way’? 6 When overtaking, you temporarily reduce your KE. Why does your KE increase again in the overtaking orbit? You may use the fact that for a stable orbit, GmM the two-body ME = (section 2.3). ME = − 2a

Extension 7 A Space Shuttle releases a communications satellite into an initial orbit of radius r = 6.66 × 106 m. Using a Hohmann transfer orbit, the satellite moves up to a geosynchronous orbit of rgeo = 4.216 × 107 m. Using the result of Question 3, calculate how long the transfer takes. (Hint: The transfer path is half an orbit.)

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3 special relativity, general relativity, inertial frame of reference, invariant, principle of relativity, fictitious forces, Maxwell’s equations, electromagnetic wave, mechanical medium, luminiferous aether, interferometry, beam splitter, aether drag, Michelson– Morley experiment, null result, postulate, simultaneity, Lorentz factor, time dilation, proper time, twin paradox, proper length, length contraction, rest mass, proper mass, relativistic mass, spacetime, mass-energy,

Seeing in a weird light: relativity Just some minor problems You may have heard it said that some physicists think that a ‘theory of everything’ is just around the corner. This attitude is not new. Many physicists thought this about what is now called ‘classical Newtonian physics’, towards the end of the late 19th century. There were just a few minor problems with understanding the way light travels through space that needed to be fixed, and then the job of physics would be finished.   Well, those ‘minor problems with light’ led to the twin pillars of modern physics: quantum mechanics and relativity. Einstein was a key player in both, especially relativity, which comes in two parts. Special relativity replaced Newton’s mechanics and the later general relativity replaced Newton’s universal gravitation. A century later, relativity still defies common sense, bending space, time and the mind, but it has not yet failed any experimental test. In this chapter we will only deal with the theory of special relativity. ‘Common sense is nothing more than a deposit of prejudices laid down by the mind before you reach eighteen.’ A Einstein

3.1 Frames of reference and classical relativity Outline the nature of inertial frames of reference.

Discuss the principle of relativity.

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Before we talk about relativistic weirdness, recall inertial frames of reference (see in2 Physics @ Preliminary section 3.3). Inertial frames of reference are required for observers using Newton’s laws or the Galilean transformation formula for relative velocity (see section 1.1). An inertial frame is any nonaccelerating (including non-rotating) reference frame. Inertial frames can move at a constant velocity relative to each other. There are no absolute velocities and there is All velocities are relative. no special absolutely stationary inertial frame. The classical Newtonian laws of mechanics and gravity are unchanged (or invariant) when transforming from one inertial frame to another (even though the values of some measurements such as velocity might change). This is called the principle of relativity. To transform a situation from one inertial frame to another, you simply apply the Galilean transformation formula to all the velocities. You’ve already seen an example of this in the gravity assist example (Figure 2.4.2).

space

I

nte

ractiv

e

Because there is no special inertial frame, no experiment purely within your own frame can detect the velocity of your frame, so absolute velocity is meaningless. You can only compare your frame’s velocity relative to others. An example of this is waiting to depart in a train, looking out the window (Figure 3.1.1) to see that a train next to you is moving slowly away, only to find a few seconds later that, in fact, relative to the station it is your train that is moving. Your acceleration (including vibrations) was negligible—you felt no effect of your uniform velocity. However, you can feel the acceleration of a non-inertial reference frame, and measure it using an accelerometer. The simplest accelerometer is a pendulum. If a pendulum hangs vertically in a car, your horizontal acceleration is zero. If you are accelerating horizontally, the pendulum will hang obliquely (Figure 3.1.2). If you are observing from within a non-inertial (accelerating) frame, Newton’s laws appear to be violated. Objects can appear to change velocity without a true net external force; in other words, you experience fictitious forces or pseudoforces (see in2 Physics @ Preliminary p 39). For example, in a car taking a corner, you experience the sensation of being ‘thrown outwards’ by a fictitious centrifugal force. If viewed from the inertial frame of the footpath, you evidently are pulled inwards by a true centripetal force. (We’ve cheated a bit. The Earth is turning, so the footpath is not strictly an inertial frame. However, Earth’s radius is so large that in most human-scale situations, fictitious forces due to Earth’s rotation are negligible.) Another view of tests for non-inertial reference frames is that they involve detecting fictitious forces. It’s a two-step process. First, analyse an object within that frame of reference and decide what true external Newtonian forces must act on the object. Then, look for apparently ‘extra’ or ‘missing’ forces—evidence of a non-inertial frame. For example, judged from the inertial frame of the ground, the downward weight mg and the upward normal force N of the seat are the only true forces on an astronaut during launch. Within the accelerating rocket, the sensation of enhanced weight (downwards) associated with g-force has the same magnitude as N but is apparently in the wrong direction and is therefore fictitious. A pendulum accelerometer hangs obliquely within an accelerating car as though there is a fictitious horizontal component of weight. In free-fall (or orbit), the apparent absence of weight is also fictitious. Your frame accelerates downwards, so true weight becomes undetectable to you, as though your true downward weight is cancelled by a fictitious upward gravity. The effects of neither ‘force’ show up separately on an accelerometer.

M o d u le

Figure 3.1.1 Who is really  moving?

Try this! Fictitious fun While sitting on a playground merry-go-round with a friend, try playing ‘catch’ with a slow moving tennis ball. The fictitious centrifugal and Coriolis forces will ‘cause’ the ball to appear to follow warped trajectories, making it difficult to catch.

5°

Worked example

T

T

5°

mg

question mg tan 5°

A Christmas decoration is hanging obliquely inside your car, 5° from vertical and pointing towards the car’s left side. Describe quantitatively the car’s motion (no skidding!).

Solution Only two true external forces act on the decoration: tension and weight (Figure 3.1.2). Because there is an angle between them, they aren’t ‘equal and opposite’, so the decoration experiences a net real force and acceleration sideways (in this case centripetal). The net force and acceleration point towards the right side of the car, so the accelerometer (and the car) is steering towards the right.

left

right

mg

Figure 3.1.2 Festive season pendulum accelerometer 59

3

Seeing in a weird light: relativity The ‘centrifugal force’ perceived by the occupants of the car to be pulling the decoration toward the left side of the car is fictitious. From Figure 3.1.2, the magnitude of the centripetal acceleration is: F ac = c = g tan 5° = 9.80 × 0.0875 = 0.86 m s–2 m

Note: This test is subjective—it requires personal judgement (hence possible bias). No measurement alone can identify a force as fictitious. For example, no pure measurement can tell the difference between true weightlessness and the fictitious weightlessness of free-fall. You can only tell the difference by looking Activity Manual, Page 21 down and seeing the Earth below; judgement says there should be gravity, but you can’t feel it. The inability of measurement alone to distinguish the effects of true gravity from the effects of g-force is what Einstein used as the starting point Perform an investigation to for his re-writing of the law of gravitation in his theory of general relativity, but help distinguish between you’ll have to wait until university physics to learn about that! non-inertial and inertial This approach to distinguishing between inertial and non-inertial frames frames of reference. relies on a classical concept of force. In Einstein’s relativity, the concept of force is more complicated and is used much less. The term fictitious force doesn’t mean the observed effects are imaginary, as the victims of a cyclone Fictitious cyclone? or astronauts who are subject to Yeah right! g-force can attest. It simply means ffects associated with that the apparent force doesn’t fit so-called fictitious Newton’s definition of a true force. forces of Earth’s rotation It is always possible to are not always negligible. re-analyse fictitious forces using an The Coriolis force is a inertial frame and to account for fictitious tangential all observed effects using only true force appearing in Newtonian forces.

PRACTICAL EXPERIENCES Activity 3.1

E

rotating frames of reference and is associated with the formation of cyclones.

Figure 3.1.3 Satellite photo of a cyclone

Checkpoint 3.1 1 2 3 4 5 6 7

60

Define an inertial reference frame. Recall the Galilean transformation formula for relative velocities. Outline why we usually treat the Earth as an inertial frame, given that it is rotating. Discuss whether or not centripetal force is fictitious. In free-fall, you don’t experience any extra apparent forces. Are you in an inertial frame? Explain. What apparatus would distinguish true weight from apparent weight due to g-force ? The values of some measurements such as velocity might change, but the laws of mechanics are the same in all frames of reference. True or False? Explain.

space

3.2 Light in the Victorian era The 19th century was a period of enormous advance in the study of electricity and magnetism. Faraday, Ampere, Oersted, Ohm and others, through theory and experiment, produced a large collection of equations and phenomena. There were hints of connections between electricity and magnetism—an electrical current can produce a magnetic field (see in2 Physics @ Preliminary section 12.3) and a changing magnetic field can induce a changing electric field or current (section 4.1). The Scottish theoretical physicist, James Clerk Maxwell (1831–1879) collected the existing equations to reduce them down to the minimum number. He reduced them down to eight equations (which expanded to 20 when he included all the x-, y- and z-components). A self-taught electrical engineer called Oliver Heaviside (1850–1925), using the newly developed mathematics of vectors, reduced Maxwell’s equations to four. We now call those four equations Maxwell’s equations. It puzzled Maxwell that his equations were almost symmetrical in their treatment of electrical and magnetic fields—almost but not quite. So he added a term to his equations, assuming that a changing electrical field can induce a magnetic field (not previously observed). When he did this, he showed that an oscillating magnetic field would induce an oscillating electric field and vice versa, resulting in a self-sustaining electromagnetic wave. From his equations he calculated the speed of that wave to be equal to the speed of light in a vacuum (which is now called c and equals 2.998 792 458 × 108 m s–1). It was either an astonishing coincidence or strong circumstantial evidence that light is an electromagnetic wave (see in2 Physics @ Preliminary p 84). Heinrich Hertz (1857–1894) experimentally confirmed the predicted speed and properties of these electromagnetic waves.

Figure 3.2.1 James Clerk Maxwell

What is light’s medium? Until then, every existing kind of wave needed a mechanical medium; for example, sound propagates through air, earthquakes through rock, musical vibrations along a violin string, ripples along water and so on (see in2 Physics @ Preliminary section 5.3). To sustain a wave, a medium needs two properties: resilience (or stiffness) and inertia (any density- or mass-related property). The higher the stiffness and the lower the inertia, the higher the wave speed. It was assumed that light also needs a medium, which was called luminiferous aether or just aether (US spelling: ether). Luminiferous means ‘light-bearing’, and ‘aether’ was the air breathed by the gods of Greek mythology. So Maxwell developed a model for aether, assigning it bizarre mechanical properties consistent with the behaviour and enormous speed of light. It needed to be far less dense than air but much stiffer than any known material. Despite its stiffness, aether was assumed to penetrate all materials effortlessly. Conversely, it needed to be able to be penetrated without resistance by all objects that move freely through space, including Earth hurtling around the Sun. If you shout with the wind blowing behind you, then, relative to you, the velocity of sound would be higher than if the air were still. This is because the velocity of sound (and other mechanical waves) is the sum of its velocity relative to the medium and the velocity of the medium itself. In other words, mechanical

Outline the features of the aether model for the transmission of light.

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3

Seeing in a weird light: relativity waves seem to obey the Galilean transformation. It was assumed that light should also obey it, so the speed of light should be affected by the motion of the aether. However, Maxwell’s equations appeared to allow only one particular value for The Galilean transformation and Newton’s the speed of light in a vacuum. laws imply it is impossible for the speed of light to appear to be the same to all observers with different relative speeds. Perhaps the speed specified by Maxwell’s equations is the speed relative to the aether only. However, this meant that the aether represented a preferred reference frame for Maxwell’s equations, which was inconsistent with the classical principle of relativity.

M and M Describe and evaluate the Michelson–Morley attempt to measure the relative velocity of the Earth through the aether. Discuss the role of the Michelson–Morley experiments in making determinations about competing theories.

Figure 3.2.2 Interference pattern in a Michelson interferometer illuminated by a mercury vapour lamp. Patterns of different shapes (such as vertical bands) are possible and depend on exactly how the interferometer is aligned.

PRACTICAL EXPERIENCES Activity 3.2

Activity Manual, Page 25

Gather and process information to interpret the results of the Michelson–Morley experiment.

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Given that the Earth was supposed to be hurtling around the Sun, through the aether at 3 × 104 m s–1, the resulting ‘aether wind’ (or aether drift) relative to Earth should affect measurements of light speed differently according to the time of day and time of year as the Earth rotated and orbited the Sun, changing its orientation relative to the aether. So in the 1880s, the experimentalist Albert Michelson (1852–1931), joined later by Edward Morley (1838–1923), attempted to measure changes in the speed of light throughout the day due to this shifting aether wind. They used a very sensitive method called interferometry (see section 21.5), which Michelson had used some years earlier to accurately measure the speed of light. Recall constructive and destructive interference (see in2 Physics @ Preliminary p 102 and p 126). If two light beams are projected onto a screen, then a bright ‘fringe’ occurs at places where the two beams are in phase (constructive interference). Where they are out of phase, destructive interference results in a dark fringe. The pattern of bright and dark fringes is called an interference pattern (Figure 3.2.2). Interference turns a pair of monochromatic (single wavelength) light beams into an extremely sensitive ruler for which the interference fringes are like magnified ruler markings one light wavelength apart. For visible light, this spacing is less than 8 × 10–7 m and corresponds to time intervals of less than 3 × 10–15 s. If the two light beams travel via different paths, then a very small change in the length of one path will change the relative phase, resulting in a detectable change in the position of fringes in the interference pattern. A change in wave speed along one of those paths should have a similar effect on phase. Michelson and Morley set up an interferometer in which the light was divided into two perpendicular beams or ‘arms’ by passing it through a half-silvered mirror or beam splitter (Figure 3.2.3). The apparatus was built on a heavy stone optical bench floating in mercury, to allow rotation and damp out vibrations. They assumed that if one interferometer arm was pointing parallel to the aether wind, the speed of light should be slightly different in the two arms. The time of flight of the light in the arm parallel to the aether wind should be slightly longer than that of light along the perpendicular arm. As the Earth (or the apparatus) rotates, this speed difference, as measured by the positions of the interference fringes (Figure 3.2.2), should change with the angle. Figure 3.2.4 summarises the classically predicted effect of aether wind on the resultant light speed in the two arms of the interferometer. Let’s calculate the expected time difference. Suppose the total distance from beam splitter MS to M1 (or M2) is L, then the round-trip for each arm is 2L.

space a

b

M1 source

Ms

aether wind

l1 l2 M2

M2

Ms

eyepiece

source

Figure 3.2.3 The Michelson–Morley interferometer drawn as (a) a simplified schematic and (b) an actual ray diagram. Multiple reflections were used to make the effective length of the arms very long hence more sensitive to changes in light speed. MS is the half-silvered beam splitter mirror.

relative to the aether, then the resultant light speed is c 2 − v 2 (Figure 3.2.4a) and the time taken for light to do a round-trip is: 2L c2 − v2

=

2L × c

1 1−

v2 c2

In the arm parallel to the aether wind (speed v), for half the trip against the wind, the speed of light would be c – v, and for the other half with the wind it would be c + v, so the time taken would be longer: L L 2L t2 = + = × c −v c +v c

a

1 1−

v2

c2 (Check that you agree that since v is smaller than c, time in the parallel arm t2 is longer than time in perpendicular arm t1.) Other factors such as thermal expansion or contraction of the apparatus could cause apparent drift in the interference pattern, but the shift due to rotation of the apparatus (or the Earth beneath it) would be a sine wave with a period equal to the rotation period of the apparatus, so any drift not due to rotation could be detected and subtracted. Michelson and Morley graphed the position of interference fringes versus rotation angle at different times of the day, but concluded that the small observed shifts could be explained as drift in the experimental apparatus. Over several years, scientists repeated the measurements, with some reports of possible changes in interference over the day; but eventually the consensus was that any observed effect was well below what was expected by the aether theory and could be explained by drift in the apparatus. George Fitzgerald and Hendrik Lorentz attempted to squeeze the Galilean transformation into Maxwell’s equations, concluding that charged particles (such as charges in atoms) moving through the aether with speed of v must

C

V

√C 2 – V 2 V

b

C V

aether wind

t1 =

aether wind

In the arm perpendicular to the aether wind (speed v), if c is light speed

M1

C

C– V

C

C+ V

V

Figure 3.2.4 Classical effect of aether wind on light velocity. (a) The resultant velocity perpendicular to the wind and (b) resultant velocity parallel to the wind. Blue = light velocity relative to aether, green = aether velocity and red = resultant light velocity

shrink in the direction of motion by a factor of 1 − v 2/c 2. The interferometer arm parallel to the aether wind would shrink just enough to compensate for the change in light speed and hence cancel the expected change in the interference pattern. 63

3

Seeing in a weird light: relativity Another reason suggested for failure to see the shift was that perhaps aether only penetrates transparent objects, so aether was trapped by large opaque mountains and valleys or buildings and dragged along by the moving Earth, similar to the way in which air is trapped in the fur of a running dog. A flea conducting scientific experiments on the dog’s skin would be unaware that outside the fur, air is whooshing backwards relative to the dog. This idea was called aether drag. If this were true, then at the tops of mountains, closer to outer space, the aether wind might be detectable. Some experimenters repeated the experiment on mountains without success (apart from a controversial partial result). Failure to detect undeniable effects of aether wind caused some physicists to question if it even existed. Maxwell’s equations only mention electric and magnetic fields. The aether is not required by the equations. Einstein assumed it didn’t exist, but said that relativity was not an attempt to explain Michelson and Morley’s negative result, but rather, he was motivated by the properties of Maxwell’s equations. However, in physics, when experiment and theory say the Today almost all physicists same thing, you’re probably on the right track. agree that there is no aether. The Michelson–Morley experiment is often called ‘the most famous failed experiment’. It was not exactly a ‘failure’. In 1907, Michelson was awarded the The result of an experiment that fails Nobel Prize for Physics for his work. to find evidence of an expected effect despite careful design and execution is more correctly called a null result. This null result was one of the most important in the history of physics, because it helped bring about a whole new way of seeing the universe.

Checkpoint 3.2 1 2 3 4 5 6 7 8

Describe Maxwell’s circumstantial evidence that light is an electromagnetic wave. Discuss why it was assumed that light required a medium or ‘aether’ to propagate in. Maxwell’s equations predicted that the speed of light should depend on the speed of the medium, but this was contradicted by the Michelson–Morley experiment. True or false? Explain. In the classical analysis of the Michelson–Morley interferometer, which arm required the longer time of flight? Is it correct to say that the Michelson–Morley experiment didn’t show any change in the interference pattern? Explain. Outline how Fitzgerald and Lorentz explained the apparent absence of evidence for aether wind. Describe aether drag. Discuss which played a greater role in motivating Einstein’s work, the work of Michelson and Morley or that of Maxwell.

Explain qualitatively and quantitatively the consequence of special relativity in relation to: • the relativity of simultaneity • the equivalence between mass and energy • length contraction • time dilation • mass dilation.

64

3.3 Special relativity, light and time Although relativity is Einstein’s theory, many of the underlying ideas or mathematical formulae were inspired or anticipated by others including Poincaré, Lorentz and Minkowski. Einstein, being a theoretician, did not conduct laboratory experiments. However, he is famous for making good use of the ‘Gedankenexperiment’ or ‘thought experiment’ to boil abstract ideas down into simple concrete ones. Theory can sometimes be derived by imagining an experiment being done, even if it is impractical. We will mention some of his thought experiments in this section.

space

Speed of light Newton regarded space and time as absolute. In practical terms, this means that the length of 1 metre, the duration of 1 second and the geometric properties of shapes would be the same to all observers everywhere. Not all physicists agreed, but the success of Newton’s laws silenced any philosophical discussion. However, Maxwell’s theory (and the Michelson–Morley experiment) pointed to the speed of light in a vacuum being constant to all observers. So Einstein said one of three things must be wrong: the principle of relativity (the invariance of laws of mechanics in all inertial reference frames), Maxwell’s equations or the Galilean transformation (the basis of all of classical mechanics). The principle of relativity seemed very fundamental to Einstein, so he didn’t reject that. In fact, he extended Galileo and Newton’s principle of relativity to include all laws of physics, not just mechanics. He called it his first postulate. Following a suggestion by Jules Henri Poincaré (1854–1912), Einstein decided that as the speed of light in a vacuum was invariant in all inertial frames, then that must also be a law of nature, which he called the postulate of the constancy of the speed of light. Maxwell’s equations accurately described electromagnetic phenomena, so Einstein didn’t want to reject them. So it must be the Galilean transformation (and hence all of classical mechanics) that was wrong. But it is difficult to see how something so simple could possibly be wrong. Suppose you are on a moving train, shining a torch towards the front of the carriage. To your eyes, the light travels the length of the carriage L. To you, its speed is the length of the carriage divided by the time t it took to get there c = L/t. To an observer at the train station, the light travelled the length of the carriage plus the distance D the carriage travelled in that time: c = (L + D)/t. The arithmetic is so laughably simple. How could both observers possibly get the same value for c? It could only be possible if you and the observer at the train station In other words, if the disagree on the lengths L or D or the time period t. speed of light is constant then length (space) and/or time are not absolute—they must depend on the state of motion of the observer. So why had no-one noticed until 100 years ago? Classical mechanics had successfully described phenomena for three centuries, but it had never been tested Classical mechanics and the for things moving at close to the speed of light. Galilean transformation are accurate approximations at speeds well below the speed of light. Only when the properties of light itself were examined, did the problems become obvious. the speed of light is the ultimate Einstein showed (in several ways), that speed limit—no observer can reach the speed of light. As a teenager, he asked ‘What would the world look like if I rode on a light beam?’ He answered as an adult with a thought experiment. A light beam is a wave of oscillating electric and magnetic fields moving at the speed c. If you were in the same reference frame as the light beam, you would observe stationary electric and magnetic fields that vary as sine waves in space, but are constant in time. This is not an allowable solution to Maxwell’s equations, so it is not possible for an observer to travel at the speed of light—it is the ultimate speed limit.

Simultaneity Einstein demonstrated that simultaneity is relative. Events apparently simultaneous to one observer are not necessarily so to all observers. Let’s use Einstein’s own

Describe the significance of Einstein’s assumption of the constancy of the speed of light. Identify that if c is constant then space and time become relative.

What’s so Special about Relativity?

E

instein called his 1905 replacement theory for Newton’s mechanics special relativity. It is a ‘special case’ in the mathematical sense of being restricted to particular conditions—to inertial reference frames. Einstein’s general relativity came 11 years later and was generalised to include non-inertial reference frames. It replaced Newton’s gravity. 65

3

Seeing in a weird light: relativity

O2

v

O1

A

B

platform

Figure 3.3.1 Lightning strikes at A and B appear simultaneously to observer O1 but not to O2.

Solve problems and analyse information using: E = mc2 l v = l0 1 − tv =

c2

t0 1−

mv =

v2

v2 c2

m0 1−

v2 c2

Analyse and interpret some of Einstein’s thought experiments involving mirrors and trains and discuss the relationship between thought and reality.

thought experiment. Observer O1 is standing on a train station platform equidistant from points A and B, which appear to O1 to be struck simultaneously by two bolts of lightning (Figure 3.3.1). Because the light travelled the same distance from both points and the light reached O1 at the same time, O1 judges that the lightning bolts struck simultaneously. Suppose observer O2 is sitting in a high speed train passing the platform without stopping. O1 calculates that at the moment the lightning struck, O2 on the train was also equidistant from A and B and so naively assumes that O2 would also see the events as simultaneous. However, by the time the light reached O1’s eyes, O2 was closer to B and had already seen the light from B but not A, and Only if two events occur simultaneously at the concluded B was struck first. same place, will all observers agree that they were simultaneous. Similar arguments can show that the order of events is relative. Since an effect must come after its cause, relativity places restrictions on possible chains of cause and effect. Because the speed of light is the universal speed limit, two events separated by distance cannot have any influence over each other over a time scale shorter than the time required for light to travel between them. Because no signal or influence can travel between a cause and its effect faster than light, apparent changes in the simultaneity or order of events based on the passage of light is more than just an optical illusion—it represents a fundamental limitation of reality.

Time dilation The relativity of simultaneity suggests that time itself is a bit rubbery. Relativity predicts that the rate of passage of time differs, depending on the velocity of the observer. Consider the following thought experiment (Figure 3.3.2). Suppose you (observer 0) are on a train, moving with speed v relative to the ground, while measuring the speed of light by shining a light pulse vertically towards a mirror on the train ceiling, a distance D from the light source. Outside on the ground is observer v who appears to you to be rushing past with horizontal speed v and watching your experiment. Both you and observer v regard your own reference you both agree on three things: (1) the speed frame as stationary. However, of light, (2) your relative horizontal speed v and (3) the height of the mirror D. You both agree on the height of the mirror because you are both in the same reference frame with respect to vertical components. Your light source also contains a detector capable of timing the interval t0 between the emission and detection of the light pulse. You do the experiment and use the speed formula c = 2D/t0 to obtain the correct speed of light. a

b

v

v mirror

mirror D source and detector

vtv source

detector

Figure 3.3.2 Measuring the speed of light on a train as seen by (a) observer 0 within the train and (b) observer v from the reference frame on the ground 66

space Observer v disagrees about what happened. She was carrying an accurate stopwatch and timed the event independently, getting a longer time interval of tv. Using Pythagoras’ theorem, she calculates that the path length of the light (Figure 3.3.2b) was not 2D, but rather: 1 2

2 D 2 + ( vt v )2 = 4 D 2 + (vt v )2

She agrees on the speed of light c so: 4 D 2 + (vt v )2

c=

tv

Square and rearrange: c 2tv2 = 4D 2 + (vtv)2  but observer 0 says c = 2D /t0 Eliminate D: Rearrange:

c 2tv2 = c 2t02 + (vtv)2 tv2(c 2 – v 2) = c 2t02 tv =

t0 1−

v2 c2

The factor 1 / 1 − v 2/ c 2 is called the Lorentz factor (abbreviated as γ). It is always larger than 1, which means that if t0 is the time between ticks on observer 0’s clock then observer v will see observer 0’s clock whiz by at speed v, ticking more slowly with a time tv between ticks. This is time dilation. The word dilation means ‘spreading out’, just like the time between clock ticks. An observer moving relative to you and observing a clock ticking (or any series of events) stationary in your frame, will judge events to happen more slowly than you observe them. Note that the dilation is only observed in clocks in other You can’t observe time dilation in clocks in your own frames of reference. frame, no matter how fast others think you’re moving. the effect is symmetrical; that is, As there is no preferred inertial frame, observers can be swapped. You both agree on your relative speed v but both insist the other observer is moving and their clock is running slow. You are both right because time is relative. A time interval observed on a clock that is stationary relative to the observer is To generalise this idea, t0 called the proper time for that reference frame. can represent the time between any two events (such as the ticks of a clock) that occur in the same place in the frame of the observer. If the events are separated in space, then extra time is required to allow for light to travel between the positions of the two events. Global positioning system (GPS) receivers estimate your position by measuring how long it takes the GPS signal to travel from the satellites to your receiver. The orbital speed of the satellites is large enough that the calculation needs to take time dilation into account. (It also takes into account a larger effect from general relativity: time runs more slowly in a stronger gravitational field.) The Lorentz factor γ approaches infinity as speed approaches the speed of light (Figure 3.3.3). This means that time in a frame of reference approaching the speed of light (relative to the observer) will come to a complete stop. In other words, nothing can be seen to happen in such a frame, which is another reason why the speed of light is the ultimate speed limit.

7 6 5 tv 4 t0 3 2 1 0

0

0.2

0.4 0.6 Speed VC

0.8

1.0

Figure 3.3.3 Plot of the ratio of a time interval in a moving reference frame to proper time (tv/t0) versus speed in units of c. Note that at speeds below ~0.1c, the ratio ≈ 1, so time behaves nearly classically. 67

3

Seeing in a weird light: relativity

Worked example

Muon and on and on

F

ast-moving muons are produced in the upper atmosphere by cosmic ray bombardment. Time dilation extends their normally short lifetimes long enough to allow many of them to make it to Earth, where they are a significant component of Earth’s background radiation.)

Figure 3.3.4 Both twins think the other’s clock is moving slower, but who is older at the end of the journey?

question A muon is like a heavy electron, and at low speed it decays with a mean lifetime of 2.2 × 10–6 s. Suppose a beam of muons is accelerated to 80% of the speed of light. What would their mean lifetime be in the laboratory reference frame?

Solution Lifetime in the muon’s frame: t0 = 2.2 × 10–6 s Speed of muon’s frame: v = 0.80c t0 Lifetime in the laboratory frame is tv: t v = 1−

v

2

=

2.2 × 10−6 2

1 − 0.80

= 3.7 × 10−6 s

2

c The twin paradox When two people pass by quickly, observing each other, they both think the other’s clock is running slower. The principle of relativity says you are both right. The twin paradox is a thought experiment in special relativity. Bill goes for an intergalactic cruise travelling at close to the speed of light (in Earth’s frame), while Phil stays on Earth (Figure 3.3.4). During the flight, they both correctly conclude that the other twin’s frame is moving, and so he is ageing more slowly. But what happens when Bill comes back home? Observations in the same frame should agree. It turns out that Bill is younger than Phil. Does this violate the principle that all inertial frames of reference are equivalent? No. Bill turned around (accelerated) to come home. The situation is no longer symmetrical. Special relativity isn’t enough to explain what Bill saw from his accelerating frame (he needs general relativity). However we have no difficulty talking about what (non-accelerating) Phil saw. By turning around and coming back, Bill left his original inertial frame and re-entered Phil’s frame, so he should agree with Phil. Phil remained in his inertial frame all along, so his conclusions (that Bill was moving and so is younger) have been consistent with special relativity throughout and, in his frame, correct. If instead Phil had hopped into another craft and caught up with Bill’s inertial frame, then Bill’s original conclusion would have been correct and Phil would have been younger. This prediction has been confirmed using highly precise, twin atomic clocks and an aeroplane.

Checkpoint 3.3 1 2 3 4 5 6 7 8

68

State Einstein’s first postulate and its alternative name. State Einstein’s second postulate and its alternative name. Outline why Newton’s classical mechanics is so successful despite a fundamental error (the Galilean transformation). Explain why the speed of light places restrictions on possible chains of cause and effect. Write the formula for time dilation. A clock moving towards you appears to slow down. If the clock were moving in the opposite direction, would it speed up? What is the name given to a time interval measured on a clock that is stationary in your frame of reference? In the twin paradox, during a period of constant relative motion, both Bill (astronaut) and Phil (earthling) observe the other twin’s watch ticking more slowly. Who’s observation is actually correct?

space

3.4 Length, mass and energy The formula for time dilation has already upset our common sense. However, once the clocks start talking to the rulers and the masses, things can only get more bizarre.

Solve problems and analyse information using: E = mc2 l v = l0 1 −

Length contraction There is a grain of truth in Lorentz and Fitzgerald’s suggestion (section 3.2)

tv =

that the arm of a Michelson interferometer contracts by a factor of 1 − v 2/ c 2 in the direction of motion. Their formula was correct, but their interpretation that it resulted from motion through the (non-existent) aether was wrong. Also, the contraction doesn’t happen in the frame of reference of the experimenter. Moreover, their hypothesis was ‘ad hoc’; it was designed only to patch a hole in the old theory without resulting in any additional testable predictions. So Einstein re-interpreted their mathematics in light of his theory of relativity. If an object is moving with speed v relative to the observer, the length of the object in the direction of that motion will be observed to be contracted according to the formula: l v = l0 1 −

v2 c2

m0 1−

v2 c2

Explain qualitatively and quantitatively the consequence of special relativity in relation to: • the relativity of simultaneity • the equivalence between mass and energy • length contraction • time dilation • mass dilation.

v2 c2

where l0 is the length judged by an observer who is stationary relative to the object (proper length) and lv is the length judged by an observer in a frame The length contraction only moving with speed v relative to the object. takes place in the dimension parallel to the motion. Just like time dilation: 1 the effect is symmetrical, which means the observers can be swapped—both insist it is the other person’s ruler that is too short 2 you cannot observe a Lorentz contraction within your own frame.

c2

t0 1−

mv =

v2

1.2 1 0.8 lv 0.6 l0 0.4

Imagine that observer 1 and observer 2 are trying to measure the length of 0.2 a rod, but all they have is a stopwatch. They already know accurately (and agree on) their relative speed v. Observer 1 is holding the rod and observer 2 is holding 0 0 0.2 0.4 0.6 0.8 1.0 the stopwatch. They whoosh past each other almost touching, both looking at Speed VC the watch. Figure 3.4.1 Plot of the ratio of length in a Observer 2 is stationary relative to the watch (Figure 3.4.2a), so he knows the moving reference frame to reading on his watch is his proper time. As the rod passes by, the watch reads proper length (lv/l0) versus zero at the start of the rod and t2 at the end, so the rod took a time t2 to pass by. speed in units of c. Note that as Therefore he calculates that the length of the rod in his frame is lv = vt2. speed approaches c, lv shrinks to zero—another reason why the Observer 1 is stationary relative to the rod (Figure 3.4.2b), so she knows that speed of light is unattainable. its length for her is the proper length l0. She agrees that the watch says t2, but the moving watch seemed to be ticking too slowly, so the number on the watch must be too small. Using the time dilation formula, she calculates that the time t1 in her frame a b was longer: t2 v t1 = v 2 v 1− 2 c Figure 3.4.2 Measuring the length of a rod using a stopwatch as seen by (a) observer 2, 60

55

60

60

5

10

50

15

45

20

40

35

30

25

5

55

10

50

15

45

20

40

35

30

25

holding the watch, and (b) observer 1, holding the rod 69

3

Seeing in a weird light: relativity

a

B

A

Observer 1 then calculates that the length of the rod is l0 = vt1 or vt 2 . l0 = v2 1− 2 c But observer 2 says that lv = vt2, so by substitution and rearrangement

b v

C' A'

B' c

C

question v

B

A'

B'

c2

Worked example

D'

C'

v2 .

l v = l0 1 −

The distance travelled by light in one year, 9.46 × 1015 m, is called a light-year (ly). The nearest star to our Sun is Proxima Centauri, 4.2 light-years away. Suppose you are travelling to Proxima Centauri at three-quarters of the speed of light. a Calculate how long it takes to get there from Earth (measured using your on-board clock). b Discuss whether this answer is a contradiction.

O

Figure 3.4.3 A fast-moving vehicle appears contracted horizontally, but also rotated away from the observer. The car is depicted when (a) stationary, (b) moving at high speed and (c) viewed from above. Corner C is normally out of sight, but at high speed, the vehicle moves out of the way fast enough to allow light reflected from C to reach your eyes at O, allowing you to see the car’s back and side at the same time. This is called Terrell–Penrose rotation.

Solution a Both you and Earth-bound observers agree on your relative speed 0.75c. In the spaceship’s frame, the distance to Proxima Centauri is contracted: l v = l 0 1−

t=

v2 c

2

= 4.2 1 − 0.752 = 2.78 ly

l v 2.78 ly = = 3.7 years c 0.75c

3.7 years is less than the 4.2 years that light takes to get there in Earth’s frame. b This is not a contradiction because in the spaceship’s frame, light would only take 2.78 years because lv = 2.78 ly. a

Earth

Proxima Centauri

v

b v

v Earth

Proxima Centauri

Figure 3.4.4 Trip to Proxima Centauri as seen by (a) earthlings and (b) the astronauts

Discuss the implications of mass increase, time dilation and length contraction for space travel.

70

Note that in the last example, the astronauts thought they experienced a short trip because the distance travelled was contracted, whereas the earthlings thought the astronauts felt their trip was short because their time had slowed.

space

Relativistic mass If you measure the mass m0 of an object at rest in your frame (rest mass or proper mass) and use the classical definition of momentum p = m0v, then in collisions, momentum is not necessarily conserved for all reference frames. However, momentum is conserved if one instead uses p = mvv where mv is the relativistic mass: m0 mv = v2 1− 2 c The relativistic mass of an object increases as its speed relative to the observer increases. As speed approaches c, the mass approaches infinity, so the force required to accelerate an object to the speed of light becomes infinite. This is yet another reason why the speed of light cannot be reached. When accelerating particles in accelerators, this increase in mass needs to be taken into account, otherwise the machines won’t work. 7 6

T

rains A and B are about to collide head-on, each with a speed 0.5c relative to the station. So, relative to train B, train A is moving at the speed of light, right? Wrong! The replacement for Galileo’s relative velocity rule in 1‑dimension is:

v (A rel. to B) =

vA − v B v v 1 − A 2B c

The speed of train A relative to train B is: 0.5c − (−0.5c ) = 0.8c 0.5c × (−0.5c ) 1− c2

5 mv 4 m0 3 2

Figure 3.4.5 Plot of the ratio of relativistic mass

1 0

Relativistic train crash

0

0.2

0.4 0.6 Speed CV

0.8

mv in a moving reference frame to rest mass m0 versus speed in units of c. As speed approaches c, the relativistic mass approaches infinity.

1

Worked example question A medical linear accelerator (linac) accelerates a beam of electrons to high kinetic energies. These electrons then bombard a tungsten target, producing an intense X-ray beam that can be used to irradiate cancerous tumours. A typical speed for electrons in the beam is 0.997252 times the speed of light. Calculate the Lorentz factor and hence the relativistic mass of these electrons, given the rest mass is 9.11 × 10–31 kg.

Solution

Lorentz factor γ =

1 1−



mv =

m0 1−

v2



v

2

=

1 1 – 0.997252

= 13.5

c2 = 9.11 × 10–31 × 13.5 = 1.23 × 10–29 kg

c2

Note: When calculating Lorentz factors close to the speed of light, use a greater number of significant figures than usual, because you are subtracting two numbers of very similar size. 71

3

Seeing in a weird light: relativity Solve problems and analyse information using: E = mc2 l v = l0 1 − tv =

c2

t0 1−

mv =

v2

v2 c2

Mass, energy and the world’s most famous equation

The kinetic energy formula K =  1 mv 2 doesn’t apply at relativistic speeds, 2 even if you substitute relativistic mass mv into the formula. Classically, if you apply a net force to accelerate an object, the work done equals the increase in kinetic energy. An increase in speed means an increase in kinetic energy. But in relativity it also means an increase in relativistic mass, so relativistic mass and energy seem to be associated. Superficially, if you multiply relativistic mass by c 2 you get mv c 2, which has the same dimensions and units as energy. But let’s look more closely at it.

m0 1−

v2 c2

PHYSICS FEATURE Twisting spacetime ... and your mind 1. The history of physics

T

here are two more invariants in special relativity. Maxwell’s equations (and hence relativity) requires that electrical charge is invariant in all frames. Another quantity invariant in all inertial frames is called the spacetime interval. You may have heard of spacetime but not know what it is. One of Einstein’s mathematics lecturers Hermann Minkowski (1864–1909) showed that the equations of relativity and Maxwell’s equations become simplified if you assume that the three dimensions of space (x, y, z) and time t taken together form a four‑dimensional coordinate system called spacetime. Each location in spacetime is not a position, but rather an event—a position and a time. Using a 4D version of Pythagoras’ theorem, Minkowski then defined a kind of 4D ‘distance’ between events called the spacetime interval s given by:   s 2 = (c × time period)2 – path length2 = c 2t 2 – ((∆x)2 + (∆y)2 + (∆z)2) Observers in different frames don’t agree on the 3D path length between events, or the time period between events, but all observers in inertial frames agree on the spacetime interval s between events.

72

Figure 3.4.6 One of the four ultra-precise superconducting spherical gyroscopes on NASA’s Gravity Probe B, which orbited Earth in 2004/05 to measure two predictions of general relativity: the bending of spacetime by the Earth’s mass and the slight twisting of spacetime by the Earth’s rotation (frame-dragging)

In general relativity, Einstein showed that gravity occurs because objects with mass or energy cause this 4D spacetime to become distorted. The paths of objects through this distorted 4D spacetime appear to our 3D eyes to follow the sort of astronomical trajectories you learned about in Chapter 2 ‘Explaining and exploring the solar system’. However, unlike Newton’s gravitation, general relativity is able to handle situations of high gravitational fields, such as Mercury’s precessing orbit around the Sun and black holes. General relativity also predicts another wave that doesn’t require a medium: the ripples in spacetime called ‘gravity waves’.

space How does this formula behave at low speeds (when v 2/c 2 is small)? mv c 2 =

m0c 2 1−

v2

 v2  = m0c 2  1 − 2  c  

−

1 2

c2 Using a well-known approximation formula that you might learn at university, (1 – x )n ≈ 1 – nx for small x:  v2  m0c 2  1 − 2  c  

−

1 2

 1 v2  1 ≈ m0c 2  1 + × 2  = m0c 2 + m0v 2 2 2 c  

1 m v 2 2 0 In other words, at low speeds, the gain in relativistic mass (mv – m0) multiplied by c  2 equals the kinetic energy—a tantalising hint that at low speed mass and energy are equivalent. It can also be shown to be true at all speeds, using more sophisticated mathematics. In general, mass and energy are equivalent in relativity and c 2 is the conversion factor between the energy unit (joules) and the mass unit (kg). In other words:

Rearrange:

mvc 2 – m0c 2 = (mv – m0)c 2 ≈

E = mc 2 where m is any kind of mass. In relativity, mass and energy are regarded as the same thing, apart from the change of units. Sometimes the term mass-energy is used for both. m0 c 2 is called the rest energy, so even a stationary object contains energy due to its rest mass. Relativistic kinetic energy therefore: m0c 2 mv c 2 − m0c 2 = − m0c 2 2 v 1− 2 c Whenever energy increases, so does mass. Any release of energy is accompanied by a decrease in mass. A book sitting on the top shelf has a slightly higher mass than one on the bottom shelf because of the difference in gravitational potential energy. An object’s mass increases slightly when it is hot because the kinetic energy of the vibrating atoms is higher. Because c 2 is such a large number, a very tiny mass is equivalent to a large amount of energy. In the early days of nuclear physics, E = mc 2 revealed the enormous energy locked up inside an atom’s nucleus by the strong nuclear force that holds the protons and neutrons together. It was this that alerted nuclear physicists just before World War II to the possibility of a nuclear bomb. The energy released by the nuclear bomb dropped on Hiroshima at the end of that war (smallish by modern standards) resulted from a reduction in relativistic mass of about 0.7 g (slightly less than the mass of a standard wire paperclip).

Evil twins

T

he most extreme mass–energy conversion involves antimatter. For every kind of matter particle there is an equivalent antimatter particle, an ‘evil twin’, bearing properties (such as charge) of opposite sign. Particles and their antiparticles have the same rest mass. When a particle meets its antiparticle, they mutually annihilate—all their opposing properties cancel, leaving only their mass-energy, which is usually released in the form of two gamma-ray photons. Matter– antimatter annihilation has been suggested (speculatively) as a possible propellant for powering future interstellar spacecraft.

Discuss the implications of mass increase, time dilation and length contraction for space travel.

Worked example question When free protons and neutrons become bound together to form a nucleus, the reduction in nuclear potential energy (binding energy) is released, normally in the form of gamma rays. Relativity says this loss in energy is reflected in a decrease in mass of the resulting atom. 73

3

Seeing in a weird light: relativity

Exploding a myth

I

t is commonly believed (wrongly) that Einstein was involved in the US nuclear bomb project. Perhaps this is because, during World War II, the nuclear physicists Leo Szilard, Eugene Wigner and Edward Teller, knowing such a bomb was possible and worried the Nazis might build one, wrote a letter to President Roosevelt suggesting the US beat them to it. They asked their friend Einstein to sign it because, being the most well-known scientist at the time, he would be taken seriously. Apart from that, Einstein did two days’ work on the theory behind uranium enrichment.

Calculate how much energy is released when free protons, neutrons and electrons combine to form 4.00 g of helium-4 atoms (2 protons + 2 neutrons + 2 electrons). At room temperature and pressure, each 4 g of helium gas is about 25 L, roughly the volume of an inflatable beach ball. Data:

Mass of proton mp = 1.672622 × 10–27 kg



Mass of neutron mn = 1.674927 × 10–27 kg



Mass of electron me = 9.11 × 10–31 kg



Mass of helium atom mHe = 6.646476 × 10–27 kg



c 2 = 8.9876 × 1016 m2 s–2

Solution Total mass of the parts:

mT = 2(mp + mn + me) = 2(1.672622 + 1.674927 + 0.000911) × 10–27 kg



= 6.69692 × 10–27 kg

Reduction in mass:

∆m = mT – mHe = (6.69692 – 6.646476) × 10–27 kg



= 5.0444 × 10–29 kg

Binding energy per He atom:

∆E = ∆mc 2 = 5.0444 × 10–29 kg × 8.9876 × 1016 m2 s–2



= 4.5337 × 10–12 J

Binding energy for 4.00 g (0.004 kg): 4.5337 × 10–12 J × 0.004 kg = 2.73 × 1012 J mHe This much energy would be released by the explosion of more than 600 tonnes of TNT.

Some physicists dislike the definition of relativistic mass mv of a moving object and prefer to talk only about the energy of an object (and its rest mass m0). There are problems with the definition, including the fact that relativistic mass doesn’t behave like a scalar, because it can be different along different directions.

Checkpoint 3.4 1 2 3 4 5 6 7 8

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Discuss why, if Lorentz and Fitzgerald came up with the correct formula for length contraction, Einstein gets the credit for explaining relativistic length contraction. Write the formula for length contraction. Would a ruler moving lengthwise relative to you appear shorter or longer? Define the term proper length. To what limit does observed length of a moving object tend as speed approaches c? Write the formula for relativistic mass. Would a mass moving relative to you appear larger or smaller? Use relativistic mass to justify the statement that the speed of light is the universal speed limit. Define all the terms in the equation E = mc  2 and explain what the equation means. Explain why an atom weighs less than the sum of its parts.

PRACTICAL EXPERIENCES

space

CHAPTER 3

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

60

20

10

0

20 30 40 0 10 50

90 8

0 70 60 50

40

30

Discussion questions 1 The principle method for detecting a non-inertial frame is measurement of acceleration. Describe an example of a non-inertial frame in which a typical accelerometer would not appear to measure an acceleration or detect extra fictitious forces. 2 Is there a test that can be performed within a frame of reference to tell if the effect measured by the accelerometer is the result of acceleration of the frame or due to an actual additional force?

80

Perform an investigation that allows you to distinguish between inertial and non-inertial frames of reference. Equipment: protractor, string, mass (50 g), tape, cardboard, chair on wheels or skateboard.

Perform an investigation to help distinguish between non-inertial and inertial frames of reference.

70

Activity 3.1: Fact or fiction: Inertial and non-inertial frames of reference

Figure 3.5.1 An accelerometer

Activity 3.2: Interpreting the Michelson–Morley experiment results Use simulations to gather data from the Michelson–Morley experiment. You will gather data as though there is and is not an aether, and then interpret the results. There are many Michelson–Morley experiment simulations available. Two web-based examples are given on the companion website.

Gather and process information to interpret the results of the Michelson–Morley experiment.

Discussion questions 1 Describe what Michelson and Morley were expecting to observe if aether were present. 2 Using the data you have gathered, explain how your observations support or refute the existence of the aether. 3 Recall the interpretation put forward by Michelson and Morley. 4 Discuss the importance of this experiment. Extension 1 Research the history of how long the belief in aether persisted in some physicists after the publication of special relativity in 1905. 2 Read the following paper, which contains a thorough review of the history of the Michelson–Morley experiment, including historical letters to and from several researchers: Shankland, R S, 1964, ‘Michelson–Morley Experiment’, American Journal of Physics, vol. 32, p 16.

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3 • •

• •

•

•

• • •

•

•

Chapter summary

Seeing in a weird light: relativity

Inertial reference frames are those that do not accelerate. Principle of relativity: The laws of mechanics are the same in all inertial reference frames. Einstein extended it to all laws of physics (first postulate of relativity). When judged within a non-inertial frame, fictitious forces are perceived. Maxwell’s equations for electromagnetism predicted only a single possible speed for light, which was assumed to be relative to a hypothetical medium called aether. Michelson and Morley failed to detect changes in speed due to aether wind, using an interferometer. Fitzgerald and Lorentz made the ad hoc suggestion that things contract when moving relative to the aether, hiding the effect of the changing relative speed of light. Einstein and others argued that aether was not required by Maxwell’s equations and was inconsistent with the principle of relativity. Second postulate of relativity: The speed of light is constant to all observers. The speed of light is the fastest possible speed. The finite speed of light means different observers disagree on the simultaneity and order of events. Only events at the same time and place are agreed by all observers to be simultaneous. 1 Lorentz factor: γ = v2 1− 2 c Proper time t0 is a time interval measured on a clock stationary in the observer’s frame.

• • •

•

•

• •

•

Proper length l0 is the length of an object stationary in the observer’s frame. Proper or rest mass m0 is the mass of an object stationary in the observer’s frame. t0 Time dilation: t v = v2 1− 2 c Clocks (and all time-dependent phenomena) evolve in time more slowly if they are moving relative to the observer’s frame. v2 Length contraction: l v = l 0 1 − 2 c Length lv of an object moving relative to the observer’s frame contracts in the direction of motion. m0 Relativistic mass: mv = v2 1− 2 c Mass of an object mv moving relative to the observer’s frame increases. Two observers in separate inertial frames will agree on their relative speed v. However, both observers will judge the other observer to be moving and, hence, subject to time dilation, length contraction and relativistic mass increase. They disagree, but both are correct because these three quantities are relative. Only when two observers are in the same frame will they agree on these. Mass and energy are equivalent: E = mc 2. A small mass is equivalent to a large energy.

Review questions Physically speaking Use the words below to complete the following paragraph:

Inertial _______________ have _______________ status in _______________ mechanics. _______________’s laws apply in these frames. If one performs measurements

Galileo, Newton, Einstein’s, Maxwell,

in _______________ , then _______________ forces might be perceived. Classical mechanics and _______________ relativity both agree that physical laws are

constancy, fictitious, change,

_______________ in _______________ frames. However, they disagree on the

non-inertial frames, length, observer, classical, value, invariant, mass, time, frames, speed, inertial, special

_______________ of the speed of light. According to _______________’s equations,

the _______________ of the speed of light does not _______________ between frames, so light doesn’t obey the transformation formula of _______________ . Because of this, measurements of _______________ , _______________ and _______________ within a reference frame moving relative to the _______________ , will depend on the _______________ of that frame.

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space

Reviewing 1 You have a priceless Elvis Presley doll hanging from your rear-vision mirror at a constant angle from vertical. Elvis’s feet lean towards the front of the car. Are you driving: A forwards at uniform speed? B backwards at uniform speed? C forwards but accelerating? D forwards but decelerating?

Solve problems and analyse information using: E = mc2 l v = l0 1 − tv =

exerted on you by the seat belt fictitious? Centrifugal force normally refers to the fictitious force you feel pushing you outwards when you steer a car. Some people have suggested re-defining centrifugal force as the outward reaction force you exert on the seat belt in response to the centripetal force it exerts on you. Re-defined in this way, is centrifugal force still fictitious? Justify your answers.

3 At the end of the 19th century, no-one was able to travel at close to the speed of light, and clocks, rulers and mass balances weren’t sensitive enough to measure relativistic changes. So why did the problems with classical physics start to become obvious then?

4 Explain why interferometry is an extremely sensitive method for measuring short differences in time or length.

5 Explain why Michelson and Morley performed their experiment at different times of the day and year.

6 If we were an entire civilisation of blind people relying on sound instead of light to decide the simultaneity of events, would our equations for relativistic length, time and mass contain c = 340 m s–1 (the speed of sound in air) instead? What’s so special about the speed of light? Discuss.

7 In Figure 3.3.2b, the dimensions of the light path have been drawn correctly. However, for simplicity, two aspects of the train’s appearance to observer v have been left out. Describe two changes that would need to be made to Figure 3.3.2b to represent these effects more correctly.

8 Suppose our relativistic twins Bill and Phil both got into spacecraft, went off in opposite directions and took journeys at relativistic speeds that were mirror images (judged from Earth). Predict and explain: a how their apparent ages will compare when they come back home b how their apparent ages will be judged by stay-athome earthlings.

mv =

c2

t0 1−

2 In a car that is cornering, is the centripetal force

v2

v2 c2

m0 1−

v2 c2

9 Prunella and Renfrew, two observers in inertial frames moving relative to each other, will always agree on their relative speed v. A third observer, Thor, standing between them, sees them both coming towards him from opposite directions, at equal speeds. Is it correct to say that relative to Thor, Prunella and Renfrew are both moving at a speed of v|2?

10 A stretch-limo drove into a small garage at near light speed. The garage attendant slammed the garage door behind the car. For a brief time the attendant saw that the relativistically shortened limo was completely contained between the closed garage door and the rear garage wall. A short time later, the stillmoving car smashed through the back wall. As far as the driver was concerned, the garage was shortened and the limo was too long for the garage so the limo was never contained between a closed door and the intact back wall. Reconcile the two differing accounts of what happened. (Hint: See section 3.3.)

11 Show that mc2 has the units and dimensions of energy. 12 In a perfectly inelastic collision, two colliding objects stick together. In a symmetrical inelastic collision between two identical objects, the final speed is zero in the frame of their centre of mass. Given that massenergy is conserved in an inertial frame, is the mass of the system the same as before the collision? Explain. (Hint: What happens to kinetic energy in an inelastic collision?)

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Seeing in a weird light: relativity

Solving Problems 13 Depending on your answer to Question 1, calculate the magnitude of your speed or acceleration if the Elvis Presley doll hangs at a constant angle of 10° from vertical.

Solve problems and analyse information using: E = mc2 l v = l0 1 −

14 The caption for Figure 3.2.3b states that increasing the length of the arms would increase sensitivity to changes in the speed of light. Justify this, using the equations given in that section.

1−

L ′ = L 1− v 2 c2 then the difference t2 – t1 between the times of flight for the two arms would be zero. Use the equations given for t2 and t1 in section 3.2.

16 In the worked example of your trip to Proxima Centauri (Figure 3.4.4), one member of the crew had a mass 80 kg at launch. Assuming his normal diet and physiology were maintained, what would you expect his mass to be during the trip: a as measured on the spaceship? b as judged from Earth?

17 Your rival in the space race plans a trip to Alpha Centauri, which is slightly further away (4.37 ly). She wants to do the trip in 3.5 years (one-way) as judged by her own on-board clock. a What speed (as a fraction of c) does she need to maintain? b How long does the trip take as judged from Earth?

19 For subatomic particles, a more conveniently sized (non-SI) unit of energy is the electron volt (eV). The conversion is E(eV) = E(J)/e where e = 1.60 × 10–19 C, the charge on an electron. A mega-electron volt (MeV) is 106 eV.

For the worked example on page 71, show that the kinetic energy of the electron in the medical linac beam is 6.4 MeV (me = 9.11 × 10–31 kg). What is the total energy of that electron?

20 Estimate the total energy (in joules) released by the

Re

78

iew

Q uesti o

n

s

v

Hiroshima bomb (∆m0 = 0.7 g).

c2

1−

v2 c2

21 In their rest frame, muons have a mean lifetime of

2.2 × 10–6 s. However, measurements (at various altitudes) of muons produced by cosmic rays indicate that, on average, they travel 6.00 × 103 m from where they are produced in the upper atmosphere before decaying. Calculate their average speed (as a fraction of c).

22 Show that if the speed of light were infinite, the following equations would revert to their classical form. a b

c

18 Calculate the total energy in the two gamma ray photons produced when an electron meets a positron (an anti-electron) (me = 9.11 × 10–31 kg).

v2

m0

mv =

that (in agreement with Fitzgerald and Lorentz’s suggestion) if the length L of the interferometer arm parallel to the aether wind shrinks to

c2

t0

tv =

15 Supposing the aether hypothesis were correct, show

v2

d

/

2 2

l v = l0 1 − v c tv =

t0

/

2 2

1−v c

mv =

m0

/

2 2

1− v c

v (A rel. to B) =

vA − vB

1 − vA vB / c 2

23 Research the history of relativity and list up to five historically important experimental confirmations of its predictions. Make a timeline of the events. Note that some experiments may pre-date relativity. For example, in 1901 W Kaufmann measured the increase in an electron’s mass as its speed increased. If possible, identify whether such examples came to Einstein’s attention before he formulated his theory. Analyse information to discuss the relationship between theory and the evidence supporting it, using Einstein’s predictions based on relativity that were made many years before evidence was available to support it.

space

PHYSICS FOCUS Can’t measure the speed of light

T

he French metric system, which evolved into the Système International d’Unités or SI units, was originally based on ‘artefact’ standards. The standard metre bar and kilogram were real objects (or artefacts) in Paris. Artefacts can degrade or be damaged, and making copies for standards labs is expensive, slow and unreliable. Artefact standards are now being replaced by fundamental physical property standards. One second is now defined as a certain number of periods of oscillation of a very stable light frequency in the spectrum of cesium-133 (in atomic clocks). Measurement standards often involve sensitive interferometry. The metre was changed in 1960 from the original bar to a certain number of wavelengths (measured interferometrically) of a colour from the krypton-86 spectrum. Being invariant, the speed of light is very useful for standards. Interferometric measurements of the speed of light became so precise that the weakest link was the experimental difficulty in reproducing the So in 1983 the krypton-86 standard metre. speed of light was fixed by definition at exactly 299 792 458 m s–1 and the standard metre was redefined as the distance travelled by light in 1|299 792 458 of a second. Now, any lab with an interferometer and an atomic clock can produce its own standard metre. Since 1983, by definition, the speed of light can no longer be measured. Traditional procedures for measuring the speed of light should now be called ‘measuring the length of a metre’. The last artefact standard, the platinum–iridium kilogram in Paris, appears to be changing mass slightly. The Avogadro project at Australia’s CSIRO is trying to develop a replacement for it with a procedure for making and testing (almost) perfect spheres of silicon that could be made in standards labs around the world without the need to copy the original sphere directly. The spheres are measured using interferometry, with the best result so far being an overall distortion from sphericity of 30 nm and an average smoothness of 0.3 nm.

2. The nature and practice of physics

3. Applications and uses of physics

5. Current issues, research and developments in physics Discuss the concept that length standards are defined in terms of time in contrast to the original metre standard.

1 Explain why standards based on fundamental physics properties are preferable to artefacts. 2 Justify (in light of relativity) the statement that the speed of light is an especially good property on which to base a measurement standard. 3 The 1960 metre standard was based on light from  krypton-86. Explain why it needed to specify the light source and why the new metre standard doesn’t. 4 Given that the value of the speed of light is now arbitrarily fixed, discuss why they didn’t just make the speed of light a nice round number such as 3.000 000 00 × 108 m s–1. 5 A single atomic layer of silicon is approximately 5.4 × 10–10 m thick. For the best silicon sphere in the Avogradro project, to approximately how many atomic layers does the reported distortion from sphericity and average smoothness correspond?

Figure 3.5.2 One of CSIRO’s accurate silicon spheres

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1

The review contains questions in a similar style and proportion to the HSC Physics examination. Marks are allocated to each question up to a total of 30 marks. It should take you approximately 54 minutes to complete this review.

Multiple choice (1 mark each) 1 Ignoring air resistance, all projectiles fired horizontally from the same height above horizontal ground will have the same: A horizontal velocity. B time of flight. C range. D final speed.

2

Which of the following orbits has a two-body mechanical energy greater than zero? A Geostationary B Elliptical C Parabolic D Non-returning comet

3

You have just rounded the top of a curve on a rollercoaster. The g-force meter you are carrying reads exactly zero. Which one of the following is true? A Your weight is the centripetal force. B Your weight is zero. C Your weight is equal and opposite to the normal force exerted on you by the seat. D Your weight is equal and opposite to the tension in your body.

4

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The Michelson–Morley experiment demonstrated that: A the aether wind was undetectable. B waves do not require a medium. C one arm of the interferometer contracted in response to the aether wind. D aether is trapped by mountains and valleys and dragged along with the Earth.

5

Observer A on the ground, watches a train (containing observer B) rush past at speed v. Both make measurements of things in each other’s frame of reference. From the following list of statements, choose the statement they disagree on. A The other observer’s frame of reference is moving with speed v. B The apparent length of my own metre ruler is longer than the apparent length of other observer’s metre ruler. C Observer B’s watch appears to run slower than observer A’s watch. D The height of the train carriage ceiling is 2.2 m above the carriage floor.

Short response 6 The escape velocity from the Earth’s surface, based

on Newton’s original concept, is 11.2 km s–1. Briefly explain two ways in which this number is not quite applicable to real Earth-surface launches. (2 marks)

7 Calculate the potential energy of a 2500 kg satellite in a geostationary orbit around the Earth. Assume a sidereal day is 23 h 56 min 4 s.  (3 marks)

8 In their rest frame, charged pions have a mean

lifetime of 2.60 × 10–8 s. A particular beam of charged pions travel an average distance of 30 m before decaying. Calculate their speed (as a fraction of the speed of light).  (4 marks)

9 Explain why if you are in a circular orbit and you briefly retro-fire your engines to slow down, you move to a faster orbit.  (3 marks)

space 10 A 9000 kg helicopter is parked at the equator and then later near the North Pole. Assuming the two locations are chosen so that the gravitational potential energy is the same at the two spots, estimate the difference in relativistic mass at the two locations. At which location will the mass be larger? (Hint: Velocity is low, so use the classical expression for kinetic energy.)  (3 marks)

Extended response 11 Critically discuss the following proposition: ‘The Michelson–Morley experiment was an embarrassment for physics because, despite a large effort, it failed to find what it was looking for and so it should be relegated to the dustbin of physics history.’ (5 marks)

12 The following formula relates the length of a

pendulum (L) to the period of its swing (T). L T = 2π g During your studies in physics you carried out an experiment to determine acceleration due to gravity. Describe and explain a method you would use to perform this measurement.  (5 marks)

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2 Context

Figure 4.0.1 A generator produces electricity in each of these wind turbines.

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Motors and Generators The first recorded observations of the relationship between electricity and magnetism date back more than 400 years. Many unimagined discoveries followed, but progress never waits. Before we understood their nature, inventions utilising electricity and magnetism had changed our world forever. Today our lives revolve around these forms of energy. The lights you use to read this book rely on them and the CD inside it would be nothing but a shiny coaster for your cup. We use magnetism to generate the electricity that drives industry, discovery and invention. Electricity and magnetism are a foundation for modern technology, deeply seated in the global economy, and our use impacts heavily on the environment. The greatest challenge that faces future generations is the supply of energy. As fossil fuels dry up, electricity and magnetism will become even more important. New and improved technologies will be needed. Whether it’s a hybrid car, a wind turbine or a nuclear fusion power plant, they all rely on applications of electricity and magnetism.

Figure 4.0.2 A simple homopolar motor

INQUIRY ACTIVITY Build your own electric motor Many of the devices you use every day have electric motors. They spin your DVDs, wash your clothes and even help cook your food. Could you live without them, and how much do you know about how they work? The essential ingredients for a motor are a power source, a magnetic field and things to connect these together in the right way. It’s not as hard as you think. All you need is a battery, a wood screw, a piece of wire and a cylindrical or spherical magnet. Put these things together as shown in Figure 4.0.2 and see if you can get your motor to spin. Be patient and keep trying. Then try the following activities. 1 Test the effects of changing the voltage you use. You could add another battery in series or try a battery with a higher voltage. 2 Try changing the strength of the magnet by using a different magnet or adding another. What does this affect? 3 Try changing the length of the screw, how sharp its point is or the material it is made from. Does it have to be made of iron?

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4

Electrodynamics: moving charges and magnetic fields Strange but true

electric current, conventional current, magnetic field, force, direct current (DC), alternating current (AC), voltage, potential difference, right-hand grip rule, right-hand palm rule, motor effect, current-carrying conductor

Some of the most surprising discoveries in science have come from the relationship between electricity and magnetism. Who would have thought that a current-carrying wire can act as a magnet? Could anyone have guessed that moving a magnet could create electricity, or that moving charges get a push from magnetic fields? All these amazing facts were discovered by people who were curious, creative and dedicated. They were interested enough to find out how things work, and now it’s your turn. Dust off your imagination and get ready to picture the concepts behind many applications of electricity and magnetism.

4.1 Review of essential concepts In this module you will develop and apply an understanding of the relationship between electricity and magnetism. First we need to review some essential concepts you encountered in the Preliminary course (see in2 Physics @ Preliminary Chapters 10 and 12). You will need to recall these concepts throughout this module and apply them to new situations.

Electric current In the presence of an electric field, charged particles and free ions will move towards a region with an opposite electric charge. In this module we will be considering the movement of electrons within metal conductors such as copper. If you connect a wire to the terminals of a battery, as in Figure 4.1.1, an electric field is set up along the length of the wire between the two ends. The free electrons in the metal wire are then attracted towards the positive terminal. The movement of these electrons is called an electric current. Protons are bound tightly inside the nuclei of atoms and the nuclei are essentially fixed in position. This means that when a wire is connected to a battery the only charges free to move inside the wire are the free electrons. 84

motors and generators An electric current is defined as the rate of flow of net charge through a region. In the wire in Figure 4.1.1, a number of electrons will flow through the area A each second, and each electron carries a charge of –1.6 × 10–19 coulombs. If the current is 1 ampere (or 1 amp), it carries 1 coulomb of charge per second through area A, which is more than 6 billion billion electrons each second. We now know that the nature of an electric current is actually a flow of electrons. However, in electric circuits we often consider currents as if they were a flow of positive particles. This type of current is called conventional current and it is in the opposite direction to the flow of the negatively charged electrons. This confusing situation has arisen because current was first thought to be a flow of positive particles. Conventional methods for determining the direction of other physical quantities, such as magnetic fields and forces, have been developed using the conventional direction of current. So we will stick with these historical conventions. It is important throughout this module to consider the direction of an electric current as the direction in which a positive charge would flow through a conductor. In Figure 4.1.1, conventional current flows from the positive terminal towards the negative terminal, as indicated by the arrow along the wire. An electrical current in which the charges only flow in one direction is called direct current (DC). This current is commonly used in small portable electronic devices and is supplied by a battery. One way of illustrating this type of current graphically is shown in Figure 4.1.2. The red line on the graph is a direct current measured by a digital ammeter. The sign of the current (+ or –) represents the direction in which the current is travelling. You can see that in this example the current has a constant value and direction over time. In contrast to direct current, an alternating current (AC) is continually changing direction. The sign of the terminal at each end of an AC circuit alternates between positive and negative over time. Each time this occurs, the electric field within the wire changes direction. This reverses the direction of the force on the charges within the wire and the current changes direction accordingly. This type of current is good for transporting electrical energy over large distances and is commonly used in larger appliances. In Figure 4.1.2, an AC current measured by a digital ammeter is shown as a blue line. As the current changes direction the blue line moves above or below the horizontal axis. The corresponding change in sign of the current indicates a change in the current’s direction.

Potential difference, emf and voltage The work done by any electrical device can be traced back to the creation of a difference in electrical potential energy. In Figure 4.1.1, a chemical reaction separates the charges inside the battery. This causes a difference in electrical potential energy between the positive and negative terminals of the The battery energy given to each coulomb battery. of charge within the battery is measured in volts and is commonly called an emf (ε). In an ideal battery it is equal to the voltage measured at the terminals when the battery is not connected to a circuit. For this type of battery this is usually about 1.5 V.

–

area A

–

–

–

+

– Figure 4.1.1 A battery creates an electric field within the wire and a current flows.

DC

+ AC Current (A)

Direct current and alternating current

copper wire

0

10 20

30

40

50

Time (ms)

–

Figure 4.1.2 A graph of AC and DC over time

Flick of a switch

T

he charges in a current-carrying wire travel along the wire much more slowly than the speed at which you normally walk. Why then does a light come on instantly when you flick the switch? All the free electrons in the wire start moving at the same time under the influence of the electric field in the wire. When you flick the switch, this field travels along the wire at close to the speed of light. Almost instantly all the free electrons are moving and losing energy in the light bulb. 85

4

Electrodynamics: moving charges and magnetic fields

Birds on a wire

H

ave you ever wondered why birds can happily sit on power lines and not get electrocuted? No, those wires are not insulated. For a current to flow through a bird on a wire, there would have to be a potential difference between its feet. If the bird could stand on the wire and touch any other object such as the ground or another wire then it would get the shock of its life.

Figure 4.1.3 There is no potential difference between the bird’s feet.

When a wire is connected to the battery’s terminals (see Figure 4.1.1), an electric field is set up within the wire. The electric field drives electrons from the negative terminal, through the wire to the positive terminal. As the electrons move through a circuit element, such as a light bulb, they collide with the ions in the metal lattice. During these collisions they lose kinetic energy to the metal lattice. The metal lattice then loses this energy as heat; in the light bulb it is also lost as visible light. The energy lost in the circuit element corresponds to a loss of electrical The difference in potential potential energy by the charges in the current. energy per unit of charge between the two points either side of a circuit element is known as the potential difference, potential drop or voltage (V). We can measure this potential difference by connecting a voltmeter to the circuit in parallel with the circuit element.

Resistance and Ohm’s law Recall that the structure of a metallic conductor is essentially a lattice of metal atoms (or ions) surrounded by a ‘sea’ of free electrons. If a potential difference V is established within the metal, these free electrons will flow as a current I. The amount of current that flows due to this potential difference is V determined by the electrical resistance of the material, which is defined by R = I and measured in ohms (Ω). A material that has a relatively high resistance will only conduct a relatively small current for a given potential difference. The resistance in a conductor is a result of the collisions of the moving charges with the ions of the metal lattice. Basically, the more collisions the free electrons have with the lattice the higher the resistance. Recall that the length, cross-sectional area, temperature and type of material within the conductor influence resistance. In many circuit components, the ratio of voltage divided by current is a This relationship is known as Ohm’s law and describes the constant. relationship in which V and I are proportional for a circuit, or circuit component, with a fixed resistance R.

Resistance and power The wire filament of the light bulb in Figure 4.1.1 has a very high resistance and, therefore, a large amount of the available energy within the circuit is lost within the bulb. This energy heats the filament in the blub to such a high temperature that it emits visible light. Whether energy in a circuit is lost as heat or in turning an electric motor, the rate at which energy is converted into another form is called electric power P : Power =

energy transferred time taken for transfer

Recall that: P = IV V2 Substituting Ohm’s law V = IR: P = I 2R or P = R where P is power in watts (W). Watts are equivalent to joules per second (J s–1), so we can determine the energy lost: Energy = Pt where energy is in joules (J) and time is in seconds (s). 86

motors and generators

Magnetic fields produced by electric currents A magnetic field exists in a region of space if a magnet at some point in that space would experience a magnetic force. From in2 Physics @ Preliminary section 12.3, you know that a current-carrying conductor produces a magnetic field. You can observe this circular magnetic field around a current-carrying wire if you place a magnetic compass needle at various locations around the wire (see Figure 4.1.4). We represent magnetic fields by solid lines, called magnetic field lines, and label these lines with the symbol B. These lines represent the places where the magnetic field has the same strength. We also place arrow heads along these By convention, magnetic field lines to indicate the direction of the field. the arrow heads indicate the direction in which the north pole of a magnet would point within the magnetic field (see the compass needles in Figure 4.1.4). You can work out the direction of the magnetic field in Figure 4.1.5 by using the right-hand grip rule. Grip the wire with your right hand and point the thumb in the direction of the conventional current along the wire. Remember, this is the direction in which positive particles would flow, from the positive to the negative terminal. Your curled fingers will now point in the direction of the magnetic field around the wire. There are several conventions we need to recall when we draw twodimensional diagrams showing currents and magnetic fields. For example, when we view the situation shown in Figure 4.1.5 from above, we represent this as shown in Figure 4.1.6. The  symbol in the centre indicates that the current is coming towards you (out of the page). When viewing the situation from below, we represent this as shown in Figure 4.1.7. The  symbol in the centre indicates the current is flowing away from you (into the page). Notice that the magnetic field lines shown in Figures 4.1.6 and 4.1.7 are drawn more closely near the wire, where the field becomes stronger. The  and  symbols indicate that the current is flowing out of and into the page respectively. You can remember this convention if you imagine that  is the head of an archer’s arrow coming out of the page at you. The crossed feathers in the back of the arrow are represented by , indicating that the arrow is pointing away from you. We also represent magnetic fields in two-dimensional diagrams as shown in To show that the magnetic field points into or out of the Figure 4.1.8. page, we use  or  • respectively. Use the right-hand grip rule to determine the

B

Figure 4.1.6 Magnetic field lines around a conventional current going out of the page

I

B

Figure 4.1.4 A straight wire carrying a current deflects the compasses around it in a circular pattern. electric current I

magnetic field B

Figure 4.1.5 The curled fingers point in the direction of the magnetic field when the thumb points in the direction of the conventional current along the wire.

B

Figure 4.1.7 Magnetic field lines around a conventional current going into the page 87

4

Electrodynamics: moving charges and magnetic fields conventional current I

B

B

Figure 4.1.8 Magnetic field lines into () and out of (•) the page for a wire carrying a conventional current upwards in the plane of the page

B –

(a)

+

direction of the magnetic field around the wire in Figure 4.1.8. You should see that the magnetic field lines would go into the page on the right-hand side of the wire (represented by ) and come out of the page on the left (represented by •). An extension of the situations you have reviewed above is the magnetic field around a current-carrying wire loop. Again we use the right-hand grip rule to determine the direction of the magnetic field around the current-carrying wire (Figure 4.1.9). Notice that the magnetic field in the centre of the loop always points in the same direction, no matter where your hand is around the loop. In the following chapters there are many situations that involve loops of wire carrying currents; therefore, it is important you are familiar with the magnetic field that is produced around them. Most applications of magnetic fields in current-carrying wire loops actually involve more than one loop. Each loop is called a turn, and many turns together are known as a solenoid (Figure 4.1.10). A solenoid is simply a long coil of wire, and the magnetic field produced is similar to that of a bar magnet, with a north The direction of the magnetic field through and south pole at each end. the centre of the solenoid is determined by using a special version of the righthand grip rule (Figure 4.1.10). In this situation, you must curl your fingers in the direction of the conventional current around the solenoid and your thumb will point in the direction of the magnetic field. Your thumb will point to the end of the solenoid that forms a north pole. A coil such as this is used to make an electromagnet or simply to produce a magnetic field.

(b)

Figure 4.1.9 (a) The right-hand grip rule can be used for a current loop. (b) Magnetic field lines around a single wire loop

N

S

Figure 4.1.10 The right-hand grip rule is used to find the direction of the magnetic field inside the solenoid.

The North Pole?

D

id you know that the Earth’s north geographic pole is actually its south magnetic pole? The north pole of a magnet is attracted towards the north geographic pole and therefore it must be a south magnetic pole. Did you also know that the Earth’s magnetic field changes direction? At irregular intervals of about 250 000 years the polarity of the Earth’s magnetic field flips and points in the opposite direction. Scientists are not sure of the effects of this flip or for how long the field disappears during each flip.

Geographic North Pole

Magnetic South Pole

S

N

Magnetic North Pole

Geographic South Pole

Figure 4.1.11 The Earth acts as though it has a south magnetic pole near the geographic north pole! The ‘magnetic north pole’ is the place to which the north end of a compass appears to point.

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Checkpoint 4.1

1 Define the nature and direction of conventional direct current. 2 Construct two-dimensional diagrams illustrating the magnetic field around a current-carrying wire from two different perspectives (end-on and side-on). 3 Sketch and compare two diagrams illustrating the magnetic field around a bar magnet and a current-carrying solenoid.

4.2 Forces on charged particles in magnetic fields We have already seen that free charged particles move when placed in an electric field, because they experience a force (Figure 4.1.1). This force is caused by the attraction and repulsion that charged particles experience within the field. Knowing why this occurs, it might seem strange to hear that a charged particle also experiences a force due to a magnetic field. There is, however, one important A charged particle only experiences a force in a magnetic field difference. when the particle is moving relative to the magnetic field, or if the strength of the magnetic field is changing. It is important to know that stationary charges or charges moving parallel to the magnetic field do not experience a force. In Figure 4.2.1a, a positively charged particle, let’s say a proton of charge q, is travelling upwards with a velocity v within a horizontal magnetic field B. The proton experiences a force F in a direction perpendicular to both the magnetic field and the direction in which it is moving. The force is given by F = qvB. The way we can tell the direction in which the force is acting is to introduce another right-hand rule. This rule is commonly called the right-hand palm a rule (or the right-hand push rule) and is illustrated in Figure 4.2.1b. To find the direction of the force that a positive particle will experience when moving through a perpendicular magnetic field: 1 Place your open right hand with the fingers pointing in the direction of the magnetic field (north to south). 2 Place your thumb at right angles to your fingers and in the direction in which the particle is travelling. b 3 The force on the positive particle will be directed out of your palm and at right angles to your hand. We can conclude that the proton in Figure 4.2.1a will experience a force out of the page at right angles to both the magnetic field and the direction in which the proton is travelling. Note that if we know the direction of any two of the three quantities represented in the right-hand palm rule, we can use this rule to determine the direction of the third quantity.

v B S

+

N

F

direction in which the particle is travelling with velocity v

B direction of the magnetic field direction of the force F on the positive particle

Figure 4.2.1 The right-hand palm rule is used to find the direction of a force acting on a positively charged particle in a magnetic field.

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Try this! Bending beams of particles Ask your teacher if they can show you a beam of electrons in a Crookes magnetic deflection tube (see Figure 4.2.2). With teacher supervision, bring the north pole of a magnet close to the front of the tube and observe the effect on the beam of electrons. By the right-hand palm rule, negative particles experience a force out of the back of your hand. Try to predict the direction in which electrons will be deflected when you place the south pole of the magnet close to the front of the tube. Now try it and then explain it to a friend. Do they agree with your explanation?

Figure 4.2.2 A Crookes magnetic deflection tube produces a beam of electrons.

Checkpoint 4.2 1 Explain why a stationary charged particle experiences a force when you move a magnet past it. 2 Identify the direction in which the proton in Figure 4.2.1a would be moving for it to experience a force into the page.

4.3 The motor effect

Identify that the motor effect is due to the force acting on a current-carrying conductor in a magnetic field.

PRACTICAL EXPERIENCES Activity 4.1

Activity Manual, Page 27

90

In the previous section we saw that charges moving in a magnetic field If a current-carrying conductor is placed in an external experience a force. magnetic field, the wire also experiences a force and this is called the motor effect. This effect occurs because the charges within the wire are travelling through the magnetic field and experience a force, just as they would if they were Remember, we are considering free charged particles (see Figure 4.2.1a). current to be a flow of positive particles. It is actually the negative electrons that experience the force from the magnetic field. It works just fine to use conventional current and consider the force is acting on positive particles to keep things simple. We can now use the right-hand palm rule we saw in section 4.2 to work out the direction of the force on a current-carrying wire in an external magnetic field To find the direction of the force that acts on a current(see Figure 4.3.1). carrying wire that is perpendicular to an external magnetic field: 1 Place your open right hand with the fingers pointing in the direction of the magnetic field (north to south). 2 Place your thumb at right angles to your fingers and in the direction in which the conventional current is flowing (from the positive to negative terminals in the circuit). 3 The force experienced by the wire will be directed out of your palm at right angles to your hand.

motors and generators When we allow current to flow through a wire within the magnetic field in Figure 4.3.1, we see that the wire moves out of the page, at right angles to both the magnetic field and the direction of the conventional current. Each positive particle of the conventional current within the wire in Figure 4.3.1 experiences a force due to its motion within the external magnetic field. Since these positive particles are within the wire, the force acts on the wire.

I S

+

N

direction in which the I current is flowing

F

B direction of the magnetic field F

Figure 4.3.1 The right-hand palm

PHYSICS FEATURE

Identify data sources, gather and process information to qualitatively describe the application of the motor effect in: – the galvanometer – the loudspeaker.

Loudspeakers

A

n excellent example of an application of the motor effect is a loudspeaker. This device is a key part of telephones, televisions and any other appliance in which an electrical signal needs to be converted into sound for us to hear. Figure 4.3.2a shows the most fundamental parts of a typical loudspeaker labelled A to D. Figure 4.3.2b illustrates its operation via the motor effect. A loudspeaker contains a current-carrying coil (C), which is commonly called the voice coil. This coil is wound around a hollow cardboard tube (B) and the tube is fixed to the cone of the speaker (A). The voice coil is suspended inside a cylindrical permanent magnet (D) that provides a uniform magnetic field at right angles to the coil. An alternating current is passed through the voice coil, causing the cone to rise and fall due to the motor effect. Each time the speaker cone pushes outwards on the air, it creates a wave of pressure that travels away from the speaker. These waves of air pressure are sound waves. By varying the frequency of the alternating current in the voice coil, the frequency (or pitch) of the sound can be varied. This means that speakers can generate a variety of sounds and reproduce sounds recorded elsewhere. At the moment in time shown in Figure 4.3.2b, the current is travelling out of the page on the left side of the voice coil. Using the right-hand palm rule

Figure 4.3.2 A loudspeaker converts electrical energy into sound. (a) A cut-away section showing the parts of the loudspeaker and (b) a simplified cross section showing the direction of the magnetic field

direction of the force on the positive particle

rule for a current

you can see that a force will be exerted upwards on the voice coil on this side. This force is exerted in the same direction around the circumference of the voice coil and causes the speaker cone to move upwards. So we can see that now we know about the motor effect and the right-hand palm rule we can explain how some everyday things work. a

A

B C

D

b

direction of force on the voice coil

N

S

A cardboard cone

N

B cardboard tube fixed to cardboard cone and wrapped in voice coil C wire coil (called the voice coil) D permanent cylindrical magnet

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Electrodynamics: moving charges and magnetic fields

Quantifying the motor effect If we place a current-carrying wire within a magnetic field (see Figure 4.3.3) the force on the conductor is given by: Solve problems and analyse information about the force on current-carrying conductors in magnetic fields using: F = BIl sin θ

F = BIl sin θ where F is magnitude of the force on the wire in newtons (N), l represents the length of the wire inside the magnetic field in metres (m), I is the size of the current in amps (A), B is the strength of the magnetic field in tesla (T) and θ is the acute angle between the magnetic field and the wire. Note that you can rearrange this equation and make any of the variables the subject to find their values.

Worked example Question If the wire in Figure 4.3.3 has a length of 5 cm within a 0.2 T magnetic field, the wire is at 30º to the field and it contains a current of 0.5 A, what is the force exerted on the wire?

I B l

S

a

SOLUTION

N

First we convert the length of the wire into the appropriate units: 5 cm = 5/100 = 0.05 m

θ sin θ =

Figure 4.3.3

a l

a = l sin θ

Then substitute:

F = BIl sin θ



= 0.2 × 0.5 × 0.05 × sin 30



= 2.5 × 10–3 N out of the page

Qualitative analysis of factors affecting the motor effect Discuss the effect on the magnitude of the force on a current-carrying conductor of variations in: • the strength of the magnetic field in which it is located • the magnitude of the current in the conductor • the length of the conductor in the external magnetic field • the angle between the direction of the external magnetic field and the direction of the length of the conductor.

92

If you have an equation that describes a relationship, the easiest way to see how the variables affect the subject of the equation is to place some numbers in the equation and see what happens. Remembering that the motor effect is the force F that a current-carrying conductor experiences in a magnetic field, let’s look at how the other variables in the equation affect the magnitude of F. As you read through this section keep referring back to the equation and check that you come to the same conclusions.

Magnetic field strength, B If the value of B was very small, then the right side of the equation would be multiplied by a very small number. Conversely, if the value of B was very large, the right side of the equation would be multiplied by a very large number. Since the magnitude of the force F is equal to the right-hand side of the equation, F is clearly directly related to B. To take our analysis one step further, consider multiplying the value of B by 2. Looking at the equation, we see that if we do this, the effect on the value of F is We can now say the force F is directly the same as multiplying F by 2. proportional to magnetic field strength B; that is, as B increases by some factor (say 2 times) F also increases by that same factor. Having understood the analysis above, it should be easy now to see the following relationships.

motors and generators Current, I As for magnetic field strength, by inspecting the formula we can see that force F will be directly proportional to current I. Length, l Similarly, F is directly proportional to the length l of the currentcarrying conductor within the magnetic field. Be particularly careful to remember that l is the total length of the wire within the magnetic field. It is noteworthy that l is regarded as a vector, but current I is not. Angle, θ When the wire is parallel to the magnetic field, the angle θ is zero. Inspecting Figure 4.3.4 you can see that if θ is zero degrees then sin θ is also zero. When you substitute zero for sin θ in the motor effect equation, you see that the This shows us the interesting situation that the force force must be zero. on a current-carrying conductor in a magnetic field is zero when the conductor is parallel to the magnetic field lines. When the current-carrying conductor is perpendicular to the magnetic field, θ is 90°. From Figure 4.3.4 you can see that sin 90 is 1. Since 1 is the maximum value for sin θ, when we substitute 1 into the equation the force F will be the maximum value it can be for each set of the other variables. This shows that the force F is a maximum value when a current-carrying conductor is perpendicular to the magnetic field B. Inspecting Figure 4.3.4 you can see that as θ increases from 0° to 90° the value of sin θ increases towards a value of 1. The rate of this increase is not So we can only say that constant (i.e. the graph is not a straight line). force F depends on θ, as it is not directly proportional to θ.

Nanotube Loudspeakers: No Magnets

A

group of Chinese researchers has developed a loudspeaker that consists only of a thin film of carbon nanotubes driven by an AC input signal. The sound generation is attributed to a thermoacoustic effect. Changes in the current flowing through the film are reflected in the film’s temperature. Those temperature changes excite pressure waves in the surrounding air and these are sound waves. The film is flexible and can be stretched and still operate unimpeded. Perhaps loudspeakers won’t have magnets in the near future!

sin θ 1

90

180

270

360 θ

–1

Checkpoint 4.3 1 Describe the relative directions of the force, the current and the magnetic field when a current-carrying wire experiences the maximum possible force due to the motor effect. 2 Compare the relationships of B, I, l and θ to F in the equation F = BIl sin θ. 3 Explain the motor effect.

Figure 4.3.4 Graph of θ versus sin θ

4.4 Forces between parallel wires In many applications of electric circuits there are wires bundled tightly together and running parallel to each other. If we want to explore the interaction of these wires, we need to bring together two of the facts we have learned so far. The first is that current-carrying conductors produce a magnetic field. The second is that a current-carrying conductor experiences a force when inside a magnetic field. We will also need to apply two of the right-hand rules we have learned to determine the direction of the magnetic fields and the forces. 93

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Electrodynamics: moving charges and magnetic fields

Qualitative analysis

Try This! the motor effect Take a piece of insulated wire about 5–10 metres long. Stretch it out between two retort stands so that there are two pieces of wire running parallel within a few centimetres of each other. Connect the ends to a 12 V battery and insert a tapping key switch at one end of the circuit. When connected briefly, the currents will run antiparallel to each other. Caution: Connect these wires for a very short time only, as they carry a large current. Predict what will happen when you press the switch. Now observe! How did you go?

Let us first consider the situation in which we have two parallel current-carrying wires with currents that are travelling in the same direction. In Figure 4.4.1a we use the right-hand grip rule to determine the direction of the magnetic fields around the wires. Now, to understand what is happening to each wire, we will consider what is happening to one wire at a time. In Figure 4.4.1b we are looking at what is happening to wire 2. The magnetic field generated by the current in wire 1 travels into the page around wire 2. Using the right-hand palm rule, we can see that wire 2 experiences a force towards wire 1. In Figure 4.4.1c we now see what is happening to wire 1. The magnetic field of wire 2 comes out of the page around wire 1. Therefore the right-hand palm rule shows that wire 1 experiences a force towards wire 2. The conclusion we can come to is that when two parallel currentcarrying conductors have currents travelling in the same direction, the two conductors are forced towards each other.

a

b

II F =k 12 l d

94

I2

F1

wire 1

wire 2

magnetic fields around parallel wires

Describe qualitatively and quantitatively the force between long parallel currentcarrying conductors:

I1

wire 1

wire 2

magnetic field due to I1

c

I1

I2

F2

wire 1

wire 2

magnetic field due to I2

Figure 4.4.1 Determining the forces on two parallel wires with currents flowing in the same direction

Now let’s consider the two parallel current-carrying wires with currents that are travelling in opposite directions. In Figure 4.4.2a the right-hand grip rule shows us the direction of the magnetic fields around the wires. To understand what is happening to each wire, we will again consider each wire in turn. Let’s look at what is happening to wire 2 first (Figure 4.4.2b). The right-hand grip rule shows the magnetic field of wire 1. This field travels into the page around wire 2. The right-hand palm rule shows that wire 2 experiences a force away from wire 1. Figure 4.4.2c shows what is happening to wire 1. The magnetic field of wire 2 goes into the page around wire 1. The right-hand palm rule then shows that wire 1 experiences a force away from wire 2. The conclusion we can now come to is that when two parallel currentcarrying conductors have currents travelling in the opposite direction, the two conductors are forced away from each other. It may be easy for you to remember the two conclusions above about the direction of forces on parallel wires, although remembering the result is generally less important than knowing how you got there. If you forget the conclusions

motors and generators a

b

wire 1

I1

wire 2

wire 1

magnetic fields around antiparallel wires

I2

wire 2

magnetic field due to I1

c

I1

I2

wire 1

wire 2

magnetic field due to I2

above and you know how to work them out yourself you can never get them wrong. You will apply similar methods in other problems later in this module. So if you are comfortable with these methods now, it will be easier later. If at any time you have trouble using your right-hand rules, come back to the relevant part of this chapter and revise. You will meet the skills you have used here several more times yet and each occasion is a chance to test your knowledge.

Figure 4.4.2 Determining the forces on two current-carrying wires with currents in opposite directions

Quantifying the relationship Using our right-hand rules, we have determined that parallel wires exert forces on each other. To quantify these forces, let’s start with an equation we have seen already. Recall the equation for the force on a current-carrying conductor: F = BIl sin θ For parallel wires, each wire is at right angles to the magnetic field of the other wire. The sin θ term in the above equation is therefore equal to 1 (see Figure 4.3.4) so the equation becomes: F = BIl Inspecting this formula we can see that the current I and length l can be measured relatively easily. To work out F, the size of the force, we now need to calculate B, the strength of the magnetic field around the wire. The strength of the magnetic field around a current-carrying conductor can be determined using the equation: I B=k d where the proportionality constant k is 2 × 10–7 N A–2, I is the current in amps, and d is the distance away from the wire in metres. Let’s consider the situation in Figure 4.4.1b. The magnetic field is being produced by wire 1, so the current I1 will be used in calculating the magnetic field strength. The force we are calculating is acting on wire 2, so the current we should use in this part of the calculation is I2. Combining the two previous equations and inserting the correct currents in each we get: F =k

Rearranging this gives:

I1 I l d 2

II F =k 1 2 l d

where F is the force on each wire in newtons and l is the length the wires are parallel in metres, so F/l is the force on each metre of wire. k is the proportionality constant 2 × 10–7 N A–2, I1 and I2 are the currents in the two wires in amps, and d is the distance between the two wires in metres. 95

4

Electrodynamics: moving charges and magnetic fields 1.5 A

1.0 A

Worked example QUESTION For the situation shown in Figure 4.4.3 calculate the magnitude of the force acting on each wire.

0.5 m

SOLUTION 2 cm

Figure 4.4.3 Two parallel current-carrying wires

Feeling the pinch

T

he piece of copper pipe shown in Figure 4.4.4 was crushed by lightning. Just like two parallel wires carrying currents in the same direction, the sides of this pipe were pulled together when a current of more than 100 000 amps was present.

Figure 4.4.4 Dramatic evidence of parallel conductors experiencing a force

From Figure 4.4.3, l = 0.5 m, l1 = 1.5 A, l2 = 1.0 A and d = 2/100 = 0.02 m. II F Use: =k 1 2 l d I1 I 2 l Rearrange to make F  the subject: F = k d Substitute:



F=

2 × 10−7 × 1.5 × 1.0 × 0.5 0.02

= 7.5 × 10–6 N

Note that as this question asked for only the magnitude of the force, you do not have to include a description of the direction. If asked, you should add that the force on each wire is directed towards the other wire.

More qualitative analysis Our last stop in our look at parallel wires is to analyse the relationships expressed in the equation: II F =k 1 2 l d If we follow the process we used in section 4.3, we can see the following relationships: • F is directly proportional to the length l. To see this easily we rearrange the formula to make F the subject. As l increases F increases. • F is also proportional to both I1 and I2. • F is inversely proportional to the distance d between the two wires. This means that as the distance increases F decreases, or as the distance decreases F increases.

Checkpoint 4.4 1 Identify the two key facts that explain the interactions of two parallel current-carrying conductors. 2 Describe the interactions of two parallel current-carrying conductors.

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PRACTICAL EXPERIENCES

motors and generators

CHAPTER 4 This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 4.1: The motor effect Observe the effect of a current-carrying wire that is placed in an external magnetic field and relate it to the mathematical formula:

F = BIl sin θ Equipment: 2 strong horseshoe magnets or ceramic magnets on an iron yoke, long wire, power supply, retort stand, clamp.

hanging from a retort stand

Perform a first-hand investigation to demonstrate the motor effect. Solve problems and analyse information about the force on current-carrying conductors in magnetic fields using: F = BIl sin θ

flexible wire

ceramic magnets on an iron yoke

low voltage DC power supply

Figure 4.5.1 Experiment set-up

Discussion questions 1 Describe the motor effect. 2 Discuss what happened when the current direction was changed.

97

4 •

•

• • •

•

•

Chapter summary

Electrodynamics: moving charges and magnetic fields

Current is the rate of flow of charge through a region, and in circuits the direction of a current is that of a positive charge, called conventional current. A current in which the charges only flow in one direction is called direct current (DC). A current in which the charges move back and forth is an alternating current (AC). Resistance is a measure of how easily a current flows and it is defined as the ratio of voltage over current. Electrical power is the rate at which energy is transferred within a circuit component. Electric currents produce a magnetic field around a wire. The direction of this field can be determined by the right-hand grip rule. A magnetic field is depicted by lines with an arrow indicating the direction in which the north pole of a magnet points within the field. The symbols × and • are used to show the direction of a magnetic field into and out of the page respectively.

• •

•

• •

•

The symbols  and  are used to show the direction of a current into and out of the page respectively. Charged particles moving in a magnetic field experience a force. When these charges are moving in a wire, the wire experiences a force called the motor effect. The right-hand palm rule relates the perpendicular directions of force, magnetic field and motion in the motor effect. A loudspeaker is a great example of an application of the motor effect. The motor effect is quantified by the equation F = BIl sin θ where the force is proportional to the magnetic field strength (B), current (I), length of the wire within the magnetic field (l ) and the angle between the wire and the magnetic field (θ). Parallel current-carrying wires experience the motor effect due to each other’s magnetic fields and this II F phenomenon is quantified by the equation = k 1 2 . l d

Review questions Physically speaking 1

Across 4 Equation to determine the force between two current-carrying wires (7, 3)

2 4

3 5

7 emf stands for this (13, 5) 9 Application of the motor effect that converts electrical energy to sound 10 Quantity related to the energy given to electrons in a circuit

6 8

7

9

11 Force experienced by current-carrying wire in an external magnetic field (5, 6)

Down 1 Electrons moving in one direction (6, 7) 2 Unit of power

10

11

98

3 The branch of electricity that deals with moving charges and magnetic fields 5 Rate of use of electrical energy 6 Opposition to the flow of electrons 8 Device that converts electrical energy to kinetic energy using the motor effect

motors and generators

Reviewing 1 2

Describe the difference between DC and AC.

3

Explain why a bird can sit on an electrical power cable and not get electrocuted.

4

Describe what is meant by metals having a ‘sea of electrons’.

5

a Recall the factors that affect the resistance of a wire. b State how they affect it.

6

Given that the definition of power is P = W /t, show that the equation for electrical power is P = VI.

7

Describe what happens to a charged particle in a magnetic field.

8

Compare the paths of two charged particles entering magnetic fields. The first is in a constant magnetic field and the second is in a magnetic field that is gradually increased.

16 Determine the direction the particle will move when it enters the magnetic field that is shown. a

Outline the journey of an electron through the circuit in Figure 4.1.1, noting the energy transformation.

9 State the motor effect in words. 10 Sketch a graph to show how changing the angle of the wire in a magnetic field changes the force experienced.

11 Explain how current can create sound in a loudspeaker.

12 Draw a labelled energy transformation diagram of a loudspeaker.

+ b

+

17 Determine the force on a current-carrying wire (I = 2 A) of length 0.5 m that is placed in a magnetic field of 3 T.

18 A physics student did an experiment to measure the force on a wire placed in an external magnetic field. The field is altered and the results are recorded below.

Force (N)

Magnetic field (T)

0.05 0.08 0.11 0.17

0.1 0.2 0.3 0.4

a Graph the results. b Determine the value of the gradient of the graph and what the gradient represents. c Given that the length of the wire is l = 0.2 m and the current is I = 2 A, comment on the accuracy of the student’s results.

solving Problems 13 Calculate the resistance in a circuit that has a battery supplying 10 V and current flow of 2.3 A.

14 Determine what happens to current in a circuit when the thickness of the wire is doubled and the voltage is increased to four times the original.

15 Using the right-hand palm rule, determine the direction of the unknown quantity B, I or F. a

iew

Q uesti o

n

s

Re

c

v

b

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5 electromagnetic induction, Faraday’s law, magnetic flux, magnetic field strength B, magnetic flux density, emf, perpendicular area, Lenz’s law, law of conservation of energy, eddy currents, induction cooktops, resistive heating, eddy current braking

Induction: the influence of changing magnetism Electromagnetic induction The discovery of electromagnetic induction was a giant step on the path to modern technology. Our understanding of this phenomenon required a great deal of new physics and involved the work of many individuals. One man, Michael Faraday, led the way with his experimental genius and intuitive diagrammatic reasoning. Faraday lacked the mathematical skills to numerically describe his discoveries, but James Clerk Maxwell took Faraday’s understanding and eventually quantified all electromagnetism. Faraday’s law and Lenz’s law provide us with the tools to explain and predict eddy currents and to understand their applications. Later they will help us in our quest to uncover the secrets of motors, generators and transformers.

5.1 Michael Faraday discovers electromagnetic induction

Outline Michael Faraday’s discovery of the generation of an electric current by a moving magnet.

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Michael Faraday (1791–1867) was born into a working-class family in London in 1791. He received little formal schooling and started work at the age of 12. At 14 he became a bookbinder’s apprentice and set about educating himself with the books he was able to access. Over this time he developed a keen interest in science and began attending scientific lectures. In 1813 he became a research assistant for the prominent scientist Sir Humphry Davy (1788–1829). In the following years Faraday became renowned as one of the greatest experimental scientists. One of his numerous experimental discoveries was the phenomenon of Electromagnetic induction is the electromagnetic induction in 1831. generation of an electric current by a changing magnetic field. Faraday showed that when he moved a magnet near a wire coil, a current flowed within the coil. He first moved a magnet into one end of a wire coil (Figure 5.1.1a). As he did this he measured a current in the coil on a galvanometer (a type of ammeter).

motors and generators He noticed that this current was only induced while the magnet was moving. He moved the magnet out of the coil, and this time he measured a current in Following many the opposite direction within the coil (Figure 5.1.1b). other experiments, he put forward his general principle of electromagnetic induction that a changing magnetic field can cause a current to be generated in a wire. This change can be caused by either the relative motion of the field and the coil or by a change in the strength of the magnetic field. In light of Faraday’s conclusion, let’s offer a simple way to understand his observations. In Figure 5.1.1 we can consider the change in the magnetic field through the coil to be represented by the number of magnetic field lines passing within the coil. In Figure 5.1.1a, notice that the magnetic field around the bar magnet is not uniform. You see that as you get closer to the poles of the magnet the magnetic field lines get closer together, indicating that the field is stronger. Therefore, as the magnet gets closer to the coil more field lines pass within the coil, indicating that the magnetic field within the coil gets stronger. So, as the magnet gets closer to the coil the strength of the field within the coil is changing and this induces the current. This gives us a general explanation for electromagnetic induction. Now let’s take things a bit further.

Faraday’s law: explaining electromagnetic induction Electromagnetic induction can be summarised by Faraday’s law. The induced emf in a coil is proportional to the product of the number of turns and the rate at which the magnetic field changes within the turns. Faraday’s law is quantified by an equation, and it is very useful to analyse the equation to understand the relationships involved. This equation is: ε = n(∆ΦB/∆t) Let us spend a little time now looking at what each of the variables in this equation means and then we can understand the relationships.

Magnetic flux ΦB Magnetic flux is a measure of the ‘amount’ of magnetic field passing through a given area. There are two variables that determine the value of magnetic flux: the strength of the magnetic field B and the area the field is passing through A. Magnetic flux ΦB is measured in weber (Wb) and can be expressed by the equation: ΦB = BA⊥ Magnetic field strength B, a quantity we are already familiar with, is also called magnetic flux density. This quantity is measured in tesla (T), or equivalently in webers per square metre (Wb m–2). This is a measure of the field strength per square metre. A ⊥ is an area that is perpendicular to the magnetic field lines. If a magnetic field passes through a circular wire loop of area A (Figure 5.1.2a) and the loop is at an angle to the field (see Figure 5.1.2b), then the field passes through an ‘effective’ area A⊥ that is perpendicular to the field and is smaller than area A (see Figure 5.1.2c). Note that flux could also be calculated using the perpendicular component of the magnetic field strength, B⊥, and the total area A.

a N

A b N

A

Figure 5.1.1 The changing magnetic field of a moving magnet can induce a current in a coil of wire.

Our sun’s magnetic influence

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arge outbursts from the Sun cause changes in the strength of the magnetic field at the Earth’s surface. This changing magnetic field induces currents in long metal pipelines and the wires of power grids, especially at high latitudes. Currents of 1000 amps have been measured in pipelines in Alaska, causing accelerated corrosion. Large currents in power grids have overloaded circuits and left millions of people in the dark for hours.

PRACTICAL EXPERIENCES Activity 5.1

Activity Manual, Page 33

Perform an investigation to model the generation of an electric current by moving a magnet in a coil or a coil near a magnet. Define magnetic field strength B as magnetic flux density. Describe the concept of magnetic flux in terms of magnetic flux density and surface area.

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Induction: the influence of changing magnetism This line is the perpendicular height of area A. B A

a

A⊥

P

b

c

Figure 5.1.2 (a) A circular coil of area A. (b) A side view of the coil at an angle to the magnetic field. (c) The ‘effective’ area of the coil perpendicular to the magnetic field viewed from point P (b)

The product of the magnetic flux density B and the area A ⊥ gives a measure of the total ‘amount’ of magnetic field passing through that area. This is the magnetic flux ΦB. The delta (∆) symbols in Faraday’s law mean a change in some quantity. So the two terms with delta symbols attached are differences between an initial value and a final value. The term ∆ΦB stands for the change in the magnetic flux and is given by: ∆ΦB = ∆ΦB final – ∆ΦB initial The ∆t term represents a period of time. This is the period of time over which the change in flux ∆ΦB is measured. We can see that the ∆ΦB/∆t term in Faraday’s law is actually the rate of change of magnetic flux, just as acceleration is the rate of change of velocity (aav = ∆v/∆t). The term ∆ΦB/∆t tells us how fast the flux is changing.

emf ε The symbol ε stands for emf, measured in volts. It is the difference in electrical potential between the two ends of a coil (X and Y in Figure 5.1.3). This emf creates an electric field within the wire of the coil and a current is established, provided the circuit is complete (X and Y are not connected in Figure 5.1.3). A current will flow as long as there is a change in the magnetic field within the coil. Qualitative analysis of Faraday’s law Let’s now look at the relationship between emf and the other terms in the equation. Recall that the equation for Faraday’s law is: ε = n(∆ΦB/∆t)

B N

S

X Y

Figure 5.1.3 A moveable magnet passes through a stationary coil with terminals X and Y. 102

The emf is proportional to the rate of change of the magnetic flux (∆ΦB/∆t). If there is a large change in flux in a small amount of time, then ∆ΦB/∆t is large; therefore, the emf produced will be large. This emf is responsible for the induced current in a closed loop of wire. If the emf is large, then so is the induced current. Another way of saying this is that if we wanted to induce a large current, we would change the magnetic field within the coil as much as possible in the shortest time possible. To do this in the example we have seen, we could move the magnet more quickly, use a stronger magnet or make the perpendicular area as large as possible for the coil. The emf is also proportional to n, the number of turns in the coil. Again, if we wanted to create a large current we would want to have as many turns in the coil as possible.

motors and generators By inspecting all the variables in Faraday’s law we can conclude that an induced potential difference, and therefore an induced current in a coil, is proportional to the rate of change of magnetic flux (∆ΦB/∆t) and the number of turns (n) within the coil. Further, we can conclude that induced currents are produced by changing the magnetic field strength B, the relative motion between B and the area A, or changing the perpendicular area A ⊥ of the coil. Our next challenge is to find the direction of an induced current, and for that we need Lenz’s law. We will cover this in section 5.2.

Try this! skipping currents

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ake a length of wire about 30–40 metres long and connect the ends to a sensitive ammeter. Lay the wire out in a large open space and swing it like a skipping rope. Can you induce a current by changing the size of the loop and using the Earth’s magnetic field? Can you explain why this should work?

Induction without relative motion Now that we have a basic understanding of electromagnetic induction, we will look at another example. In one of his early experiments Faraday wound two coils of wire around an iron ring (Figure 5.1.4). He noticed that a current was induced in coil B for a short time after the current in coil A was switched on or off. To explain this induced current, recall that a current travelling through a wire produces a magnetic field around the wire. We can then follow similar reasoning to that in the different situation described earlier. The list of events below explains why Faraday observed induced currents for a short time after he connected or disconnected coil A. The steps below are indicated in Figure 5.1.5, which shows the currents in both coils and the magnetic field produced in coil A. 1 Coil A is connected by closing the switch and a current begins to flow. 2 The current in coil A produces a magnetic field around the iron ring (see Figure 4.1.9). This field gets stronger as the current increases to its maximum value. The changing magnetic flux from coil A causes a changing magnetic flux in coil B, which induces an emf and therefore a current in coil B. The rate of change of magnetic flux power in coil B is positive, rapid at first but then slows down (Figure supply 5.1.5), so the induced current is positive, high at first but decreases rapidly. 3 The current and magnetic field in coil A both reach their maximum value. The rate of change of magnetic flux in coil B is now zero, so the induced emf and current are also zero. 4 Coil A is disconnected by opening the switch. The current in coil A decreases rapidly, but does not stop immediately because there is normally a brief spark in the switch that allows current to continue flowing briefly. 5 The current in coil A, and therefore the magnetic field it produces, rapidly decreases. The changing magnetic flux from coil A causes a changing magnetic flux in coil B that, in turn, induces an emf and therefore a current. The rate of change of magnetic flux in coil B is negative, initially high but rapidly decreases. Therefore, the induced current is negative, high at first but decreases rapidly to zero. The example above illustrates our previous conclusions that it is the change in the magnetic field (more precisely, the changing magnetic flux) passing through a wire coil that induces a current, and the induced current is proportional to the rate of change of magnetic flux. Now we can see that the direction of the induced current is determined by whether the magnetic flux is increasing or decreasing.

Describe generated potential difference as the rate of change of magnetic flux through a circuit. layers of copper coil interwound with cotton and calico

soft iron ring

switch

+

G

–

coil B coil A

Figure 5.1.4 Basic set-up of Faraday’s experiment

IA

BA

IB 2 1

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Figure 5.1.5 The behaviour of currents and magnetic fields in Faraday’s iron ring experiment 103

5

Induction: the influence of changing magnetism

Checkpoint 5.1 1 Outline Faraday’s discovery of induction by a moving magnet and summarise his conclusion. 2 Define magnetic flux in terms of magnetic flux density. 3 Using the term magnetic flux, explain why removing a magnet quickly from a coil induces a relatively large current.

5.2 Lenz’s law Heinrich Lenz (1804–1865) independently made many of the same discoveries as Faraday. He also devised a way to predict the direction of an induced current This method in a closed conducting loop due to a changing magnetic field. is called Lenz’s law and it states that an induced current in a closed conducting loop will appear in such a direction that it opposes the change that produced it. This means that the induced magnetic field from a wire loop will oppose the change in magnetic flux that causes the induced current. Let us look at the example shown in Figure 5.2.1 to understand this better. a

direction of movement

b

S

direction of movement

N

S N N

B

I

Figure 5.2.1 (a) Bar magnet moves towards a wire loop. (b) The magnetic field due to the induced current

S

As the north pole of the magnet gets closer to the wire loop in Figure 5.2.1a, the magnetic flux passing through the coil increases. This change in flux induces a current in the wire loop. Now let’s find the direction of the current to agree with Lenz’s law. The north pole of the magnet is coming towards the coil, so the magnetic flux pointing downwards through the coil is increasing. Lenz’s law says that the magnetic field produced by the induced current should oppose this change in flux, so the induced flux should point upwards through the coil. Using the right-hand grip rule for a wire coil or solenoid, we see that the current would need to flow in the direction shown in Figure 5.2.1b, to produce an upwardspointing magnetic field within the loop. This field is pointing in the opposite direction to the changing magnetic flux, so it reduces the changing magnetic flux from the approaching magnet. Simply, you can think of the interaction of these two magnetic fields as if the north poles of two bar magnets are facing each other. The two north poles repel each other, and in this way the induced field acts to minimise the increasing magnetic flux within the coil. 104

motors and generators Now, let’s look at what would happen when you move a magnet away from the loop with its north pole facing the loop (Figure 5.2.2). In this case the magnetic flux in the coil is decreasing and pointing downwards. The induced magnetic field should therefore be also pointing downwards to add to the reducing field and try to minimise the change. Using the right-hand grip rule, we see that we need a current as shown to produce a downwards-pointing induced magnetic field. Again, you can think of the interaction of these two magnetic fields as the interaction of two magnets in which a north pole is facing a south pole. The two ends of the magnets are attracted to one another and in this way the induced field acts to add to the decreasing flux within the coil. These examples leave us with Lenz’s law as a tool to determine the direction of an induced current in a wire coil due to a changing magnetic flux.

The law behind Lenz’s law: the law of conservation of energy While we were learning about Lenz’s law you may have wondered why the current needs to produce a magnetic field to oppose the change in flux? Well the This law states that answer lies in the law of conservation of energy. energy cannot be created or destroyed, only converted from one form to another. In the case of Lenz’s law, it is the fact that energy cannot be created that is important. If the induced current in Figure 5.2.1 was in a direction that added to the changing flux through the coil there would be an attractive force on the magnet. This would mean that the magnet’s motion would cause it to be pulled through the coil. The amount of energy you could get from the induced current (heat from electrical resistance) and the induced magnetic field would be much more than that put in initially to move the magnet and change the flux through the coil. This would mean you would be getting something (energy) for free (without doing any work), which is not possible. Energy must be ultimately converted from the work done to move the magnet into heat energy from the electrical resistance within the wire coil. There must be a balance between the energy that goes into a system and the energy that comes out. So, whether you push the magnet towards the loop or pull it away from the loop, you will always experience a force that resists the motion. This force is an attraction or a repulsion between the magnetic fields of the magnet and the wire loop. So far in this chapter we have explained the cause of magnetic induction using Faraday’s law, used Lenz’s law to find the direction of an induced current, and the right-hand grip rule to find the direction of this current’s magnetic field. Keep these ideas in mind, as we will apply them in the next two chapters.

direction of movement

S N

Figure 5.2.2 Induced current due to a decreasing magnetic field in accordance with Lenz’s law

Account for Lenz’s law in terms of conservation of energy and relate it to the production of back emf in motors.

Checkpoint 5.2 1 2 3

Define Lenz’s law. Describe the induced current and magnetic field in Figure 5.2.2 if the south pole of the permanent magnet is pointing towards the wire coil. Justify the current and magnetic field shown in Figure 5.2.1b in terms of the law of conservation of energy.

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Induction: the influence of changing magnetism

5.3 Eddy currents Explain the production of eddy currents in terms of Lenz’s Law.

We have seen that charges moving in a magnetic field experience a force in accordance with the right-hand palm rule (sections 4.2 and 4.3). This effect occurs for free charges and charges within conductors. When a current-carrying wire experiences a force within an external magnetic field, we call this the motor effect. When charges within a wire experience a force within a changing magnetic field, inducing a current, we call this electromagnetic induction. Many conductors that experience a changing magnetic field and produce an induced current are much larger than a wire. We call these induced currents eddy currents. Eddy currents can be produced by the relative motion of a conductor and a magnetic field. These eddy currents are small loops of current within the conductor. They are the same as induced currents in wires subjected to changing magnetic flux, except that the currents are not confined to a loop of wire. These currents are set up in accordance with Lenz’s law and produce magnetic fields that act to minimise the change in magnetic flux within the path of the current. Figure 5.3.1 shows a square piece of copper sheet swinging like a pendulum through a magnetic field. When this piece of metal moves through the magnetic field, we notice that there is a braking effect, slowing its swing. After a few swings it comes to rest. It stops much more quickly than it does when we remove the magnetic field. To explain this situation we can use the right-hand rules or approach the problem in terms of Lenz’s law. When we use Lenz’s law, we use the right-hand grip rule for solenoids or coils to predict the direction of magnetic fields and eddy currents. As the copper square swings into the magnetic field on the left of Figure 5.3.1 let’s consider what happens to a positive charge (as shown) on the leading edge of the square. If this positive charge was within a piece of wire it would experience a force F1 upwards as shown. If this was a square loop of wire, this force would generate a conventional current moving anticlockwise around the loop. As this charge is not confined to a wire, the charge moves upwards initially and then loops around to form a complete circuit (an eddy current). Using the right-hand grip rule, we can see that the eddy current (I1) shown would produce

direction of swing

I1

F2 causing braking effect

F1

N

+

F3

S

I2

+ F4 causing braking effect

Figure 5.3.1 A square metal sheet is swung through a uniform magnetic field between two bar magnets. A braking effect is observed due to induced currents and their magnetic fields, in accordance with Lenz’s law.

106

motors and generators a magnetic field out of the page (indicated by the N for the north pole of the current’s magnetic field). The flow of positive charges in the direction of F1 is a current. This current experiences a force due to the uniform magnetic field. This force F2 opposes the motion of the copper, acting as a braking effect. As the copper square leaves the magnetic field (on the right in Figure 5.3.1) Lenz’s law tells us that the eddy current should produce a force to slow the square’s departure from the field. The positive charge shown experiences a force F3 upwards, as shown. This causes a flow of positive charges in the direction of F3. This current experiences a force due to the uniform magnetic field. This force F4 opposes the motion of the metal sheet, again acting as a braking effect. If you have access to the equipment that demonstrates the situation in Figure 5.3.1, then try observing it for yourself. You may also be able to observe the effect of cutting slots through the piece of metal swinging in the field. The slots limit the size of the eddy currents that can be produced and therefore the size of the induced magnetic fields, and the braking effect is considerably less. We will meet this idea of reducing the size of eddy currents again in chapters 6 and 7.

Try this! Racing magnets Find two identical magnets. Get a piece of copper or aluminium sheet and a sheet of a non-metal, such as glass, with a surface similar in smoothness to the surface of the metal. Place the two sheets at the same angle (say 60º) to the table surface and place the magnets at the same height on each sheet (see Figure 5.3.2). Now predict which magnet is going to win the race and why. Now race! Did everyone agree? Explain your observations to a friend. aluminium or copper

glass

60°

Figure 5.3.2 Which magnet will win?

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PHYSICS FEATURE

PRACTICAL EXPERIENCES Activity 5.2

Gather, analyse and present information to explain how induction is used in cooktops in electric ranges.

Activity Manual, Page 39

Induction cooking

I

nduction cooktops are a great example of a growing application of eddy currents. The main appeal of induction cooking is its efficiency and fast heating. Heat is not transferred to the pan from a hot plate or flame in induction cooking. The heat is generated within the pan itself and then flows into the food being cooked. This means that minimal heat is lost to the air before reaching the food, making this much more efficient than other cooking methods. The operation of an induction cooktop is illustrated in Figure 5.3.3. A rapidly changing strong magnetic field is generated in a large wire coil, using an alternating current. Both the intensity and direction of this field change continuously over very short periods of time. The resulting rapidly changing magnetic flux within the base of the frying pan induces strong eddy currents, causing resistive heating. Resistive heating by eddy currents occurs when the charges flowing within the metal collide with the ions in the metal lattice. Kinetic energy is transferred to the metal ions as vibrations, and this increases the temperature of the metal. The amount of heat Q produced by resistive heating is proportional to the resistance of the material R   and the square of the current I  , as shown by Joule’s law:

Q = Pt = I 2Rt where Q is in joules, power P is in watts or joules per second (J s–1), I is in amps A, R is in ohms (Ω) and t is in seconds (s).

eddy currents produced in base of frypan ceramic surface coil supplied with high frequency AC

AC

rapidly changing magnetic field B

Figure 5.3.3 The frypan on an induction cooktop up heats due to eddy currents.

From Joule’s law we can see that a large current and relatively high resistance would result in a large amount of resistive heating. This explains why specialised cookware is required to gain maximum efficiency from induction cooking. Another method of heat production within induction cookware is a process called magnetic hysteresis losses. When a magnetic field is applied to, and then removed from, a magnetic material such as iron, a permanent magnetic field remains within the material. If a magnetic field is then introduced in the opposite direction to this remnant field, some energy is expended reducing the remnant field to zero before the field can build in the other direction. Energy is dissipated in this process as heat in the material and therefore also raises the temperature.

Checkpoint 5.3 1 2

108

Explain the formation of eddy currents in a small, flat, square metal sheet that falls between the poles of a magnet. Describe how Lenz’s law can be used to predict the formation of eddy currents.

PRACTICAL EXPERIENCES

motors and generators

CHAPTER 5

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 5.1: Generating electric current Using the equipment listed, write up an investigation that will allow you to generate alternating current. Once you have produced alternating current, investigate how changing the distance between a coil and a magnet, the strength of the magnet and the relative motion between the coil and the magnet will affect the electric current produced. Equipment: coil of wire (transformer coil), galvanometer, magnet (either electromagnet or a series of permanent magnets). Discussion questions 1 Describe the relationship between the distance between the coil and the magnet and the electric current produced. 2 Determine how the strength of the magnet affects the current produced. 3 What effect does making the magnet move instead of the coil and vice versa have on the current produced?

Activity 5.2: Making use of Eddy currents Research induction cooktops and check to see if advertised claims about their efficiency are true. Look at the use of eddy currents in braking. What forms of transport use it and where could it be applied? Discussion questions 1 Explain why AC and not DC must be used for an induction cooktop to work. 2 Discuss the efficiency claims of induction cooktops in comparison to traditional cooktops. 3 List the advantages and disadvantages of eddy current braking. (Hint: See Physics Focus on page 112.)

Perform an investigation to model the generation of an electric current by moving a magnet in a coil or a coil near a magnet. Plan, choose equipment or resources for, and perform a first-hand investigation to predict and verify the effect on a generated electric current when: • the distance between the coil and magnet is varied • the strength of the magnet is varied • the relative motion between the coil and the magnet is varied. Plan, choose equipment or resources for, and perform a first-hand investigation to demonstrate the production of an alternating current. Gather, analyse and present information to explain how induction is used in cooktops in electric ranges. Gather secondary information to identify how eddy currents have been utilised in electromagnetic braking.

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Chapter summary

Induction: the influence of changing magnetism

Michael Faraday discovered that a current can be generated by a changing magnetic field when moving a magnet within a wire coil. Magnetic field strength B in tesla (T) is equivalent to magnetic flux density in webers per square metre (Wb m–2). Magnetic flux ΦB is a measure of the magnetic field passing through a certain area. This is equal to the magnetic flux density B multiplied by the perpendicular area A⊥ through which the field is passing. An emf produced in a coil is proportional to the rate of change of magnetic flux and the number of turns in the coil.

•

Induced currents are produced by changing the magnetic flux due to relative motion, changing the flux density or changing the perpendicular area. Lenz’s law and the right-hand grip rule can be used to predict the direction of an induced current. Lenz’s law states that an induced current in a closed conducting loop will appear in such a direction that it opposes the change that produced it. Lenz’s law can be explained in terms of the law of conservation of energy. The production of eddy currents can be explained in terms of Lenz’s law. Applications of eddy currents include induction cooktops and eddy current braking.

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Review questions Physically speaking The theme of this word search is induction. There is a twist; there is no list provided, so you have to work out the words that have been included. Find the 10 hidden words that have to do with induction, list them and write their definitions.

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motors and generators

Reviewing 1 Outline Faraday’s experiment that led to the discovery

11 A conductive wire is placed on rails of an electrical circuit and forced to move to the right, as shown in the diagram below.

of electromagnetic induction.

2 Recall the factors that affect the size of the induced

wire

emf that is created by induction.

bulb

3 Magnetic flux is a measure of the magnetic field passing through a certain area. Use this statement to explain how magnetic flux can be altered.

rails

4 Outline how emf can be used to produce current. 5 Describe the place of relative motion in inducing emf. 6 Explain how Lenz’s law supports the law of

a Determine which is the positive and which is the negative end of the wire. b Determine the direction of the current in the circuit.

conservation of energy.

7 Compare induced current in a wire and eddy currents. 8 Explain why a solid square piece of copper swinging through a magnetic field will slow more quickly than one with slits in it.

12 Explain how eddy currents can be a problem. 13 Give examples of how eddy currents can be of use. 14 Predict the direction of the eddy currents in the following examples. a

9 Compare and contrast the use of eddy currents in induction cooktops and electromagnetic braking. (You may need to refer to Physics Focus on page 112.)

square metal sheet

Solving problems

rotating metal disc

15 Justify the claim that induction cooktops need

10 Using Lenz’s law, predict the direction of current

special cookware.

in the following situations. Sketch these diagrams, showing the induced currents. a c S

N

N

A

d

X

B

expanding wire loop Y

iew

Q uesti o

n

s

v

b

A

Re

S

b

111

5

Induction: the influence of changing magnetism

PHYSICS FOCUS Eddy Currents Stop Me in My Tracks

a

B

rakes are used in vehicles when it is necessary to reduce speed quickly. When a vehicle brakes, friction is usually used to convert kinetic energy into heat. Typically this involves a brake pad made of a heat-resistant material being forced against a metal disc attached to a wheel. This leads to wear on brake components and requires the ongoing cost of replacement. Some high-speed trains and modern roller-coasters now use eddy current braking. This method requires no physical contact between brake components and therefore no wear and tear. Some roller-coasters use stationary magnets to induce eddy currents in the moving roller-coaster. High speed trains in Europe have electromagnets fixed to the train, to induce eddy currents in the rails beneath, as shown in Figure 5.4.1.

N S

S N

N S

S N

N S

S N

coils

N S

S N

top of rail

Figure 5.4.1 An eddy current brake in action A series of current-carrying coils are suspended a few millimetres above the top of the rail. Their magnetic fields are produced with a polarity that is in the opposite direction to the neighbouring coil (Figure 5.4.1) by alternating the direction of the direct current within the coil.

112

b

B

top of stationary rail

– coil wound around soft iron core

location of coil

+ I rail

direction of train’s motion

Figure 5.4.2 (a) Simplified diagram of eddy current braking system. (b) Induced eddy currents in rail seen from above

The operation of these brakes is outlined in Figure 5.4.2. A direct current is passed through the coil, producing a magnetic field and the soft iron core intensifies the field. These fields extend into the iron rail and, as the train moves, the top of the rail experiences a changing magnetic flux. In accordance with Lenz’s law, this changing flux is opposed by the production of eddy currents and their associated magnetic fields (see Figure 5.4.2b). The braking effect is derived from the force that each current experiences due to the magnetic field of the other. For example, the eddy current in the rail experiences a force due to the magnetic field of the coil and an equal and opposite force is experienced by the coil due to the magnetic field of the eddy current. The kinetic energy of the train is converted to heat through resistive heating and magnetic hysteresis losses. A graph of typical rail heating is shown in Figure 5.4.3. The adoption of this technology is not yet widespread, as there are environmental and structural problems associated with excessive rail heating. Use is limited to rail lines with rails that can withstand the temperature changes without affecting their performance.

Change in rail temperature (°C)

motors and generators

Extension

30

6 Research other methods of applying eddy currents to braking and compare these methods to the one outlined above. 7 The kinetic energy lost when braking can be recycled by regenerative braking systems. Research these systems and outline their operation.

25 20 15 10 5 0 0

20

40

60

80

100

120

140

160

Braking force (kN)

H4. Assesses the impacts of applications of physics on society and the environment

Figure 5.4.3 Graph of rail heating 1 Identify and describe the function of eddy currents in train eddy current brake systems. 2 Calculate the amount of kinetic energy Ek that must be dissipated as heat to stop a train with a mass m of 5 × 104 kg travelling at a velocity v of 300 km h–1, using Ek = ½mv 2. (Remember to convert velocity to the appropriate units.) 3 If typical braking forces of 100 kN are applied, identify the associated change in track temperature. 4 Assess the environmental impact of an excessive rise in rail temperature if eddy current braking was used on a high speed train in Australia. 5 Identify further research that could be conducted to allow the widespread adoption of train eddy current brakes.

H5. Identifies possible future directions of physics research

Gather secondary information to identify how eddy currents have been utilised in electromagnetic braking.

113

6

Motors: magnetic fields make the world go around The magic of motors

DC motors, coils, armature, rotor, stator, electromagnets, torque, commutator split‑ring, commutator brushes, current‑carrying loop, galvanometer, back emf, supply emf, single-phase, three-phase, AC induction motor, ‘squirrel cage’ rotor, shaded-pole induction motor

Much of the activity and infrastructure that makes our lives of convenience possible is hidden from our view. Electric motors contribute significantly to our modern existence but remain a mystery. To most they are magical devices that convert electricity into motion with the flick of a switch. What better way to appreciate our comfortable lives than to understand one of the things that drives it. This chapter will equip you to contemplate how one invention, the electric motor, contributes to your quality of life in the age of technology.

6.1 Direct current electric motors Direct current (DC) motors transform electrical potential energy into rotational kinetic energy. They can be powered by relatively lightweight batteries and are therefore easily portable. The most common type of DC motor found in battery-operated toys is shown in Figure 6.1.1. These motors typically draw a current from a number of batteries. This current flows armature or rotor commutator brushes through a wire coil within an external magnetic field. The coil experiences a force we know as the motor effect, which causes the motor to rotate. The essential components that allow a DC motor to operate are summarised in Table 6.1.1.

stator (curved magnets)

commutator contacts

coils

Figure 6.1.1 Parts of a simple DC motor 114

motors and generators Table 6.1.1  Parts of a DC motor Part

Description and role

Coils

Many loops of wire that carry a direct current. These wires experience a force (due to the motor effect) that causes the motor to turn when current is flowing. This is the part of a motor that contains the main current-carrying coils or windings. For DC motors this is the rotor, but for AC motors it is usually the stator. In a DC motor, the rotor consists of coils of wire wound around a laminated iron frame. The frame is attached to an axle or shaft that allows it to rotate. The iron frame is laminated to reduce heating and losses due to eddy currents. The iron itself acts to intensify the magnetic field running through the coils. Stationary permanent magnets (electromagnets in large DC motors) that provide an external magnetic field around the coils. Permanent magnets are curved to maximise the amount of time the sides of the coil are travelling perpendicular to the magnetic field, to maintain maximum torque. A device with metal semicircular contacts that reverse the direction of the current flowing in each coil at every half rotation. This reversal of the current makes continuing rotation possible in a DC motor. Conducting contacts (generally graphite or metal) that connect the commutator to the DC power source.

Armature

Rotor

Stator

Commutator split-ring

Commutator brushes

Describe the main features of a DC electric motor and the role of each feature. Identify that the required magnetic fields in DC motors can be produced either by current-carrying coils or permanent magnets.

Brushless DC motors

T

he designs of many modern DC motors are more complex than the simple example we study in this chapter. The brushless DC motor has numerous significant advantages, and is used commonly in computer cooling fans (Figure 6.1.2). The cylindrical permanent magnets in the rotor have alternating poles around the circumference. Small electronics switch the direction of the current in the stator coils. The magnetic field from the stator coils rotates and this causes the rotor to rotate.

S

N S N

Figure 6.1.2 The brushless DC motor of a computer cooling fan

Now that we have seen the basic structure of a DC motor, we need to turn our attention to understanding and explaining its operation. In section 4.3 we encountered the motor effect, but now we will apply it to rotating a currentcarrying coil. Figure 6.1.1 shows that motors contain many coils of wire. For simplicity we will begin by looking at a single coil and how it would act as a motor.

Let’s torque about rotating coils A torque is the turning effect (or turning moment) of a force. The force that causes an electric motor to turn is the motor effect, so we must see how to calculate a torque before we can begin to perform calculations for motors. The idea of torque is illustrated in Figure 6.1.3a, which shows a force F due to a person sitting on a see-saw. In this case the force is acting perpendicular to the The magnitude (τ) of the torque involved is calculated using see-saw. the equation: τ = Fd

Define torque as the turning moment of a force using:  τ = Fd

115

6

Motors: magnetic fields make the world go around

a

pivot

F

d

where torque in newton metres (Nm) is given by the distance d  from the pivot If the force is not acting in metres multiplied by the force F in newtons. at right angles to the see-saw (Figure 6.1.3b), it is the perpendicular component F⊥ of the force that is used. As F⊥ = F cos θ, the formula for torque becomes: τ = F⊥d = F cos θd

b d

F θ

F⊥

F cos θ = ⊥ F

θ

Notice that the angle θ between F⊥ and F is the same as the angle θ the see-saw makes with the horizontal. We will use this fact again soon. Note that you could also use τ = Fd⊥.

Worked example QUESTION

Figure 6.1.3 A force acting on the end of a see-saw exerts a torque. (a) Force at right-angles to the see-saw and (b) force angled to the see-saw.

Identify that the motor effect is due to the force acting on a current-carrying conductor in a magnetic field.

a

axis of rotation

B

Y

X

FYZ

FWX

l

Z

W

Calculate the torque exerted on the see-saw in Figure 6.1.3b if a force of 980 N was acting at a distance of 4.0 m from the pivot at an angle θ of 30°.

SOLUTION

τ = F⊥d = F cos θd

= 980 × cos 30 × 4.0



= 3390 Nm

Quantifying torque on a coil When a current-carrying loop (or coil) is placed in a magnetic field, it is the force we know as the motor effect that applies a torque on the coil. Figure 6.1.4a If we know the shows a current-carrying loop within a magnetic field. direction of the current in the coil, we can use the right-hand palm rule to give us the direction of the forces on each side of the coil. Recall from our previous study of the motor effect that charges moving parallel to the magnetic field do not experience a force. This explains why only the forces FWX and FYZ act on the coil between points W and X, and Y and Z respectively, when it is in the position shown in Figure 6.1.4a. During the coil’s rotation, forces are experienced by sides XY and WZ, but they essentially cancel each other out and the coil is not free to move in these directions. Now let’s apply our new understanding of torque to this coil. The forces FWX and FYZ apply a torque to the coil, and at the moment shown in Figure 6.1.4a the magnitude of each torque (τ) is given by:

A

τ = F ⊥d

b FWX

θ F⊥

WX

d

rotation

axis of rotation

θ d

YZ

FYZ

Figure 6.1.4 (a) The current-carrying coil is able to rotate about the axis of rotation (dashed line). (b) The same coil a short time later has rotated slightly, as seen from point A along the axis of rotation. 116

We will use the wire between points W and X as our example and consider the situation in Figure 6.1.4b in which the coil has rotated. We must now account for the fact that force FWX is not acting perpendicular to the coil, so we use the formula seen previously: τ = F⊥d = FWX cos θd As the force FWX is the motor effect, we recall the equation: FWX = BIl sin θ Note that θ in this equation is the angle we met in section 4.3. It is the angle of a current-carrying wire to the magnetic field through which it passes. This is not the same angle as in the cos θ term of the previous equation. Since the

motors and generators two sides of the coil that experience a force (wx and yz in Figure 6.1.4) are always perpendicular to the field, θ = 90°. This makes sin θ = 1 and therefore: FWX = BIl Combining this equation with τ = FWX cos θd (using the other θ!) gives: τ = BIl cos θd where l is the length W to X (WX) and d is half the length of X to Y (XY/2). Now we consider the whole coil. The two forces FWX and FYZ are equal and both make the coil rotate in the same direction. The total torque on the coil is: τtotal = 2BIl cos θd The product 2ld is the area of the coil A. This gives: τ = BIA cos θ Motors have more than one coil, so we make one final adjustment to this expression, giving: τ = nBIA cos θ where n is the number of turns in the coil. This formula quantifies the torque (turning effect) on a rotating coil in newton metres (Nm). It is important to note that the angle θ in this equation is the angle of the coil relative to the magnetic field, as shown in Figure 6.1.4b.

Solve problems and analyse information about simple motors using:   τ = n BIA cos θ

Worked example QUESTION

axis of rotation

The coil in Figure 6.1.5 contains 50 turns and is carrying a 6.0 × 10–2 A current in a 5.0 × 10–3 T magnetic field. Calculate the torque on the coil if the coil has rotated 30° relative to the field.

= 4.5 × 10–5 Nm

S

cm

θ

I

N

cm



5

6

SOLUTION Using τ = n BIA cos θ and converting the lengths to metres: τ = 50 × 5.0 × 10–3 × 6.0 x 10–2 × (0.06 × 0.05) × cos 30

F B

F

Figure 6.1.5

Operation of a simple DC motor Let’s follow the rotation of a coil within a simple DC motor to understand how motors work. We will analyse the situation shown in Figure 6.1.7. In Figure 6.1.7a the coil of a DC motor is connected to a battery via a split-ring commutator at A. At the moment shown in Figure 6.1.7a, the two curved metal contacts of the commutator split ring direct the conventional current around the coil through points W, X, Y and Z in turn. Figure 6.1.7b shows the coil in Figure 6.1.7a as seen from the point A along the axis of rotation. The coil is seen end-on, and we can see the direction in which the current flows along sides These sides experience a force F that imposes a torque τ WX and YZ. on the coil, causing it to turn. The magnitude of the torque on the coil at positions 1–5 is shown on the graph in Figure 6.1.7c. Let’s now follow this coil through half a rotation and see what happens to the current and torque on the coil. Recall that the torque τ on the coil in Figure 6.1.7 is given by the equation:

PRACTICAL EXPERIENCES Activity 6.2

Activity Manual, Page 48

Describe the forces experienced by a currentcarrying loop in a magnetic field and describe the net result of the forces.

τ = nBIA cos θ 117

6

Motors: magnetic fields make the world go around In this example the values of all the variables on the right-hand side of the The variation in the cos θ equation remain constant, except for the angle θ. term accounts for the variation in the magnitude of the perpendicular force F⊥ experienced by the sides WX and YZ. This force determines the amount of torque acting on the coil, and therefore the torque varies accordingly as the coil rotates. • At position 1 in Figure 6.1.7b, the perpendicular force F⊥ on the coil is at a maximum and therefore the torque is at a maximum. This occurs when θ is zero and cos θ is 1. • At position 2, the coil has rotated 45° with respect to the magnetic field. The magnitude of the perpendicular force (F⊥) acting on the coil is less than at position 1. This means that the torque acting on the coil has reduced. At this point, cos θ is approximately 0.7. Note that real motors have curved magnets in the stator to ensure that each coil maintains θ = 0 as long as possible. This provides maximum torque for the longest time possible during each rotation. In the example in Figure 6.1.7, the magnets in the stator are not curved, so only at positions 1 and 5 is θ = 0. • At position 3, the perpendicular component of the force F⊥ has dropped to zero. Consequently, there is no torque acting on the coil at this instant. Here θ is 90° and cos θ is 0, so torque is zero. At this point the brushes and split ring within the commutator have broken contact momentarily and no current flows in the coil. The current needs to be reversed in the coil at this point, so that the motor continues to turn for the next half rotation. • Between position 3 and 4, the commutator has reconnected the coil and reversed the direction of the current flowing in the coil (now flowing through points Z, Y, X and W in turn). The opposite sides of the split-ring commutator are now connected to the terminals of the battery. If the current had remained in the original direction once the motor rotated past point 3, the forces on the coil would have been in the opposite direction to the initial rotation. This would stop further rotation. • As the motor moves through position 4, the perpendicular component of the force F⊥ is increasing. This means the torque is increasing and will reach a

Try this! Model a simple motor

T

o get a better understanding of the forces on a coil as it rotates, make a model of a simple motor. You need some coat-hanger wire, a few bamboo skewers, Blu-Tack, two small magnets, a cork and paper. Construct a coil as shown in Figure 6.1.6. Here the green skewer indicates the magnetic field. The red skewer points on the magnets represent the direction of the force due to the motor effect. Work with a friend and model the forces on a coil as it rotates.

Figure 6.1.6 A simple model of an electric motor b axis of rotation

a X F

Y

1

I

F

YZ

WX

F

F

B

S

c Position

Z

2

F

N

θ = 0°

F⊥

WX

W

Torque 0 max.

F

F⊥

θ

YZ F

θ = 45°

WX

+ –

3

A

Figure 6.1.7 (a) A simplified model of a DC motor. (b) The coil as seen from point A in part a, showing forces and current over a half rotation. (c) Torque on the coil over a half rotation 118

θ = 90°

YZ F

4

F⊥ YZ F

5

YZ

F

WX F

F⊥

θ = 45°

WX F

θ = 90°

Time

motors and generators maximum again at position 5. At position 5 the coil has performed half a rotation and the pattern we have seen from positions 1 to 5 will repeat until the coil is back to its original orientation at position 1.

PRACTICAL EXPERIENCES Activity 6.1

PHYSICS FEATURE

Activity Manual, Page 42

Galvanometers

T

he galvanometer was developed in the 1800s to measure the relative strength and direction of electrical currents. Its descendant is the analogue ammeter (Figure 6.1.8), which is calibrated in units of amperes and its basic structure is shown in Figure 6.1.9. In an ammeter, the coil is connected in parallel to a low-resistance wire within the ammeter. When a current is passed into the ammeter, only a small amount of the total current flows through the coil and this is proportional to the total current. When a current passes through the coil shown in Figure 6.1.9, it experiences a force due to the motor effect. This force exerts a torque on the coil, causing it to rotate around the pivot. A spring provides resistance to the torque, and the needle comes to rest when the torque on the coil equals the torque from the spring. The torque on the coil is proportional to the current passing through the coil. Therefore the amount the coil and needle move indicates the size of the current. To ensure that the needle moves (or deflects) by the same amount with each ampere of current being measured, the magnets are curved around the coil and iron core. This ensures that the magnetic field is always perpendicular to the flow of current along the sides of the coil. A uniform maximum torque will then be experienced by the coil through its whole range of movement. This means that the scale you read on the ammeter can be uniform (i.e. the scale has the same-sized divisions throughout).



Identify data sources, gather and process information to qualitatively describe the application of the motor effect in: • the galvanometer • the loudspeaker.

Figure 6.1.8 A ‘moving coil’ ammeter

pointer

force

permanent magnet

spring

coil

N

pivot

moving coil

S

soft iron core magnetic field

Figure 6.1.9 Cross-section of a ‘moving coil’ galvanometer

Checkpoint 6.1 1 2 3

Explain how each part of a DC motor contributes to its operation. Define torque and explain how torque varies during the rotation of a DC motor. Compare the features of a DC motor and a galvanometer. 119

6

Motors: magnetic fields make the world go around B

C

D

Current (A)

A

motor current 0 0

Time (ms)

Figure 6.2.1 Graph of net current for a DC motor averaged over many cycles

Account for Lenz’s Law in terms of conservation of energy and relate it to the production of back emf in motors.

6.2 Back emf and DC electric motors We now know that when a current is applied to the coil of a motor, the coil experiences a force. This force exerts a torque on the coil and the rotor begins to rotate. If we continue to apply the same current to the coil, the net force on the coil continues increasing the motor’s speed (recall Newton’s second law). This may make you wonder why a motor doesn’t just keep speeding up. We know from experience that they don’t because our toys with motors don’t accelerate forever. The main reason DC motors reach a maximum operating speed is that a Back emf is a back emf is generated in the coil as the motor rotates. potential difference across the terminals of the motor created by the changing magnetic flux passing through the wire coils within the motor (see section 5.1). Look back at Figure 6.1.7 for a moment. At position 1 the magnetic flux ΦB through the coil is zero. At position 2 in Figure 6.1.7b, the magnetic field is pointing right to left through the coil and the magnetic flux is increasing. This tells us that there should be an emf induced that would produce a magnetic field pointing from left to right to reduce the increasing magnetic flux. This field would be produced by a current flowing through Z, Y, X and W in turn. We can see from Figure 6.1.7a that this would be a current that opposes the one generated by the supply emf (through W, X, Y and Z in turn). The result of this would be that the net current in the coil would be less than the supply emf could generate. Another way to think about it is to recall the equation for Faraday’s law: ε = n(∆ΦB/∆t)

Try this! a back emf Connect a DC motor to a battery and place an ammeter in the circuit to measure the current flowing through the motor. Predict how the amount of current flowing will change if you apply a significant load on the motor. Try it! How did you go?

Explain that, in electric motors, back emf opposes the supply emf.

120

We can see that this changing flux ∆ΦB over time ∆t would generate a potential difference or emf ε. This potential difference (back emf ) is in the opposite direction to the applied potential difference (supply emf) that causes the rotor to turn. As the speed of the motor increases the back emf increases, as ∆ΦB/∆t increases. Eventually this potential difference cancels out most of the applied potential difference and virtually no current flows through the coil. At this point there is no net torque acting on the rotor and it turns at a constant speed. Let us analyse this situation using Figure 6.2.1. At time A on the graph, the motor is connected to a DC power source. The applied potential difference causes a current (blue line) to flow in the motor’s coil and this builds quickly to its maximum value. Once the coil starts turning, a back emf is generated due to the changing flux within the coil. This back emf is in the opposite direction to the applied potential difference and therefore reduces the net potential difference, which in turn reduces the current flowing in the coil as the speed of rotation increases. At time B, the motor has reached its maximum rotational speed. Here most of the supply emf has been cancelled out by the back emf. If the motor has no load attached to it, only a small current continues to flow. This residual current is required to overcome any friction within the motor and any voltage drop due to losses such as resistive heating. There is no net torque on the coil between times B and C and the motor operates at a constant speed.

motors and generators At time C, a large load is applied to the motor, such as the motor turning a wheel to move a toy car. The motor slows down quickly under this load and the amount of back emf is reduced. This means that the applied potential difference is greater and therefore a larger current flows through the coil. At time D, a larger current continues to flow through the coils. If the motor is not designed to handle the resistive heating produced by this larger current, the motor may ‘burn out’. Burn out occurs when insulation melts at high temperature, and may cause other components to melt. The motor will then cease to operate efficiently.

Back emf measures motor speed

I

Checkpoint 6.2 1 2 3

Define the term back emf. Describe the relationship between back emf and supply emf during the operation of a motor. Analyse the production of back emf in terms of Lenz’s law.

n some motors back emf is used to measure their speed. The input voltage is turned off for a short amount of time and the back emf is measured. Since the back emf is proportional to the speed of the motor, a calibration can be applied and the speed can be calculated.

6.3 Alternating current electric motors Many large appliances in your home contain motors that use single-phase 240 V alternating current (AC). Single-phase AC was illustrated in Figure 4.1.2 as a blue curve, showing a current changing direction many times a second. To connect an appliance to the mains power, you insert a plug (Figure 6.3.1) into the wall socket. This allows single-phase AC to be supplied by the active pin and the circuit to be completed by the neutral pin. Some industrial motors require more power and torque than can be supplied by single-phase AC. These motors are typically supplied with threephase 415 V AC by a plug with four or five pins. Figure 6.3.2 shows the AC signal available for three-phase equipment. This involves three alternating currents that can be applied to different coils within a motor at any one time.

active

neutral

earth

Current

+ 0

Time

–

Figure 6.3.1 A typical plug for an electrical appliance in Australia

Figure 6.3.2 Each of the three AC in three-phase power is 120° out of phase with the others.

121

6

Motors: magnetic fields make the world go around AC motors have the same basic components as DC motors but lack the need for others. They all contain a rotor and stator. Their magnetic fields can be generated by current-carrying coils. They utilise the motor effect to transform electrical potential energy into rotational kinetic energy. Let’s explore examples of the AC induction motor that exist both in industry and in the home.

Three-phase AC induction motors PRACTICAL EXPERIENCES Activity 6.3

Activity Manual, Page 51

The simplest induction motor is the three-phase AC induction motor (Figure 6.3.3). These motors are used in industry for their efficiency and In Figure 6.3.5a, reliability. Three-phase AC is fed to the coils in the stator. each pair of magnetic poles in the stator is fed one phase of the AC signal. The peak current of each phase is reached sequentially around the stator (Figure 6.3.5b), creating a magnetic field (the stator field) that rotates.

electromagnets

rotor

Figure 6.3.3 In this cutaway image of a three-phase AC induction motor, you can see the stator, consisting of electromagnets arranged to form a hollow cylinder. Within the stator sits the rotor, which is mounted on the motor’s shaft.

Universal motors

T

he most common type of motor in appliances around the house is the universal motor. This type of motor is capable of using both AC and DC. Figure 6.3.4 shows that coils act as electromagnets in the stator of these motors. These provide stronger fields and are lighter than permanent magnets. The coils are connected in series with the rotor coils and use a single-phase alternating current. These motors are found in many household appliances in which a variable speed is required, such as drills and blenders.

armature coils

carbon brushes shaft

stator coils segmented commutator

spring-loaded brush holders

Figure 6.3.4 A typical universal electric motor, showing the main components. Some motors would have additional stator coils. The commutator feeds current to the armature coils in the position where most torque will be experienced.

122

motors and generators a

b

conducting bars

Current

+ 0

Time

– i rotor

ii

1

iii 2

end ring

3

Figure 6.3.5 (a) In a three-phase motor, as the current in each pair of opposite

1

2

coils peaks, the field appears to rotate, dragging the rotor around with it. (b) A ‘squirrel cage’ rotor. The rotor is made of iron laminations to cut down undesirable eddy currents. The induced currents flow lengthwise in copper or aluminium rods which are joined at the ends (as in a squirrel cage).

3

stator pole

These motors contain a ‘squirrel cage’ rotor (Figure 6.3.5b) that does not Around the circumference of these require the input of an external current. rotors are a number of parallel conducting bars. These bars are joined at the ends by an end ring that allows current to flow from one bar to another. The rotating stator field induces a current in these bars and a magnetic field is induced in accordance with Lenz’s law. This induced magnetic field interacts with the rotating field from the stator and the resulting forces cause a torque on the rotor. This causes the rotor to spin without the need for a commutator as in a DC motor. As these motors do not need brushes and therefore have fewer moving parts, they are more efficient and more reliable. Another way to understand the operation of an AC induction motor is to consider a positive particle within one of the conducting bars in the stator. Let’s consider the bar marked A in Figure 6.3.6a. As the rotating magnetic field moves upwards past bar A this is equivalent to the bar moving downwards in a stationary magnetic field. Figure 6.3.6b shows this equivalent situation in which bar A moves relative to the magnetic field. Using the right-hand palm rule (see section 4.2) we see that a positive particle in bar A would experience a force into the page. This is equivalent to a current being induced into the page in bar A. Now we must use the right-hand palm rule (see section 4.3) to deduce the direction of the motor effect on a current-carrying conductor. This indicates that a force is exerted upwards on bar A and this is in the same direction as the rotating field in Figure 6.3.6a. This force on bar A is the same as the force experienced by each bar as the magnetic field rotates. These forces, in the same direction as the rotating stator field, exert a torque on the rotor and are responsible for its rotation. a

stator

b

2

force on bar A due to motor effect

B ‘squirrel cage’ conducting bars

A

rotating magnetic field

B I motion of bar A relative to the magnetic field

2

Figure 6.3.6 (a) The magnetic field due to one phase of an AC inductor motor. (b) The force acting on squirrel cage rotor bar A in part (a) due to the rotating magnetic fields 123

6

Motors: magnetic fields make the world go around

Single phase AC induction motors

Try this! Induction exerts forces Suspend a small piece of a lightweight conductor (e.g. aluminium foil) from fishing line. Use your right-hand palm rule to predict the force acting on the foil as you move a strong magnet vertically past the foil. Can you observe any movement? If not, use the scientific method to discover why it’s not working and try again.

Describe the main features of an AC electric motor. Gather, process and analyse information to identify some of the energy transfers and transformations involving the conversion of electrical energy into more useful forms in the home and industry.

A more complicated, but widely used, AC induction motor is the ‘shaded-pole’ AC induction motor shown in Figure 6.3.7. These motors require only a single phase of AC and can be found in most household electric fans. In these motors the alternating current is passed through a coil wrapped around the soft iron casing. The stator field induced by the alternating current passes through the casing and through the squirrel-cage rotor. Figure 6.3.7a shows the rotor removed and leaning on the motor. A cross-section of a squirrel cage rotor is also shown (Figure 6.3.7b) and clearly shows the conducting bars within the rotor. Four small copper shading rings can be seen within the stator in Figure 6.3.7a. These are inserted into the stator on each side of the rotor on opposite poles. The currents induced in these shading rings in accordance with Lenz’s law act to delay the magnetic flux passing through the rotor. This produces an asymmetric magnetic field passing through the rotor shown in Figure 6.3.7c. This leads to a changing magnetic field in each cycle of the alternating current that sweeps across each pole of the stator. The sweeping change in magnetic field strength across the rotor is essentially the same as the rotating magnetic field we studied in the three-phase AC induction motor. This rotating magnetic field causes a torque on the rotor in the same way as outlined for the three-phase induction motor. b

a

conducting bar

squirrel cage rotor shading rings

c

rotor shading rings delay the phase of part of the field to produce a rotating field

thick, copper shading ring

Figure 6.3.7 (a) A shaded-pole AC induction motor taken from a small household fan, with (b) a cross-section of the rotor. The conducting bars can be seen clearly within the laminated rotor. (c) The principle of a simple single-phase, ‘shaded pole’ induction motor. The distorted (or ‘shaded’) field causes the rotor to turn in one direction in preference to the other.

124

motors and generators

The right motor for the job Now that we have seen a few examples of some common electric motors, let’s consider for a moment why they are chosen for their common applications. The initial price and operating cost are key factors in making a decision. Operating costs depend on factors such as energy efficiency and replacement of parts due to wear and tear. The amount of torque required and how often a motor will be put under load are also critical factors. These will determine how much current is needed, the strength of the magnetic field required and numerous other parameters. Of course there are other constraints including portability, size and weight to consider. All these factors lead to a lot of homework for an engineer who is trying to design a machine or appliance driven by a motor. A summary of the characteristics of the different types of motors discussed in this chapter is provided in Table 6.3.1. Can you see why they are used in their common applications? Table 6.3.1  Characteristics of motors Type

Advantages

Disadvantages

Common applications

Simple (brushed) DC motor

Efficiency (%) 40–90

Low cost, battery powered, speed easily controlled

Toys, power tools, treadmill exercisers, automotive starters

Brushless DC motor

30–90

AC universal motor

40–60

Three-phase AC induction motor

70–90

Single-phase (shadedpole) AC induction motor

20–35

Long working life, low maintenance, high efficiency High starting torque, compact design, high running speeds High starting torque, high power, high efficiency, good power to weight ratio Inexpensive, long working life, high power, multi-speed

Short working life, high maintenance (brushes), sparking and ozone production High cost of some designs, requires a controller Less efficient than equivalent DC motor Requires three-phase power

Inefficient, low starting torque

CD and DVD players, computer hard drives Blenders, vacuum cleaners, hair dryers, portable power tools, sewing machines Industrial machinery, pumps and compressors Fans

Checkpoint 6.3 1 2 3

Compare the features of an AC induction motor and a simple DC motor. Construct a flow chart to account for the operation of a three-phase AC induction motor. Justify the choice of a three-phase AC induction motor for use in industrial machinery.

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Motors: magnetic fields make the world go around

PRACTICAL EXPERIENCES CHAPTER 6

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 6.1: Applications of the motor effect Identify data sources, gather and process information to qualitatively describe the application of the motor effect in: • the galvanometer • the loudspeaker.

Discussion questions 1 Outline how the motor effect is used to make music in a loudspeaker. 2 Explain how a loudspeaker differs from a motor in its use of the motor effect. 3 Determine the difference between the way in which the motor effect is used in a loudspeaker and in a galvanometer.

armature

N

S

Part A: Make a loudspeaker and determine how the motor effect is used to make it work. Equipment: two horseshoe magnets, strong sticky tape, thin insulated wire, cardboard, power supply, alligator clips and wires. Part B: Make a working galvanometer and determine the differences in this application of the motor effect. Equipment: PVC-covered copper wire (150 cm) with bare ends, wooden base board, armature block, magnets, split pins, knitting needle, rivets, wire strippers, drinking straw, rheostat (10–15 ohms, rated at 5 A or more).

brush

Activity 6.2: Motors and torque commutator

Figure 6.4.1 A simple motor

Solve problems and analyse information about simple motors using:  τ = nBIA cos θ

Make a motor like the one shown and note what factors change its performance. Calculate the torque of your motor. Equipment: insulated wire, magnets, magnetic field sensor and data logger (if available), paperclips, Blu-Tack, connecting wires with alligator clips, power supply. Discussion questions 1 Investigate the factors that determine the effectiveness of the motor. 2 Calculate the amount of torque in your motor and list ways in which torque can be increased.

Activity 6.3: AC induction motors Perform an investigation to demonstrate the principle of an AC induction motor.

Using the equipment supplied, make a model of an AC induction motor and relate each part to the parts in a real AC motor. Equipment: aluminium foil, fishing line, retort stand and clamp, ceramic magnet. Discussion questions 1 Outline how the metal is made to move. 2 Explain why the AC induction motor is so efficient.

126

Chapter summary •

•

•

•

•

The main features of a DC motor are the rotor (coils, shaft and frame), stator (permanent or electromagnets), commutator split ring and commutator brushes. The roles of these components are summarised in Table 6.1.1. Torque τ is the turning effect of a force F. τ = Fd where d is the distance from the pivot to the point where the force is applied. A current-carrying loop in an external magnetic field experiences forces due to the motor effect that generate a torque. The torque τ on a current-carrying coil can be quantified by the relation τ = nBIA cos θ where n is the number of turns in the coil, B is the strength of the magnetic field, I is the strength of the current in the coil, A is the area of the coil and θ is the angle between the plane of the coil and the magnetic field lines. The torque on a coil in an external magnetic field varies as the coil rotates. The torque is at a maximum when the plane of the coil is parallel to the magnetic field

•

•

•

•

motors and generators

and zero when the coil is perpendicular to the magnetic field. The current-carrying coil of a galvanometer experiences a torque due to the motor effect. This torque is balanced by a spring and this causes the needle to deflect by an amount proportional to the current flowing. This allows the determination of the magnitude and direction of the current being measured. When a DC motor rotates, its coils experience an induced emf (back emf ) set up in accordance with Lenz’s law. This back emf opposes supply emf and reduces the current flowing through the motor’s coils. AC induction motors contain a squirrel-cage rotor (conducting bars, shaft and frame) and stator electromagnets. AC induction motors generate a rotating magnetic field that induces currents in the squirrel-cage rotor. These current-carrying conductors in the rotor experience a force due to the motor effect, which exerts a torque on the rotor.

Review questions Physically Speaking

Reviewing

The terms in the following list belong to two distinct groups. Group these terms into the two groups and add the definition of each term. Create a diagram to display the relationship between them.

1 a Identify the type of motor in Figure 6.4.2.

• • • •

Split-ring commutator Squirrel cage Brushes Stator

• • • •

b Identify each of the labelled features. c Construct a table to list the parts you have identified and the role each plays.

2 Explain why radial (curved) magnets in a motor allow

Fan Bearings Armature Magnets

for greatest efficiency.

3 Describe the differences and similarities in the way permanent and current-carrying coils produce magnetic fields within a DC motor.

D E

4 Define a galvanometer and outline what it is used for. 5 A galvanometer has a spring attached to the centre of it, distinguishing it from a simple DC motor. Give reasons for its presence.

6 Recall Lenz’s law and explain how Lenz’s law accounts for the conservation of energy.

A

Figure 6.4.2

B

C

7 Describe what is meant by back emf. 8 Determine how back emf is produced in a motor. 9 Explain what a manufacturer does to a motor to account for back emf.

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Motors: magnetic fields make the world go around

10 a Label the parts of the AC motor in Figure 6.4.3. b Explain what you could do to this motor to make it into a DC motor. B

C

18 Students undertook to measure the torque produced by a simple DC motor.

The motor contained 100 turns and the square armature was 0.03  m in length. The motor is attached to a piece of string holding a mass (Figure 6.4.4).

D A

motor

E

motor shaft

r edge of table string

F

mass

Figure 6.4.3

11 12 13 14

Figure 6.4.4

Give examples of where an AC motor would be used. Compare and contrast DC and AC motors.

The motor was turned on and allowed to wind up the mass until it stalled and stopped. At this point the radius of the windings r was recorded. The current supplied to the motor was gradually increased and the process repeated.

Explain how an AC induction motor works. Explain how a single-phase induction motor gets started.

15 Distinguish between situations in which AC universal motors and AC induction motors would be best suited.

The table of results is shown below.

solving Problems 16 Calculate the maximum torque that is generated by a force of 460 N applied to an object at a distance of 3 m from its axis of rotation.

17 Calculate the torque in a square coil with sides of length 3 cm. The current in the coil is 2 A and it is placed in a magnetic field of 0.3 T.



Re

128

Q uesti o

n

s

v

a b c d

iew

τ = F × d = mg × d

Mass = 0.5 kg

Torque (Nm)

Current a

Radius (m)

0.27

0.1

0.055

0.54

0.2

0.11

0.81

0.3

0.165

1.08

0.4

0.22

1.35

0.5

0.276

1.62

0.6

0.331

Draw a graph of torque versus current. Determine the gradient of the line. What quantity does this value represent? From this value, determine the magnetic field in which the armature is spinning.

motors and generators

PHYSICS FOCUS Linear motors

F

igure 6.4.5 is of a maglev train—a train that floats on its rail and moves at very high speeds. You will study the floating mechanism in detail in Module 3 ‘From Ideas to Implementation’, but let’s have a look at how the train actually moves forward. The maglev train operates solely on electric power. The propulsion method is via an electric motor—but one with a difference—it is a linear motor. This is, in principle, an electric motor that has been unwrapped and flattened. Magnetic fields on the train and rail are continually created to attract and repel each other. It is these interactions that apply the forces to propel the train forward. Linear motors produce motion by moving in a straight line rather than the traditional rotational motion. There are two main parts: the stator unrolled (the primary) and laid flat on the rail, and the secondary, which is the glider on the train that floats over the rail.

3. Applications and uses of physics

Why use this type of motor? Reasons include: • It has no moving parts so there is no wear and tear. • The train rides on an air cushion, so less energy is lost due to friction. • Electromagnets are used for braking, so the train is a lot quieter. 1 Outline how a DC motor works. 2 Outline how an AC induction motor works. 3 Compare the uses for an AC induction motor with that of an AC universal motor. 4 Draw a diagram of a DC motor and explain what can be done to it to make it an AC motor. 5 Determine how torque is calculated in a motor. 6 Justify the use of linear motors in such applications as the maglev train. 7 Evaluate the cost associated with maglev trains and standard trains.

Research 8 Find out exactly how a linear motor works. 9 Compare the torque produced in a standard motor with the linear force in a linear motor.

Figure 6.4.5 A linear motor propels this maglev train.

129

7

Generators and electricity supply: power for the people Technology that changed our lives

generator, transformer, step-down transformer, step-up transformer, flux leakage, magnetic hysteresis, power stations, substation, transmission towers, insulators, lightning protector

Widespread access to affordable electricity has arguably been the single greatest catalyst for change in modern society. It has had an impact on every aspect of our lives, from our health to our wealth and even what we do in our leisure time. Some key developments that enabled this revolution were the invention of the AC generator and the transformer. Their significance is far beyond the reaches of the power lines they service and, as we shall see, they involve more than meets the eye.

7.1 AC and DC generators

Permanent magnets or electromagnets provide an external magnetic field.

In chapter 6, we saw that the real value of electric motors was that they convert electrical potential energy into rotational kinetic or mechanical energy. A logical question that follows is ‘how do we get the electrical potential energy that turns these motors?’. For many of our household applications, the answer is ‘we use a generator’. When you switch on an electrical appliance at home, you may not realise where the energy comes from. In some distant power plant there is a huge rotating machine as big as a building that looks like a huge electric motor. The difference is that this generator is producing electricity instead of using it. The simplest AC motor design is the synchronous AC motor (Figure 7.1.1). If this motor was supplied with 50 Hz AC from the power point, it would spin at 50 revolutions per second in synchronisation with the electrical signal. This S design has no practical application, but more complicated designs are used in clocks and tape drives, due to their constant speed.

N

AC supply Slip-ring commutator continuously connects the rotating coil to the AC supply.

130

Figure 7.1.1 A synchronous motor uses a slip-ring commutator to feed AC to the motor.

motors and generators Although it lacks applications as a motor, this design would produce an electric current if you turned the coil with your hands. In this way it is acting as When we turn the motor ourselves, its coil experiences a a generator. changing magnetic field and an emf is induced (recall Faraday’s law). If the loop is connected by a circuit, a current flows, and we have turned the motor into a generator. The parts of a generator are essentially the same as those of a motor (see Table 7.1.1). The difference is that we physically turn a generator and it produces electrical potential energy. So the operation of a generator is the opposite of that of a motor. In this section we will only consider simple generators; we will see a more complicated version in section 7.3.

Describe the main components of a generator.

Table 7.1.1  The parts of a simple generator and their function Part

Description and role

Armature

The armature is the part of a generator that contains current-carrying coils. These carry an induced current caused by a changing magnetic field. For simple generators these are the coils in the rotor, but in other generators the armature is in the stator. These are the many loops of wire that carry electrical current. In many generators there are two sets of coils. One set is the electromagnets that provide the magnetic field (in simple generators this field is provided by permanent magnets). The other set is in the armature and these electromagnets carry the current produced by the generator. The rotor generally consists of coils of wire wound around a laminated iron frame. The frame is attached to an axle or shaft that allows it to rotate. The iron frame is laminated to reduce heating and losses due to eddy currents. The iron itself acts to intensify the magnetic field passing through the coils. In simple generators the stator is the stationary permanent magnets or electromagnets that provide an external magnetic field around the rotor. These magnets are curved to maximise the amount of time the sides of the rotor coil are travelling perpendicular to the magnetic field. In some generators the stator contains the armature coils and a magnet turns as part of the rotor to produce a changing magnetic field. A simple DC generator contains semicircular metal contacts (a split ring) that reverse the direction of the current flowing out of the rotor coil every half rotation. This reversal of the current ensures that the current being produced is DC. A simple AC generator has two circular metal contacts. Each slip ring is connected to one end of the coils in the rotor. These provide an alternating current that changes direction every half rotation. In more complicated examples these provide a current to an electromagnet in the rotor and a current is produced in the stator. Brushes are conducting contacts (generally of graphite or metal) that connect the commutator to the external circuit.

Coils

Rotor

Stator

Commutator split ring Commutator slip ring

Commutator brushes

Figure 7.1.2a shows a typical hand-operated generator found in schools. This generator has the features of both an AC and a DC generator. With the flick of a switch you can connect to the components for either type of generator and produce a current by winding the handle. If you have access to one of these make sure you try it out and try to identify all its parts.

A simple AC generator

PRACTICAL EXPERIENCES Activity 7.1

Activity Manual, Page 54

Figure 7.1.3a shows a simple model of an AC generator. This generator therefore contains a slip-ring commutator that simply connects each end of the rotating coil to an external circuit. The rotor of the generator shown is being turned by hand. Let’s follow a full rotation of this generator as shown in Figure 7.1.3b and use the graph (Figure 7.1.3c) to gain an understanding of its operation. Figure 7.1.3c shows the amount of flux ΦB passing within the coil WXYZ. It also shows the induced emf ε caused by the rate of change of flux ∆ΦB/∆t in accordance with Faraday’s law (see section 5.1). This emf would result in a current flowing within the coil, if the output terminals were connected. 131

7

Generators and electricity supply: power for the people

a

b

slip-ring commutator to produce AC

slip-ring commutator to produce DC

S coils or windings

N

rotor and armature

stator magnets

output terminals

Figure 7.1.2 (a) A typical AC and DC generator found in high schools and (b) a close-up showing the main parts of this generator (see Table 7.1.1 for details)

• At position 1 (Figure 7.1.3b), the plane of the coil WXYZ lies parallel to the magnetic field. This means that no magnetic field lines are passing within the coil, so the amount of magnetic flux ΦB within the coil is zero. Recall Faraday’s law:

Try This! Modelling generators Use the model of a simple motor suggested in section 6.1 to model the operation of a generator. Add slip-ring and split-ring commutators by sticking pieces of aluminium foil around the cork. Ensure you can describe and explain the current that would be produced during a full rotation.

I

ractiv

e

nte

M o d u le

132

ε = n(∆ΦB/∆t) which shows that the emf ε induced in the coil is proportional to the rate of change in magnetic flux ∆ΦB/∆t passing within the coil—not the flux value itself. At position 1, the emf is at a maximum because the rate of change of the magnetic flux is at its maximum. This can be seen in Figure 7.1.3c in which the tangents to the curve (illustrating the slope of the curve) for the magnetic flux ΦB represent the rate of change of magnetic flux ∆ΦB/∆t. The tangent at position 1 is at the maximum positive value and, therefore, so is the induced emf, according to Faraday’s law. • If the output terminals of the generator are connected, the induced emf in the coil will cause a current to flow. At position 1, this occurs in the direction marked (i.e. through Z, Y, X and W in turn). To determine the direction of this current, let’s consider a positive charge in the wire between points W and X at position A. The right-hand palm rule shows that a charge moving upwards at position A, as the coil rotates within the magnetic field, will experience a force towards point W. You could also determine the direction of the current using Lenz’s law. Review Lenz’s law and then give it a try. • At position 2 the coil has rotated 45º. The amount of flux passing through the coil has increased from zero and will continue to increase until position 3. The rate of change of magnetic flux ∆ΦB/∆t at position 2 has decreased and will continue to decrease until the coil reaches position 3. Around position 3 the flux is changing relatively slowly. At position 3 the flux reaches a maximum and then starts to decrease. Since the rate of change of flux ∆ΦB/∆t is zero (as shown by the slope of the tangent) at position 3, the emf must also be zero according to Faraday’s law. This means there will be no current flowing through the coil at this point.

b a

F

rotation X

S

I

B A W

B

1

Y

I

v

2

Position

Z

WX

3

ε ΦB

YZ YZ YZ

5

+max.

WX

4

F

ε and ΦB

–max. 0

YZ

WX

N

c

motors and generators

WX

tangents show rate of change

WX

ΔΦB

YZ

Δt

YZ

6

WX

YZ

7 WX

YZ

8 axis of rotation

P

output terminals

WX

9

WX

YZ

Time

Figure 7.1.3 (a) A simplified model of an AC generator and (b) the coil as seen from point P in part (a), showing the direction of the current flowing over one rotation. (c) A graph of magnetic flux and induced emf over one rotation

• Once the coil rotates past point 3, the flux ΦB begins to decrease and the rate of change of flux ∆ΦB/∆t is increasing. We can see this as the slope of the curve for ΦB gets bigger (as shown by a tangent to the curve). When the flux ΦB is decreasing, ∆ΦB = ∆ΦB final – ∆ΦB initial is a negative value and so is ∆ΦB/∆t. This means that according to Faraday’s law the induced emf now has the opposite sign to the sign it had before position 3. This means that the current produced by this emf is in the opposite direction to the one that would have flowed before position 3. This explains why we get an AC current, because at positions 3 and 7 in Figure 7.1.3b the current will change direction. • Between positions 3 and 5 the flux decreases and is zero at position 5. The rate of change of flux increases to a maximum at position 5. This means that the emf increases to a maximum at point 5 and, therefore, so does the induced current. Notice that in Figure 7.1.3b the current flowing along the side of the coil WX is now flowing into the page. You can use the right-hand palm rule to show that the current is now flowing through points W, X, Y and Z in turn. This further illustrates that the current has reversed direction at point 3 and we are producing an AC current at the output terminals. • Follow the rotation through to position 9 and you will see that the current changes direction again at position 7 and the full cycle is completed at position 9. Each rotation induces an emf in the coil that is in one direction for half a rotation and in the opposite direction for the other half of a rotation. This alternating emf can produce an alternating current and our explanation of the operation of AC generators is complete.

A simple DC generator Now that we have seen how an AC generator works, let’s look at a DC generator. Figure 7.1.5a shows a simple model of a DC generator. The obvious difference is the split-ring commutator connecting the coil to the external circuit. We have seen this situation in a simple DC motor (see Figure 6.1.7) but now it is acting as a DC generator. Note that the graph for emf ε is the emf that would be measured at the output terminals. This is therefore affected by the inclusion of a split-ring commutator that is fundamental to the functioning of a DC generator. Let’s focus on what is different about the operation of a DC generator.

A Linear Generator magnet motion of magnet

coil

Figure 7.1.4 A torch without batteries

W

ith energy shaping up to be the single biggest issue for humans in the future, there are many new devices available to help us save energy. A torch without batteries is just one example. These torches contain a permanent magnet that can slide back and forth inside a coil of wire. The current generated in the coil is stored in a capacitor and is then ready for use in the torch.

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7

Generators and electricity supply: power for the people • As before, we see that the change in the rate of change of magnetic flux ∆ΦB/∆t in the coil causes a changing induced emf (Figure 7.1.5c). The operation of this generator is the same as the AC example until we get to position 3. At position 3 the brushes of the commutator reverse the contact between the coil and the external circuit. • The current flowing in the coil WXYZ is in the same direction as for an AC generator (see Figure 7.1.3b). The difference is that now this current only flows in one direction in the external circuit. As the current in the coil changes direction at position 3, the split-ring commutator reverses the connection. This causes the current to flow in the same direction in the external circuit throughout the entire rotation. • From the graph of emf in Figure 7.1.5c you can see that the emf induced in the external circuit rises and falls but is always in the same direction. This produces a current that also rises and falls and is always in the same direction. This is DC. We usually see a DC (or emf ) represented on a graph as a straight line with one value (e.g. the red line in Figure 4.1.2). This might be the case with a current from a battery, but not for the output direct from a DC generator. • At position 7 we see that again the split-ring commutator reverses the contact to the external circuit. A current in this external circuit would continue to flow in the same direction, even though the current in the coil again changes direction.

Regenerative braking

A

n energy-efficient way to slow a moving vehicle is to convert the kinetic energy of the vehicle into electricity. This energy can then be stored and reused to get the vehicle moving again. Traditionally braking was achieved by using friction and the energy was lost as heat. Modern trains and electric hybrid automobiles now use these systems to increase efficiency. The electric motors on the wheels of the vehicle are used as generators when the brakes are applied, and energy is stored in batteries or a capacitor.

Position

b 1

2 a X S

I

B

axis of rotation

+max.

WX

ΦB

YZ

3

N

YZ

tangents show rate of change ∆ΦB

WX

4

∆t

YZ Z

W

0

WX Y

A

ε and ΦB –max.

YZ

WX

F

I

c

5

6

WX YZ

ε

YZ WX YZ

P

7 output terminals

WX YZ

8 WX

9

WX

YZ Time

Figure 7.1.5 (a) A simplified model of an DC generator and (b) of the coil as seen from point P in part (a), showing one complete rotation. (c) A graph of magnetic flux and induced emf over this rotation

134

motors and generators

Comparing AC and DC generators The differences in the structure and operation of simple AC and DC generators The two styles of commutator are essential to the should now be clear. production of either AC or DC and are the main difference. As DC generators have more complicated commutators, they are generally less reliable. The continual electrical reconnection and the extra mechanical wear in the split-ring commutator reduces the service life of a DC generator and increases running costs. Electrical arcing between the split ring and the brushes also produces electromagnetic radiation. This can cause interference in other electronics and to radio communication. When we consider the operation of simple generators, we see that both styles of commutator connect the rotating coils to an external circuit. The commutators in a DC generator perform the additional function of reversing the direction of the connection to the outside circuit each half revolution. This acts to maintain the current in the external circuit flowing in one direction, even though the current within the rotating coil is changing direction. We will see in section 7.3 that more complicated generators produce their current in the stator. We will learn that this is the case in a generator used in a large-scale AC power plant. This configuration is best for generating high currents and is designed to produce three-phase electricity.

PHYSICS FEATURE Comparing motors and generators

A

Describe the differences between AC and DC generators. Gather secondary information to discuss advantages/ disadvantages of AC and DC generators and relate these to their use.

Compare the structure and function of a generator to an electric motor.

dynamotor. These can be used to convert DC electricity to different voltages. This could occur if the set contained a DC motor and DC generator. As we will see in the next chapter, there is a handy device that can easily convert AC to different voltages, but it is more difficult with DC. Motor-generators can also be used to convert AC to DC, DC to AC and AC to AC at another frequency. Modern electronics have now allowed all these conversions to be made without the use of a motor-generator, but there are still some applications where electronics are not practical.

comparison between motors and generators is now relatively easy. First, take a look at Tables 6.1.1 and 7.1.1 to identify the similarities and differences. You may have noticed that we had already seen the principles behind the operation of a generator when we studied back emf in motors (section 6.2). The induced emf that opposes supply emf in a motor would produce a current if there was no supply emf— and you would have a generator. Figure 7.1.6 shows the AC voltage is applied to the combination of a motor and a motor coil through slip rings generator. This illustrates the conversion of mechanical energy to AC voltage output is through slip rings and brushes electrical energy in the generator and back again in the motor. It also N shows the similarity in construction mechanical energy input is converted of both an AC motor and an AC into electrical energy S generator. This combination of a generator motor and a generator is commonly known as a motor-generator set or Figure 7.1.6 A motor-generator set or dynamotor

mechanical energy output

N

S

motor

back emf generated by turning of motor voltage applied to motor

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Checkpoint 7.1 1 2 3

Use labelled diagrams to outline the structure of simple AC and DC motors. Compare the structure and function of motors and generators. Describe and contrast the structure and function of AC and DC generators.

7.2 Transformers Discuss why some electrical appliances in the home that are connected to the mains domestic power supply use a transformer. Describe the purpose of transformers in electrical circuits.

Our everyday use of electricity requires a large variety of electrical currents and voltages. With DC, this usually means that we have the right battery for the job. With AC, it’s another story altogether. When you plug in your laptop computer, it typically requires an AC input of about 18.5 V, but the average value of the voltage from the power point is about 240 V. (Note: the Australian standard is now 230 V.) The conversion of AC to the correct voltage occurs, with a range of +10% to –6%, in a small box on the laptop’s power cable, which contains a device called a transformer. changing magnetic flux

Try This! exploring electromagnetism To observe the magnetic flux produced by a changing current, place a wire with many loops near or wrapped around a magnetic compass. Connect a battery to your coil and observe the compass needle as you connect and disconnect the circuit.

340

200

0

22

32

0

W

0 280 3 0 26 00

24

14

40

0

20

160

S

N

60

80

100

E

0

12

Figure 7.2.2 A current through a coil induces a magnetic flux.

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AC supply

voltage Vp

voltage Vs

primary coil with np turns

load

secondary coil with ns turns

Figure 7.2.1 In an ideal transformer, the iron core ensures that all the flux generated in the primary coil also passes through the secondary coil.

Transformers are devices that can easily convert alternating currents to higher or lower voltages. They consist of two coils of insulated wire wound around an iron core (see Figure 7.2.1). You may notice that they have the same basic construction as Faraday’s iron ring we studied Faraday observed earlier (see Figure 5.1.4). As we saw in Chapter 5, an induced current in a secondary coil only when there was a changing current in the primary coil, and this is illustrated in Figure 5.1.5. The changing current in the primary coil created a changing magnetic flux that passed through the secondary coil. This changing flux induced an emf in It is the secondary coil and an induced current was measured. important to note that when a direct current was applied to the primary coil it did not produce a continually changing magnetic field, so transformers are only practical for AC. To quantify the voltage conversions performed by transformers, we recall Faraday’s law from Chapter 5: ε = n(∆ΦB/∆t)

motors and generators Let’s consider applying a 50 Hz input voltage Vp across the primary coil of the transformer in Figure 7.2.1, which has np turns (or loops) of wire. If we assume there are no losses, this will produce the same changing magnetic flux ∆ΦB/∆t within both coils. This changing magnetic flux will produce a 50 Hz alternating voltage Vs across the secondary coil with ns turns. Substituting into Faraday’s law gives:  ∆Φ B   ∆Φ B  Vp = np  and Vs = ns   ∆t   ∆t 

Dividing one expression by the other gives: Vp Vs

=

np ns

This expression shows us the relationship between the number of turns and the voltages across each coil of a transformer. If we analyse the expression above, we can see that if the primary coil has more turns (loops) than the secondary coil (i.e. np/ns > 1), the voltage Vp across the primary coil will be greater than the voltage Vs across the secondary coil. This is called a step-down transformer because it produces AC with a lower voltage. When the secondary coil has more turns than the primary coil (i.e. np/ns < 1), the voltage Vs measured across the secondary coil is higher than the This is called a step-up transformer because it produces input voltage Vp. an AC output with higher voltage. Figure 7.2.3 is a simplified diagram showing the change in magnitude of the output voltage Vs compared to the input voltage Vp for these two types of transformers.

Identify the relationship between the ratio of the number of turns in the primary and secondary coils and the ratio of primary to secondary voltage. Vp

t

Vs

t

Worked example Figure 7.2.3 The simplified graphs show

QUESTION

the AC output of a step-up transformer (blue line) and a step-down transformer (red line) in the secondary coil of a transformer. The input voltage in the primary coil is shown for comparison (black line).

Calculate the number of turns required in the secondary coil of a transformer to produce 18.5 V AC from 230 V AC if there are 100 turns in the primary coil.

SOLUTION Using:

Vp Vs

=

np ns

V  ns = np  s   Vp  18.5 Substituting Vp = 230 V, Vs = 18.5 V and np = 100 gives: ns = 100 × 230 = 8 turns and rearranging to make ns the subject gives:

Transformer efficiency and design The law of conservation of energy states that energy cannot be created or destroyed, but can be converted from one form to another. This means that the amount of energy you put into a transformer must equal the amount you get out. This is true for all transformers, but some of the energy that went into a transformer is always converted to heat. It is for this reason that transformers are never 100% efficient.

Compare step-up and stepdown transformers. Solve problems and analyse information about transformers using:

Vp Vs

=

np ns

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Generators and electricity supply: power for the people Efficiency is often expressed as a percentage and can be determined using the following equation: useful power output Efficiency (%) = × 100 total power input The losses of energy that occur within a transformer cause the useful electrical power output to be less than the power input. We saw in Chapter 4 that power is the rate at which energy is converted and can be determined using the following relationships: V2 P = IV, P = I 2R and P = R where P is electrical power in watts (W), I is current in amps (A), V is potential difference (or voltage) in volts (V) and R is resistance in ohms (Ω). In formulating the equation: Vp np = Vs ns we assume that there are no losses; that is, that transformers are ideal. We will continue this assumption whenever we do calculations for transformers, but it is important to consider transformers in a more realistic way, to understand their losses and design considerations. Using the equation above, we can determine another relationship for a transformer that illustrates the changes in current. In an ideal transformer the power input in the primary coil (Pp) and the power output in the secondary coil (Ps) would be: Pp = IpVp and Ps = IsVs

Neon lights

N

eon lights are one very visible application of transformers. Typically about 25 000 volts are required to excite the gas in a neon tube, so a step-up transformer is used to convert the voltage from 230 V AC. In Figure 7.2.4, the red light is produced by neon gas and the violet-blue light is produced by argon.

Figure 7.2.4 Neon and argon discharge tubes

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Rearranging and substituting into the relationship above gives: I s Pp np = I p Ps ns Pp = Ps for an ideal transformer, so: I s np = I p ns

This illustrates that both the voltage and the current output of a transformer change with respect to the input. We have therefore gained another insight into the workings of transformers. By inspecting both equations: Vp np I s np = and = I p ns Vs ns

we can see that for a step-up transformer (ns > np), the transformer increases for a the voltage but decreases the current of AC electricity. Alternatively, step-down transformer (ns < np) the voltage decreases but the current increases. Let’s see how this situation satisfies the law of conservation of energy. Since energy (in joules) must be conserved and power (in joules per second) is the rate at which energy is converted, then power must also be conserved if no energy is being stored in a transformer. As power is equal to the product of the current and voltage: Pp = Ps and IpVp = IsVs

motors and generators As we already know, the voltage is changed in a transformer and, for the above relationship to hold, the change in current must be inversely proportional to the change in voltage. We have already shown that this is the case by deriving the two expressions: Vp np I s np = and = I p ns Vs ns Combining these expressions gives: I s Vp = I p Vs

This expression shows that current and voltage are inversely proportional. For example, in a step-up transformer Vs is larger than Vp so Vp/Vs must be less than 1. In this case, Is/Ip is also less than 1. This means that Is is less than Ip. We now see that transformers obey the law of conservation of energy when they alter the current and voltage of AC electricity. For an ideal transformer, the input power is equal to the output power and this is maintained by the inverse relationships between input and output current and voltage. These findings explain the design features evident in Figure 7.2.5, which shows a step-down transformer. The increase in current in the secondary coil is the reason a larger diameter wire is used for the secondary coil. As discussed in Chapter 5, larger currents produce greater losses due to resistive heating. The amount of heat Q produced by resistive heating is proportional to the resistance of the material R and the square of the current I as shown by Joule’s law: Q = Pt = I2Rt where Q is in joules, power P is in watts or joules per second (J s–1), I is in amps, R is in ohms and t is in seconds (s). So we see that larger currents produce more heat. The resistance of a metallic conductor is described by the following equation: ρl R= A where ρ is resistivity, l is length and A is cross-sectional area. This equation shows that resistance in a wire is proportional to its length and inversely proportional to its cross-sectional area. So, by increasing the diameter of the wire in the secondary coil of the transformer (Figure 7.2.5), the resistance is decreased and this minimises resistive heating. resistive heating is one of the mechanisms We have now identified that responsible for energy losses in all transformers. The transformer shown in Figure 7.2.5 is a very common style of transformer in household appliances because it minimises another loss mechanism—flux leakage. The transformer in Figure 7.2.5 has a central iron core around which the primary and secondary coils are The overall shape of the iron core acts to contain and direct the wound. magnetic flux from the primary coil through the secondary coil. This ensures that the maximum amount of flux passes through the secondary coil and therefore maximises induction in the secondary coil. If a significant amount of flux did not pass through the secondary coil, then energy would not be transferred and could be lost by other mechanisms.

Explain why voltage transformations are related to conservation of energy.

secondary coil primary coil

Figure 7.2.5 A step-down transformer in which the coils are wound around a central laminated iron core

Gather, analyse and use available evidence to discuss how difficulties of heating caused by eddy currents in transformers may be overcome.

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a primary coil

cross soft iron core section I

B

increasing magnetic flux Φ

I

secondary coil

b solid core

increasing magnetic flux Φ eddy currents cross section

c laminated core

eddy current cross section

Figure 7.2.6 (a) A simple model of a transformer. (b) Large eddy currents are set up in the core. (c) Smaller eddy currents are set up in a laminated core.

The Sounds of AC

I

f you ever get a chance to see a large industrial transformer you will probably hear it hum. This is the sound of the laminations in the core vibrating. As the direction of 50 Hz AC output alternates back and forth, the magnetic field within the core also changes direction. The alternating magnetic flux exerts forces on the core laminations and causes them to vibrate.

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The remaining mechanisms for losses in transformers are sometimes referred The first of these is loss due to magnetic hysteresis as the to as core losses. magnetic field continually changes direction. We were introduced to this effect in our discussion of induction cooktops (see Chapter 5 Physics Feature p 108). In induction cookware hysteresis served a useful purpose in heating food, but in a transformer we wish to minimise the effect. This can be done by making the core of a transformer from a magnetically ‘soft’ iron that does not stay magnetised once an external magnetic field is removed. Magnetically ‘soft’ iron is therefore used in transformer cores to reduce heat loss through magnetic hysteresis. This soft iron is also used in the rotors of motors and generators and in the stators of AC motors to reduce losses due to magnetic hysteresis. The other type of core loss in transformers is resistive heating due to eddy currents. To reduce this effect, the soft iron cores of transformers are made from thin laminated sheets that are electrically insulated from each other. Figure 7.2.6a shows a simple set-up of two coils that would act as a transformer. This diagram shows a moment in time in which the current in the primary coil is producing an increasing magnetic flux in the direction indicated. In accordance with Lenz’s law, currents would be induced in both the secondary coil and as eddy currents in the iron core (Figure 7.2.6b). To minimise the energy lost as heat due to these eddy currents, the core is made from thin laminations (Figure 7.2.6c) that are electrically insulated from one another. This minimises the magnitude of the eddy currents and therefore the amount of heat lost through resistive heating. We have now seen that transformers come in many shapes and sizes, but they all have the same basic structure. They contain a magnetically soft, laminated iron core and two coils of insulated wire. Large transformers can be up to 99% efficient, but there are always some losses as heat. These losses, although small in proportion, can affect the performance of a transformer, if allowed to build up. In small transformers, air removes heat by convection, but in large transformers oil is used (Figure 7.2.7). The oil is passed through cooling tubes and heat is dissipated to the air. This type of transformer is used in electrical substations and at power plants as part of the distribution of electricity to the community. We will explore this more in the next section.

low voltage terminals

high voltage terminals oil tank

laminated core low voltage coils

high voltage coils

cooling tubes

Figure 7.2.7 An industrial transformer used in the transmission of electricity

motors and generators

Checkpoint 7.2 1 2 3 4 5 6

Identify a device that uses a transformer and outline the role the transformer plays in the operation of this device. Define step-up and step-down transformers. Explain why transformers do not work with DC electricity. Calculate the voltage produced by a transformer for which np/ns is 0.01 and the input voltage is 12 V. Recall the law of conservation of energy and analyse its application to the operation of a transformer. Identify the losses that occur in transformers and outline the design features that minimise these losses.

7.3 Electricity generation and transmission PHYSICS FEATURE

PRACTICAL EXPERIENCES Activity 7.2

Activity Manual, Page 58

a

b

Edison and Westinghouse

I

n the late 1800s a rivalry developed in the US over the type of electricity that should be generated. Thomas Edison (1847–1931) had established a system based on DC electricity. It was primarily used for household lighting and Edison had patents on much of the technology involved. One of the major drawbacks of this system was that it had to be distributed at the voltage used in households, as transformers could not be used with DC to easily alter the voltage and current. Transmission therefore required large currents, and this limited the distance DC could be transmitted because of the huge power losses as a result of resistive heating. This meant that power stations had to be dotted all over a large city and distribution was not practical in rural areas. When Edison was establishing his system there was no viable alternative to DC. Nicola Tesla (1856–1943), an employee of Edison at one stage, eventually developed AC motors and generators. The patents for these were promptly purchased by George Westinghouse (1846–1914). Alternating current had the advantage that voltage and current could be changed using transformers. It could therefore be generated at low voltages and converted to high voltages for transmission using low currents. This allowed transmission over long distances

Figure 7.3.1 (a) Thomas Edison and (b) George Westinghouse with acceptable energy losses, and made AC more economical. Edison embarked on a campaign to discredit the use of AC and promote it as unsafe. To do this he used an AC generator to publicly electrocute animals, including an elephant. The showdown between Edison’s DC and Westinghouse’s AC came when a proposal was put forward to generate hydroelectricity at Niagara Falls. Several designs were considered, including those from Westinghouse and Edison. Westinghouse was awarded the contract and, due to its many advantages, AC electricity was eventually adopted as the standard worldwide. Gather secondary information on the competition between Westinghouse and Edison to supply electricity to cities.

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AC power generation and delivery

Figure 7.3.2 A coal-fired power station a

C B

three phase output

stator N

A

A S

B

C

neutral

Current or EMF

b + 0V

A

B 10

C

20

30

40

–

Time (ms)

Figure 7.3.3 (a) A three-phase generator. The DC output is fed to the rotating magnet via slip rings (not shown). The ends of each of the three stator coils are connected together—this becomes the ‘neutral’. (b) The other ends carry the high voltages, which are one-third of a cycle apart. Explain the role of transformers in electricity substations.

142

Today, more than 99% of the world’s electricity is generated as AC. Most of the electricity generated in New South Wales is produced by burning coal in a power plant like the one shown in Figure 7.3.2. Heat from the burning of coal in these facilities is used to produce steam. This steam is used to turn turbines, which are connected to large generators that produce electricity. These generators typically produce three-phase 23 kV AC at 50 Hz. Figure 7.3.3 shows the main features of a three-phase generator used in a power station. Once generated, electricity is transmitted to consumers through a vast network, such as the one summarised in Figure 7.3.4. The most common use of this energy is in our homes. It arrives there in wires above your street or below your feet in underground cables. In order to get there efficiently, electricity goes through the following steps on the way to your home. The 23 kV 50 Hz output voltage from a power station’s generators is passed through step-up transformers to produce 330 kV or even 500 kV AC. High voltages mean low currents and this reduces power losses during transmission due to the resistance of the lines. Large transmission lines are used to transport this electrical energy over long distances to a terminal substation. Fuel sources for power stations are typically a long way from major population centres, so this long-distance transmission is essential. At the end of the transmission line, step-down transformers in a terminal substation convert the voltage to 132 kV or 66 kV for transmission to zone substations (see Figure 7.3.5). Higher voltages are less important for these shorter transmission distances. As the voltages decrease, the effort required to insulate each wire from the others decreases and so does the cost involved. At zone substations this power is again stepped-down using transformers. Most commonly it is converted to 11 kV AC and distributed into communities. Typically, transformers on power poles finally step-down the AC to 400 V and 230 V for industrial and household use respectively. These voltages are low enough so that discharges cannot occur through air within appliances, and the insulation of wires from each other is easy and economical. A typical power pole in a suburban street resembles the one shown in Figure 7.3.6. Three of the four main wires shown carry the three phases produced in the power plant’s generator (see Figure 7.3.3). The other wire is the neutral that completes the circuit. The potential difference between neutral and a single phase is 230 V. The difference between any two of the three phases is 400 V. A single phase and the neutral are connected to each household and all four wires deliver three-phase power to industry (see Figure 7.3.4).

motors and generators

step-up transformers GENERATION

POWER STATION 23 kV 50 Hz

330 kV or 500 kV AC transmission

HIGH VOLTAGE TRANSMISSION LINES TRANSMISSION

TERMINAL SUBSTATION 132 kV or 66 kV AC transmission

step-down transformers 132 kV & 66 kV AC SUB-TRANSMISSION

ZONE SUBSTATION

11 kV AC distribution step-down transformers 11 kV AC DISTRIBUTION

400 V and 230 V AC distribution

pole step-down transformer

CONSUMER

household single-phase 230 V AC

N

3 2

1

230 V 400 V

factory three-phase 400 V AC

Figure 7.3.4 A typical electricity supply network in New South Wales 143

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Generators and electricity supply: power for the people

four wires carrying three-phase AC and the neutral wire

insulator

Figure 7.3.5 An electricity substation

Figure 7.3.6 A power pole in Marrickville, Sydney. Note that the four main wires are not insulated. They are spaced far enough apart to stop discharges occurring. They are insulated from the pole by small ceramic insulators.

PHYSICS FEATURE Transformers in the Home

T

he use of transformers does not end when electricity is fed into your home. The components of many household appliances require higher or lower voltages than the 230 V available from the wall socket. A good example is the microwave oven. The component within a microwave oven that produces the microwaves typically requires thousands

of volts, while the control circuits and control panel on the front only use small voltages. This means that a microwave oven would require both step-up and step-down transformers to supply the power for its components. Discuss why some electrical appliances in the home that are connected to the mains domestic power supply use a transformer.

Power losses during transmission and distribution Discuss the energy losses that occur as energy is fed through transmission lines from the generator to the consumer.

In order to deliver electrical power efficiently across NSW the power loss between The main the generator and the customer must be as small as possible. losses during transmission result from the resistance in the transmission wires and the induction of eddy currents. First, let’s look at losses due to resistance. Using a relationship we have seen before: P = I 2R we see that power loss P in transmission lines is proportional to the resistance R and the current I squared. Therefore, if we double the resistance in transmission wires we would double the power loss. But if we double the current we increase So obviously, we want the resistance in the power loss by a factor of four. transmission wires as low as possible but, even more importantly, we want to use a relatively small current. This is achieved using transformers to step-up AC to as

144

motors and generators high a voltage as practical. During this process the current is reduced to a minimum. Another way to illustrate why we want to reduce transmission current comes from another relationship we saw earlier: ρl R= A where ρ is resistivity, l is length and A is cross-sectional area. This equation shows that the resistance R of a wire is proportional to its length l . This means that, as the distance of transmission increases, the resistance also increases. A typical resistance for one type of high-voltage transmission line is about 0.4 ohms per kilometre. We can see that if large currents are used, transmission will quickly become uneconomical over the hundreds of kilometres typically required. A large proportion of the energy generated would be lost as heat due to resistive losses. The best solution is again to minimise the current being transmitted. The two previous paragraphs highlight the importance of transformers in the distribution of electricity. Transformers can readily step-up voltages and in doing so they minimise the magnitude of the current involved. They can also step-down the voltage when required to a value suitable for the consumer. This ability of transformers has changed modern society enormously, and we will take a closer look at this soon. ρl also shows us that the resistance of a wire is inversely The equation R = A proportional to its cross-sectional area A. So making the diameter of the wire larger will reduce its resistance. Unfortunately, thick copper wires are very heavy and require larger structures to support them. Larger structures cost more money, so a compromise is required. Aluminium is used as a conductor in high-voltage transmission lines. Although it has a higher resistance than copper, it is much lighter, so the wires have a larger diameter to reduce resistance without being too heavy. Losses during AC transmission also occur due to the formation of eddy currents. The constantly alternating current in transmission lines produces a constantly changing magnetic flux. This can induce eddy To currents in nearby conductors and energy will be lost. minimise this, transmission lines are held at a distance from metal transmission towers by insulators. Losses due to eddy currents also occur in the cores of transformers. Recall that these losses are minimised by constructing these cores from thin laminations of magnetically soft iron.

Gather evidence and analyse secondary information to discuss the need for transformers in the transfer of electrical energy from a power station to its point of use.

PRACTICAL EXPERIENCES Activity 7.3

Activity Manual, Page 62

lightning protector insulators

Transmission structures A typical high-voltage AC transmission tower is shown in Figure 7.3.7. The three sets of wires carry the three phases generated in the power station. As with lower voltage transmission (Figure 7.3.6), high-voltage transmission lines are not coated with insulation. These three sets of wires must be kept at large distances from each other and from the metal tower. This not only minimises losses due to eddy currents, it also minimises the chance of current flowing through the air (electrical discharge) between conductors or to the ground through the tower. In ideal conditions, 500 kV

high voltage transmission lines

Figure 7.3.7 A high-voltage transmission tower 145

7

Generators and electricity supply: power for the people Gather and analyse information to identify how transmission lines are: • insulated from supporting structures • protected from lightning strikes.

a

transmission lines must be approximately half a metre away from the tower that supports them. In conditions in which the insulators are wet and covered in pollution deposits, this distance must be greater. Typically, therefore, these wires are held more than a metre from their supporting tower. Figure 7.3.8 shows that insulators used for lines are typically long chains of ceramic components. These components are disc shaped and have deep corrugations on their underside. These features increase the distance any current would have to travel in a discharge. The corrugations also minimise the chance of pollution deposits settling on the surface and keeps some of the insulator dry in wet weather.

High-Voltage DC Transmission

M

odern electronics have made the task of converting high-voltage AC to high-voltage DC (HVDC) relatively simple, and HVDC transmission is now economically viable for distances of more than several hundred kilometres. These systems lose less power due to eddy currents induced in metal support structures and require only two transmission lines. For short distances, these savings do not offset the cost of the extra electronics, but if the transmission distance is long enough, the savings are significant. HVDC is the method of choice for transmitting electricity by underwater cables due to the excessive losses involved. The Victorian and Tasmanian electricity grids are connected across Bass Strait via a 400 kV DC link called Basslink. This provides a twoway link along which electricity flows according to the demand at either end. In this way, either state can make up for a shortfall of electricity in the other.

b

The impact of AC generators and transformers on society

insulator arcing horn cement ball

conductor

Figure 7.3.8 (a) A technician services insulators on a power line. (b) Insulators used are typically long chains of ceramic components.

146

cup porcelain

The development of AC generators and transformers in the 20th century enabled the widespread availability of electricity. AC generators efficiently convert the energy from sources such as coal into electricity. Transformers enabled the efficient long-distance transmission of this electricity to the households of whole nations. These developments have arguably been the single greatest catalysts for change in modern society. Before the availability of household electricity, a significant part of the daily ritual was the lighting of fires for cooking and heating. Refrigeration was generally not available, so many perishable foods could only be eaten fresh or when in season. None of the electrical appliances that make our lives so convenient had been developed, so much more time was devoted to chores and manual tasks. As you can imagine, these differences would have resulted in a very different life for the average citizen in a society without electricity. The availability of electricity has also had many positive and negative effects on society. • The ability to transmit electricity efficiently over long distances has allowed more people in large cities to live further from the

motors and generators city centre. It is not uncommon in modern society for people to commute large distances to work. • People in most rural areas enjoy the same access to electricity as city dwellers. This has improved their quality of life and allowed many people to remain in country areas, who might have otherwise decided to move to the city. • The quality of life improved when electricity was affordable in every home. This changed the way we live through the many appliances available. Many of us find it hard to conceive a life without a refrigerator, television, DVD player or computer. • Tasks performed by electrical machinery decreased the amount of unskilled labour required and increased unemployment in certain parts of society.

Assess the effects of the development of AC generators on society and the environment. Discuss the impact of the development of transformers on society.

The widespread generation of electricity has also had a significant effect on the environment. • When industry was able to move away from the power plants in city centres this took pollution away from many people’s homes. This improved the quality of their environment and improved their health. • The availability of electric power for heating has reduced the need to burn wood or coal in houses, significantly improving air quality in large cities. • The use of hydro-electricity generation required the construction of large dams. These destroyed the habitat of many plants and animals. It also displaced many people from their homes, and this continues today as our demand for electricity grows. Lightning protection • The construction of long-distance power lines also required igure 7.3.7 shows a typical high-voltage the destruction of habitat and increased rates of erosion transmission tower. The two wires at the very where vegetation has been removed. top of the tower are called lightning protectors or • Radioactive waste from nuclear power plants requires shield conductors. These wires are placed long-term storage and poses a threat to the environment if above the tower and transmission lines to not contained. significantly reduce the chance of lightning • The burning of fossil fuels for the production of electricity has led to a significant increase in atmospheric carbon striking the transmission lines. At regular dioxide levels. Scientists believe this will have significant intervals these wires are connected directly to consequences in the near future through global climate the ground. These connections act to transfer change. huge currents from lightning strikes directly to • The burning of fossils fuels has also produced pollutants earth. This protects the transmission system such as sulfur dioxide. This contributes to the formation of from dangerous spikes in current and voltage, acid rain, which can damage forests, aquatic life and manwhich can damage substation transformers and made structures. cause blackouts.

PHYSICS FEATURE

F

Checkpoint 7.3 1 2 3 4 5

Explain how transformers allow the transmission of AC power over long distances. Identify the role of transformers in household appliances. Outline the energy losses in high-voltage transmission lines and the steps taken to minimise them. Describe how transmission lines are insulated from supporting structures and protected from lightning strikes. Assess the impact of transformers and AC generators on society and the environment. Justify your answer.

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PRACTICAL EXPERIENCES CHAPTER 7

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 7.1: AC and DC generators Gather secondary information to discuss advantages/ disadvantages of AC and DC generators and relate these to their use.

Research how AC and DC generators work then place the information that you have found into a table outlining the advantages and disadvantages of each. Use this information to create a list of applications that would benefit from using each. Discussion questions 1 Compare the structure of AC and DC motors. 2 List applications for which AC and DC generators are preferred and explain your choice.

Activity 7.2: Edison and Westinghouse Analyse secondary information on the competition between Westinghouse and Edison to supply electricity to cities.

Research the competition between Westinghouse and Edison to dominate supply of electricity. Use this information to present a 5 minute speech on the topic. Discussion questions 1 Outline the benefits of both DC and AC electricity supplies. 2 State reasons that Edison used to show that AC was more dangerous.

Activity 7.3: Power to the people Gather and analyse information to identify how transmission lines are: – insulated from supporting structures – protected from lightning strikes. Perform an investigation to model the structure of a transformer to demonstrate how secondary voltage is produced. Gather, analyse and use available evidence to discuss how difficulties of heating caused by eddy currents in transformers may be overcome. Gather and analyse secondary information to discuss the need for transformers in the transfer of electrical energy from a power station to its point of use. Gather, process and analyse information to identify some of the energy transfers and transformations involving the conversion of electrical energy into more useful forms in the home and industry.

148

In this investigation, you will gather first-hand data to see how a transformer works. You will then gather information from secondary sources in order to look into how heating due to eddy currents are overcome in a transformer. Equipment: transformer with removable coils, power supply, ammeter, voltmeter, connecting wires, light bulb. Discussion questions 1 Outline the difference between a step-up and step-down transformer. 2 Explain why the first-hand data does not fit the equation for a transformer. 3 Outline methods used to overcome heating in transformers due to eddy currents.

Chapter summary •

•

• • •

•

•

•



The difference between motors and generators is that motors use electricity to produce mechanical energy and generators produce electricity from mechanical energy. The main components of a generator are the armature, coils, rotor, stator, split-ring or slip-ring commutator and commutator brushes. A current is produced in the rotating coil of a generator due to the changing magnetic flux within the coil. An AC generator contains a slip-ring commutator; a DC generator contains a split-ring commutator. The operation of a transformer relies on the changing magnetic flux from the primary coil passing through the secondary coil. This changing flux then induces an emf and produces a current in the secondary coil. The law of conservation of energy states that energy cannot be created or destroyed, only converted from one form to another. Transformers alter the voltage and current of AC electricity. There is an inverse relationship between the voltage and current produced by a transformer. If the voltage is increased the current is decreased and vice versa. This is in agreement with the law of conservation of energy. For an ideal transformer, the relationship between the number of turns in the coils of a transformer and the voltages in either coil is: Vp np = Vs ns

•

•

•

•

•

•

•

motors and generators

A transformer that increases the voltage of AC electricity is called a step-up transformer; one that decreases the voltage is called a step-down transformer. Loss mechanisms in transformers include flux leakage, resistive heating by eddy currents and hysteresis loses. These are minimised by using a suitably shaped and magnetically ‘soft’ laminated iron core. The increase in AC voltage and subsequent decrease in current in transformers allows the efficient transmission of AC power over long distances. Household electrical appliances contain components that require various voltages and it is transformers within these appliances that provide these voltages. The main losses in the transmission of electricity are due to resistive losses and eddy currents. These losses are minimised through appropriate isolation of conductors and minimising resistance. Transmission lines are isolated and insulated from their supporting structures by insulating components called insulators. The higher the voltage, the larger the length of the insulator required to minimise the occurrence of electrical discharge. Transformers and AC generators have enabled the widespread distribution of AC electricity. This has had an enormous impact on society and the environment. These impacts include changes in where many people live, their quality of life, the nature of their employment, their health and that of our environment.

Review questions Physically speaking Copy and complete the table below by adding a definition for each key term.

Term

Definition

Step-up transformer Generator armature Generator Substation High-voltage insulator

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7

Generators and electricity supply: power for the people

Reviewing A

F

X B

Look at Figure 7.4.1. Identify each of the parts that have been labelled.

3 4

Explain how you could tell the difference between an AC and DC motor.

5

Outline methods of reducing energy lost along the path from generation to consumer.

6

List the applications that would be best suited to using an AC generator and those that would be more appropriate using DC generators.

7

State the features of AC and DC generators that make them appropriate to their use.

8

a List the effects that generators have had on society and the environment. b Assess these effects.

9

Give reasons why transmission lines need to be insulated from supporting structures and describe how it is done.

Y

I

I

Z

W

C

P

1 2

D

Figure 7.4.1 An AC generator

B

Create a table to compare and contrast the parts and function of an electric motor and a generator. a Identify where energy is lost during the distribution of AC from the time the electricity is generated to when it reaches the household consumer. b Describe the energy that is lost.

10 Explain how transmission lines are protected from lightning strikes. 11 a Identify parts A–E of the transformer in Figure 7.4.2. b Identify the function of each of these parts. c If np = 250 and ns = 750, identify if this is a step-up or a step-down transformer. A changing magnetic flux

B E

voltage Vp

voltage Vs

C

D

Figure 7.4.2 Parts of a transformer

12 13

Explain how current is produced in the secondary coil of a transformer. Discuss how the number of turns in the coils of a transformer determines the voltage in the secondary coil.

14 Show that Vp Vs

=

np is true for a transformer that is 100% efficient. ns

15 Outline the energy transfers for a transformer that is not 100% efficient. 16 Determine how currents in primary and secondary coils of a transformer are related.

17 Explain why transformers are needed in order to bring electricity effectively from a power station to the consumer.

18 Many appliances that are used in the home have a transformer attached to their power cord. Explain the presence and need for these.

150

motors and generators 19 Give examples of how transformers have impacted on the way in which we live.

20 Discuss the production of eddy currents in transformers and the problems that arise from them.

21 Outline methods of reducing the effects of eddy currents in transformers.

Solving Problems 22 Calculate the voltage in the secondary coil of a transformer that has 20 turns in the primary coil and 300 turns in the secondary coil if the input is 240 V.

23 Determine the efficiency of a transformer that has a turns ratio of 10:500, an input voltage of 50 V and output of 200 V.

24 Complete the following table. Coils in primary

Primary voltage (V)

100

6

320

240

50 000

Coils in secondary

Secondary voltage (V)

Step-up or step-down

200 50 30 500

240

25 Complete the following table by calculating the unknown quantities.

6 240

0.5

Current in secondary coil (A)

Voltage in secondary coil (V)

Turns ratio

240 2

12

0.1

1000

iew

Q uesti o

n

s

v

5

Voltage in primary coil (V)

Re

Current in primary coil (A)

151

2

The review contains questions in a similar style and proportion to the HSC Physics examination. Marks are allocated to each question up to a total of 30 marks. It should take you approximately 54 minutes to complete this review.

Multiple choice

4

Look at Figure 7.5.2.

(1 mark each) 1 Two current-carrying wires of different lengths are

III

placed side by side as shown in Figure 7.5.1.

0.5 cm

The length of the 2 m wire and the separation between the two wires are doubled. Identify the magnitude of the new force acting on each wire. A F B 2F C 4F D F/2

3

152

N

Figure 7.5.1

0.8 m

2

IV

S

2m

Identify the answer that has the correct part matched to the correct role.

Part

Role

a b

Armature

Provides magnetic field

Brushes

c

Commutator

d

Magnet

Allow current direction to change within the motor Maintains electrical contact without tangling wires Part of motor or generator that contains current-carrying coils or windings

Determine the output voltage from a household transformer for which the ratio of windings is 40:500. A 3000 V B 19 V C 0.05 V D 240 V

I



Figure 7.5.2  A generator

II

Identify the answer that lists its parts correctly.

I

II

III

IV

a

Axis of rotation

Force

Output terminals

Current

b

Force

Current

Axis of rotation

Output terminals

c

Axis of rotation

Output terminals

Force

Current

d

Output terminals

Axis of rotation

Current

Force

5

A series of substations can be found in suburban streets. The role of these substations is mainly to: A step up the voltage. B step down the voltage. C boost the power being transmitted. D add extra branches of wires to serve more houses.

motors and generators

Extended questions 6

Describe the main features of an AC induction motor and their roles.  (3 marks)

7

Outline the procedure you followed in order to investigate three factors that affect the generation of an electrical current.  (3 marks)

8

Discuss the how competition between Edison and Westinghouse led to the efficient supply of electricity to cities.  (2 marks)

9

Explain how eddy currents are used in an induction cooktop to cook a steak.  (2 marks)

10

A wire is placed in an external magnetic field as shown in Figure 7.5.3. a Determine the direction of the force on the wire. b Calculate the size of the force experience by the wire.   (3 marks)

12

Compare and contrast how the motor effect is used in a galvanometer and a loudspeaker.  (3 marks)

13

From the four chapters in this module, identify some of the energy transfers and transformations involving the conversion of electrical energy into more useful forms in the home and industry.   (4 marks) Gather, process and analyse information to identify some of the energy transfers and transformations involving the conversion of electrical energy into more useful forms in the home and industry.

0.1 T magnetic field within the shaded area

1 cm

B

0.5 A

Figure 7.5.3  A wire in a magnetic field

1 cm

11

A simple electric motor has dimensions as shown in Figure 7.5.4. The single coil carries a 0.1 A current and sits within a 0.05 T magnetic field at 40º to the vertical. a Explain why the magnets in a real motor would be curved. b Determine the force generated by the motor effect on each side of the motor. c Calculate the torque in the motor.   (5 marks) axis of rotation m

4c

4 cm 40°

I N

S

Figure 7.5.4  A simple electric motor

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3 Context Figure 8.0.1 Semiconductor integrated circuits have produced laptop computers that have much more computing power than even the most powerful early military computers that used thermionic devices.

154

from ideas to Implementation In just over 100 years we have gone from not knowing what constitutes matter to understanding its subatomic structure. From the discovery of the mysterious cathode rays came an understanding that these are fundamental particles that hold solid matter together. Technology, such as X-rays, sprang from this discovery and showed that their use in medical imaging enabled better patient diagnosis. An understanding of the way electrons move in space and solids gave rise to the plethora of electronic devices that have transformed our technology and way of life. The control of electron motion in a vacuum resulted in early thermionic devices (‘valves’), such as the triode, to control the flow of current in electronic circuitry. From these came television, radio, radar and the start of the modern electronic age. The study of electron motion in solids resulted in miniature semiconductor devices that replaced thermionic devices and enabled faster, more efficient and much more sophisticated electronics, such as integrated circuits, to be made. These devices are the backbone of all modern-day electronics, and include mobile phones and computers, as well as electronics and instrumentation used in homes, hospitals, industry, and in space. The accidental discovery of superconductors (solid material through which electrons travel unimpeded) may one day transform our technology yet again to produce much faster computers and more efficient transportation, energy production and transmission.

Figure 8.0.2 CDs and DVDs can be used as spectrometers.

Inquiry activity CDs: windows to viewing photons Neon signs and most street lights use discharge tubes to create the many different-coloured lights. In all of these, an electrical current is passed through a gas at reduced pressure. The light from the glowing gas is characteristic of the type of gas in the tube. A spectrometer is a device that separates the colours of the spectrum of the gas by using the same principle of interference as early research into X-ray diffraction. You can make your own spectrometer of visible light at home simply by looking at the light reflected from the back of a CD or DVD. Ordinary light (from the Sun and an incandescent bulb) will show you the continuous spectrum of white light that makes up the colours of the visible spectrum. Wait until it’s dark and find lighting that does not use incandescent lamps. Fluorescent tubes or bulbs are commonly available lighting that uses electrical discharges. These are filled with mercury vapour and an inert gas. 1 Compare the spectrum of colours from a fluorescent lamp with that of an incandescent lamp. Does it come in discrete colours or is it continuous like the light from the incandescent lamp? 2 Use your spectrometer under the street lighting. Can you tell if this lighting has the same gas as your fluorescent lamp or is it different? Or does it use a mixture of gases?

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8

From cathode rays to television Mysterious rays

fluorescence, Geissler tubes, cathode rays, cathode ray tubes, discharge tubes, Aston dark space, cathode glow, Crookes dark space, negative glow, Faraday dark space, positive column, anode glow, cyclotron motion, oscilloscope, CRO, timebase, sawtooth, blanking, shadow mask

Try This! Human-powered lighting You can make a fluorescent light tube temporarily give out flashes of light by rubbing it quickly with a piece of fur or wool in a darkened room.

Figure 8.1.1 The high voltage produced in static electricity will temporarily light up a fluorescent tube. 156

In today’s terminology, ‘cathode rays’ are collimated beams of electrons in an evacuated vessel. They were given this label at a time when not even the structure of the atom was known. The discovery of these ‘mysterious’ rays led to the discovery of the electron, and paved the way not only for understanding the atom but has also led to technology such as television, radar, electronics, the oscilloscope and lighting.

8.1 Cathode ray tubes Physicists began passing electrical current through air at lower pressure as soon as the vacuum pump was invented by Otto von Guericke in 1650. Among the most notable physicists to experiment with this was Michael Faraday, who, in 1838, reduced the pressure of air in a glass vessel and applied a high voltage to electrodes imbedded at each end of the tube. He noted that a narrow channel of light was produced between the negative (cathode) and the positive (anode) electrodes. Today we use this phenomenon to produce lighting in all shapes and sizes (such as neon signs) and are generally known as discharge tubes. In 1855 Heinrich Geissler (1815–1879), a master glassblower and skilled Geissler, in collaboration instrument maker, improved the vacuum pump. with the physicist Julius Plücker (1801–1868), reduced the pressure in the tubes to the point where the colours and patterns of the gaseous discharge disappeared and were replaced by a green glow (also known as fluorescence) that came from the glass. Moreover, Geissler was able to seal the tubes and maintain the very low pressure such that a vacuum pump was no longer needed. Julius Plücker called these tubes Geissler tubes.

from ideas to Implementation near vacuum

We now know that this green glow is caused by electrons leaving the cathode and striking the glass at high speeds, causing it to glow, but during Geissler and Plücker’s time this was a complete mystery. The name cathode rays was first used by Eugene Goldstein, a German physicist, who coined the name because it appeared that unknown beams were being emitted by the cathode. The tubes displaying this property came to be known as cathode ray tubes. The next great challenge for physicists at the time was to find the composition of these cathode rays from their behaviour.

cathode

The cathode ray debate: charged particles or electromagnetic waves?

Figure 8.1.2 A high voltage across

A variety of cathode ray tubes were made by a number of physicists in order to study the behaviour of cathode rays and hopefully determine their nature. In about 1875, the most prominent of these physicists Sir William Crookes placed a piece of metal in the shape of a Maltese cross in the path of the cathode rays as shown in Figure 8.1.3. This produced a sharp shadow on the glass behind the cross, indicating that cathode rays travel in a straight line. A magnetic field at right angles to the direction of the beam caused it to deflect in the same way that negatively charged particles would deflect (see Figure 8.1.4). Crookes concluded that cathode rays consisted of negatively charged particles. A moveable paddle wheel struck by the cathode rays started to rotate and roll along the tube (see Figure 8.1.8). This implied that cathode rays carried energy and momentum. Crookes also went on to show that the properties of cathode rays did not depend on the type of cathode material. From all of these experiments, the British physicists (such as Crookes) concluded that cathode rays were beams of negatively charged sub-microscopic particles. However, German physicists such as Heinrich Hertz (1857–1894) carried out experiments that seemed to contradict the British viewpoint. Hertz was unable to deflect the cathode rays with an electric field between two parallel plates placed inside the tube. This suggested that cathode rays did He also found that cathode rays could pass through very not have a charge. thin gold foil without damaging it. Thus Hertz and other German physicists concluded that cathode rays had no mass or electric charge, and were possibly a form of electromagnetic radiation. There were also objections to the paddle wheel experiment because it was thought that the cathode ray heated one side of a paddle, causing the gas in contact with it to heat up and expand thus causing the paddle, to move. The German physicists knew that electrical current can cause a magnetic field, but they were unable to detect such a field around the cathode rays. These contradictions strengthened their view of the wave nature of the cathode rays. From a modern perspective, we can easily resolve the debate between the British and German physicists. Cathode rays pass through a thin gold foil because the space between atoms is much larger than the size of an electron, and so a few atomic layers of gold allow the electrons to pass without much chance of a collision with a gold atom. However, the atomic structure of matter was unknown at the time. It was difficult to measure a magnetic field due to the cathode rays because the current was extremely small—a very sensitive magnetic field probe is needed to make such a measurement.

anode

–

+ glass tube

switch

A

high voltage (variable)

electrodes at each end of an evacuated glass tube causes the glass near the positive electrode (anode) to glow. Explain that cathode ray tubes allowed the manipulation of a stream of charged particles. Crookes tube (cathode ray tube) cathode

anode

mask holder

Figure 8.1.3 A Maltese cross placed in the path of cathode rays produces a sharp shadow on the glass indicating that cathode rays travel in a straight line.

cathode –

+

collimator S

anode

N

magnet

Figure 8.1.4 Cathode rays are deflected by a magnetic field in the same way that negatively charged particles are deflected. Explain why the apparent inconsistent behaviour of cathode rays caused debate as to whether they were charged particles or electromagnetic waves.

157

8

From cathode rays to television

PHYSICS FEATURE Anatomy of a discharge tube

E

anode dark

Aston Crookes space nergy-efficient lighting, such as fluorescent Faraday tubes and neon signs, are known as discharge – + tubes. These are evacuated glass vessels filled with negative gas at approximately 1% of atmospheric pressure cathode positive anode glow and two electrodes at opposite ends of the tube. glow column glow A large potential difference between these electrodes causes an electrical current to flow Figure 8.1.5 A discharge tube showing the typical bright and dark spaces through the gas in the tube, resulting in colours that depend on the type of gas and its pressure. Bright regions in the discharge are areas where the Figure 8.1.5 illustrates an operating discharge electrons have sufficient energy to ionise the gas. tube with the characteristic bright and dark spaces. Some of the ions recombine with electrons, which The Aston dark space next to the cathode is very thin results in the emission of light. Not all atoms will be and may go unnoticed. This is followed by the first ionised. Some will simply have their electrons gain luminous region called the cathode glow. Next is the energy while remaining bound to the atom or Crookes dark space and the negative glow (a luminous molecule; this is known as excitation. All excited region). The Faraday dark space follows this and is the electrons will fall back to their previous energy state largest of all the dark spaces. The largest luminous and, in doing so, give out light. A minimum energy is region, known as the positive column, follows and is required to excite or ionise an atom or molecule. the most prominent feature of the discharge. The Many discharge characteristics are related to the positive column may display periodic regions of bright average distance that electrons travel between and dark spaces known as striations. Finally, there colliding with the gas atoms or molecules. An increase is the anode dark space and then the anode glow in gas pressure leads to an increase in the density of adjacent to the anode. gas particles, and results in an electron travelling a An operating discharge tube consists of a mixture shorter distance between collisions, and vice versa. of ions, electrons and neutral gas atoms. The At low pressure, electrons have sufficient distance discharge starts because energetic particles (such as to accelerate and reach the required ionisation energy; electrons and protons) continually stream from outside however, they may strike the sides of the vessel more the Earth and strike the surface and atmosphere. often than they strike the gas. A minimum number of This so called ‘cosmic radiation’ can strike gas atoms ionisations are required to sustain the avalanche of and produce free ions and electrons in the discharge electrons that maintain the discharge. Below a tube. An electric field causes the free electrons to minimum pressure, it becomes difficult to start an gain sufficient energy to ionise gas atoms and electrical discharge along the tube. produce further free ions and electrons, which in turn will accelerate to produce more ionisation. This avalanche of ionisation is the way an electrical + Figure 8.1.6 Ions that strike the discharge is started. cathode liberate – To maintain the current through the discharge, secondary electrons, e– electrons (known as secondary electrons) are which help sustain – e the discharge. continually ejected from the cathode surface by the + bombardment of ions that are attracted to it. These secondary electrons cause further ionisation, which PRACTICAL results in further ion bombardment of the cathode EXPERIEN CES surface, and the process repeats.

Activity 8.1

158

Activity Manual, Page 68

from ideas to Implementation

Anatomy of a discharge tube (continued) At higher pressures there is a large density of gas particles, so there is a high probability that electrons will strike them. However, too high a pressure results in very frequent electron collisions such that they do not reach sufficient energy to ionise gas particles. Thus, the electric field strength must be increased for them to reach this energy. There is an optimum pressure for starting a discharge at a relatively low electric field. Electrons in dark regions have not reached sufficient energy to ionise or excite the gas. Figure 8.1.7 shows the potential difference and the magnitude of the electric field strength between the cathode and anode during a discharge. The largest potential difference occurs between the negative glow and the cathode. This is due to the ions being attracted to the cathode and thus crowding around it. Any electron outside this region will see an almost equal amount of positive charge (from the ions) and negative charge (from the cathode), resulting in a much lower electric field between the negative glow and the cathode. Let us now trace the path of electrons from cathode to anode. Electrons ejected from the cathode are accelerated by the high electric field until they have sufficient energy to cause the cathode glow. This depletes the electron energy gained, so they must accelerate once more through the cathode dark space until they have sufficient energy to cause the negative glow. Once again this process depletes their energy, and they must accelerate through the Faraday dark space until they gain sufficient energy to cause the ionisation and excitation in the positive column. Striations may result in the positive column for the same reasons that other bright regions occur. However, the electric field is more uniform in this region, so the bright and dark regions (the striations) are uniformly spaced. Again, electrons gain sufficient energy to ionise and excite the gas, which causes a luminous region and depletes the energy of the electrons. The electrons then accelerate through a region (a dark space) until they gain sufficient energy to excite and ionise the gas, and so on.

V

E –

Distance along the tube

+

Figure 8.1.7 The potential difference V and the magnitude of the electric field strength E along the axis of a discharge tube

Clear luminous and dark regions will be produced if electrons that leave the cathode remain in step. This is not possible in practice. Some electrons will strike gas particles before others and have their energy depleted sooner. This results in a spread of energies as electrons move along the tube. Thus, the luminous regions are diffuse and do not have well-defined boundaries. If electrons become too out-of-step with each other, then the striations will start to merge into each other and will not be distinguishable from the dark spaces between them. This may happen at too high a pressure. As the pressure is lowered, striations become more widely spaced because there is a lower gas particle density and the electrons will travel longer, on average, before striking a gas particle. The converse happens at higher pressures. Electrons strike gas particles more frequently, so the striations will be more closely spaced. Finally, electrons are collected by the anode, which also repels any ions, thus forming the anode dark space. This is a region of high electric field that causes the electrons to accelerate to the anode and to excite or ionise the gas around it, causing the anode glow. As a rule, the positive column is less bright than the negative glow and differently coloured. Helium results in a red cathode glow, a green negative glow, and a reddish-purple positive column. Neon produces yellow, orange and red in the respective regions. Nitrogen emits pink, blue and red. Each gas has a characteristic set of colours, depending on the degree of excitation that a bound electron experiences.

159

8

From cathode rays to television

cathode

anode

+

collimator

– paddle wheel

Figure 8.1.8 Cathode rays caused a paddle wheel to turn, implying they carried both momentum and energy. – +

+ + ++ +

–

––––– + cathode rays are not deflected, and behave in the same way as light

Finally, the apparent lack of deflection of the cathode rays by an electric field arose because the cathode ray tubes were not in a total vacuum— there was residual air in the tube, albeit at a low pressure. A fraction of the electrons in the beams (cathode rays) struck air molecules and ionised them. As shown in Figure 8.1.9, the positively charged ions are attracted to the negatively charged plate and the electrons are attracted to the positively charged plate. This essentially neutralises the charge on the plates and results in zero electric field between them. Therefore, there can be no deflection of the cathode ray. A resolution to the debate started with the French physicist Jean Perrin (1870–1942), who showed that a metal plate acquired a negative charge when it was struck by cathode rays. British physicist, Joseph John Thomson (1856–1940) was able to definitively show that cathode rays can be deflected by an electric field simply by producing an even lower pressure in the tube—thus reducing the number of ions. Moreover, he measured the charge to mass ratio of electrons. To quantitatively examine his work, we must first revise the motion of electric charges in electric and magnetic fields.

Figure 8.1.9 The cathode ray is not deflected by oppositely charged plates because ions and electrons neutralise the charge on the plates.

Checkpoint 8.1 1 2 3 4 5 6 7 8

+

a

Activity Manual, Page 72

Describe Faraday’s work with cathode ray tubes. Outline the contribution of Geissler and Plücker to the development of discharge tubes. Recall the different types of cathode ray tubes that were developed to understand cathode rays. Create a table that lists the observations and inferences made from each of the tubes. List the evidence that led Hertz to believe that cathode rays were not particles. Give a reason for not being able to detect the magnetic field from cathode rays. Explain why Hertz could not detect a deflection when he applied an electric field across the cathode rays. Describe how the debate about whether cathode rays were particles or waves was finally resolved.

–

b

Figure 8.2.1 The electric field lines for (a) a positive point charge radiate outwards, and (b) a negative point charge radiate inwards.

Discuss qualitatively the electric field strength due to a point charge, positive and negative charges and oppositely charged parallel plates.

160

PRACTICAL EXPERIENCES Activity 8.2

8.2 Charges in electric fields Any region in space in which there is an electric force on a charged object is said to contain an electric field. The field points in the direction of the force on a very small positive charge, known as a test charge. The magnitude of the field on the test charge—the electric field strength—increases in proportion to the force. A way of visualising the electric field is to draw lines that indicate the direction of the force on the test charge. Closely spaced lines mean higher electric field strength and vice versa. Equally spaced lines indicate it is a uniform field. Electric field lines are not real lines. They are used to give a qualitative description of the field. We can only draw a finite number of lines. The field is actually continuous—it exists everywhere in space. The electric field lines around a point positive or negative charge are shown in Figure 8.2.1. Although this is a two-dimensional drawing, the lines actually radiate outwards for a positive charge and inward for a negative charge in three

from ideas to Implementation dimensions. The greatest repulsive or attractive force is in the region of the lines with the closest spacing—near the charge in this case. There are some guidelines for drawing electric field lines for two or more charges: • The lines must begin on a positive charge and/or end on a negative charge. • Larger charges have more lines starting or ending on them. • Lines cannot cross. • Lines are always at right angles to a conducting surface. The electric field lines for a pair of opposite and equal point charges are shown in Figure 8.2.2. The number of field lines leaving the positive charge is equal to the number of lines ending on the negative charge. The lines are radial at very close distances to the charges. These lines are more closely spaced near the charges to indicate higher electric field strength in that region. The electric field lines between two equal charges q of the same sign are shown in Figure 8.2.3. The region at the centre, between the charges, is where the electric field strength is zero, because the electric fields of the two charges cancel each other. The field lines between two charged parallel metal plates of opposite sign but Equally spaced and parallel equal magnitude are shown in Figure 8.2.4. lines indicate that the field is equal in magnitude and direction and is said to be uniform. The field lines start to curve near the edges and become unevenly spaced, indicating a non-uniform field (called the edge-effect). b

a

+

+

–

+

–

Figure 8.2.2 The electric field lines for two point charges of equal magnitude but opposite sign

Identify that charged plates produce an electric field.

negative

–

positive

Figure 8.2.3 Electric field lines around (a) two equal positive charges, and (b) two equal negative charges

Electric field strength between parallel plates A side-view of a uniform field between two plates is shown in Figure 8.2.5 (we have ignored edge effects). The electric field strength E was defined as the ratio of the force F on a small positive charge q given by: F E = q (see in2 Physics @ Preliminary section 10.6). It has units of force per unit charge, which in SI units is newtons per coulomb or N/C or N C–1. However, in practice neither the force between the plates nor the charge on them is easily measured. We need a more convenient expression in terms potential difference V  between the plates. The equal and opposite charges on the plates were produced by applying a potential difference V  between them, using a power supply or a battery. Energy from the power supply moved the electrons from one plate to the other, resulting in equal and opposite charges on the plates. Recall that it was shown that the

Figure 8.2.4 The electric field lines between two oppositely charged parallel metal plates – – – – – – – – – – – negative +

d

E + + + + + + + + + + + positive

Figure 8.2.5 Parallel plates with a uniform electric field between them

Describe quantitatively the electric field due to oppositely charged parallel plates.

161

8

From cathode rays to television energy required to move an object in the direction of the force is known as work and is defined as: Work = force × distance = W = F × d (see in2 Physics @ Preliminary section 4.3). In this case, the displacement d is the separation of the parallel plates. Rearranging this formula as follows: W F = d and substituting this expression for force in the expression for electric field given previously, we obtain: W E= q×d Recall also that the work done on charges is related to the potential difference V  by the following expression: W=q×V (see in2 Physics @ Preliminary section 10.8). Substituting this into the previous expression for electric field strength, we obtain: q ×V E= q×d E=

Thus, the electric field strength E is easily calculated from the plate separation d and the potential difference between them V. From this equation, you can see that an alternative unit for the electric field strength is volts per metre (V m–1).

Solve problem and analyse information using: F = qvB sin θ F = qE and V E= d



Worked example Question

– – – – – – – – – – – d

e– E + + + + + + + + + + + L

Figure 8.2.6 Two parallel plates with a uniform electric field

V d

Two parallel plates are separated by d = 1.0 cm and have a side length of L = 2.0 cm (Figure 8.2.6). A potential difference of V = 10 V is applied between them. An electron enters halfway between the plates with a velocity of v0 = 2.7 × 106 m s–1 parallel to the plates. Calculate: a the electric field strength E  between the plates b the vertical force F  experienced by the electron as it travels between the plates c the acceleration of the electron d the time it takes for the electron to travel the full length of the plates e the vertical displacement of the electron just as it exits the plates f the change in kinetic energy of the electron between its entry to and exit from the plates. Note that the electron charge e = – 1.6 × 10–19 C, and its mass me = 9.11 × 10–31 kg.

Solution a The electric field strength E is given by: V 10 E= = = 1000 V m−1 (or N C−1 ) d 0.010 162

from ideas to Implementation b The vertical force is given by: F = e × E = 1.6 × 10–19 × 1000 = 1.6 × 10–16 N upwards c The acceleration can be obtained from Newton’s second law (see section 3.4 of in2 Physics @ Preliminary): a=

F 1.6 × 10−16 = = 1.76 × 1014 m s–2 = 1.8 × 1014 m s–2 downwards me 9.11 × 10−31

d The motion of the electron is similar to projectile motion (see Module 1 ‘Space’). There is only acceleration in the vertical direction. We are concerned with the horizontal component of the velocity, which remains constant. Therefore, the time t to travel the length L of the plates is given by: t=

L 0.020 = = 7.4 × 10−9 s v 0 2.7 × 106

e The electron only accelerates vertically, so we use the acceleration from part c and one of the SUVAT equations (see in2 Physics @ Preliminary section 1.3) to work out the vertical displacement s given by: 1 s = ut + at 2 2

where u is the initial vertical velocity = 0 m s–1, t is the time to travel the length of the plates (from part d), and a is the vertical acceleration = 1.8 × 1014 m s–2 (from part c). Therefore, the displacement is given by:



1 s = 0 + × 1.8 × 1014 × (7.4 × 10−9 )2 = 0.0049 m = 4.9 mm 2 f The change in kinetic energy is due to the vertical deflection of the electron by the electric field. This is work done by the field on the electron, and is given by: W = e × ∆V

where ΔV is the change in potential that the electron experiences due to its vertical deflection s. The relationship between potential difference and the displacement s is given by: ∆V = E × s



Combining this with the above expression for work gives: W = e × E × s = 1.6 × 10–19 × 1000 × 0.0049 = 7.8 × 10–19 J

Checkpoint 8.2 1 2 3

Explain the meaning of the arrows and spacing of lines when drawing electric field lines. Calculate the electric field strength at the location of a charge of 1.28 × 10–18 C that experiences a force of 1.1 × 10–18 N. Sketch the electric field lines around: a a point positive charge b two oppositely charged parallel metal plates.

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From cathode rays to television

B

8.3 Charges moving in a magnetic field

r F + +q

v

Figure 8.3.1 A positively charged particle enters into a magnetic field directed into the page and undergoes anticlockwise circular motion. Identify that moving charged particles in a magnetic field experience a force. Describe quantitatively the force acting on a charge moving through a magnetic field: F = qvB sin θ



A charged particle moving in a magnetic field will experience a force, as discussed in Module 2 ‘Motors and generators’. Consider a magnetic field directed into the pages (see Figure 8.3.1) and a positively charged particle that enters the field The particle will experience a force as given by the rightfrom the left. hand palm rule (refer to Module 2 ‘Motors and generators’). The force F on the particle changes direction every time the particle changes direction, such that the resulting motion is circular. This circular motion is known as cyclotron motion. A negatively charged particle will undergo circular motion in the opposite sense. The magnitude of the force F on a charge q that moves with a speed v at right angles to a magnetic field B is given by: F = qvB The speed of the charged object remains constant even though its direction changes. This means that the magnetic force simply changes the direction of the particle without adding or subtracting energy, provided that the magnetic field is constant. In general, the magnitude of the force on a charged particle with a velocity at an angle θ with respect to the magnetic field is given by: F = qvB sin θ Note that a particle that is parallel to the field (θ = 0°) does not experience a force. A charged particle that enters a magnetic field at an angle other than θ = 0° or 90° travels in a spiral along the magnetic field (also known as helical motion), as shown in Figure 8.3.2. However, the component of the velocity at right angles to the field is still circular. A particle of mass m with charge q enters a uniform magnetic field B at an angle θ and a speed v. Recall from the Module 1 ‘Space’ that the force directed towards the centre of circular motion is known as the centripetal force. In this case, the centripetal force is the magnetic force; that is: centripetal force = magnetic force

Solve problem and analyse information using: F = qvB sin θ F = qE and V E= d

a

mv 2 = qvB sin θ r where r is the radius of curvature of the particle. Rearranging this expression, we obtain the following expression for the radius: mv r= qB sin θ c

b +

F

θ

v sin θ

v + B

+

v v cosθ

B

B

Figure 8.3.2 (a) A proton enters a magnetic field at an angle θ. (b) The component of the velocity at right angles to the field undergoes circular motion. (c) The total motion is helical along the magnetic field.

164

from ideas to Implementation Worked example Question Protons are occasionally ejected from the Sun at high speeds towards the Earth. These can be caught in the Earth’s magnetic field and are trapped along magnetic field lines in a region known as the Van Allen belt. A proton travelling at a speed of v = 1.0 × 107 m s–1 strikes the Earth’s magnetic field at an angle of 30° at a distance of 3000 km above the surface where the magnetic field B = 3.5 × 10–5 T. Assume that the Earth’s magnetic field at this distance is uniform. The charge on a proton is 1.6 × 10–19 C and it has a mass of 1.67 × 10–27 kg. a Calculate the magnitude of the force on the proton. b Calculate the radius of curvature of the proton. c Determine if the cyclotron motion of the proton will cause it to come in contact with the Earth.

Solution a The magnitude of the force is given by:

F = qvB sin θ = 1.6 × 10−19 × 1.0 × 107 × 3.5 × 10−5 sin 30° = 2.8 × 10 –17 N

b The radius of curvature r is given by: r=

mv 1.67 × 10−27 × 1.0 × 107 = 6.0 × 103 m = qB sin θ 1.6 × 10−19 × 3.5 × 10−5 sin 30°

c No, the proton’s cyclotron motion will not cause it to come in contact with the Earth’s surface, since the cyclotron radius is 6 km and the proton is 3000 km above the surface.

Checkpoint 8.3 1 2

Outline what will happen to a negative charge moving to the left at right angles to a magnetic field that is directed out of the page. Calculate the magnitude of the force experienced by an electron that travels at right angles to a magnetic field of 2 T at a speed of 3 m s–1.

8.4 Thomson’s experiment In 1897 Joseph John (JJ) Thomson not only provided a definitive resolution to the debate about whether cathode rays were particles or electromagnetic waves, but he also measured the charge to mass ratio of the main constituent of cathode rays—the electron. Thomson’s cathode ray tube is shown schematically in Figure 8.4.1. Cathodes rays (electrons) were accelerated from the cathode to the anode, which consisted of two anodes aligned along the axis of the tube and separated by a small distance. Each anode had a horizontal slit cut into it so that cathode rays could pass through. The separation of the anodes produced a flat beam that passed between two parallel plates and struck the end of the tube, which had a fluorescent screen painted on the inside surface. The beam produced a well-defined narrow horizontal line on the screen.

Outline Thomson’s experiment to measure the charge/mass ratio of an electron.

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From cathode rays to television magnetic field electromagnet cathode rays

large voltage

Figure 8.4.1 Thomson’s cathode ray tube apparatus for measuring the charge to mass ratio of electrons

fluorescent screen

cathode anodes

electromagnet

charged plates

The parallel plates deflected the beam vertically upwards by placing a positive charge on the top plate. A magnetic field (using an electromagnet outside the tube) was applied at right angles to the electric field and the direction of the beam. The direction of the magnetic field was such that it deflected the beam downwards. Adjusting the magnetic and electric forces such that they were equal and opposite resulted in the beam passing through undeflected. We now quantitatively describe this experiment and how it leads to the charge to mass ratio of the electron. Equating the electric to the magnetic force on a particle with a charge q and speed v in a magnetic field B and an electric field E, we obtain the following expression: Magnetic force = electric force   qvB = qE

From this we obtain an expression for the speed: E v= B Recall that the relationship between the magnetic force and centripetal force (see section 8.3) is given by: mv 2 = qvB r This enables an expression for charge to mass ratio to be obtained: q v = m rB Substituting the expression for v into this equation: q E = 2 m rB

Thomson was able to calculate the radius of curvature r from the deflection of the beam on the fluorescent screen when the electric field was switched off and magnetic field switched on. He could calculate the magnitude of the electric field E because he knew the spacing d of the plates and the potential difference V  between them, and made use of the relationship E=V/d. Finally, knowing the number of turns in the wires of the electromagnet and the current flowing through it, he was able to calculate the magnitude of the magnetic field B. Thomson found that the charge to mass ratio always came to: q = 1.76 × 1011 C kg −1 m regardless of the cathode material, indicating that a fundamental particle was being emitted. This, in essence, marks the discovery of the electron. In 1891, the Irish physicist George Johnstone Stoney (1826–1911) suggested that the fundamental unit of electricity be called an electron—6 years before Thomson’s publication of his now famous experiment. 166

from ideas to Implementation

Checkpoint 8.4 1 2

Explain the purpose of Thomson’s experiment. Describe how Thomson’s experiment obtained the charge to mass ratio of the electron.

8.5 Applications of cathode rays A cathode ray oscilloscope (or simply oscilloscope or CRO) is a device used to measure the variation of voltage in time across an electrical component. Both the CRO and TV have elements in common with Thomson’s original cathode ray tube apparatus. Here we are not referring to the new plasma or LCD television sets, but the older style scanning electron beam sets.

Cathode ray oscilloscope (CRO) The display screen of a cathode ray oscilloscope, as seen by the user, is shown in Figure 8.5.1. Despite the many knobs and switches on its front control panel, the CRO simply displays a real-time graph of voltage (y-axis) versus time (x-axis). The oscilloscope has had many uses by scientists and Figure 8.5.1 The display screen and the control technicians in the design and operation of electronic and panel of a cathode ray oscilloscope electrical equipment. The way this real-time graph of voltage versus time was achieved can be understood by looking at the path of electrons horizontally electron gun main components inside the CRO. A schematic diagram of a deflecting plates bright spot cathode anode CRO (see Figure 8.5.2) consists of an electron gun, horizontal on screen and vertical deflection plates and a fluorescent screen. where heater electrons hit The electron gun (see Figure 8.5.3) produces and accelerates the electron beam. It consists of a heater, cathode fluorescent screen and anode. The heater is a wire filament with a large enough current flowing through it so that it reaches a high temperature. vertically deflecting plates Electrons in the wire then acquire enough energy to escape the wire and enter the region between cathode and anode. They accelerate toward Figure 8.5.2 A cathode ray oscilloscope deflects an electron beam with vertical and the anode and pass through a hole at its centre. The shape of the anode and its horizontal electric fields between separation from the cathode enables a narrow and focused beam of electrons to parallel plates. travel to the fluorescent screen. Some electron guns can become more sophisticated than the description given here, to achieve better focusing. The horizontal deflection plates cause the beam to sweep horizontally across the fluorescent screen (from left to right) by periodically changing electric field Outline the role of: between the plates—this is the time axis. The speed at which the beam sweeps • electrodes in the electron gun across the screen is controlled by the timebase dial on the front control panel of • the deflection plates or coils the CRO. This changes the frequency at which the beam sweeps across the • the fluorescent screen screen. The waveform of the potential difference across the horizontal plates is in the cathode ray tube of shown in Figure 8.5.4 and is commonly known as a sawtooth waveform. conventional TV displays and oscilloscopes. Electronics in the CRO blocks the beam on its way back to the left of the screen so that it doesn’t retrace itself (known as blanking).

167

8

From cathode rays to television anode (positive)

electrons 'boil' off the heated cathode

collimator

electron beam

heater cathode (negative) electrons attracted to the positive anode

Figure 8.5.3 The components of an electron gun used in both cathode ray oscilloscopes and CRT televisions

V

Time

sawtooth voltage for timebase

V

Time sinusoidal vertical voltage

Figure 8.5.4 A sawtooth voltage waveform on the horizontal deflection plates of a CRO sweeps the electron beam across the screen to display the sinusoidal waveform on the vertical deflection plates.

The vertical deflection plates cause the beam to move up or down in synchronisation with an input voltage. For example, a sinusoidal voltage will display a sinusoidal waveform (known as a trace) on the screen.

Television electron gun electron beam

magnetic coils

fluorescent screen

Figure 8.5.5 A television picture tube showing the electron gun, deflection coils and fluorescent screen

168

Cathode ray tube (CRT) television sets used the principles of the cathode ray tube for most of the 20th century. These are now being superseded by plasma and liquid crystal display television sets, which use different operating principles and allow a larger display area with a sharper image. However, the CRT television holds quite a significant historical place in this form of communication. A schematic diagram of a colour CRT television set is shown in Figure 8.5.5. Its basic elements are similar to those of the CRO. The main difference is the method of deflecting the electrons. Magnetic field coils placed outside the tube produce horizontal and vertical magnetic fields inside it. The magnitude and direction of the current determine the degree and direction of electron beam deflection. Recall your right-hand palm rule for the force on charged particles in a magnetic field. The vertical magnetic field will deflect the electrons horizontally; the horizontal field will deflect them vertically. The picture on the screen is formed by scanning the beam from left to right and top to bottom. The electronics in the television switches the beam on and off at the appropriate spots on the screen in order to reproduce the transmitted picture. However, to reproduce colour images, colour television sets need to control the intensity of red, blue and green phosphors on the screen. Three separate electron guns are used, each one aimed at one particular colour. The coloured dots on the screen are clustered in groups of red, blue and green dots that are very close to each other and generally cannot be distinguished by eye without the aid of a magnifying glass. For this reason a method of guiding the different electron beams to their respective coloured dots was devised. A metal sheet, known as a shadow mask (Figure 8.5.6) and consisting of an array of holes, is placed behind the phosphor screen. Each hole guides the three beams to their respective coloured phosphor as the beams move horizontally and vertically. Black and white television sets did not need the shadow mask since they had only one beam.

from ideas to Implementation glass

deflecting coils

mask

electron guns

blue beam red beam

R G B

vacuum

mask

phosphor dots on screen

fluorescent screen

green beam

focusing coils fluorescent screen

Try this! Do not adjust your horizontal! If you have access to an old black and white TV set or an old style monochrome computer monitor, try holding a bar magnet near the front of the screen and watch how the image distorts. This occurs because the magnetic field deflects the electrons that strike the screen. DO NOT do this with a colour TV set. This can magnetise the shadow mask and cause permanent distortion of the image and its colour. You can move a bar magnet near the back of a colour TV set to deflect the electrons from the electron gun and therefore distort or shift the image without causing permanent damage to the TV set.

holes in mask

electron beams

Figure 8.5.6 A colour CRT television set has three electron guns that will only strike their respective coloured phosphor dots with the aid of a shadow mask.

Can an oscilloscope be used as a television set?

T

he similarity between the cathode ray oscilloscope (CRO) and CRT television suggests that a CRO can be used as a television set. In fact, there have been some devices that have made use of the CRO as you would a computer monitor. So, in principle, it can be used as a television. One is then forced to ask ‘why did they need to deflect the beam in a television set with magnetic fields rather than with electric fields as in the CRO?’ In principle all television sets could be made in the same design as a CRO; however, it is much easier and cheaper to deflect the beam with a magnetic field on the outside of the tube rather than embed electrodes in the glass and inside the vacuum—this is a little trickier. So now another question arises: ‘why not deflect the beam of the CRO using magnetic fields, wouldn’t it result in cheaper CROs?’ Cathode ray oscilloscopes are precision instruments. The horizontal sweep rate must be able to be increased to very high frequencies in order to detect signals that change very quickly. Electric fields can be made to change very quickly without significant extra power requirements. However, a magnetically deflected system requires higher and higher voltages with increasing horizontal and vertical deflection frequencies in order to maintain the same current in the coils, and therefore, the same angle of beam deflection – thus having a significantly greater power requirement. Cathode ray tube television sets, however, only operate at fixed and relatively low scanning horizontal and vertical frequencies. Thus it is simpler and cheaper for the mass market to deflect with a magnetic field.

Checkpoint 8.5 1 2 3 4

Outline the purpose of a CRO. List the main parts of a CRO. Describe the role of each of these parts in the CRO. State the similarities and differences between the cathode ray tube CRO and CRT TV. 169

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From cathode rays to television

PRACTICAL EXPERIENCES CHAPTER 8

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 8.1: Changing pressure of discharge tubes Perform an investigation and gather first-hand information to observe the occurrence of different striation patterns for different pressures in discharge tubes.

Connect a set of discharge tubes that are at different pressures to an induction coil or high voltage power supply and observe the different striation patterns. The patterns are hard to see unless the room is very dark. Equipment: induction coil, connecting wires, discharge tubes at different pressures, DC power supply. Discussion questions 1 Draw a labelled set of diagrams showing the distinct patterns that occur during the evacuation of the tube. 2 Describe the change in the striation pattern with changing pressure.

Activity 8.2: Cathode ray tubes evolve Perform an investigation to demonstrate and identify properties of cathode rays using discharge tubes: • containing a Maltese cross • containing electric plates • with a fluorescent display screen • containing a glass wheel. Analyse the information gathered to determine the sign of the charge on cathode rays.

170

Connect the induction coil or high voltage power supply to each of the cathode ray tubes. Observe the behaviour of the electron beam under the influence of a magnetic or an electric field. Note the result of placing a Maltese cross or a paddle wheel in the path of the electron beam. Equipment: induction coil, connecting wires, cathode ray tubes in the following set-ups: Maltese cross, magnetic field (you can use a permanent magnet), electric field, paddle wheel, DC power supply. Discussion questions 1 List which experiment supported the idea that cathode rays were waves and which supported the particle theory. State the observation that led to the particle or wave conclusion. 2 Identify the properties of cathode rays that were determined from these experiments.

Chapter summary • • •

• •

• •

Cathode rays are electron beams in an evacuated vessel. Cathode ray tubes are the evacuated glass vessels in which cathode rays travel. A discharge tube consists of an anode and a cathode at the ends of a glass vessel filled with a low-pressure gas that glows when a high voltage is applied between the anode and the cathode. Geissler tubes are sealed discharge tubes that do not require a vacuum pump. Cathode rays were thought to be negative particles due to their direction of deflection when subjected to a magnetic field. Cathode rays moved a paddle wheel, implying they carried momentum. Initial inconclusive experiments led to the debate between British and German scientists about whether cathode rays consisted of particles or were a form of electromagnetic wave.

•

•

• •

•

from ideas to Implementation

Thompson’s experiment provided definitive proof of the particle nature of cathode rays from their motion in electric and magnetic fields. Moreover, he measured the charge to mass ratio of the electron. Cathode rays (or electron beams) are used in the cathode ray oscilloscope (CRO) and cathode ray tube (CRT) television. The CRO is used to display a changing voltage in time. The CRO deflects the electron beam with electric fields, while the CRT television deflects them with magnetic fields. Colour CRT TV sets have three electrons beams— one for each of the coloured phosphors (red, green, blue) on the screen.

Review questions Physically Speaking

Reviewing

The items in the columns are not in their correct order. Copy the table and match each of the definitions to the apparatus.

1 State the origin of the name cathode rays. 2 State one inconsistency of the behaviour of cathode

Apparatus

Definition

Oscilloscope

Shows that electrons travel in a straight line by the shadow cast on the tube

Discharge tube

Uses magnetic fields to deflect electron beams

CRT TV

An instrument used to measure a time-varying voltage

Maltese cross

A glass vessel containing gas at low pressure that can be ionised by high voltage

rays and how it was solved.

3 Explain the significance of Jean Perrin’s experiment. 4 Describe the different components that constitute a discharge tube.

5 Construct a table that names the features in a discharge tube, their characteristics and positions.

6 Briefly explain how a discharge is started and maintained in a discharge tube.

7 Describe how light is produced inside a discharge tube, in terms of the electrons in an atom.

8 Explain what would happen when a positive charge is placed in an electric field.

9 Estimate the difference in the force experienced by a charge placed in an electric field when the charge is tripled.

10 Describe how the force on a charge in a magnetic field varies as the angle of entry changes.

11 Compare and contrast the method of producing an image in a CRO and a CRT TV.

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From cathode rays to television

solving Problems

16 Calculate the acceleration of the electron while in an electric field strength of 1000 V. The mass of an electron is 9.11 × 10–31 kg and its charge is 1.60 × 10–19 C.

12 Calculate the electric field needed for an electron to

CRT a force of 3.2 × 10–16 N. phosphor experience SED electron phosphor electron 13 Calculate the work done in order to move an object gun emitter through 10 cm with a force of 3 N.

14 Consider Figure 8.6.1.

deflecting yoke

17 Using the information from Question 16, calculate the final velocity of an electron initially at rest, when it had travelled a distance of 1.00 cm.

spacer

18 Calculate the force on an electron that enters a magnetic field of 0.1 T at a speed of 3.2 × 106 m s–1.

100 V



+

10 mm



0V

Figure 8.6.1 A positive charge between two plates a Calculate the work needed to move a positive charge from the bottom plate to the top plate. b Calculate the electric field between these two plates. c Determine the direction of the force experienced by the charge. d Calculate the magnitude of the force experienced if the charge is a proton.

15 As a charge moves through an electric field it gains

Re

iew

172

Q uesti o

n

s

v

kinetic energy. Obtain an expression for the kinetic energy gained in terms of the electric field strength E and the distance s travelled by the charge q.

19 a Calculate the radius of curvature of a proton that

enters a magnetic field 0.1 T at 2.1 × 106 m s–1. b Determine the magnitude and direction of the force that is experienced by the proton.

20 Using the value that Thomson determined for charge to mass ratio 1.76 × 1011 C kg–1 and the electron charge of 1.6 × 10–19 C, determine the mass of an electron.

from ideas to Implementation

PHYSICS FOCUS Where to from here?

S

cience has come a long way since the first experiments with discharge tubes. The applications range from a CRO in scientific fields to TVs for the general public. But development has continued. TVs have changed; more and more households now have either an LCD or plasma screen TV. There are also surface-conduction electron-emitter display TVs (SED-TVs), which are still at the planning stage. These use the traditional cathode ray tube method but instead of having one electron gun for each colour (red, blue, green), these sets have a surfaceconducting electron emitter (SCE) for each coloured phosphor dot on the screen—there may be up to a million of them! Each SCE is made of two carbon layers with a gap between them; one with a negative electrode, the other with a positive electrode. A voltage of only 10 volts is needed to make an electron appear at one side of the gap. These cross the vacuum to strike the different phosphor dots lining the glass, resulting in a glow. The big advantage with this system is that the matrix that controls the SCEs allows each pixel to be activated simultaneously rather than in the traditional method of one row at a time.

luminescence black matrix

colour filter glass substrate

electrode Va

phosphor

metal back film electron beams electron emitter glass substrate Vf

Va

field emission nanogap

scattering several nm

Figure 8.6.2 A diagram of a surface-conduction electron-emitter display TV

H3. Assesses the impact of particular advances in physics on the development of technologies

H4. Assesses the impacts of applications of physics on society and the environment

H5. Identifies possible future directions of physics research

But development is not only on the entertainment front. Lighting has benefited from discharge tube technology. Commercial fluorescent lights have been around since 1938. They consist of a vacuum tube containing mercury vapour, filled with an inert gas such as argon and coated with phosphor powder. Neon lights are similar but are filled with inert gases such as neon, argon and krypton. These glow with different gas-dependent colours, when an electrical current is passed through them. The most recent and economical is the compact fluorescent light. Set to eventually replace all incandescent lights in Australia, these lights are basically a twisted fluorescent tube. 1 List the parts that are needed to make a discharge tube. 2 Outline the necessity for having low pressure inside the tube. 3 Compare and contrast the Geissler tube and the fluorescent light. 4 Determine why fluorescent lights are more efficient than incandescent lights. 5 List advantages and disadvantages of SED TVs. Suggest why a design that seems to encompass all the features wanted by consumers is not yet being made commercially. 6 Explain the need for magnetic coils in CRT TV sets. 7 Give reasons why magnetic coils are not needed in the SED TV.

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9

Electromagnetic radiation: particles or waves? The wave nature of light

electromagnetic wave, transmitter, receiver, standing waves, nodes, anti-nodes, refraction, polarisation, hertz, black body, classical theory, ultraviolet catastrophe, quanta, photons, spectrometer, Planck’s constant, photoelectric effect, photocathode, photoelectrons, cut-off frequency, work function, stopping potential, photocell, dynodes, photomultiplier tube, positron emission tomography (PET), gamma photons

There was considerable debate about whether light was a wave or a stream of particles, long before there was a similar debate about cathode rays. However, by the 19th century there was growing experimental evidence that light had wave properties, but the nature of these waves was not known.

9.1 Hertz’s experiments on radio waves

PRACTICAL EXPERIENCES Activity 9.1

Activity Manual, Page 75

174

A suggestion about the nature of light waves came from the Scottish physicist James Clerk Maxwell (1831–1879). From his studies of Michael Faraday’s experiments with electricity and magnetism, he derived four fundamental Remarkably, these equations that linked electricity and magnetism. equations predicted that oscillating electric charges should produce a wave that travels through space at the speed of light. This wave consisted of oscillating electric and magnetic fields at right angles to each other, and was called an electromagnetic wave. Although Maxwell’s equations did not conclusively prove that light was an electromagnetic wave, he strongly suspected that it was, because the predicted speed of these electromagnetic waves was the same as the speed of light. Moreover, his equations predicted that all electromagnetic waves travelled at the speed of light, regardless of their frequency. Sadly, he died at age 48 before his theory could be tested experimentally. In 1887 Heinrich Hertz (1857–1894) was the first to experimentally verify Maxwell’s electromagnetic wave theory. Hertz’s apparatus was essentially a radio wave transmitter and receiver. Today, almost all of us carry out a more

from ideas to Implementation sophisticated version of Hertz’s experiment when we tune in to a radio station. More importantly, we are able to select different radio stations by their Hertz needed to make an antenna that transmitted transmitting frequency. at a specific frequency and a receiver that was tuned to that frequency. This was not such an easy task when electronics were not available. Hertz used a number of different transmitters and receivers during the course of his experiments. Each successive apparatus was refined to give more accurate results. The basic principle of Hertz’s apparatus is shown in Figure 9.1.1. The transmitter consisted of a pair of metal rods placed end to end with a small gap between them. He used an induction coil to place charges of opposite signs on these rods at very large potential difference, causing a spark to jump across the gap. This caused electrical current to oscillate back and forth across the gap and along the rods, thus producing an electromagnetic wave. The frequency of the oscillation was determined by the dimensions of the rods. His first apparatus produced a frequency of 50 million cycles per second, but he had no way of The mathematics for calculating the frequency was known measuring that. during Hertz’s time and so he was able to calculate this oscillation frequency. The electromagnetic wave emitted by the transmitter was detected by one of several receivers. The receiver shown in Figure 9.1.1 is a loop of wire with a gap. The natural oscillating frequency of this loop had to match the frequency of the transmitter. The dimensions of the loop and gap determined this frequency and were accurately calculated by Hertz. The receiver showed that a spark jumped across the gap even if it was placed many metres away from the transmitter, thus indicating that an electromagnetic wave has been transmitted in air.

Hertz measures the speed of radio waves Hertz was able to obtain a fairly accurate measurement of the speed of his radio waves by a number of methods. His most famous paper for measuring their speed in air relied on the phenomenon of standing waves. Recall that two waves of the same frequency, wavelength and amplitude can form standing waves if they overlap while travelling in opposite directions (see in2 Physics @ Preliminary section 7.4). Figure 9.1.2 shows a standing wave produced by a rope tied to a wall. The nodes are points of zero amplitude and no motion (thus the term standing wave). The points of maximum amplitude are known as anti-nodes.

a

plate rod

spark gap

to induction coil

b

metal loop

Figure 9.1.1 General schematic representation of all of Hertz’s apparatuses, which consisted of (a) a transmitting dipole antenna and (b) a receiver of various designs Outline qualitatively Hertz’s experiments in measuring the speed of radio waves and how they relate to light waves.

TRY THIS! Any experiment in a thunderstorm

A

A N

A N

N

Figure 9.1.2 A standing wave on a rope (A = anti-node, N = node)

Hertz formed a standing wave of electromagnetic radiation by reflecting the radio waves from a large flat zinc plate, as shown in Figure 9.1.6, in which only a standing wave of the electric field is shown. Hertz then moved his receiver coil along this wave. A spark in the gap was produced at the anti-nodes but not at the nodes. The distance travelled between nodes or anti-nodes is half a wavelength. Doubling this

You can reproduce Hertz’s experiment at home. The next time there is a thunderstorm, turn on your radio to the AM band (FM does not work for this experiment). Turn the tuning dial to a frequency at which no radio station is transmitting and listen. You will hear crackles and hisses of thunder near and far. Incredibly, you can hear more crackling than the number of lightning flashes that you see, because not all lightning flashes are large enough to be visible. This is the essence of Hertz’s experiment— electromagnetic waves produced by the motion of charge in lightning travel to your radio receiver at the speed of light and are detected. 175

9

Electromagnetic radiation: particles or waves?

PHYSICS FEATURE The evolution of Hertz’s experiment Hertz’s first apparatus produced an electromagnetic wave with a wavelength of 6 m. There are difficulties with precision experiments when using such a long wavelength. You need to be far enough away from the transmitter so that you are not simply detecting a spark in your Figure 9.1.3 Heinrich Hertz was a receiving antenna due to skilled experimental physicist. purely electrostatic effects as a result of being too close to the high-voltage induction coil. Moreover, the spark in the detecting antenna becomes very weak if the distance between it and the transmitter is too great. You have to contend with reflections from the walls of the room and the objects within it. You must also face the problem of tuning the transmitter and receiver to the same frequency when there is no instrument that can help you. Heinrich Hertz overcame all of these issues. Figure 9.1.4 shows the evolution of the different types of transmitters and receivers used by Hertz. The first transmitter consisted of zinc spheres at the ends of the two rods (see Figure 9.1.4a). The antenna could be tuned by sliding the spheres along the rods. The receiver was a square loop of wire with a gap. The second type of antenna, was tuned by changing the area of the flat plates at the end of the rods (see Figure 9.1.4b). The receiver was a circular loop with a gap. The first two antennas produced a wavelength of 6 m. The next generation of transmitter and detector (Figure 9.1.4c) was a smaller antenna and a similar receiver (no loop) for a wavelength of 0.66 m.

176

transmitters a

receivers

λ 6m

b

6m

c

0.66 m

Figure 9.1.4 The different types of antennas and receivers used by Hertz include (a) a dipole transmitting antenna with spheres and a square loop detector for 6 m wavelength, (b) a transmitting antenna with square plates and a circular loop receiver for 6 m wavelength and (c) a small dipole antenna and similar dipole receiver for 0.66 m wavelength. parabolic reflector

to induction coil

transmitting antenna

receiving antenna

Figure 9.1.5 Hertz’s transmitting and receiving antennas used parabolic reflectors to direct the electromagnetic wave without much loss of energy with distance.

Because these antennas were smaller, they were placed in parabolic reflectors (Figure 9.1.5) to direct and detect the beam without much loss. This configuration produced the most accurate results.

from ideas to Implementation distance gave the wavelength λ. Hertz was able to calculate the frequency f of the transmitted radio waves. The speed of these waves c was then determined using the following well-known wave speed formula: c = f λ Hertz used a different frequency and found that the speed of the waves remained the same. Although this didn’t prove that the speed of these radio waves was the speed of light, it was strong supporting evidence of Maxwell’s theory 1m 2m 3m 4m 5m 6m 7m 8m Distance from the wall that the speed of all electromagnetic radiation was the same. Moreover, the speed of these waves was exactly the measured Figure 9.1.6 Standing electromagnetic waves produced by reflection speed of light. from a large zinc plate enabled Hertz to measure the Hertz also showed that the path of these waves speed of light. could be bent in the same way as light (refraction) by passing them through a large asphalt prism. (See in2 Physics @ Preliminary section 8.3.) Hertz showed that the electric and magnetic fields Heinrich Hertz: ‘Oops’ of the radio waves had a unique direction in space, known as polarisation. The electric field in Figure 9.1.6 points vertically. ertz initially published his estimate of the speed of electromagnetic waves as When the detecting loop was at an anti-node and the direction 2 × 108 m s–1. This relied on the accuracy of his of the gap aligned with the electric field (vertically), then a calculation of the oscillation frequency of the spark jumped across the gap. When the loop was rotated so transmitter. It was an embarrassing moment when that the gap was at right angles to the electric field, there was Poincaré, a great French mathematician, wrote him no spark. a letter pointing out that he had not included a It took great skill for Hertz to show that polarised factor of the square root of 2 in his frequency electromagnetic waves exist and have a finite speed equivalent calculation. As a result, his actual measured speed to the speed of light. Moreover, these waves shared other was 2.8 × 108 m s–1, which is only 7% off the properties with light such as reflection and refraction. This set actual value of 3 × 108 m s–1. So don’t feel so bad the scene for the emergence of radio communications and next time you make a numerical mistake—even hence the modern field of telecommunications—mobile phone great scientists do it. technology is a sophisticated version of Hertz’s experiment. Hertz did all of this before his life was cut short at the age of 36. One of the greatest honours that can be bestowed upon a scientist is to name a unit of measurement after them. The international unit of frequency is no longer called cycles per second—it is known as the hertz.

H

Checkpoint 9.1 1 2 3 4 5 6 7 8 9

Describe Maxwell’s contribution to understanding the connection between electricity and magnetism. Describe an electromagnetic wave. Recall the two predictions that Maxwell made about electromagnetic waves. Describe how Hertz produced and controlled the frequency of electromagnetic waves. Outline the requirements needed of the receiver in order for it to detect any given frequency. Recall the definition of a standing wave. Sketch a diagram of a standing wave in a rope, labelling a node and anti-node. Outline how Hertz manipulated the equipment in order to determine the wavelength of the electromagnetic wave produced. Define polarisation and explain how Hertz showed that electromagnetic waves were polarised. 177

9

Electromagnetic radiation: particles or waves?

9.2 Black body radiation and Planck’s hypothesis

cause these metals to radiate different colours from dull red to yellow.

PRACTICAL EXPERIENCES Activity 9.2

Activity Manual, Page 78

TRY THIS! Sun power Use a magnifying glass to focus the Sun’s rays to a point on a wad of tissue paper—it should catch fire. Now colour another tissue paper with black ink, say from a felt-tip pen or by dipping it in black ink. Let it dry. Focus the Sun’s rays onto this blackened tissue. You should find that the paper catches fire much more quickly. This is because black absorbs more of the incoming radiation and therefore heats up more quickly.

ultraviolet visible

infra-red

10

8 Intensity I (arb.units)

Figure 9.2.1 Increasing temperatures

The success of Maxwell’s theory of electromagnetic waves led other physicists to apply it to the long-standing problem of radiation from hot objects. As an object such as a piece of metal is heated, its temperature rises and it emits colours from a dull red to orange and then to yellow, eventually becoming a bluish-white (if the object has not melted). This is shown in Figure 9.2.1 for metal parts at increasing temperatures. The brightness or intensity (power radiated per unit area) of radiation depends on both wavelength and temperature, as shown by the plot in Figure 9.2.2. The maximum intensity of the curve shifts to smaller wavelengths as the temperature is increased. This simply means that as an object becomes hotter, its colour changes The challenge for physicists was to mathematically from red towards blue. predict the exact form of this curve using electromagnetic wave theory. An object that absorbs all the wavelengths of the spectrum is referred to as a black body absorber. In a strange twist of terminology, an object that emits all the wavelengths of the spectrum is called a black body emitter. The curves shown in Figure 9.2.2 are known as black body radiation curves. The Sun is close to being a black body emitter and so its temperature can be determined from this plot of intensity versus wavelength. An object that is painted black is not usually an ideal black body, as some radiation is always reflected from its surface. An ideal black body is generally pictured as a cavity with a small hole (see Figure 9.2.3). Any radiation that enters the hole becomes trapped inside the cavity and undergoes many reflections before eventually being absorbed. Similarly, a heated cavity with a small hole acts as a black body emitter. A small fraction of the radiation from within the cavity exits through the hole, so the intensity of this radiation can be measured (see Figure 9.2.2). A number of physicists attempted to calculate the mathematical relationship for the black body radiation curve on the basis of electromagnetic wave theory and thermodynamics (the physics of the movement of heat). We will call this

6 λmax

T = 6000 K

4 5000 K

λmax

4000 K

2 3000 K

0

0

1.0

Wavelength λ (μm)

2.0

Figure 9.2.2 The intensity of light from a hot object is dependent on wavelength. 178

3.0

from ideas to Implementation

These assumptions were radical ones since classical physics said that objects can have any energy. Max Planck’s theory defied this assumption. Incredibly, Planck himself resisted this theory wholeheartedly for many years and thought of it as nothing more than a mathematical convenience. However, this idea of the quantised nature of energy heralded the start of quantum physics, which, together with relativity, became the foundations of the modern physics we use today.

Einstein’s contribution to quantum theory: the photon Planck originally restricted his concept of energy quantisation to emitters in the walls of a black body cavity. He still believed that the electromagnetic energy radiated as a wave. However, in 1905 Einstein called into question this view of He proposed that electromagnetic radiation and of light in particular. radiant energy is made of concentrated bundles of energy that later came to be called photons. That is, Einstein had proposed that light is made of particles and is not a wave, as shown in Figure 9.2.5. Einstein assumed that such an energy bundle is initially localised in a small volume of space, and that it remains localised as it moves away from the source with velocity c (the speed of light). Ironically, we still express the speed of photons in terms of the wave properties of frequency f and wavelength λ, as follows:

Figure 9.2.3 A spherical cavity with a hole acts as a black body absorber of radiation.

Identify Planck’s hypothesis that radiation emitted and absorbed by the walls of a black body cavity is quantised.

classical theory Radiance

the classical theory approach. They assumed that the walls of the cavity were made from tiny oscillators that emit electromagnetic waves—just as Hertz had Although this theory was able to assumed for his transmitting antenna. reproduce the shape of the graph for large wavelengths (see Figure 9.2.4), it failed badly at the shorter wavelengths. Here the calculated intensity increased towards infinity, which violated the law of conservation of energy. This was called the ultraviolet catastrophe, because this started to occur at the ultraviolet end of the radiation spectrum. Max Planck (1858–1947) solved this problem in 1900 and was able to mathematically reproduce the black body radiation curve by making the following radical assumptions: • The emitters in the walls of the cavity can only have energies E given by E = nhf where f  is the emitted frequency in hertz, h is a constant (now called the Planck constant = 6.63 × 10–34 J s) and n is an integer. The energy E is measured in joules. • The emitters can absorb or radiate energy in ‘jumps’, or quanta. Two consecutive energy states of an emitter differ by hf.

experiment and Planck theory

0

1

2 3 Wavelength (μm)

4

Figure 9.2.4 The classical theory of black body radiation could only explain the long wavelengths, Planck’s theory could explain all wavelengths.

c = f  λ a

b

y E

y

B z

photon with energy hf

c x z

x c

Figure 9.2.5 Light can be shown as (a) an electromagnetic wave and(b) a collection of particles called photons.



Identify Einstein’s contribution to quantum theory and its relation to black body radiation. Explain the particle model of light in terms of photons with particular energy and frequency. Identify the relationships between photon energy, frequency, speed of light and wavelength:   E = hf and c = f  λ

179

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Electromagnetic radiation: particles or waves?

PHYSICS FEATURE Max avoids a catastrophe

D

etailed understanding of Max Planck’s mathematical manipulations is beyond the scope of this book. Even Planck, at first, looked on his solution as nothing more than a mathematical act of desperation. But we can gain some qualitative insights into his theory by adopting a view that is slightly different from that of his original work. Imagine that the electromagnetic wave emitters in the walls of the cavity produce standing waves, as shown in the simplified one-dimensional representation in Figure 9.2.7 in which the walls on either side of the waves represent the walls of the cavity. The intensity at a particular wavelength of the radiation from the cavity is related to the energy in the wave at that particular wavelength. Both the classical theory and Planck’s theory assumed that the energy in the wave is equal to the energy from the emitter. However, according to classical theory, the average energy for each emitter in the walls is kT, where k is a known constant (the Boltzmann constant) and T is the temperature of the walls in kelvin. Therefore, the energy per standing wave is also kT. An understanding of the way intensity is measured is important to understanding the difference between the ultraviolet catastrophe and Planck’s radiation law. An instrument known as a spectrometer is used to measure the intensity of the black body radiation. All spectrometers take in a small wavelength range simultaneously—they never measure just a single wavelength. For example, a typically good spectrometer cannot distinguish between wavelengths of 6000.0 × 10–10 m and 6000.1 × 10–10 m. Note from Figure 9.2.7 that the difference in wavelength between two consecutive standing waves becomes progressively smaller with increasing frequency. This implies that at higher frequencies more wavelengths can enter the spectrometer. Since each wavelength contributes an equal amount of energy, and more of them are entering the spectrometer with increasing frequency, this will register as an increasing intensity since the total energy entering the spectrometer increases. As the wavelength becomes shorter, and 180

Figure 9.2.6 Max Planck’s radical hypothesis avoided the ultraviolet catastrophe and started a new field of physics called quantum physics.

Figure 9.2.7 A simplified representation of the standing waves inside a metal cavity

therefore the frequency increases, the intensity will increase towards infinity. This is the origin of the ultraviolet catastrophe, and which, of course, does not happen. Planck restricted the energy of the emitters in the cavity to integer multiples of hf, which is the same as only allowing integer multiples of a certain frequency f. This restricted the number of standing waves that can occur. He also used a well-established law of thermal physics that states that a particle is more likely to have a lower than a higher energy. That means there is a dramatic decrease in the number of emitters with higher energies. As the frequency becomes very high, and therefore the wavelength very short, these two restrictions ensure that the number of emitters with these higher frequencies becomes very small and approaches zero. This registers as a diminishing intensity in the spectrometer and, thus, reproduces the actual black body radiation curve.

from ideas to Implementation Einstein assumed that the energy E of the photon is related to its frequency f  by the equation: E = hf where h is Planck’s constant = 6.63 × 10–34 J s. In classical theory, the intensity was determined by the electric field of the electromagnetic wave. In the new quantum theory, it is the number of photons that determines the intensity. He then used this idea to explain peculiar properties of metals when they are irradiated with visible and ultraviolet light, known as the photoelectric effect.

Solve problems and analyse information using:  E = hf   and c = f  λ

Worked example question A laser pointer emits light with a wavelength of 6.50 × 10–7 m. The power of the laser is 1.00 × 10–3 W. Assume that Planck’s constant h = 6.63 × 10–34 J s, and the speed of light c = 3.00 × 108 m s–1. a Calculate the energy in each photon of the laser beam. b Calculate the number of photons emitted each second.

Solution a The energy E in a photon is given by E = hf

where h is Planck’s constant and f is the frequency of the light. We can calculate the frequency from the following: c = f λ



Therefore: E =h



c 3.00 × 108 = 6.63 × 10−34 × = 3.06 × 10 –17 J λ 6.50 × 10−9

The energy per photon is 3.06 × 10–17 J.

b The laser has a power of 1.00 × 10–3 W. This means it emits 1.00 × 10–3 joules per second. The number of photons per second is given by: number of photons per second =

power 1.00 × 10−3 = energy per photon 3.06 × 10−17

= 3.27× 1013 photons per second

Checkpoint 9.2 1 2 3 4 5 6

Describe what happens to the wavelength of light emitted as the temperature of an object is increased. Define a black body. Outline how classical theory described black body radiation. Explain the ultraviolet catastrophe. Outline the assumptions Planck used to solve the problem. Explain Einstein’s revolutionary thoughts on light.

181

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Electromagnetic radiation: particles or waves?

9.3 The photoelectric effect Describe Hertz’s observation of the effect of a radio wave on a receiver and the photoelectric effect he produced but failed to investigate.

light electrons

–

+

evacuated quartz tube V

R

A

Figure 9.3.1 Electrons ejected from the cathode register a current on the ammeter, when the cathode is struck by ultraviolet light.

0

c

gst en pla tin um

1

tun

2

zin

pot ass sod ium ium

Maximum kinetic energy (eV)

3

f0 f

–W 0.5

1.0

1.5

Frequency ( 1015 Hz)

Figure 9.3.2 The energy of photoelectrons increases with increasing frequency. No electrons are ejected below the cut-off frequency f0. 182

In his experiments on radio waves, Hertz noticed that the sparks were easier to produce in the gap of the detector loop whenever it was directly exposed to the ultraviolet light from the spark of the transmitting antenna. Placing an ordinary glass plate between the transmitter and the detector reduced the brightness of the spark, because ordinary glass blocks ultraviolet light. Replacing the glass plate with a quartz glass plate caused the spark to be more easily produced, because quartz glass allows ultraviolet light to be transmitted. He concluded that ultraviolet light has an effect on the receiving loop, but he was not in a position to explain it. However, he realised that it was an extremely important phenomenon and stopped his wave research in order to study it. Hertz had discovered the photoelectric effect. The photoelectric effect is the ejection of electrons from the surface of a polished metal when light is shone on it. The metal on either side of the gap of Hertz’s detector loop emitted electrons when struck by ultraviolet light, which assisted in producing the spark. Ironically, the results of this experiment were used by Einstein to prove the particle nature of light, in contrast to Hertz who wanted to prove its wave nature. The photoelectric effect was studied by other physicists who used a cathode ray tube, as shown in Figure 9.3.1. The tube either had a quartz window or was made of quartz, which allowed ultraviolet light to pass through and strike the cathode, known as a photocathode. This caused electrons, known as photoelectrons, to leave the cathode and be collected by the anode, thus registering a current on the ammeter. The photoelectric effect was considered to be strange because: • No electrons are ejected from the metal below a certain frequency no matter how intense the light. This is called the cut-off frequency f0. This is in contrast to classical theory, which says that all electromagnetic waves have energy and, if you wait long enough, all electrons can gain energy and leave the surface. • The kinetic energy of the electrons increases as the frequency of the incident light increases (that is, going from red to blue to ultraviolet and beyond) as shown in Figure 9.3.2. However, there is no change in the electron energy if the frequency of the incident light is constant but the intensity is increased. Increasing the intensity simply increases the number of electrons but their energy remains the same. From classical wave theory, an increase in intensity should result in an increase in the energy of the electrons. • There is no delay between the time the light is shone on the surface and the time the electrons are emitted, no matter how dim the light source. From classical theory, electrons would require a length of time to gain enough energy from a low-intensity light source in order to escape the surface of the metal. Einstein explained the photoelectric effect by assuming that an electron is ejected when it absorbs a photon that has energy E = hf. Although electrons involved in electrical conduction in a metal are free to move around, they are still bound to the metal as a whole. Energy is needed to separate them from the metal. This energy is called the work function W. The value of W depends on the type of metal. The photoelectric effect is mainly a surface phenomenon, so the surface must be free of oxide films, grease or other surface contaminants. Einstein then simply conserved energy by stating that:

from ideas to Implementation

where Kmax is the maximum kinetic energy of the ejected electrons. Rearranging this equation, we get: Kmax = hf – W

I

nte

ractiv

e

Photon energy = energy of ejected electron + energy needed by the electron to leave the metal. This is encapsulated in the equation: hf = Kmax + W

M o d u le

PRACTICAL EXPERIENCES Activity 9.3

Current

Activity Manual, Page 81 From this we can see that the kinetic energy is zero (that is, no ejected electron) when hf = W. This accounts for the cut-off frequency. In addition, the kinetic energy increases as the frequency increases. Increasing the light intensity PRACTICAL merely increases the number of photoelectrons and therefore increases the current EXPERIENCES emerging from the surface, but does not change the kinetic energy of the electrons. Activity 9.4 Activity Manual, Page 88 Einstein’s theory also accounts for the absence of a time delay, because all of the energy is supplied in a concentrated bundle; it is not spread uniformly over a large area, as in wave theory. In 1921 Einstein won the Nobel Prize in Physics for his explanation of the photoelectric effect. The graph in Figure 9.3.3 shows the dependence of the photoelectron current on the applied potential difference between photocathode and anode. There are two interesting features of this graph. The first is that there is a current even when there is no potential difference. At this point, all the energy to produce the current comes from the incoming photons. The second, and more interesting, feature is –Vstop 0 Potential difference that the current is stopped by a certain negative potential difference placed on the photocathode. Not surprisingly, this potential is called the stopping potential Figure 9.3.3 Current of photoelectrons versus Vstop and is used to determine the maximum kinetic energy of the electrons. The applied potential difference ammeter will read zero current when eVstop equals maximum electron kinetic energy. between cathode and anode

Worked example question A zinc photocathode is irradiated with ultraviolet light. A voltage of –2.0 V is required to stop the photoelectrons. a Calculate the maximum kinetic energy of the photoelectrons. Use the electron charge e = –1.6 × 10–19 C. b Calculate the wavelength of the ultraviolet light striking the photocathode, given the work function of zinc W = 6.88 × 10–19 J and Planck’s constant h = 6.63 × 10–34 J s.

Solution a The maximum photocathode kinetic energy Kmax is given by: Kmax = eVstop = 1.6 × 10–19 × 2.0 = 3.2 × 10–19 J b Use Einstein’s photoelectric effect formula:  Kmax = hf – W

where Kmax is obtained from part a, h is Planck’s constant and the zinc work function W = 6.88 × 10–19 J. In order to calculate the wavelength we first require the frequency, which is obtained by manipulating Einstein’s formula: f=



K max +W 3.2 × 10−19 + 6.88 × 10−19 = = 1.52 × 1015 Hz −34 h 6.63 × 10

We can now use the relationship between frequency and wavelength: c 3.00 × 108 λ= = = 1.97 × 10−7 m f 1.52 × 1015 183

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Electromagnetic radiation: particles or waves?

Checkpoint 9.3 1 2 3 4

Define the photoelectric effect. Outline Hertz’s observations regarding the photoelectric effect. Describe the significance of the cut-off frequency. Complete the table below by listing the observation of the photoelectric effect that classical theory predicted incorrectly. Observation

Classical prediction

Electrons are not emitted from the surface of a metal no matter how intense the light if the frequency of the light is below the cut-off frequency.

5 6

Explain how the work function helps explain the measured energy of photoelectron. Outline the significance of the stopping voltage.

photomultiplier outputs an electrical impulse to electronic circuits

anode

dynode

incoming light photon

photo cathode

electron

Figure 9.4.1 The photomultiplier tube uses the photoelectric effect and can multiply the number of electrons produced by millions of times. Identify data sources, gather, process and present information to summarise the use of the photoelectric effect in photocells.

184

9.4 Applications of the photoelectric effect The photoelectric effect presents a convenient way of producing an electrical signal from light. The apparatus used to produce the traditional photoelectric effect (a vacuum tube with a photocathode and anode), called a photocell (photoelectric cell), was not of practical use outside the classroom for demonstrating the photoelectric effect. The current that was produced was far too small for most applications. A modification to this tube was made by placing a series of specially coated surfaces, called dynodes, between the photocathode and the anode (see Figure 9.4.1). This is known as a photomultiplier tube. The dynodes are at successively increasing positive voltage with respect to the cathode, so they attract the electrons emitted by the photocathode. Each incident electron causes many other electrons to be emitted from the dynode surface. This effect is multiplied at the next dynode, and so on until a large number of electrons reach the anode and register a relatively large current. This makes the photomultiplier tube extremely sensitive to low levels of light, such that it is possible to detect individual photons. The material from which the photocathode is made has been improved and it can eject electrons from incident light ranging from the ultraviolet to the infra-red range of the electromagnetic spectrum. This high sensitivity of the photomultiplier tube makes it suitable for the detection of small changes in light intensity. This has opened up many applications in astronomy, nuclear physics, blood testing and medical imaging. One method of medical imaging is called positron emission tomography (PET). One use of this method is the imaging of tumour cells within the body. The patient is injected with a short-lived radioactive chemical that contains molecules that attach themselves to particular tissue types such as tumour cells. This chemical emits two very high energy photons (known as gamma photons) that exit the body in opposite directions and strike a ring of photomultipliers around the patient (see Figure 9.4.2). Each photomultiplier has a material in

from ideas to Implementation front of it, known as a scintillator, that gives out flashes of light when the gamma photons strike it. The flashes of light are converted to electrical signals by the photomultiplier tubes. Computers calculate the time of arrival of the individual gamma photons and are able to locate their point of origin in the body. In this way, an image of the tumour cells can be generated from the different pairs of gamma photons arriving from different locations. Another use of the photoelectric effect is in night-vision devices used by the military to see in almost total darkness by using the few photons coming from stars. If we broaden our definition of the photoelectric effect to mean any process by which light is converted to electricity, without the need to eject an electron from a surface, then we can include many devices made from semiconductor materials. Although the electron stays within the solid, it becomes free from the atom to which it was bound, when struck by a photon. However, one important use of this type of photoelectric effect is the conversion of sunlight into electrical energy in solar cells. More detail will be given on this topic in Chapter 10 ‘Semiconductors and the electronic revolution’.

a

detectors

positron-electron collision

gamma rays created

b

Checkpoint 9.4 1 2 3

Explain why the classic photocell is not used in commercial applications. Outline how a photomultiplier tube works. Give examples of applications in which the photoelectric effect is used.

Figure 9.4.2 (a) A positron emission tomography scanner uses photomultiplier tubes to obtain (b) an image of tumour cells.

PRACTICAL EXPERIENCES

from ideas to Implementation

CHAPTER 9

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 9.1: Receiving radio waves Produce sparks with the induction coil. Tune the portable radio to a position on the AM band at which there is no radio station. Move around the classroom and listen to the clicking sounds on the radio that are in step with the sparking from the induction coil. Equipment: induction coil, power supply, radio.

Perform an investigation to demonstrate the production and reception of radio waves.

Discussion questions 1 Explain how your experiment differs from the one carried out by Hertz. 2 Describe what you notice as you move further away from the coil. 185

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Electromagnetic radiation: particles or waves?

PRACTICAL EXPERIENCES Activity 9.2: Black body radiation

Identify data sources, gather, process and analyse information and use available evidence to assess Einstein’s contribution to quantum theory and its relation to black body radiation.

Gather information to assess Einstein’s contribution to quantum theory and black body radiation. Present your information in the form of a newspaper article relating what you have found to people without scientific backgrounds. Discussion questions 1 Define the UV catastrophe. 2 Outline how Planck explained the difference between observations and theory. 3 State Einstein’s contribution to black body radiation.

Activity 9.3: Photocells Identify data sources, gather, process and present information to summarise the use of the photoelectric effect in photocells. Solve problems and analyse information using: E = hf and c = f λ

Part A: Gather information about how the photoelectric effect is used in photocells. Part B: Set up an electrical circuit to investigate the energy that a photon of light gives to an electron. You can carry out the experiment first hand or use a computer simulation. Discussion questions 1 List devices that can be classified as photocells and explain how their working principle is related to the photoelectric effect. 2 Use the graph of electron energy (y-axis) and photon frequency (x-axis) to calculate the Planck constant.

Activity 9.4: Einstein versus Planck Process information to discuss Einstein and Planck’s differing views about whether science research is removed from social and political forces.

Gather information about the differing views held by Planck and Einstein and use it to write an informed paragraph on your views. Discussion questions 1 Outline Einstein’s and Planck’s view of scientific research. 2 List political pressures on Planck and Einstein during the times of their main discoveries, and assess the influence of these pressures on Planck’s and Einstein’s research.

Chapter summary • • • •

•

186

Maxwell’s theory predicted that all electromagnetic radiation travels at the same speed. Maxwell strongly suggested that light is an electromagnetic wave. Hertz performed a number of experiments to demonstrate the propagation of electromagnetic waves. Hertz also measured the speed of electromagnetic waves by causing them to become standing waves, then measuring the wavelength, calculating the frequency and using c = f λ. A black body is an object that is capable of absorbing or emitting all wavelengths of the electromagnetic spectrum.

•

•

•

•

Classical physics could not reproduce the measured curve of intensity versus wavelength of radiation emitted from a black body. The ultraviolet catastrophe is a prediction by the classical electromagnetic wave theory that the radiation intensity from a black body becomes infinite at very short wavelengths, which is false. Max Planck was able to predict the black body radiation curve by assuming that the energy of the oscillators in the black body walls is quantised in integer steps of hf, where h is Planck’s constant and f is the frequency. Planck thought that the quantisation of energy was only a mathematical trick and not related to reality.

from ideas to Implementation •

• •

Einstein realised that this quantisation was real and that light is made of particles, each with a bundle of energy given by hf. The particle of light is called a photon. The photoelectric effect is the emission of electrons from the surface of a metal that has been struck by light.

• •

Einstein explained the photoelectric effect in terms of the photon model of light. The photoelectric effect has applications in all forms of light detection, medical applications and the conversion of light into electrical energy.

Review questions Physically speaking

15 Discuss the significance of the work function in

Create a visual summary of the concepts in this chapter by constructing a mind map incorporating the following terms and equation:

16 What determines the maximum kinetic energy of

hertz, radio waves, quanta, photon, Maxwell, photoelectric effect, black body, UV catastrophe, photomultiplier, work function, Einstein, kinetic energy, threshold, frequency, polarisation, E = hf

Reviewing 1 State the significance of Maxwell’s equations. 2 Outline the purpose of Hertz’s experiments. 3 Outline the procedure followed by Hertz to determine the speed of radio waves.

4 Describe how a standing wave is set up. 5 Explain how Hertz used standing waves to determine the wavelengths of radio waves.

6 Outline how Hertz’s results support Maxwell’s predictions.

7 Describe the results that Hertz obtained to support the idea of polarisation.

8 Recall the relationship between wavelength of radiation and its intensity.

9 Outline how classical theory predicted the ultraviolet catastrophe.

10 Explain the idea of quanta. 11 Explain how Einstein expanded on Planck’s idea of

understanding the behaviour of photoelectrons. a photoelectron?

17 What is a photomultiplier tube and how does it work? 18 Describe how the photoelectric effect can be used in PET scans to detect tumours.

Solving problems 19 Calculate the energy of a photon of red light with a wavelength of 656 nm.

20 Figure 9.5.1 is a graph of voltage versus frequency of light. a Convert the voltage scale to energy of photoelectrons. b Determine the cut-off frequency. c Use information from the graph to determine the work function of the material. d Explain what will happen to the graph if the intensity of light is increased. e Explain changes in the graph if the cathode was made of a different material. f Calculate the Planck constant from the graph. V0 (V) 3 2 1

quantisation of oscillators in a black body cavity.

13 Justify the use of quartz glass for the photoelectric effect experiment. the kinetic energy of the emerging photoelectrons.

0.75

1.0

f (1015 Hz)

–1

Figure 9.5.1 Frequency versus voltage of light

iew

Q uesti o

n

s

14 Relate the frequency of light on a photocathode to

0.25 0.50

v

of the photoelectric effect.

0

Re

12 Discuss Einstein’s contribution to the understanding

187

10

Semiconductors and the electronic revolution Electronics transforms the world

valence electron, valence level, conduction level, valence, conduction bands, energy gap, band gap, forbidden energy gap, solid state physics, electron volt (eV), hole, doping, n-type semiconductor, donor, p-type semiconductor, acceptor, donor energy level, acceptor energy level, extrinsic, intrinsic, microprocessor, diode, diffusion, depletion region, forward bias, reverse bias, photovoltaic cells, thermionic devices, transistors, plate, triode, electronics, bipolar transistor, emitter, collector, base, field-effect transistors (FET), source, drain, gate, MOSFET, integrated circuits

Figure 10.0.1 An old valve computer weighed several tonnes but had much less computing power than your mobile phone. 188

Electronics has transformed almost every aspects of modern life, from television, computers and mobile phones to medical diagnostics and treatment. The discovery of cathode rays led to devices such as radios, televisions and computers. The earliest computers, which used cathode ray tube devices, weighed several tonnes and filled very large rooms. Today, a basic mobile phone has much more computing power than the most powerful early computer. This reduction in size (and power consumption) came with the discovery of semiconductors. The working parts of semiconductor devices are so small that a microscope is needed to see their structure. An example of semiconductor circuitry is the microprocessor, the brains of your computer. It is no larger than a coin but contains millions of circuits.

from ideas to Implementation

Figure 10.1.1 An abstract energy level diagram helps visualise the ionisation energy of an atom, which would produce free charges for an electrical current.

Energy

a

b

Energy

Electrical conduction occurs as a result of the flow of charge. Gases, which are normally electrical insulators, are made to conduct large currents during lightning strikes, due to the movement of ions and electrons created by the ionisation of air atoms. Ionisation is the removal of the outer electron of an atom, which is known as a valence electron. The energy required to remove this electron is usually supplied by the electric field during lightning. In general, we are not interested in the exact motion of the charges in a current, we just want to know whether there will be electrical conduction. A simple and abstract way of visualising whether there will be conduction in gas atoms is shown in Figure 10.1.1. The vertical axis is energy and the horizontal lines indicate the position an electron can have. The lower line (or level), known as the valence level, is the energy it has while being bound to the atom. The upper level, known as the conduction level, is the energy it must acquire to become free and contribute to electrical conduction. Placing two individual atoms together will result in two pairs of these levels Note that the levels are not at exactly the same energy (see Figure 10.1.2a). but become shifted slightly vertically with respect to each other. This odd behaviour can be explained by the laws of quantum physics, which are beyond the scope of this section. Basically, energy levels of electrons in close proximity to each other are forbidden from being the same, so placing five atoms close together results in five pairs of these lines, as shown in Figure 10.1.2b. To form a solid we must place many atoms, say 1023, close together, and so It is impossible to we have 1023 pairs of energy levels (see Figure 10.1.3). draw these individual levels because they are very close to each other and too numerous. We simply represent them by two shaded areas called energy bands. The lower and upper bands are labelled as valence and conduction bands respectively. The energy gap between them is referred to as the band gap or forbidden energy gap. Electrical conduction in a solid occurs because some electrons have gained enough energy to be in the conduction band and therefore become free They are no longer localised to a particular ion but are shared by electrons. all ions in the lattice, since they are free to move throughout the solid. Unlike conduction in a gas, the ions in solids are not free to move, so the only charge carriers are the electrons. Quantum physics determines that no electron can have an energy in the energy gap. Moreover, only a fixed number of electrons are allowed per energy level in the energy bands. When an energy level is occupied by the maximum number of electrons allowed by quantum physics, we say that the ‘level is filled’. Any further electrons will start to occupy the level above it until this new level is filled, and so on until a whole energy band is filled. Ordinary experience tells you that not all solids conduct electricity. All metals are good conductors; however, materials such as glass and plastic are insulators. There is also a third type of solid known as a semiconductor, such as silicon and germanium, which conduct electricity but much less than metals. The energy band diagram, shown in Figure 10.1.4, can be used to illustrate the difference in the conductivities of these three types of materials.

Energy

10.1 Conduction and energy bands

Figure 10.1.2 Energy level diagrams for (a) two atoms and (b) five atoms in close proximity

conduction band energy gap

valence band

Figure 10.1.3 The energy band structure of solids

Identify that some electrons in solids are shared between atoms and move freely. Describe the difference between conductors, insulators and semiconductors in terms of band structures and relative electrical resistance.

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Semiconductors and the electronic revolution a

b

c conduction band

overlap

energy gap

valence band

Figure 10.1.4 Energy band diagrams for (a) a conductor, (b) a semiconductor and (c) an insulator

Compare qualitatively the relative number of free electrons that can drift from atom to atom in conductors, semiconductors and insulators.

The type of atom, the bonding between atoms and the number of valence and inner electrons of an atom are some factors that determine the type of energy band diagram for the resulting solid. This is the area of physics known as solid state physics, which relies heavily on quantum physics. Figure 10.1.4a shows an energy band diagram for conductors. There is no clear energy gap and therefore no distinct conduction or valence band. We can say that the conduction band is partially filled and therefore the solid contains free electrons. The energy band diagrams for a semiconductor and an insulator are similar The much smaller energy gap of the semiconductor (Figure 10.1.4b, c). distinguishes it from an insulator. The valence band of an insulator is full and, due to the relatively large size of its energy gap, none of the electrons in the valence band have enough energy to reach the conduction band. However, some electrons in the valence band of the semiconductor can acquire enough thermal The energy to jump across the small gap into the conduction band. number of conduction electrons that result is much less than for a conductor, so a semiconductor is a much poorer conductor than a metal but a better conductor than an insulator such as plastic or glass.

Checkpoint 10.1 1 2 3 4 5

Describe how gases, which usually have insulating properties, can be made to conduct electricity. Describe why the energy levels of two neighbouring atoms are not exactly the same. State why only electrons are charge carriers in solids. Define the terms electrical conductor and insulator. Describe how it is possible for electrons in the valence band of semiconductors to reach the conduction band.

10.2 Semiconductors The size of the energy gap determines whether a solid is a semiconductor or an insulator. A convenient unit for measuring the energy gap is the electron volt (eV). This is a unit of energy and is related to the joule: 1 eV = 1.6 × 10–19 J Most semiconductors have an energy gap less than 5 eV. Insulators have greater energy gaps. Examples of energy gaps for semiconductors and insulators are given in Table 10.2.1. The most common semiconductors are silicon (Si) and germanium (Ge). 190

from ideas to Implementation Some electrons in the valence band of a semiconductor will gain enough energy from the ambient thermal (heat) energy in the solid to jump up to the conduction band. This occurs readily at normal room temperature. Heating a semiconductor further increases the number of electrons jumping from the valence band to the conduction band, thereby making the solid more conducting. For example, heating a piece of silicon will lower its resistance. At absolute zero (T = 0 K), there is no thermal energy available for the electrons, all of which remain in the valence band; therefore, at absolute zero the semiconductor behaves as an insulator.

Table 10.2.1 Energy gaps in electron volts of common semiconductors at room temperature

Semiconductor crystal

TRY THIS!

Energy gap (eV)

Si Ge InP GaP GaAs CdS CdTe ZnO ZnS Diamond

1.14 0.67 1.35 2.26 1.43 2.42 1.45 3.2 3.6 5.4

Semiconducting pencil Graphite is a form of carbon that is used in ‘lead’ pencils (there is no lead in pencils, just graphite and clay, but we will refer to it as the lead of the pencil). Graphite also behaves as a semiconductor. Obtain a length of pencil lead (about 10 cm) and attach each end to the probes of a digital ohm meter. It should measure units of ohms. Now heat it (with, say, a match) and observe that the resistance decreases. The resistance may decrease by up to 0.5 Ω and it will be more for thinner leads. This decrease in resistance is due to an increase in the number of electrons jumping from the valence band to the conduction band. The resistance will increase once more when the lead cools.

Electrons and holes When an electron is excited into the conduction band, it leaves a vacancy in the The hole valence band. This vacancy is known as a hole (see Figure 10.2.1). behaves like a positively charged particle and moves in the opposite direction to the electron. In reality, the other electrons in the valence band move to fill the vacancy, but in doing so they leave behind another vacancy, resulting in the apparent motion of the hole. It is much simpler to treat the hole as a positively charged particle than to track the motion of all the electrons that move in the opposite direction. The magnitude of the charge on an electron and a hole are Therefore, the the same at 1.6 × 10–19 C, but they are opposite in sign. motion of both electrons and holes contribute to the motion of electrical current.

Identify absences of electrons in a nearly full band as holes, and recognise that both electrons and holes help to carry current.

v

conduction band

v

narrow energy gap v

v

Doping At normal room temperature (300 K) the average kinetic energy of an electron is about 0.026 eV. However, the energy gap of silicon is 1.14 eV. It would appear that the electrons don’t have enough energy to jump to the conduction band, so why is silicon, or any of the other semiconductors listed in Table 10.2.1, slightly conducting at room temperature? The answer is that 0.026 eV is an average energy. This means there are some electrons that have a much lower energy and

valence band

applied electric field electrons

holes

Figure 10.2.1 Holes are created in the valence band of a semiconductor when electrons are excited up to the conduction band. 191

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Semiconductors and the electronic revolution

A hole in a bottle

A

way of picturing the concept of a hole in a semiconductor is to consider a bottle that is almost full of water with a small air bubble underneath its closed cap. Turning the bottle upside down will cause the bubble to move upwards. The question is: ‘did the bubble move up or did the water move down?’ The answer is that both are true, but it is much simpler to follow the motion of the bubble than the more complex motion of the water. In the same way, we describe the motion of the hole in a semiconductor rather than the more complex motion of all the valence electrons.

a +4

+4

+4

+4

+4

+4

+4

+4

+4

+4

+4

+4

+4

+5

+4

+4

+4

+4

+4

+4

+4

+4

+3

+4

+4

+4

+4

b

c

electron

192

hole

some that have a much higher energy. It is these higher energy electrons that make it across the gap, but there are not many of them. Physicists have found that the conductivity of semiconductors can be improved by introducing impurities into the crystal lattice, a process that is called doping. In order to understand how an impurity atom can improve the conductivity, we must examine what happens at the crystal lattice level. We will use silicon as an example, but similar explanations can be applied to all other semiconductors. Silicon belongs to group 4 of the periodic table, which means that the silicon atom has 4 electrons available for bonding with other silicon atoms (see Figure 10.2.2a). Silicon can be doped with phosphorus P by replacing a silicon atom with a phosphorus atom. Phosphorus is in group 5 of the periodic table and has 5 electrons available for bonding, but it can only bond with 4 atoms in the silicon crystal lattice. This leaves one electron unbonded (see This semiconductor is called an n-type semiconductor. Figure 10.2.2b). (n represents the negative charge of the unbonded electron.) Impurities that produce unbonded electrons are called donor impurities. These unbonded electrons are weakly attached to their atoms and can easily be moved into the conduction band by ambient heat, and thus can contribute to an electrical current. Similarly, silicon can be doped with boron, which belongs to group 3 and therefore has 3 electrons available for bonding with 3 atoms of silicon. There is no electron available for the fourth silicon atom to bond with and, as a result, a vacancy in charge has been created, which we call a hole. An electron from a neighbouring silicon atom can now jump across to fill this vacancy (see Figure 10.2.2c). This, in turn, creates a vacancy for the atom it left behind, and so can be viewed as the movement of the hole. As already mentioned, a hole behaves like a positive charge since it moves in the opposite direction to the electron. The semiconductor behaves as though it has positive charge carriers and is therefore called a p-type semiconductor. Impurities that produce a charge vacancy are called acceptor impurities. Note that both n- and p-type semiconductors remain electrically neutral even though there is an unbonded electron or a vacancy is created. The neutral impurity atoms were added to an already neutral solid, so the net charge remains zero. Both n- and p-type doping improve the conductivity of a semiconductor by creating an excess of negative and positive charge carriers respectively. The unbonded electron in n-type semiconductors needs a small amount of energy to move to a different location. Similarly, very little energy is required to cause a neighbouring electron to occupy the vacancy of a p-type semiconductor. The energy-band diagram of an n-type semiconductor is shown in Figure 10.2.3. A new energy level has been created in the energy gap just below the conduction band. This level is occupied by the unbonded electrons of the impurities and is called the donor energy level. This level is very close to the conduction band. The difference in energy between the two levels is usually 0.026 eV or less, and so most of the electrons in the donor level are able to jump up to the conduction band at room temperature, making the semiconductor much more conducting than the undoped crystal. Figure 10.2.2 The silicon lattice (a) has no doping, (b) has been doped with phosphorus and (c) has been doped with boron.

from ideas to Implementation The energy-band diagram for a p-type semiconductor in Figure 10.2.4 shows that there is an extra energy level, called an acceptor energy level, near the valence band. This level is full of vacancies (holes), allowing the valence electrons to jump up and occupy these vacancies and therefore contribute to an electrical current. The gap between the acceptor level and the donor (valence) level must be equal to or less than 0.026 eV for conduction to take place at room temperature. A semiconductor is labelled as extrinsic if conduction is dominated by donor or acceptor impurities. Otherwise, it is known as an intrinsic semiconductor.

conduction band

Identify differences in p- and n-type semiconductors in terms of the relative number of negative charge carriers and positive holes. Describe how ‘doping’ a semiconductor can change its electrical properties.

conduction band

donor impurity energy level

holes acceptor impurity energy level

valence band

valence band

Figure 10.2.3 Energy-band diagram of an

Figure 10.2.4 Energy-band diagram of

n-type semiconductor

a p-type semiconductor

PRACTICAL EXPERIENCES Activity 10.1

Activity Manual, Page 91

Checkpoint 10.2 1 2 3 4 5 6

Distinguish between a semiconductor and an insulator in terms of the size of the energy gap. Describe what is meant by a hole. Describe doping in semiconductors. Explain the difference between n-type and p-type semiconductors. Identify elements that can be used to make n-type and p-type semiconductors. Distinguish between acceptor and donor energy levels.

10.3 Semiconductor devices Combinations of p- and n-type semiconductors can be used to make all modern electronic devices, ranging from the simple remote control to the sophisticated computer microprocessor. The basis of this technology is a result of the properties of the junction between p- and n-type materials.

The p–n junction A p–n junction is a p-type material joined onto an n-type material. In reality, the p–n junction is made from a continuous single crystal in which the concentration of impurities has been made to change abruptly from p-type to n-type as we cross the junction. However, to clarify the physics of the junction region, we will assume that the p-type and n-type crystals are initially separated and then brought together. The p–n junction is used as the most basic electronic device, the diode. It allows current to move in only one direction, as shown in Figure 10.3.1.

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Semiconductors and the electronic revolution V

a

p

I

n

forward bias

A +

b p

c

n

I

forward bias O

V

reverse bias

Figure 10.3.1 (a) A forward-biased diode and (b) its symbol. (c) The graph shows the current versus voltage characteristics of a diode.

electric field

depletion layer

p-type

n-type E E free electrons

free holes fixed negative ions

fixed positive ions

Figure 10.3.2 The electric field in the depletion region of the p–n region stops further diffusion of charge.

194

The conventional current easily flows from p- to the n-type material when the positive terminal of a power supply is connected to the p-type end of the device. Reversing the connections of the power supply (positive connected to n-type and negative to p-type) results in a very tiny reverse current, the current is essentially stopped. The symbol for the diode is an arrow that indicates the direction of the conventional current when the positive and negative terminals of the power supply are connected to the p- and n-type materials respectively (see Figure 10.3.1b). A key to understanding the physics of a p–n junction is to first examine the concept of diffusion. Free particles are in continuous random motion and move from areas of high to areas of low concentration. For example, a drop of ink placed in a beaker of water results in the ink particles slowly spreading outwards so that the water becomes uniform in colour. This occurs because the point of origin of the free particles is very small (drop of ink) compared with the rest of the medium (beaker of water). So it is more likely that the particles will move to the rest of the medium. This is diffusion and it is a property of all randomly moving particles. The energy for the movement of these particles (kinetic energy) comes from the ambient thermal energy. At a temperature of absolute zero there will not be any diffusion. Similarly, the holes in the p-type material and the electrons in the n-type material are in constant random motion due to their thermal energies, resulting in a uniform density of the charge carriers across their respective materials. When the two crystals are brought into contact, electrons from the n-type crystal will naturally want to diffuse into the p-type crystal, because the electron density of its conduction band is lower. Conversely, the holes in the p-type region want to This results in an diffuse into the n-type crystal, as shown in Figure 10.3.2. electrical current, which is quickly stopped because an electric field builds up from the n to the p region, due to this charge separation, and stops the further flow of charge. A large part of this diffusion region results in the combination of electrons with holes, leaving it depleted of charge. This is known as the depletion region. However, there continues to be an excess of electrons and holes on the p and n sides respectively. A depletion region is like two parallel plates with equal and opposite charges on them, resulting in an electric field in the space between them, which contains no charge. The energy that created the electric field essentially came from the thermal energy of the charge carriers. At a temperature of absolute zero (0 K), there will not be enough energy for diffusion to occur and therefore no electric field and no depletion region will be created. Electrons that diffused across and combined with holes cannot easily drift back since they lost energy as they fell into the holes. Similarly, holes that diffused combine with electrons from the n-type material. The trapping of the diffused holes and electrons has resulted in the conversion of thermal energy into the electrostatic energy stored in the electric field region. As shown in Figure 10.3.3a, connecting the positive and negative terminals of a battery to the p and n sides of the junction respectively creates an electric field opposing that in the depletion region. This lowers the junction electric field strength and therefore allows further charge to diffuse across the junction. A thin p–n junction will enable the excess diffused charge to reach the

from ideas to Implementation metal terminals on either side of the junction, thus resulting in the flow of current in the outside circuit. This type of connection to the p–n junction is known as forward bias. This also has the effect of reducing the width of the depletion region. Conversely, connecting the negative and positive terminals of a battery to the p and n sides respectively, as shown in Figure 10.3.3b, increases the electric field strength in the junction region. This will cause some of the diffused electrons to travel back to the n region, setting up a small reverse current. This type of connection is known as reverse bias. It also has the effect of increasing the width of the depletion region.

a

Rectifiers

b

The p–n junction or diode is used to convert AC to DC electricity in chargers or power supplies for electronic devices such as mobile phones and computers. A simple circuit, but not one that is used often, is shown in Figure 10.3.4. A sinusoidal alternating voltage, say from the AC mains supply, is applied to the p end of the diode. The output from the n end is the positive part of this signal. The negative part of the AC voltage is blocked. A component consisting of parallel plates (called a capacitor) at the output stores the positive charge and helps smooth the output signal so that a DC voltage will result. V

input

output C

hole flow p-type

electron flow

n-type

E electron flow +

temporary hole flow

–

electron flow

temporary wide depletion layer electron flow

p-type

n-type

–

E +

Figure 10.3.3 (a) Forward-biased and (b) reverse biased p–n junctions

voltage smoothed

0

narrow depletion layer

t

Figure 10.3.4 A rectifier circuit can be used as a power supply.

Light-emitting diodes (LED) and laser diodes A current that passes through a forward-biased diode will lead to some recombination of electron and hole pairs. This recombination occurs when a conduction electron drops from the conduction to the valence band to occupy a vacancy (the hole), as shown in Figure 10.3.5. This change in energy can appear as heat given to the solid, or as the emission of light. The energy of the photon hf is equal to the energy lost by the electron, which is the energy gap Eg such that: hc E g = hf = λ where Planck’s constant h = 6.63 × 10–34 J s, f is the frequency, c is the speed of light and λ is the wavelength. Manufacturers of LEDs can control the wavelength of the emitted light by controlling the size of the energy gap. Most remote-control units around the home have an LED on the end that you point towards a device. The LED looks like a very tiny plastic light blub, but works very differently. The wavelength of these LEDs is usually in the infra-red range of the spectrum and so is not visible to the eye. However, it is visible to most types of digital cameras such as those in mobile phones and video cameras.

PRACTICAL EXPERIENCES Activity 10.2

Activity Manual, Page 94

light emission conduction band E = hƒ valence band

hole flow

hole and electron recombine

p-type

electron flow n-type

+

–

electron flow

E = hƒ electron flow

Figure 10.3.5 A forward-biased light-emitting diode (LED)

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Semiconductors and the electronic revolution

PRACTICAL EXPERIENCES Activity 10.3

PHYSICS FEATURE

Activity Manual, Page 96

Photovoltaic cells—solar cells

S

region is made thin enough so that most of the olar cells use the p–n junction to convert light incoming light can be transmitted to the p-type region, directly into electricity. They have the more formal name of photovoltaic cells (or PVs). Very little greenhouse where photons are absorbed to create the conduction electrons. These are swept up into the n region by the gas emissions are associated with PVs, which makes electric field at the p–n junction and are collected by them a strong candidate to replace coal-fired power the front contact metal grid, which can be connected to stations for future power generation. Currently they have an external circuit. The size of the voltage created by many applications that include the supply of electricity solar cells depends on the potential difference across to remote locations or homes, and satellites. the depletion region. In a PV, a valence band electron absorbs a photon The engineering of solar cells has advanced greatly and is excited up to the conduction band, thus in recent years so that greater efficiency can be contributing to an electrical current. In practice, PVs obtained at lower cost. For example, an antireflective are simply p–n junctions. The photons absorbed in the coating is placed on top so that very few photons are p region produce conduction electrons near the lost due to simple reflection. The size of the energy gap depletion region. These electrons are swept to the n determines the wavelength range that will be absorbed side of the junction by the electric field (see Figure to produce electrons—some parts of the sunlight 10.3.6). If the depletion layer is thin enough, then spectrum deliver more energy than others. The most of these electrons can reach the n region without maximum efficiency of 23% has been achieved in recombining with a hole. The electrons in the n region recent years but not on a commercial scale. have a much higher lifetime before recombining with a hole. Electrodes placed on the ends of energy the semiconductor collect this newly electrical from Sun transmission created current and deliver it to an system external circuit. Electrons will travel transparent around a circuit and back to the p region, adhesive where they will recombine with holes. antireflection The actual structure of a solar cell is coating solar cover glass shown in Figure 10.3.7. The n-type front arrays contact

n-type layer (semiconductor) light absorption

conduction band

v

absorbed photon

substrate

junction p-type layer (semiconductor)

valence band new hole and electron created p-type

hole flow

electron flow

current

n-type

electron flow

load resistor

electron flow (current)

electron flow

Figure 10.3.6 A solar cell is a p–n junction that absorb photons to create conduction electrons. 196

back contact

freed electrons

holes filled by freed electrons

Figure 10.3.7 How a solar cell operates

from ideas to Implementation

Try this! seeing infra-red Grab a remote control that you use for the TV set or DVD player. Look at the LED at the end of the remote control while pressing any button (for example, the volume or channel button). You should not see anything happening. Now point the remote control at your mobile phone camera or any other digital camera. While looking at the screen of the camera, press any button on the remote control. You should now see the LED flashing, because the camera is sensitive to the infra-red light from the LED but your eye is not.

Figure 10.3.8 The infra-red light from the LED of a remote control has been made visible by the digital camera that took this photograph.

Checkpoint 10.3 1 2 3 4 5 6 7

Describe what is used to make a diode. Outline how a p–n junction is connected to a power supply in order for electricity to flow. Describe what happens to the distribution of electrons and holes when p- and n-type semiconductors are brought into contact. Outline how a rectifier works. Explain why light is emitted in a LED. Describe the different parts of a solar cell. Explain why the n-type layer in a solar cell needs to be thin.

10.4 The control of electrical current We have seen that the diode (or rectifier) enables current to be conducted in one direction, for example, to convert AC to DC. Diodes are one of the most basic electronic components, but they have limited control over current. Remarkably, the first type of radio receiver used only a diode as a means of detecting a radio signal and obtaining the audible sound without the need for batteries or any other kind of power supply. This was called a crystal radio set and is still made today by enthusiasts to detect AM radio stations. A great deal more control over current is required for the wide variety of electronic devices we use every day. For example, mobile phones have to transmit an electromagnetic wave at a precise frequency by making electrical current oscillate through an antenna. The phone also needs to detect a very weak electromagnetic wave, then amplify it (make it larger) and decipher the information contained within it such as voice, SMS, etc.

Figure 10.4.1 Thermionic devices (valves) were replaced by the much smaller transistors in modernday electronics. 197

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electrons

A number of discoveries and inventions led to components that were used to accurately control the direction and magnitude of electrical current. The first of these were vacuum tubes known as thermionic devices or valves, which were later replaced by semiconductor technology such as diodes, transistors and integrated circuits.

anode electrons

+

cathode heater

Thermionic devices

electrons A

Figure 10.4.2 A thermionic diode enables current to flow in one direction.

anode electrons grid cathode heater heater (filament)

cathode grid

anode

Figure 10.4.3 A triode allows the electron current to be controlled electrically.

+150 V

The transistor

output

0V

Figure 10.4.4 A small AC signal is amplified by a triode circuit.

198

During his research on incandescent lamps, Thomas Edison (1847–1931) discovered that a current can be made to flow between a heated filament and a metal electrode in a vacuum by applying a potential difference between them. The filament and electrode were the cathode and anode respectively. Heating the filament gives some electrons enough energy to escape the surface. These electrons are then accelerated to the anode. John Ambrose Fleming (1849–1945) applied this effect to the construction of the first thermionic rectifier (diode) or valve, which allowed current to flow in only one direction. The name valve comes from an analogous mechanical device that allows fluids or gases to flow in one direction, such as the valve (or nozzle) on a bicycle or car tyre that enables air to be pumped in but does not let it out. A thermionic diode consists of three components: the filament (or heater), the cathode and the anode (see Figure 10.4.2). The heater is used to heat the cathode, which releases electrons that are attracted to the anode (the plate). Lee De Forest (1873–1961) inserted a metal grid between the cathode and anode of a thermionic diode. The grid allows electrons to pass through (see Figure 10.4.3). This device is known as the triode. De Forest found that the electron current could be stopped by placing a negative voltage on the grid, or it could be allowed through by using a positive voltage. A voltage between these two extremes could be used to control the amount of current flowing through. One application of the triode is the amplifier (see Figure 10.4.4). If the voltage on the grid is negative to the point that it only allows a very tiny current through, then a small AC signal applied to the grid will cause current flowing to the anode to vary in the same way. However, because of the large potential difference between anode and cathode, this current is larger than that in the original signal; that is, the triode has acted as a current amplifier. Moreover, this current could be passed through a resistor, resulting in a varying voltage across it. In this way, small voltage signals could be amplified. The invention of the triode was in part responsible for the revolution in electronics, such as in radio and television, in the early 20th century. Transistors are semiconductor devices that control and amplify electrical current; they have largely replaced thermionic devices. Just as the p–n junction has replaced the thermionic diode, the transistor has replaced the thermionic triode. The transistor was invented in 1947 by John Bardeen and Walter Brattain, who with their boss William Shockley received the Nobel Prize for it in 1956. In the early days of the transistor, germanium was used as the semiconductor because the methods existed for growing germanium crystals of high purity. Although silicon is more abundant and retains its semiconducting properties at higher temperatures, it needed a crystal of higher purity than could be achieved at the time. Germanium diodes and transistors were replaced with those made from silicon as soon as techniques for growing crystals improved and resulted in higher purity silicon.

from ideas to Implementation The most common transistor is the bipolar (or junction) transistor. A schematic diagram of a bipolar transistor is shown in Figure 10.4.5. This transistor works in a similar way to the original transistor of Bardeen and Brattain, but its structure is very different. The bipolar transistor (Figure 10.4.5) consists of a p-type semiconductor sandwiched between two n-type semiconductors. A lead (or wire) is connected to each layer of the material. The two outer layers are called the emitter and the collector; the central layer is called the base. This is often called an n–p–n transistor. The width of the base has been exaggerated for clarity, but is usually very thin. An electron current will only flow from the emitter to the collector if there is a potential on the base with respect to the emitter (see Figure 10.4.6). Recall that in a p–n junction there is an electric field that points from n to p. Applying a positive voltage on the base of an n–p–n transistor with respect to the emitter reduces the size of this field to almost zero. This then allows electrons from the emitter to move into the base region. These electrons are then accelerated into the collector region by the electric field at the junction between base and collector. The electrons that moved into the collector can then flow into an external circuit. Note that no current will flow without applying the potential to the base. The current to the base is much smaller than the current that it allows to flow from emitter to collector. Thus the transistor, like the triode, is a current amplifier. The circuit given in Figure 10.4.7 has a small AC voltage applied to the base of an n–p–n transistor, which causes a much larger current to flow into an external resistor between the emitter and the collector. The voltage across the resistor is larger than the input signal but varies in the same way. This circuit is a basic amplifier. Bipolar transistors also can be made in the p–n–p configuration.

Identify that the use of germanium in early transistors is related to lack of ability to produce other materials of suitable purity.

electrons electrons n collector

p base

collector

n

electrons emitter

small current

n

emitter

p

base

n

collector

emitter base

Figure 10.4.5 A schematic diagram of a bipolar transistor

+ Vbase

n

large current

+ Vcollector

Figure 10.4.6 An electron current flows through an n–p–n transistor.

p

load resistor

n

amplified output

input +

+ Vbase

Vcollector

Figure 10.4.7 This small signal amplifier uses an n–p–n transistor.

A comparison of solid state and thermionic devices Semiconductor devices such as diodes and transistors do the same job as their equivalent thermionic devices (the valve and the triode), but they have many The semiconductor devices are much smaller than their advantages. thermionic counterparts and use much less power. They are mechanically robust, whereas thermionic devices are made of glass and so are fragile, and semiconductors have longer operating lifetimes due to their lower temperature of operation.

Describe differences between solid state and thermionic devices and discuss why solid state devices replaced thermionic devices.

199

10

Semiconductors and the electronic revolution

PHYSICS FEATURE Integrated circuits

A

nother type of transistor, called the field-effect transistors (FET), shown in Figure 10.4.8, consists of two n regions embedded in a larger and lightly doped p substrate. The two n regions are called the source and the drain. An insulating layer made from an oxide of the semiconductor (such as silicon oxide) covers these, and a metal electrode (called the gate) is placed on top of this layer. Because the substrate is lightly doped, it is not very conducting, so that no current will flow naturally between the source and the drain. When a positive potential is applied to the gate, it causes electrons in the substrate to be attracted to the region between the two n regions and forms a conducting channel. In this way, current can be controlled by the voltage on the gate. This metal–oxide–semiconductor field-effect transistor (MOSFET) is not as common as the bipolar transistor, but its construction makes it easy to place many MOSFETs on a single wafer of silicon. Combinations of MOSFETs, resistors and capacitors form very complicated circuits known as integrated circuits (IC). Integrated circuits containing hundreds of thousands of transistors may be as small as a few millimetres square. The IC is commonly known as the silicon chip, and is used in all modern-day electronics such as computers, televisions, mobile phones, etc. There are many processes that go into producing an integrated circuit and they vary between different manufacturing plants and depend on the specifications of the required IC. The following description simply gives an indication of the many-step processes that go into making an IC.

An n-type layer is sandwiched between a lightly doped p-type silicon layer (the substrate) and a layer of silicon dioxide (SiO2) deposited on it. Photographic, chemical and ion-beam bombardment techniques are used to dissolve (etch) unwanted areas and deposit or dope other areas for the required patterns of transistors, resistors and capacitors.

gate (small voltage) source (–)

drain (+) insulating layer

n

n few electrons can pass substrate (p)

Figure 10.4.8 A schematic diagram of a metal–oxide–semiconductor field-effect transistor (MOSFET)

silicon dioxide film n-type

p-type substrate

p

Figure 10.4.9 A wafer of p-type silicon with n-type and insulating layers is the starting point in the manufacture of an integrated circuit.

Checkpoint 10.4 1 2 3 4 5 6 200

p

Describe the purposes of a diode and a triode. Referring to Figure 10.4.2, explain the need for the heater in a thermionic diode. Explain how a triode works. Outline how an amplifier works. Recall the reasons why germanium was originally used in semiconductor devices. Identify problems associated with the use of germanium in these devices.

PRACTICAL EXPERIENCES

from ideas to Implementation

CHAPTER 10

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 10.1: Electrons and holes Using the computer simulation, investigate the behaviour of p-type and n-type semiconductors, and a p–n junction. Discussion questions 1 Describe the changes in the energy diagram when the semiconductors are doped with p-type or n-type dopants. 2 Use your observations of the simulator diode to explain what happens inside a diode.

Activity 10.2: History of communication Gather information about the history of the transistor and the development of telecommunications, and consider how the materials and technology available at the time influenced what could be developed. Discussion questions 1 Explain how the use of semiconductors changed telecommunications. 2 Identify areas in which the introduction of the transistor has influenced social behaviour.

Activity 10.3: Solar cells Solar cells are devices that can convert sunlight directly to electrical energy. Carry out an experiment to determine the effect of sunlight on a solar cell. Equipment: solar cell, fan or small motor, digital multimeter.

Perform an investigation to model the behaviour of semiconductors, including the creation of a hole or positive charge on the atom that has lost the electron and the movement of electrons and holes in opposite directions when an electric field is applied across the semiconductor.

Gather, process and present secondary information to discuss how shortcomings in available communication technology lead to an increased knowledge of the properties of materials with particular reference to the invention of the transistor. Identify data sources, gather, process, analyse information and use available evidence to assess the impact of the invention of transistors on society with particular reference to their use in microchips and microprocessors.

Identify data sources, gather, process and present information to summarise the effect of light on semiconductors in solar cells.

Discussion questions 1 Determine the angle (with respect to the direction of the Sun) of the solar cell that delivers the greatest amount of power to the motor, or the greatest voltage. 2 Determine the effect on power output of different wavelengths of light by placing, for example, coloured cellophane in front of the solar cell.

201

10 • • • •

• • •

•

Chapter summary

Semiconductors and the electronic revolution

The conduction of electrons can be described with energy band diagrams. Electrons in the valence band are not free to move and do not contribute to conduction. Electrons in the conduction band are free to move and enable conduction to take place. Insulators have a large energy gap between the conduction and valence bands and not many valence electrons have enough energy to jump up to the conduction band. Metals have no band gap and, therefore, there are always electrons in the conduction band. The energy-band diagram of semiconductors is similar to that of insulators but with a much smaller band gap. Semiconductors can be made more conducting by exciting electrons from the valence band to the conduction band by heat energy or a photon. The conductivity of semiconductors can be improved by introducing impurity atoms into the crystal lattice.

•

• • •

•

• •

A semiconductor with impurities that introduce extra electrons into the lattice is known as an n-type semiconductor. A semiconductor with impurities that introduce a vacancy in the bonding is known as a p-type semiconductor. The introduction of impurities into the semiconductor crystal lattice is known as doping. The impurities in semiconductors introduce extra energy levels in the energy gap region and so improve the conductivity of semiconductors. Semiconductor devices can be made by using the properties of the junction between p- and n-type semiconductors, and include the diode, the solar cell and the transistor. Transistors replaced thermionic valve triodes used in the control of electrical current. Transistors have the advantage of being faster, smaller and more energy efficient than thermionic valve triodes, and thousands can be produced on a small silicon wafer to make complicated integrated circuits.

Review questions Physically speaking

Reviewing

Create a visual summary of the concepts in this chapter by constructing a mind map incorporating the terms and equation in the following table.

1

Outline the difference between energy levels and energy bands.

2

Describe the difference in terms of energy gaps between insulators, conductors and semiconductors.

3

Explain why it is wrong to say n-type semiconductors are negatively charged or that p-type semiconductors are positively charge.

4

Explain why the depletion zone occurs when p- and n-type semiconductors are placed in contact with each other.

p-type

n-type

Semiconductor

Energy bands

Depletion zone

Conductor

Valence band

Forbidden gap

Insulator

Thermionic devices

Solid state

Triode

5

Explain why the depletion zone will not form at a temperature of absolute zero.

Diode

Transistor

Solar cell

6

Describe in terms of electric field strength why forward and reverse bias occurs.

Holes

Electrons

Doping

7 8

Describe the path of an electron through a solar cell.

9

Explain in terms of the electric field at a p–n junction how current can flow in only one direction.

202

Give reasons why thermionic devices were not energy efficient.

from ideas to Implementation 10 Explain how a triode can be used to control the amount of current flowing in a circuit.

11 Compare and contrast a triode with a transistor. 12 Explain why silicon was not originally used in making semiconducting devices.

13 Explain the significance of integrated circuits on society.

Solving problems 14 A 1 milliwatt (10–3 W) red laser pointer outputs

a wavelength of 650 nm (nm = 1 × 10–9 m). a Calculate the energy of a photon emitted from the laser. b Calculate the number of photons being emitted every second. (Hint: Planck’s constant h = 6.63 × 10–34 J s.)

16 A photon of wavelength 3.35 μm (1 μm = 1 × 10–6 m) has just enough energy to raise an electron from the valence band to the conduction band in a lead sulfide crystal. Calculate the energy gap between these bands in lead sulfide. (Hint: Planck’s constant h = 6.63 × 10–34 J s.)

17 Figure 10.5.1 shows a sinusoidal input voltage Vi as a function of time t for the given circuit. Draw a similar graph of the output voltage Vo as a function of time. Vi +v Vi 0

R

Vo

t

15 The electrical conductivity of undoped silicon can

iew

Q uesti o

n

s

v

Figure 10.5.1

Re

be increased by irradiating it with photons. This has the effect of exciting valence electrons into the conduction band. Given that the energy band gap of silicon is 1.14 eV, calculate the longest wavelength of a photon that can excite a valence electron to the conduction band. (Hint: Planck’s constant h = 6.63 × 10–34 J s.)

203

11

Superconductivity Surprising discovery

crystal, constructive interference, destructive interference, path length, diffraction grating, Bragg law, phonons, critical temperature, type-I superconductors, type-II superconductors, critical field strength, vortices, flux pinning, BCS theory, Cooper pair, coherence length, energy gap, spin

Just as an improved understanding of the conducting properties of semiconductors led to the wide variety of electronic devices, research into the conductivity of metals produced quite a surprising discovery called superconductivity. This is the total disappearance of electrical resistance below a certain temperature, which has great potential applications ranging from energy transmission and storage to public transport. An understanding of this phenomenon required a detailed understanding of the crystal structure of conductors and the motion of electrons through them.

11.1 The crystal structure of matter A crystal is a three-dimensional regular arrangement of atoms. Figure 11.1.1 shows a sodium chloride crystal (ordinary salt also called rock salt when it comes as a large crystal). The crystal is made from simple cubes repeated many times, with sodium and chlorine atoms at the corners of the cubes. Crystals of other materials may have different regular arrangements of their atoms. There are 14 types of crystal arrangements that solids can have. The regular arrangement of atoms in crystals was a hypothesis before Max Von Laue and his colleagues confirmed it by X-ray diffraction experiments. William and Lawrence Bragg took this method one step further by measuring the spacing between the atoms in the crystal. Let us first look at the phenomenon

Figure 11.1.1 Crystal structure of sodium chloride. The red spheres represent positive sodium ions, and the green spheres represent negative chlorine ions.

204

from ideas to Implementation of interference of electromagnetic radiation, and examine how this was applied to crystals using X-rays. Then we will see how the BCS theory of superconductivity made use of the crystal structure of matter.

Try This!

Checkpoint 11.1

Crystals in the kitchen

Explain what is meant by the crystal structure of matter.

Look at salt grains through a magnifying lens. Each grain is a single crystal that is made from the basic arrangement of sodium and chlorine atoms shown in Figure 11.1.1. Although the grains mostly look irregular due to breaking and chipping during the manufacturing process, occasionally you will see an untouched cubic or rectangular prism that reflects the underlying crystal lattice structure.

11.2 Wave interference The wave nature of light can be used to measure the size of very small spaces. Recall that two identical waves combine to produce a wave of greater amplitude when their crests overlap, as shown in Figure 11.2.1a (see in2 Physics @ Preliminary sections 6.4 and 7.4). The overlapping waves will cancel to produce t=0s a resulting wave of zero amplitude when the crest of one wave coincides with the trough of the other (Figure 11.2.1b). This addition and subtraction is called constructive and destructive interference respectively and is a property of all wave phenomena. t=1s As an example, two identical circular water waves in a ripple tank overlap (see Figure 11.2.2). The regions of constructive and destructive interference radiate outwards along the lines as shown. Increasing the spacing between the sources t = 3 s (Figure 11.2.2b). causes the radiating lines to come closer together a

b

t=0s

t=4s

Figure 11.2.1 Two identical waves (red, green) travelling in opposite directions can add (blue) t=1s

(a) constructively or (b) destructively.t = 5 s

The interference of identical waves from two sources can also be represented by outwardly radiating transverse waves (see Figure 11.2.3). The distance that a twave = 3 s travels is known as its path length. t = 6 s Constructive interference occurs when the difference in the path length of the two waves is equal to 0, λ, 2λ, 3λ, 4λ or any other integer multiple of the wavelength λ. Destructive interference occurs when the two waves are half a wavelength out of step. This corresponds to t=4s t=7s a path length difference of λ/2, 3λ/2, 5λ/2 etc.

t=5s

waves in phase

lines of destructive interference

lines of constructive interference

b

Figure 11.2.2 Interference of water waves for two sources that are (a) close together and (b) further apart

constructive interference

t=6s

destructive interference constructive interference

t=7s

a

Figure 11.2.3 Constructive and destructive interference between identical transverse waves from two sources 205

11

Superconductivity

Light also has wave properties and produces interference effects when it is passed through two narrow and closely spaced slits (see Figure 11.2.4). Each slit acts as a source of light waves that are in step with each other. The resulting constructive and destructive interference pattern appears as bright and dark bands on a screen. The position of the bright bands can be determined by applying the condition that the path length difference between the two waves must be an integer multiple of wavelengths (see Figure 11.2.5). This can be expressed by the following equation:

screen double slit

mλ = d sin θ where m = 1, 2, 3 …, d is the spacing between the slits and θ is the angular position of the bright bands. The central bright band corresponds to m = 0. The first band on either side of it corresponds to m = 1, and so on. Interference also occurs when many slits are constructive used (see Figure 11.2.6). The bright bands interference become narrower and sharper on the edges, with y increasing number of slits. In practice, the slits θ are parallel straight lines scratched, etched or moulded onto a glass plate. This arrangement of multiple slits is known as a diffraction grating. A very good grating for visible light can screen have 2400 lines per millimetre. Such a grating provides very sharply defined bands whose

Figure 11.2.4 An interference pattern is formed by light passing through two narrow slits.

S1 L S2

slit 1 d

θ θ

slit 2 d sin θ

waves in phase

Figure 11.2.5 The path difference between waves that produce constructive interference constructive interference

waves in phase

TRY THIS! Diffraction grating in my stereo A commonly available diffraction grating is the humble CD or DVD. It has thousands of tiny and closely spaced pits that cause ordinary white light to break up into the colours of the rainbow because there is an angle for constructive interference for each wavelength contained in white light. A more clearly defined diffraction pattern can be made by shining a low power (less than 1 mW) laser pointer on the CD or DVD so that the beam is reflected to a nearby wall. You should notice that there are fainter dots near the reflected dot. These fainter dots arise from constructive interference of the laser light.

waves in phase

destructive interference

constructive interference

Figure 11.2.6 Interference from multiple slits in a diffraction grating 206

from ideas to Implementation position can be accurately determined. The angular positions of the bright bands Knowledge are determined by the formula given above for the double slit. of the wavelength can be used to determine the slit separation and vice versa. To obtain an interference pattern, the distance between slits must be close to the wavelength of the light falling on them. The interference pattern from a diffraction grating is also called a diffraction pattern.

Checkpoint 11.2 1 2 3

Recall the conditions for constructive interference. Describe how these differ from those for destructive interference. Compare the interference pattern produced by light and that produced by two interfering water waves.

11.3 X-ray diffraction X-rays are high-frequency electromagnetic waves first discovered by Wilhelm Conrad Röntgen (1845–1923). He was experimenting with cathode ray tubes and noticed that a fluorescent screen, some distance away from the tube, began to glow each time the tube was operated. Röntgen had discovered a new type of radiation, which he called X-rays to indicate that their nature was unknown. Röntgen also found that X-rays had great penetrating power and could be used for medical applications. Figure 11.3.1 shows one of the first X-ray pictures of a hand taken by Röntgen. Several years of experimentation by scientists after Röntgen’s discovery showed that X-rays were electromagnetic waves like light but with an extremely short wavelength. A schematic diagram of an X-ray tube is shown in Figure 11.3.2. A beam of electrons strikes a metal target. The rapid deceleration of electrons on striking the target causes X-rays to be emitted. The wavelength of the most intense X-rays depends on the material from which the target is made. A single narrow slit can also be used to obtain a blurry diffraction pattern. Using this crude method, rough estimates of 10–10 m were obtained for the wavelength of X-rays. It wasn’t physically possible for a proper diffraction grating with a well-controlled slit spacing of 10–10 m to be made, so accurate estimates of X-ray wavelength were not possible. Max von Laue (1879–1960) learned of the hypothesis that crystals were regular arrangements of atoms with a spacing between atoms of about 10–10 m. He realised that crystals were the gratings he needed for X-ray diffraction. In 1912 W. Friedrich and P. Knipping followed von Laue’s suggestions and passed a narrow beam of X-rays through a zinc sulfide crystal. A photographic plate placed behind the crystal produced a regular pattern of spots (Figure 11.3.3), which was the interference pattern of the crystal which acted as a diffraction grating.

Figure 11.3.1 X-ray picture of a hand taken by Röntgen cathode (electron source) electron beam

anode

vacuum X-rays

heavy metal target metal rod (removes heat and electrons)

cooling fins

Figure 11.3.2 An X-ray tube 207

11

Superconductivity

a

Checkpoint 11.3

X-rays

1 2

Recall Röntgen’s observations that led to the discovery of X-rays. Identify why X-rays are emitted when electrons strike a metal surface. Explain the significance of crystals in determining the wavelength of X-rays.

crystal

3 lead collimator

photographic film

Figure 11.3.3 Laue’s suggestion for X-ray diffraction

Outline the methods used by the Braggs to determine crystal structure.

11.4 Crystal structure The Laue diffraction patterns qualitatively showed that a crystal indeed consisted of a regular array of atoms. The idea of using the diffraction pattern to measure the atomic spacing came from the Australian-born physicist Sir William Lawrence Bragg (1890–1971), who, working with his father William Henry Bragg (1862–1942), realised that crystals can be considered as consisting of many planes oriented along different directions (see Figure 11.4.1). Bragg realised that X-rays could penetrate the crystal structure and be reflected from a set of parallel planes (see Figure 11.4.2). Constructive In 1912, Bragg interference is produced for certain reflection angles. showed that the wavelength λ of X-rays was related to the spacing d between planes by modifying the double slit constructive interference equation as follows: nλ = 2d sin θ where n is an integer and θ is the angle of incidence measured between the X-ray beam and the crystal plane (see Figure 11.4.3). This is the Bragg law that governs all modern X-ray diffraction. This pioneering work has now become a standard method of determining the crystal structure of materials.

Figure 11.4.1 The different planes of a cubic crystal

Figure 11.4.2 Crystal lattices consist of many parallel planes in many directions that can reflect X-rays.

incident angle

θ

θ

reflection angle

θ θ λ

d

d sin θ

208

Figure 11.4.3 The parallel planes reflect X-rays to produce constructive interference according to the Bragg law.

from ideas to Implementation

Which came first: λ or d ?

From measurements of the diffraction angle θ, the Braggs were able to determine the wavelength λ of X-rays. This needed an accurate knowledge of the spacing d between planes in the crystal. However, this spacing was not known accurately, since there was no method available to measure it. How can you measure λ without knowing d and vice versa? The Braggs solved this dilemma by using the diffraction pattern to determine the arrangement of the atoms, without the need to know the space between them. For example, it is possible to know that a crystal has a cubic arrangement from a Laue diffraction pattern, without the need to know the spacing of atomic planes Knowledge of the lattice geometry, the density or the wavelength of X-rays. of the crystal (mass/volume) and the mass (in grams) of the individual atoms enables the spacing between atoms to be calculated accurately. This calculated spacing d can now be used with the X-ray diffraction angle θ, to determine the wavelength λ from the Bragg diffraction equation. Today, we simply use the database of known crystal lattice spacings to determine X-ray wavelengths.

Checkpoint 11.4 1 2 3

Outline the Braggs’ contribution to the understanding of X-rays. State the Bragg law. Outline how the Braggs determined the spacing of the atoms on crystals.

11.5 Electrical conductivity and the crystal structure of metals X-ray diffraction has shown that the atoms of most metals exist in one of three types of crystal lattices (see Figure 11.5.1). The crystal structure of metals can be viewed as a lattice of positive ions surrounded by a ‘sea’ of nearly free electrons, which makes metals such good electrical conductors. The binding mechanism in metals is the attractive force between the positive ions and the electron gas. Remarkably, quantum physics predicts that there should be little or no resistance to the motion of electrons in a perfect crystal lattice since electrons behave like waves propagating through it. That is, the perfect regularity of the crystal enables the electrons to travel unimpeded, as a wave through the crystal.

a

b

Identify that metals possess a crystal lattice structure. Describe conduction in metals as a free movement of electrons unimpeded by the lattice. Identify that resistance in metals is increased by the presence of impurities and scattering of electrons by lattice vibrations.

c

Figure 11.5.1 The structure of most metallic crystals can be (a) body-centred cubic, (b) face-centred cubic, or (c) hexagonal close-packed. 209

11

Superconductivity

Anything that disturbs the regularity of the lattice results in electron collisions and contributes to the electrical resistance of the crystal. In practice, metals do show electrical resistance, as evidenced by the increase in temperature of the metal wire in an electric heater or the heating element on a stove. Resistance in metals originates from collisions of electrons with irregularities in the crystal lattice. These can be caused by lattice vibrations, or impurities (a foreign atom substituted for one in the crystal lattice) and defects of the lattice (such as a missing atom) (see Figure 11.5.2). The crystal lattice of all metals above a temperature of  0 K consists of waves of lattice vibrations known as phonons. A phonon colliding with an electron causes it to lose energy and thus contributes to electrical resistance. Real metal wires consist of many small crystals joined together and separated by irregular boundaries. The boundaries also serve as places where electron collisions take place and thus contribute to electrical resistance. a

b

c

Figure 11.5.2 Electrical resistance is caused by electron collisions due to crystal lattice (a) impurities, (b) defects and (c) vibrations.

The Kelvin temperature scale

T

he kelvin temperature scale differs from the Celsius scale by the following expression: kelvin = Celsius + 273.15

  The lowest possible temperature is 0 kelvin (or simply 0 K), which is –273.15°C. The kelvin is named after the British physicist William Thomson (1824–1907), who later was given the title Lord Kelvin. In 1900 he is reported to have said, ‘there is nothing new to be discovered in physics now. All that remains is more and more precise measurement.’ This was before the discovery of superconductivity, relativity, quantum physics, and all of the modern physics that has led to a radical transformation of our society through technology and our understanding of nature and the universe. Even great scientists can be short sighted.

Figure 11.5.3 William Thomson (Lord Kelvin)

Checkpoint 11.5 1 2 3 4

210

Outline the significance of X-ray diffraction to the structure of metals. Explain the effect on electrical resistance of irregularities that are introduced into a crystal. Give reasons why resistance in metals does not match the near-zero resistance predicted for crystal structures. Outline the role of a phonon in electrical resistance.

from ideas to Implementation

11.6 The discovery of superconductors 0.15 Resistance (Ω)

The phenomenon of superconductivity, in which the electrical resistance of certain materials completely vanishes at low temperatures, is one of the most interesting and sophisticated in condensed matter physics. It was first discovered by the Dutch physicist Heike Kamerlingh Onnes (1853–1926), who was the first to liquefy helium (which boils at 4.2 K at standard pressure). In 1911 Kamerlingh Onnes discovered the phenomenon of superconductivity while studying the resistance of metals at low temperatures. He studied mercury because very pure samples could easily be prepared by distillation. The historic measurement of superconductivity in mercury is shown in Figure 11.6.1. The electrical resistance of mercury decreased steadily when it was cooled, but dropped suddenly to zero at 4.2 K. Soon after this discovery, many other elemental metals were found to have zero resistance when their temperatures were lowered below a certain temperature that is characteristic of the material. This is called the critical temperature Tc, some of which are given in Table 11.6.1.

Hg 0.10

0.05

0.00 4.1

4.2 4.3 Temperature (K)

4.4

Figure 11.6.1 The resistance of mercury as measured by Kamerlingh Onnes

Table 11.6.1  Some superconductors, their critical temperatures and critical magnetic fields Material

Critical temperature (kelvin)

Critical magnetic field strength (tesla)

Type I

Tungsten

0.02

0.0001

(elements)

Titanium

0.4

0.0056

Aluminium

1.18

0.0105

Tin

3.72

0.0305

Mercury (α)

4.15

0.0411

Lead

7.19

0.0803

Type II

Nb–Ti alloy

(compounds and alloys)

Nb–Zr alloy

10.8

11

PbMo6S8

14.0

45

V3Ga

16.5

22

10.2

Process information to identify some of the metals, metal alloys and compounds that have been identified as exhibiting the property of superconductivity and their critical temperatures.

12

Nb2Sn

18.3

22

Nb3Al

18.9

32

Nb3Ge

23.0

30

Type II

YBa2Cu3O7

92

(high-temperature ceramic compounds)

Bi2Sr2Ca2Cu3O10

110

Tl2Ba2Ca2Cu3O10

125

HgBa2Ca2Cu3O8

135

Too high to measure, typically ~200 (estimated)

Superconductivity was an unexpected phenomenon. As shown in Figure 11.6.2, it was expected that by cooling a conductor the lattice vibrations (phonons) will be gradually reduced in amplitude. The reduction in lattice vibrations also reduces the number of collisions of electron with the crystal lattice and therefore reduces the electrical resistance. One would expect the resistance to gradually decrease to very low values at a temperature of 0 K. This is why it was surprising to find that the resistance dropped to zero at a relatively high temperature.

Electrical resistance

Class

superconductor

normal metal

0

TC

Temperature (K)

Figure 11.6.2 The resistance of a normal conductor and a superconductor 211

11

Superconductivity

Checkpoint 11.6 1 2 3

Recall how superconductivity was discovered. Define critical temperature. Compare the expected measurements of resistance as temperature is reduced with the experimental results.

11.7 The Meissner effect T Tc Ba 0

Ideal conductor

Superconductor

Figure 11.7.1 A thought experiment that illustrates superconductors are not the same as perfect conductors

In 1933, Walter Meissner and Robert Oschenfeld discovered that superconductors expel magnetic fields from their interiors in a way that is different from the behaviour expected of hypothetical perfect conductors. Figure 11.7.1 illustrates a thought experiment that highlights this difference. Imagine that both the ideal conductor and superconductor are above their critical temperature Tc; that is, they both are in a normal conducting state and have electrical resistance. A magnetic field Ba, is then applied, which penetrates both materials. Each sample is then cooled to below its critical It is found that temperature, so that they both have zero resistance. the superconductor expels the magnetic field from inside it, while the ideal conductor maintains its interior field. Note that the energy needed by the superconductor to expel the magnetic field comes from the superconducting transition, which is exothermic. Switching off the magnetic field induces currents in the ideal conductor that prevent changes in the magnetic field inside it, as stated by Lenz’s law (Module 2 ‘Motors and generators’). However, the superconductor returns to its initial state; that is, it has no magnetic field inside or outside it.

Checkpoint 11.7 Explain what happens to a magnetic field passing through an ideal conductor and a superconductor when the conductors are cooled to below their critical temperatures.

11.8 Type-I and type-II superconductors High magnetic fields destroy superconductivity and restore the normal conducting state. Depending on the character of this transition, we may distinguish between type-I and type-II superconductors. The graph in Figure 11.8.1 illustrates changes in the internal magnetic field strength Bi (the field inside the superconductor) with increasing applied magnetic field. It is found that the internal field is zero (as expected from the Meissner effect) until a critical magnetic field Bc is reached, at which a sudden transition to the normal state occurs. This results in the penetration of the applied field into the interior. Superconductors that undergo this abrupt transition to the normal state 212

Checkpoint 11.8 1 2

Describe the significance of internal magnetic fields and critical magnetic fields. Distinguish between type-I and type-II superconductors.

Bc External field Ba

Figure 11.8.1 Type-I superconductors abruptly become normal conductors at field strengths above a critical magnetic field.

Internal field Bi

above a critical magnetic field strength are known as type-I superconductors. Most of the pure elements listed in Table 11.6.1 tend to be type-I superconductors. Type-II superconductors, on the other hand, respond differently to an applied magnetic field (see Figure 11.8.2). These superconductors have two critical field strengths, Bc1 and Bc2. As field strength is increased from zero, there is no change in the internal magnetic field of the superconductor until Bc1 is reached. At this field strength, the applied field begins to partially penetrate the interior of the superconductor. However, the superconductivity is maintained at this point. Superconductivity vanishes above the second, much higher, critical field Bc2. For applied fields between Bc1 and Bc2, the applied field is able to partially penetrate the superconductor, so the Meissner effect is incomplete and the superconductor is able to tolerate very high magnetic fields. Type-II superconductors are the most technologically useful because the second critical field can be quite high, enabling high field strength electromagnets Most compounds listed in Table to be made out of superconducting wire. 11.6.1 are type-II superconductors. Wires made from, for example, niobium–tin (Nb3Sn) have a Bc2 as high as 24.5 tesla, though in practice it is lower. There is a misconception among some non-specialists that the term type II refers to the copper oxide based high-temperature superconductors discovered in the late 1980s. Although these are type-II superconductors, so are many superconductors discovered before that time.

Internal field Bi

from ideas to Implementation

Bc1

Bc2

External field Ba

Figure 11.8.2 Type-II superconductors have

11.9 Why is a levitated magnet stable? Figure 11.9.1 shows a spectacular demonstration of the Meissner effect in which a small permanent magnet floats on top of a high critical temperature superconductor (YBa2Cu3O7) cooled with liquid nitrogen (at 77 K). It demonstrates the repulsion of the magnetic field by the superconductor and thus the levitation of the magnet. Eddy currents are created on the surface of the superconductor, and, consistent with Lenz’s law, these essentially produce a magnetic field that mirrors the field of the magnet, resulting in the repulsion and subsequent levitation of the magnet (see Figure 11.9.2). In reality, when the magnet is first placed over a small piece of superconductor it is unstable and falls off to the side. This is because the magnet will float over its ‘mirror image’ provided that image can keep moving with it. The superconductor is small and cannot produce a satisfactory magnetic field image near its edge, which results in ineffective repulsion. So why does a levitating permanent magnet remain stable on top of a small superconductor? Even a little nudge causes the magnet to spring back to its original position as if somehow tied by invisible springs to that point. To explain

a partial penetration of the magnetic field between two critical fields.

Figure 11.9.1 A permanent magnet levitates above a superconductor due to the Meissner effect. 213

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induced shielding current

permanent magnet high-temperature superconductor image of permanent magnet

Figure 11.9.2 Eddy currents on the surface of the superconductor essentially create a mirror image of the magnet resulting in repulsion and levitation.

this, we need to expose a little secret used when demonstrating this levitation experiment. If the magnet is lightly placed over a newly cooled high-temperature superconductor, you should find that the magnet does not stay levitated for long and falls off very quickly. To get the magnet to stay, hold the magnet over the superconductor and, rather than letting it go, thrust it slightly towards the superconductor. This is a subtle movement and usually goes unnoticed by the audience. Now release the magnet and it will remain there stably. Incredibly, if the magnet is then removed and dropped back over the superconductor, it levitates stably without the need to thrust the magnet towards the superconductors. It is as if the superconductor has ‘remembered’ that the magnet was there. Moving the magnet back and forth parallel to the surface of the superconductor or allowing the superconductor to warm up above Tc and then cooling it down again will make the levitation of the magnet unstable once more. The magnet must again be thrust towards the superconductor to achieve stability. This is explained in the following section.

Vortex states and flux pinning

a

Ba

b

600 Å

Figure 11.9.3 (a) Illustration and (b) photograph of a regular array of normal conducting regions (dark areas) in a type-II superconductor where the field penetrates the material. 214

Stable levitation of a permanent magnet above a small flat superconductor only occurs with type-II superconductors. Certainly levitation occurs when using type-I superconductors but with a type-II superconductor the levitation is particularly stable and robust. The answer lies in the properties of type-II superconductors for an applied magnetic field between the two critical fields Bc1 and Bc2. Recall that for type-II superconductors, there is partial penetration of the magnetic field at field strengths between Bc1 and Bc2. This partial penetration is in the form of a regular array of normal conducting regions, as illustrated in Figure 11.9.3a. Techniques have been developed to photograph these regions, which are shown as a regular array of dark areas in Figure 11.9.3b. These normal regions allow the penetration of the magnetic field in the form of thin filaments, usually called vortices. The vortices are aptly named because each is a ‘vortex’ or swirl of electrical current that is associated with this state (see Figure 11.9.3a). While in the vortex state, the material surrounding these normal regions can have zero resistance and partial flux penetration. Vortex regions are essentially filaments of normal conductor (non-superconducting) that run through the sample when an external applied magnetic field exceeds the lower critical field Bc1. As the strength of the external field increases, the number of filaments increases until the field reaches the upper critical value Bc2; the filaments then crowd together and join up so the entire sample becomes a normal conductor. One can view a vortex as a cylindrical swirl of current surrounding a cylindrical normal-conducting core that allows some flux to penetrate the interior of type-II superconductors. Thrusting a permanent magnet towards a type-II superconductor will cause the applied magnetic field at the superconductor to be within the region of the two critical fields, which creates the vortex state. In principle, the motion of a levitating permanent magnet will cause these vortices In practice, real materials (such as high critical temperature to move. superconductors) have defects (missing or misplaced atoms, impurity atoms) in their crystal lattices. They are also composed of many crystals, all bound together, resulting in many crystal boundaries. The crystal defects and boundaries stop the motion of the vortices, which is known as flux pinning. This provides the stability of a levitating magnet. Pinning the motion of its magnetic field lines also means stopping the motion of the magnet. Flux pinning can only occur in type-II superconductors.

from ideas to Implementation

Checkpoint 11.9 1 2 3 4

Explain what effect is being demonstrated by Figure 11.9.1. Explain the contribution of eddy currents to the levitation of a magnet over a superconductor. What are vortices? Explain how thrusting the magnet towards the superconductor increases the stability of the levitation.

11.10 BCS theory and Cooper pairs According to classical physics, part of the resistance of a metal is due to collisions between free electrons and the crystal lattice’s vibrations, known as phonons. In addition, part of the resistance is due to scattering of electrons by impurities or defects in the conductor. As a result, the question arose as to why this doesn’t happen in superconductors. A microscopic theory of superconductivity was developed in 1957 by John Bardeen, Leon Cooper and J. Robert Schrieffer, and is known as the BCS theory (after their initials). The central feature of this theory is that two electrons in the superconductor are able to form a bound pair called a Cooper pair if they somehow experience an attractive interaction between them. At first this notion seems counterintuitive since electrons normally repel one another because of their like charges. An explanation of the formation of Cooper pairs relies heavily on quantum physics; but here we present a classical picture of their formation (shown in An electron passes through the lattice Figure 11.10.1) and an explanation. and at some point the positive ions are attracted to it, causing a distortion in their nominal positions. The second electron (the Cooper pair partner) is attracted by the positive charge of the displaced ions. This second electron can only be attracted to the lattice distortion if it comes close enough before the ions have had a chance to return to their equilibrium positions. The net effect is a weak delayed attractive force between the two electrons. This short-lived distortion of the lattice is sometimes called a virtual phonon because its lifetime is too short to propagate through the lattice like a wave, as a normal phonon would. From the BCS theory, the total linear momentum of a Cooper pair must be zero. This means that the electrons travel in opposite directions, as shown in Figure 11.10.1. In addition, the nominal separation between the Cooper pair (called the coherence length) ranges from hundreds to thousands of ions! If electrons in a Cooper pair were too close, such as only a couple of atomic spacings apart, then the electrostatic (coulomb) repulsion would be much larger than the attraction from the lattice deformation and they would repel each other, A current flowing in a and there would be no superconductivity. superconductor just shifts the total moment slightly from zero so that, on average, one electron in a Cooper pair has a slightly larger momentum magnitude than its partner. They do, however, still travel in opposite directions. The interaction between electrons in a Cooper pair is transient. Each electron in the pair goes on to form a Cooper pair with another electron, and this process continues with the newly formed Cooper pairs so that each electron

Describe the occurrence in superconductors below their critical temperature of a population of electron pairs unaffected by electrical resistance. Discuss the BCS theory.

a e–

b + + e– + +

c + + + +

e–

Figure 11.10.1 Classical description of the coupling of a Cooper pair. (a) The first electron approaches a section of the lattice and (b) deforms part of the lattice electrostatically. (c) A second electron is attracted to the net positive charge of this deformation.

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PHYSICS FEATURE High-temperature superconductors: The exceptions to the rule

I

n 1986 a class of materials was discovered by Bednorz and Müller that led to the superconductors we use today on a bench top with liquid nitrogen to cool them. Bednorz and Müller received the Nobel Prize in 1987 for this work (the fastest ever recognition by the Nobel committee). The material we use mostly in school science labs is the yttrium–barium– copper oxide compound YBa2Cu3O7, otherwise known as the 1-2-3 superconductor. It is classified as a high-temperature (Tc) superconductor. The critical temperatures of some high-temperature superconductors Figure 11.10.2 Bednorz and Müller discovered high-temperature superconductors. are given in Table 11.6.1. Critical temperatures as high as 135 K have been achieved. This has made experiments on superconductivity more accessible, since these need only to be cooled by liquid nitrogen (with Cu-O chains a boiling point of 77 K), which is cheap and readily available. This is in contrast to the expensive and bulky equipment that uses liquid helium for cooling the traditional types of superconductors. The crystal lattice structure of YBa2Cu3O7 is shown in Figure 11.10.3. Cu(2) Unlike traditional superconductors, conduction mostly occurs in the planes Y containing the copper oxide. It has been found that the critical temperature Ba CuO2 is very sensitive to the average number of oxygen atoms present, which can Cu layer O vary. For this reason the formula for 1-2-3 superconductor is sometimes given as YBa2Cu3O7 – δ where δ is a number between 0 and 1. The nominal distance between Cooper pair electrons (coherence length) in these superconductors can be as short as one or two atomic spacings. Cu(1) As a result, the electrostatic repulsion force will generally dominate at these distances, causing electrons to be repelled rather than coupled. For this reason, in these materials it is widely accepted that Cooper Figure 11.10.3 The crystal structure of YBa2Cu3O7, pairs are not caused by a lattice deformation, but may be associated with a high-temperature the type of magnetism present (known as antiferromagnetism) in the copper superconductor oxide layers. This means that high-temperature superconductors cannot be explained by the BCS theory, since that mainly deals with lattice deformations mediating the coupling of electron pairs. The research continues into the actual mechanism responsible for superconductivity in these materials.

216

from ideas to Implementation goes on to form Cooper pairs with other electrons. The end result is that each electron in the solid is attracted to every other electron, forming a large network of interactions. Causing just one of these electrons to collide and scatter from atoms in the lattice means the whole network of electrons must be made to collide into the lattice, which is energetically too costly. The collective behaviour of all the electrons in the solid prevents any further collisions with the lattice. In this case, the Nature prefers situations that spend a minimum of energy. minimum energy situation is to have no collisions with the lattice. A small amount of energy is needed to destroy the superconducting state and make it normal. This energy is called the energy gap. In addition to having a linear momentum, each electron behaves as if it is spinning. This property, not surprisingly, is called spin. (The electron is not The BCS theory requires actually spinning, but behaves as though it does.) that the spins of Cooper pair electrons be in opposite directions.

Checkpoint 11.10 1 2 3

Describe how classical physics explains resistance in metals. Outline how Cooper pair electrons form. Describe what is meant by ‘the total linear momentum of a Cooper pair must be zero’.

11.11 Applications of superconductors The main advantage of superconductors is that there is no heat lost when passing current through them, which means that currents can be made to persist indefinitely. For example, a superconducting electromagnet is made such that external power is applied for only a very short time. The electromagnet is then formed into a closed loop that enables the current (and field) to persist as long as the superconductor stays below its critical temperature; that is, the external power supply can be switched off! The main disadvantage of superconductors is that they must be maintained at very low temperatures. This requires specialised vessels known as cryostats, which contain the cooling fluids such as liquid nitrogen and liquid helium. The brittle nature of high-temperature superconductors has limited their applicability. As a result, most superconducting applications today still use the more traditional superconductors such as niobium-tin (Nb3Sn), which can be made into flexible wires. The disadvantage is that they require cooling with liquid helium, which is much more expensive than liquid nitrogen and must be contained in more sophisticated cryostats.

Discuss the advantages of using superconductors and identify limitations to their use.

Medical applications The first large-scale commercial application of superconductivity was in magnetic resonance imaging (MRI). This is a non-intrusive medical imaging technique that creates a two-dimensional picture of, for example, tumours and other abnormalities within the body or brain. This requires a person to be placed inside a large and uniform electromagnet with a high magnetic field. Although normal electromagnets can be used for this purpose, their resistance would

Figure 11.11.1 Patients are located inside the bore of the superconducting magnet of an MRI machine. 217

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dissipate a great deal of heat and have large power requirements. Superconducting magnets, on the other hand, have almost no power requirements apart from that required for cooling. Once electrical current flows in the superconducting wire, the power supply can be switched off because the wires can be formed into a loop and the current will persist indefinitely, as long as the temperature is kept below the transition temperature of the superconductor. Superconductors can also be used to make a device known as a superconducting quantum interference device (SQUID). This device is extremely sensitive to small magnetic fields and can detect magnetic fields from the heart (10–10 tesla) and even the brain (10–13 tesla). For comparison, the Earth’s magnetic field is about 10–4 tesla. As a result, SQUIDs are used in non-intrusive medical diagnostics of the brain. Process information to discuss possible applications of superconductivity and the effects of those applications on computers, generators and motors and transmission of electricity through power grids.

Figure 11.11.2 ITER, a proposed test reactor for future clean energy production using nuclear fusion

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Scientific research The traditional use of superconductors has been in scientific research requiring high magnetic field electromagnets. One application of powerful superconducting electromagnets is in high-energy particle accelerators, such as the Large Hadron Collider at CERN (see section 15.4), in which beams of protons and other particles are accelerated to almost the speed of light and made to collide with each other to create more elementary particles. It is expected that this research will answer fundamental questions such as those about the origin of the mass of matter that makes up the universe. A future use of superconducting electromagnets is in nuclear fusion energy generation using plasmas. A plasma is a fully ionised gas that is obtained by heating it to millions of degrees and trapping it inside a toroidal structure known as a tokamak by large electromagnets. The nuclei of the ions fuse together, producing energy. The operating gases of such reactors are deuterium and tritium, the isotopes of hydrogen. Deuterium is abundant in water, but tritium will be made inside the tokamak as a by-product of fusion reactions. There is no long-term radioactive waste with this process, which is why it is known as clean nuclear energy. Currently an international research plasma reactor—the International Thermonuclear Experimental Reactor (ITER)—is being built. The aim of the project is to demonstrate that energy production is possible with this method. A diagram of the projected reactor is shown in Figure 11.11.2. The magnetic field coils on the ITER will be made from superconductors and will need to only be powered once; the current through them will be sustained indefinitely, as long as the coils are kept cool.

Levitating trains Magnetic levitation (maglev) trains have been built that use powerful electromagnets made from superconductors. The superconducting electromagnets are mounted on the train and kept cool with liquid helium. As shown in Figure 11.11.3, normal electromagnets on a guideway beneath the train repel (or attract) the superconducting electromagnets to levitate the train while pulling it forwards. The superconducting electromagnets rely on the conventional ‘like-pole’ repulsion—not the Meissner effect—to achieve levitation.

from ideas to Implementation Although such a superconducting maglev train has been built and has demonstrated a top speed of 581 km h–1, there are several issues that limit its widespread commercial use. A strong magnetic field inside the train will exclude passengers with pacemakers or devices that have magnetic data storage including computers and credit cards. The powered conventional electromagnets on the guideway that levitate and propel the train are expensive to run over long distances, so alternative propulsion schemes may have to be used.

Power generation, transmission and storage Superconductors have the potential to make electricity generation and transmission more efficient, which will reduce energy costs and the emission of greenhouse gases. Electricity generators used in coal-fired, nuclear and hydroelectric power plants use electromagnets, which heat up due to the resistance of their wires. Replacing these with superconducting wires will at least halve the amount of power lost in the electromagnets, even when the energy cost of making the liquid helium or nitrogen to keep the superconducting electromagnets in the superconducting state is taken into account. The transmission of power to homes and businesses is carried out by high-voltage transmission lines. A high voltage enables a small current to be passed through the transmission lines, to minimise the amount of resistive heating in the wires. Nevertheless, there is some heating of the transmission lines and a substantial energy loss associated with it. Superconducting transmission lines will not suffer from such heating losses. Moreover, problems associated with high-voltage leakage of power by ionisation of the air can be overcome by reducing the voltage on the transmission lines and increasing the current through them. Energy production at, say, coal-fired power plants varies, depending on the anticipated demand for electricity. At the moment, any overproduction of energy is stored by pumping water to higher levels in a large reservoir. Releasing this water into hydroelectric generators enables this stored energy to be retrieved as electricity. The problem with this method is that it is highly inefficient. A third of the excess energy is needed to operate the pumps to the reservoir. Superconductors offer the possibility of storing an electrical current indefinitely in superconducting rings. This current could be retrieved at any time, provided the rings remain in the superconducting state. This technology may also solve problems such as the variability of supply by solar energy generation. Night-time power could be obtained from daytime storage of excess solar power.

superconducting levitation magnet

superconducting propulsion magnet

guideway propulsion magnet vehicle

gliding skid guidance and braking levitation and propulsion magnet

armature windings (iron core) guideway

Figure 11.11.3 Levitation of a train using onboard superconducting magnets on a guideway that propels it with conventional electromagnets

PRACTICAL EXPERIENCES Activity 11.1

Activity Manual, Page 100

Checkpoint 11.11 1 2

List the advantages and disadvantages of superconductors. State an application of superconductors and explain why it is an improvement on existing technology.

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PRACTICAL EXPERIENCES CHAPTER 11

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual. Perform an investigation to demonstrate magnetic levitation. Analyse information to explain why a magnet is able to hover above a superconducting material that has reached the temperature at which it is superconducting. Gather and process information to describe how superconductors and the effects of magnetic fields have been applied to develop a maglev train. Process information to discuss possible applications of superconductivity and the effects of those applications on computers, generators and motors and transmission of electricity through power grids.

Activity 11.1: Applications of superconductivity Magnetic levitation is the ability to control magnetic repelling forces in order to balance objects above each other. You will demonstrate that stable levitation of a permanent magnet can be achieved. Equipment: 2 circular magnets with holes in the centre, superconductor kit (optional), liquid nitrogen. Discussion questions 1 Describe the orientation of the poles of the magnets in order for levitation to occur. 2 Describe your attempts at achieving levitating magnets without the supporting rod through the centre of the magnets. Discuss how this might differ for a levitating magnet over a superconductor.

Figure 11.12.1 Set-up to have magnets levitate

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Chapter summary •

•

• • •

• • •

An X-ray diffraction pattern results from the interference of X-rays to form a pattern of dots on photographic film that can then be used to deduce the crystal structure of a solid. The Bragg law relates the X-ray wavelength λ to the spacing d between crystal planes and the diffraction angle θ by the equation: nλ = 2d sin θ where n is an integer. Resistance in metals originates from collisions of electrons with irregularities in the crystal lattice. Crystal lattice vibrations are known as phonons. Superconductivity is the total disappearance of electrical resistance below a critical temperature Tc that depends on the material of the superconductor. The Meissner effect is the expulsion of the magnetic field from the interior of a superconductor. Magnetic fields can destroy superconductivity and restore the normal conducting state. Type-I superconductors undergo an abrupt transition to the normal state above a single critical magnetic field.

•

•

•

•

• •

•

from ideas to Implementation

Type-II superconductors have two critical magnetic fields. The transition to the normal state occurs above the higher critical field. Type-II superconductors form regions of normal conductivity around which there are circulating electrical currents, known as vortices. Magnetic field lines penetrate the interior of a type-II superconductor through the normal regions of the vortices. The BSC theory explains superconductivity by the coupling of electron pairs, known as Cooper pairs, through an interaction with a lattice phonon. The energy gap is the minimum energy required to destroy the superconducting state. Applications of superconductors include electromagnets for MRI machines, and superconducting quantum interference devices (SQUIDs) for the detection of very small magnetic fields. Superconductors have potential applications in fusion energy research, levitating trains and power generation and transmission.

Review questions Physically Speaking

Reviewing

The items in the columns are not in their correct order. Match each of the key concepts with its closest definition.

1 Outline the work carried out by the Braggs in

Concept

Definition

Superconductivity

The point below which there is zero resistance The array of dots on a photographic film created by X-ray diffraction A burst of lattice vibrations The state of matter in which electrical resistance is zero Allows a magnetic field to penetrate while maintaining the superconducting state The mathematical relationship between X-ray wavelength and crystal lattice spacing The temporary attraction between electrons mediated by a lattice deformation Expels magnetic fields from the interior The regular arrangement of atoms in a solid

Crystal lattice Laue pattern Phonon Bragg law

Tc

Meissner effect

Type-II superconductor Cooper pairs

understanding the wavelength of X-rays.

2 Describe the property of crystals that make them useful in understanding X-rays.

3 Explain the difference between constructive and destructive interference.

4 Outline how the approximate wavelength of X-rays was found. State the value achieved.

5 State the prediction by quantum physics of resistance in crystal structures and explain why these are not always seen.

6 Describe the role of phonons in superconductivity. 7 Outline the contributions of Kamerlingh Onnes to the understanding of superconductivity.

8 Explain how the drop in temperature allows for zero resistance.

9 Describe the Meissner effect.

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10 State the difference between a conductor and a superconductor when a magnetic field is applied and the temperature is reduced beyond the critical temperature.

Solving Problems 19 Calculate the separation of diffraction grating slits when a helium–neon laser (λ = 633 nm) is shone on it. The angular position of the second-order bright band is 30°.

11 Explain why the superconducting transition is exothermic.

12 Describe why a magnet levitates above a superconductor that is below its critical temperature.

13 Explain why there is the formation of eddy currents on the surface of superconductors.

14 Explain what is meant by a Cooper pair and describe

20

a Identify the critical temperature of each metal in Figure 11.12.2. b Identify which of the two plots (diamonds or squares) shows a material with easily obtainable superconducting properties. State your reasons.

15 Outline an energy argument as to why electrons travel through a superconductor unimpeded when it is below its critical temperature.

16 Outline the BCS theory. 17 Create a table to compare the advantages and

Resistance (Ω)

how they are formed.

1000 800 600 400 200 0

0

disadvantages of superconductors.

Re

222

iew

Q uesti o

n

s

v

18 List applications of superconductors in everyday life.

50

100

150

Temperature (K)

Figure 11.12.2

200

250

from ideas to Implementation

PHYSICS FOCUS

H3. Assesses the impact of particular advances in physics on the development of technologies

Semiconductors to Superconductors

H5. Identifies possible future directions of physics research

1950s

1960s

1970s

1980s

Silicon transistor

TTL Quad gate

8-bit Microprocessor

32-bit Microprocessor

1 transistor

16 transistors

4500 transistors

275 000 transistors

1990s

2000s

32-bit Microprocessor 64-bit Microprocessor

3 100 000 transistors

592 000 000 transistors

Figure 11.12.3 Moore’s law states that the number of transistors on a chip doubles every 2 years. Since the introduction of the commercial transistor in the late 1940s, the size and processing ability of computer chips has changed dramatically. According to Moore’s law, the number of transistors on a chip doubles every 2 years. But there is a limit—the physical size of atoms. In the distant future, a new generation of computers that make use of quantum physics may overcome the size limitations of current (classical) computer technology. Quantum computers allow for processes to happen simultaneously rather than sequentially as in classic computers. For example, to factorise a 400 digit number would take a quantum computer a few minutes, but it would take current computers billions of years to do the same calculation. Another distant technology that may speed up our current limitations on computation speed are the superconducting switching devices known as Josephson junctions. Although not the only technology being researched for future computers, this is one to watch.

1 Define what a semiconductor is. 2 Explain the significance of doping semiconductors. 3 State the significance of semiconductors in the computing industry. 4 Outline the advantages of silicon over the previously used germanium. 5 Explain why the p–n junction is important to modern electronics. 6 Discuss the impacts the p–n junction has had on society. 7 Analyse the implication of a limit to the size of silicon chips. 8 Discuss the need to involve quantum mechanics in the next generation computers. 9 Describe why quantum computers are so much faster than standard computers.

Research 10 Research and outline how the Josephson junction works. 11 Research and describe the possible technologies that can take over from semiconductors. 12 Outline the role of nano-carbon tubes in speeding up processing ability.

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3

The review contains questions in a similar style and proportion to the HSC Physics examination. Marks are allocated to each question up to a total of 25 marks. It should take you approximately 45 minutes to complete this review.

Multiple choice (1 mark each) 1 Predict the direction of the electron in Figure 11.13.1 as A B C D

I

II

III

Conductor Insulator Insulator Semiconductor

Insulator Conductor Semiconductor Conductor

Semiconductor Semiconductor Conductor Insulator

The graph in Figure 11.13.3 shows how the resistance of a material varies with temperature. Identify each of the parts labelled on the graph.

A B C D

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Figure 11.13.1

An electron in a magnetic field

The diagrams in Figure 11.13.2 represent semiconductors, conductors and insulators. The diagrams show the conduction and valence bands, and the energy gaps. Which answer correctly labels each of the diagrams?

A B C D

3

–

I

II

III

Critical temperature Superconductor material Critical temperature Normal material

Superconductor material Critical temperature Normal material

Normal material

Superconductor material

I

Figure 11.13.2

Normal material Superconductor material Critical temperature

II

II

I

Figure 11.13.3

III

Energy bands

Resistance (Ω)

2

it enters the magnetic field. Straight up Left Right Down

III

Temperature (K)

Resistance varies with temperature

from ideas to Implementation 4

5

Experimental data from black body radiation during Planck’s time showed that predicted radiation levels were not achieved in reality. Planck best described this anomaly by saying that: A classical physics was wrong. B radiation that is emitted and absorbed is quantised. C he had no explanation for it. D quantum mechanics needed to be developed. Figure 11.13.4 shows a cathode ray tube that has been evacuated. Which answer correctly names each of the labelled features?

III

Figure 11.13.4  An evacuated cathode ray tube

C D

6

Explain, with reference to atomic models, why cathode rays can travel through metals.  (2 marks)

7

Outline how the cathode ray tube in a TV works in order to produce the viewing picture.   (2 marks)

8

Give reasons why CRT TVs use magnetic coils and CROs use electric plates in order to deflect the beams, given that both methods work.  (2 marks).

9

In your studies you were required to gather information to describe how the photoelectric effect is used in photocells. a Explain how you determined which material was relevant and reliable. b Outline how the photoelectric effect is used in photocells.  (3 marks)

10

Justify the introduction of semiconductors to replace thermionic devices.  (4 marks)

11

Magnetic levitation trains are used in Germany and Japan. The trains in Germany use conventional electromagnets, whereas the one in Japan uses superconductors. Compare and contrast the two systems.  (3 marks)

12

a b

II

I

A B

Extended response

I

II

III

Striations Faraday’s dark space Crooke’s dark space Cathode

Cathode Striations

Anode Cathode

Anode

Faraday’s dark space Striations

Faraday’s dark space

Determine the frequency of red light, which has a wavelength λ = 660 nm. (Speed of light c = 3.00 × 108 m s–1) Calculate the energy of a photon that is emitted with this wavelength. (Planck’s constant h = 6.63 × 10–34 J s)  (4 marks)

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4 Context

Figure 12.0.1

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Bubble chamber tracks formed by the passage of ionising particles through a liquid that is kept at a temperature above its boiling point. The curved trajectories are the result of charged particles interacting with a magnetic field, and allow information concerning the charge and mass of the particles to be determined.

Quanta to Quarks We all talk about ‘quantum leaps’, but did you know that Max Planck made the first quantum leap in 1901 when he introduced the idea of the quantisation of energy? At the beginning of the 20th century, scientists were confronted by an accumulation of experimental observations and explanations that lacked unification. Black body radiation, the photoelectric effect, radioactivity and the emission of sharp spectral lines by atoms in a gas discharge tube could not be adequately explained within the framework of Newtonian classical physics. A new physics, quantum physics, was born. The story has some inspiring characters and storylines: how Niels Bohr synthesised the works of Planck, Einstein and Rutherford and proposed the now commonly recognised Rutherford–Bohr atomic model, and so provided an explanation for spectral lines; and how Louis de Broglie in 1924 took Planck’s idea, reversed it and proposed a ‘totally crazy idea’ that all matter has wave properties, which, in turn, gave birth to quantum mechanics. The investigations into radioactivity led to the artificial manufacture of elements and the dawn of the atomic age. The development of particle accelerators often referred to as ‘atom smashers’ in the media, led to the discovery of a ‘particle zoo’, as a plethora of new subatomic particles were identified. Today, the quest to understand the building blocks and forces of nature continues, with the building of the Large Hadron Collider (referred to as the LHC), which is designed to produce conditions that mimic the environment present just after the birth of the universe.

Figure 12.0.2 The blue glow in the core of

Inquiry activity A chain reaction

a water-cooler nuclear results from the radiation emitted when energetic charged particles travel faster than light through water.

Obtain some scrap paper! Gather together a crowd of people—your class will do— the more the better! And make sure you have safety glasses for all! Screw up the paper into hundreds of balls, each about the size of a ping-pong ball. Make sure everyone has at least six balls of paper and then gather everyone in close together. These are the rules of the chain reaction game. 1 Safety glasses on at all times! 2 If you are hit by a paper ball you throw two balls of paper high up into the air. Your teacher can lob in the first paper ball. It makes a great video clip! You can also vary the rules. For example, try throwing one paper ball when you are hit rather than two. Have fun!

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12

From Rutherford to Bohr Pieces of a jigsaw!

Rutherford, Bohr, spectra, orbit, spectrum, quantum number, Planck’s constant, quanta, photon, absorption spectra, emission spectra, Balmer series, Rydberg’s constant, transition, stationary state, Zeeman effect

By the beginning of the 20th century a large number of experimental facts had accumulated that could not be explained by existing theories: • the discovery of ordered series in atomic spectra • the photoelectric effect • radioactivity • evidence that the atom had internal structure.

12.1 Atomic timeline

Ernest Rutherford

O

nce, a distinguished stranger, amazed by his unscholarly accent and appearance, mistook him for an Australian farmer.’   At the University of Manchester, Rutherford would proclaim to his recruits that: ‘all science is either physics or stamp collecting.’ ‘

Figure 12.1.1

228

Ernest Rutherford is most well known for his alpha-particle scattering experiments. He was awarded the Nobel Prize for Chemistry in 1908 for his work in investigations into the disintegration of the elements, and the chemistry of radioactive substances.

In the early 20th century many prominent scientists doubted the existence of atoms. Today atomic theory is widely accepted and taught throughout science curricula, forming the foundation upon which scientists, technologists and engineers understand the properties of matter. The ideas underpinning atomic models have changed over time, driven on by the interplay between available technology, theoreticians and experimentalists. The history of the ‘atom’ originates in Greece more 2000 years ago and the quest to reveal its inner structure continues today. The atomic age was born, ushered in by the development and construction of atomic weapons, nuclear reactors and particle accelerators.

quanta to quarks Table 12.1.1 

Atomic timeline

Democritus (c 460 bce–c.370 bce)

He was a Greek philosopher who proposed that there was a limit to how small one could divide matter—the smallest indivisible particle was called an atom (atomos, Greek meaning ‘without slices’ or ‘indivisible’).

Aristotle (384 bce–322 bce) John Dalton (1766–1844)

He criticised Democritus, and proposed a model based upon four elements—earth, air, fire and water. His view held for some 2000 years.

Henri Becquerel (1852–1908) JJ Thomson (1856–1940)

In 1896 he discovered that certain elements emitted radiation and decayed, suggesting that the atom was divisible.

Ernest Rutherford (1871–1937)

In 1911 he proposed the Rutherford planetary model of the atom, based upon the results of Geiger and Marsden’s scattering experiments at the Cavendish Laboratories.

Niels Bohr (1885–1962)

In 1913 Bohr proposed the Rutherford–Bohr model also commonly called Bohr’s model. Bohr provided a set of three postulates to address the issues raised by Rutherford’s earlier model, and this led to the development of a mathematical model to account for the spectra of the hydrogen atom.

Louis de Broglie (1892–1987)

In 1924 de Broglie introduced the concept of matter waves. This concept provided a mechanism for electrons to inhabit a stable orbit by having an integral number of wavelengths fitting around the circumference of the orbit, forming a standing wave.

James Chadwick (1891–1974)

In 1932 he reported the discovery of the neutron. Rutherford some 12 years earlier had proposed its existence, and this discovery completed the constituents of the basic atomic model with which most people are familiar today.

Experimental surprise

R

A Scottish teacher, Dalton in 1801 proposed his atomic model, based upon his studies in chemistry that: • Matter is composed of small indivisible atoms. • Elements contain only one type of atom. • Different elements contain different atoms. • Compounds contain more than one type of atom.

utherford’s radium produced alpha particles, and these massive particles travelled at approximately 1.6 × 10+7m s–1. Rutherford in a later lecture described the extraordinary backscattering of alpha particles as ‘… the most incredible event that has ever happened to me in my life. It was almost as incredible as if you had fired a 15-inch shell at a piece of tissue paper and it came back and hit you’.

In 1904 he proposed the ‘plum pudding’ model of the atom in which electrons were embedded in a positive sphere like ‘plums’ in a pudding. This model was based upon Thomson’s experimental work and his discovery of electrons in 1897.

Checkpoint 12.1 Outline the main atomic models proposed between ancient times and 1913.

~10–10 m

12.2 Rutherford’s model of the atom In 1907, New Zealander Ernest Rutherford (1871–1937) moved to the University of Manchester in England where, with Johannes Geiger (1882–1945) and Ernest Marsden (1889–1970), he continued his earlier work on firing alpha particles at metal foils. They were shocked to find that approximately one alpha particle in every 8000 was deflected by the platinum and gold foils through angles greater than 90º. The Thomson (plum pudding) model of the atom (Figure12.2.1) predicted only small scattering because the atom had no large concentrations of charge or mass to deflect the massive and fast-moving alpha particles. A new atomic model was needed.

–

–

electron

– – – –

Figure 12.2.1

positively charged material

Thomson’s plum pudding atomic model

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Structure

Discuss the structure of the Rutherford model of the atom, the existence of the nucleus and electron orbits.

The alpha particle experiments posed many questions. Rutherford hypothesised that for alpha particles to be deflected as observed, it would require a massive, but tiny, positively charged ‘charge centre’ (nucleus), approximately 10–15 m in diameter with a set of orbiting electrons (like planets orbiting the Sun). From Einstein’s analysis of Brownian motion, the radius of an atom was approximately 10–10 m, meaning that the tiny nucleus contains 99.9% of the mass and the atom is mostly empty space. By 1910 Rutherford had formalised his atomic model with mathematical equations and directed Geiger and Marsden to thoroughly test his model. The series of experiments between 1908 and 1911 provided clear evidence that Thomson’s model was not correct. Interestingly, the scientific community was not persuaded or even interested. The two people who brought Rutherford’s model into mainstream science were Niels Bohr (1885–1962) and a young member of Rutherford’s team Henry Moseley (1887–1915).

Limitations Despite the success of Rutherford’s planetary atomic model in explaining the scattering of alpha particles, the model failed to explain other important questions: • What is the nucleus made of? • How are the orbits of the electrons arranged around the nucleus? • What keeps the negatively charged electrons from losing energy and spiralling into the positive nucleus?

a +–+ – – – + + + –

α α

b nucleus α α α

+

Figure 12.2.2

Scattering deflections predicted by (a) Thomson’s and (b) Rutherford’s atomic models

a

Rutherford proposed that electrical attraction provided the centripetal force to keep electrons in orbit. However, orbiting electrons are accelerating, and Maxwell’s classical electromagnetic theory predicted that accelerating electrons would radiate away their energy in a fraction of a second and spiral into the nucleus, therefore Rutherford’s atomic model was not stable. Additionally, Rutherford’s planetary model could not explain the observed line spectra of excited gases. viewing screen

α–pa

rticle

source containing radon

s

b metal foil

+

α–particle

+ nucleus

Figure 12.2.3

230

The components used by Geiger and Marsden to measure the deflection of alpha particles fired at thin metal-foil targets

quanta to quarks

Checkpoint 12.2 1 2

Explain why Thomson’s ‘plum pudding’ model only predicted small deflections in alpha particles passing through a thin metal foil. Outline the significance of Marsden and Geiger’s scattering experiments to the development of the Rutherford atomic model.

12.3 Planck’s quantised energy Max Planck (1858–1947) in 1901 proposed a theory to model the spectrum of a black body (see section 9.2). This theory dictated that the energy of oscillations of atoms or molecules cannot have just any value; they can only possess a discrete amount of energy that is a multiple of the minimum value related to the frequency of oscillation by the equation:

Discuss Planck’s contribution to the concept of quantised energy.

E = hf Planck’s assumption suggests that the energy of any vibration could only be a whole number multiple of hf: E = nhf where n = 1, 2, 3…, E is energy (in joules), n is called a quantum number, h is Planck’s constant (with a value of 6.63 × 10–34 J s) and f is the frequency in hertz (s–1). This would mean that energy was not a continuous quantity as had been believed, but rather that it was quantised into discrete packets. Interestingly, at the time of this work Planck considered this idea more of a mathematical device to obtain the ‘right answer’ rather than an actual physical reality of nature. In 1905, Einstein extended the concept of quantisation and proposed a new theory of light. Einstein took Planck’s suggestion that the vibrational energy of atoms or molecules in a radiating object was quantised with energy. He argued that the vibrational energy of the atoms or molecules could only change by a multiple of hf and therefore proposed that light would be emitted in discrete packets (quanta) also obeying the equation E = hf. Gilbert Lewis (1875–1946) in 1926 named these discrete packets of light photons (from photos, Greek meaning ‘light’).

ramp

stairs

Figure 12.3.1

This simple analogy shows the difference between continuous and quantised energy states. On the ramp the box can have any amount of potential energy, but the box on a staircase can only have discrete amounts of potential energy.

Worked example Question A photon has an energy of 2.8 eV. a Calculate its energy in joules.

b Calculate the frequency of the photon.

Solution a You will recall from section 10.2 that 1 eV = 1.602 × 10–19 joules

So: 2.8 × 1.602 × 10–19 = 4.5 × 10–19 joules

b Now rearranging E = hf and substituting in the energy and Planck’s constant we obtain: f=

E 4.5 × 10−19 = = 6.8 × 1014 Hz h 6.63 × 10−34

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Quantised Energy?

W

hy don’t we notice that energy comes in discrete packets or quanta in our everyday life? The value of Planck’s constant is very small; therefore, for large everyday objects such as a cricket ball or a car, the energy appears to be a ‘continuous’ quantity. At the molecular and atomic level, however, the quantisation of energy becomes very apparent when energy is absorbed or emitted.

Checkpoint 12.3 1 2

Outline the idea Planck proposed to explain black body radiation. Explain the difference between the use of E = hf  by Planck and Einstein.

12.4 Spectral analysis

spectroscopes

I

n a spectroscope, light from a source passes through a slit and enters the collimator tube. The lens at the end of the collimator makes the light parallel before it illuminates the grating. The diffraction grating is viewed through a small telescope mounted on a rotatable platform so that the angle of the observed colour or spectral line can be measured.

When light is passed through a prism or diffraction grating, the constituent colours present in the light are revealed in the spectrum produced. The prism separates the light using the property of refraction, whereas the diffraction grating uses the property of interference. The traditional device used by scientists to examine spectra is called a spectroscope. Modern instruments are called spectrometers. There are two types of spectra: absorption spectra and emission spectra. • Absorption spectra can be produced by passing white light (a continuous spectrum) through a cool gas. The atoms or molecules in the gas will absorb certain specific wavelengths (colours) of light. The atom that absorbed the light is now in an excited state and will spontaneously emit a photon of light, usually in a different direction. Therefore the original beam of light will now have certain wavelengths depleted, and these will appear as a series of dark lines when observed through a spectroscope. • Emission spectra can be produced when a gas is excited. This can be achieved by heating the gas or by passing an electrical current through a low pressure gas. The light produced when viewed with a spectroscope will often be made up of a series of bright coloured lines. Hydrogen absorption spectrum

source

collimator

slit

lens

grating θ telescope

eye

Figure 12.4.1

232

A spectroscope showing the major components

Hydrogen emission spectrum

400 nm

Figure 12.4.2

700 nm

Absorption and emission spectra of hydrogen

H alpha line 656 nm Transition n = 3 to n = 2

quanta to quarks Ancient peoples have observed the emission colours of a range of materials. A sprinkling of common salt (sodium chloride) into a flame produces an intense golden flame, and during the smelting of copper ore the flames in the furnace are often coloured an intense vivid green. Spectra are a ‘window’ into the hidden atomic structure. Each element has its own unique spectral ‘fingerprint’. Chemists and students of chemistry still use the flame test to identify the constituent elements present in chemical samples. A continuous emission spectrum can be produced by a hot object, forming a rainbow or continuum. For example, if you look at the light emitted from a white-hot piece of metal through a prism or diffraction grating you will see a full rainbow of colours. Anders ångström (1814–1874) was the first to make detailed measurements of the visible spectrum of hydrogen, and in 1885 Johann Balmer (1825–1898) commenced a detailed study of the visible emission spectrum for hydrogen, which is now referred to as the Balmer series. Table 12.4.1  The Balmer series for hydrogen Johannes Rydberg (1854–1919) generalised Balmer’s Spectral line Balmer’s generalised equation equation:  1 1 1 =R  2 − 2  ⎛ 1 1 1⎞ Name Colour λ (nm) λ  n f ni  = R⎜ 2 − 2⎟ λ ⎝ nf ni ⎠ nf ni Hα

Red

where λ is the wavelength of the spectral line, R is Rydberg’s Blue-green Hβ constant (1.097 × 107 m–1), nf is the final state of the electron Blue-violet Hγ and ni is the initial state of the electron. Violet Hδ The emission spectrum in Figure 12.4.2 is the Balmer series emission spectrum for hydrogen corresponding to the values of nf = 2, and ni = 3, 4, 5, 6, 7 in the generalised Balmer equation. The improvements in experimental techniques and technology for detecting infra-red and ultraviolet radiation led to the discovery of additional spectral lines and other spectral series. For example, Theodore Lyman (1874–1954) discovered the first spectral line in the Lyman series in 1906 and took another 8 years to detect the remaining lines. The generalised Balmer equation could be used to calculate accurately the wavelength of the spectral lines for all the newly discovered series.

Worked example Question Using the generalised form of Balmer’s equation, calculate the wavelength of the emitted photon for a transition between n = 5 and n = 3.

Solution Using the generalised form of Balmer’s equation to calculate

1 : λ

 1 1 1  1 1 = R  2 − 2  = 1.097 × 107  2 − 2  = 7.80 × 105 m–1 3 5  λ  n f ni 

Now take the inverse to determine the wavelength of the emitted photon: λ = 1.28 × 10–6 m

656.3

2

3

486.1

2

4

434.0

2

5

410.2

2

6

Solve problems and analyse information using:  1 1 1 = R  2 − 2 λ  n f ni 

The test of time

R

ydberg’s constant R was initially determined empirically from spectroscopy, and then theoretically by Niels Bohr, who used a mixture of classical and quantum ideas. Later its value was again calculated using more fundamental constants in terms of quantum mechanics: R∞ =

mee 4 8εo2h 3c

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E=0 −0.85 −1.5 −3.4

n=4

ionised atom (continuous energy levels)

n=5

n=3 Paschen series

n=2

excited states

Balmer series

Energy (eV)

−5

−10

Figure 12.4.3

Energy level diagram for the hydrogen atom with the Lyman, Balmer and Paschen series emission transitions and associated spectra for the Balmer series

−13.6

ground state

n=1

−15

Lyman series

Table 12.4.2  Spectral series associated with the hydrogen atom Series

Spectral region

nf

ni

Lyman Balmer Paschen Brackett Pfund

Ultraviolet Visible and UV Infra-red Infra-red Infra-red

1 2 3 4 5

2, 3, 4, 5, 6,

Discovered 3, 4, 5, 6, 7,

4, 5, 6, 7, 8,

5 6 7 8 9

… … … … …

1906–1914 1885 1908 1922 1924

Worked example Question a Predict the values for nf and ni for the transition of the second longest wavelength emission in the Lyman series. b Calculate the wavelength for the second-longest wavelength in the Lyman series.

Solution a The second-longest wavelength in the Lyman series will correspond to the secondlowest energy transition. In the Lyman series we know that nf is always 1. Thus ni = 2 will correspond to the lowest energy transition and ni = 3 will correspond to the second-lowest energy transition. Therefore nf =1 and ni = 3. 1 b Using the generalised form of Balmer’s equation to calculate : λ  1 1 1 1 7 1 = R  2 − 2  = 1.097 × 10  2 − 2  = 9.75 × 106 m–1 1 3  λ n n  f



i

Now take the inverse to determine the wavelength of the emitted photon for the transition between ni = 3 and nf = 1. λ = 1.03 × 10–7 m

234

quanta to quarks Worked example Question Using the information in the table, construct an energy level diagram for n = 1 to n = 5 for the hydrogen atom. a On your diagram, draw and label the Lyman, Balmer and Paschen series. b Identify the values of ni and nf for a transition of an electron between energy levels that would absorb the highest frequency photon for the Paschen series, based on the energy values provided.

Principal quantum number (n )

Energy (eV)

1 2 3 4 5

–13.6 –3.4 –1.51 –0.85 –0.54

c Identify the values of ni and nf for the transition of an electron between energy levels that would emit the highest energy photon for the Balmer series.

Solution a Refer to Figure 12.4.3 and construct a similar diagram for the given values n = 1 to n = 5. b From your diagram for part a, we see that for the Paschen series ni = 3 to nf = 5 is the greatest possible jump in energy, therefore this transition would absorb the highest energy photon. c From your diagram for part a, we see that for the Balmer series ni = 5 to nf = 2 is the greatest possible jump in energy, therefore this transition would emit the highest energy photon.

Checkpoint 12.4 1 2 3 4 5 6

Recall the two types of spectra. List the names of five spectral series. Recall the basic process that produces the spectral series. Explain the relationship between the energy of an emitted photon and quantum number n. Calculate the wavelength of a photon emitted by an electron jumping from n = 3 to n = 2. Outline the spectral series that corresponds to a transition from n = 3 to n = 2.

12.5 Bohr’s model of the atom Niels Bohr (1885–1962), a Danish physicist, spent a short time at the Cavendish Laboratory, Cambridge, working with JJ Thomson before joining Rutherford’s team at the University of Manchester in 1912. Bohr focused his attention on the planetary model and was equally impressed by both its successes and its obvious limitations. Bohr set to work and built upon Rutherford’s model by synthesising a mixture of Planck’s quantum theory and classical physics.

Bohr’s postulates In 1913 Bohr announced the revised planetary model of a hydrogen atom based upon the quantisation of energy and angular momentum of the electron. His new Rutherford–Bohr model included a set of three postulates to address the identified limitations of Rutherford’s model.

Figure 12.5.1

Niels Bohr

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From Rutherford to Bohr

Analyse the significance of the hydrogen spectrum in the development of Bohr’s model of the atom. Define Bohr’s postulates.

A Square peg in a Round Hole?

D

espite Bohr showing all the signs of becoming a theorist in an experimentalist laboratory, Rutherford responded to comments by saying, ‘Bohr’s different. He’s a football player!’

Bohr’s postulates for his atomic model 1 Electrons exist in stable orbits. An electron can exist in any of several special circular orbits with no emission of radiation. These orbits are called stationary states. 2 Electrons absorb or emit specific quanta of energy when they transition between stationary states (orbits). In contradiction to classical electromagnetic theory, a sudden transition of an electron between two stationary states will produce an emission or absorption of quantised radiation (a photon), described by the Planck–Einstein relation. 3 Angular momentum of electrons is quantised. An electron in a stationary state (orbit) has a quantised angular momentum that can take only values of nh where n is the principal quantum number. 2π

Structure of the Rutherford–Bohr model The Rutherford–Bohr model has a small, positively charged nucleus that contains most of the atom’s mass. The electrons orbit the nucleus in classical circular paths. The electrons do not radiate energy continuously as predicted by Maxwell’s classical electromagnetic theory (i.e. accelerating charges will radiate electromagnetic waves, which would result in the electron spiralling into the nucleus), due to the quantisation conditions of energy associated with each electron orbit. When an electron jumps to a higher or lower orbit it will absorb or emit a quantum of energy in the form of a photon. Despite being a hybrid theory that spanned both classical and the quantum physics, the Rutherford–Bohr model proved to be extremely successful in explaining many experimental observations.

Checkpoint 12.5 1 2

Explain the importance of stationary states to Bohr’s model. Summarise the role of Planck and Einstein’s concept of quantisation to the development of Bohr’s atomic model.

12.6 Bohr’s explanation of the Balmer series Diagrammatic illustration Let’s consider the Balmer series emission spectra for a hydrogen atom. When an excited electron in a stationary state (orbit) of ni > 3 jumps down to the stationary state (orbit) nf = 2, a photon is emitted. The energy of this photon will be equal to the energy difference ∆E between these two stationary states: Energy of emitted photon = ∆E = Ei –Ef Process and present diagrammatic information to illustrate Bohr’s explanation of the Balmer series.

236

This photon will have a characteristic frequency and wavelength, which can be calculated using the relationships E = hf and v = f λ (where v is c the speed of light) rearranged in the forms: E c f = and λ = h f

quanta to quarks Balmer series

Figure 12.6.1 illustrates the Rutherford–Bohr atomic model electron transitions for the Balmer spectral series.

Bohr’s mathematical model Bohr’s model was not simply a diagrammatic representation of an atom. He backed up his model with a mathematical framework that was based upon both classical and quantum ideas. Most importantly, the model was supported by experimental observations.

n=1

nucleus

n=2 n=3

Quantised energy of the stationary states of the Bohr atom

n=4 n=5 n=6

The expressions for the quantisation of the angular momentum and radii can be used to derive a quantised expression for energy: 1 E n = 2 E1 n Figure 12.6.1 Diagrammatic representation of Bohr’s explanation

Theoretically determining spectral lines

of the Balmer line series for the hydrogen spectrum

Now recall Bohr’s second postulate concerning the emission or absorption of a specific quantum of energy when an electron transitions between stationary states (orbits). This obeys the Planck–Einstein relation hf = Ei – Ef. 1 If we now substitute in E n = 2 E1 we obtain: n 1 1 ∆ E between stationary states = hf = 2 × E 1 − 2 × E 1 ni nf If we take out the common factor of –E1 and make f the subject, we obtain the expression: − E1 ⎛ 1 1⎞ f = ⎜ 2 − 2⎟ h ⎝ nf ni ⎠ c Using the relationship v = f λ where v is c the speed of light, we have f = . λ Substituting this we obtain the expression: c − E1 ⎛ 1 1⎞ = ⎜ 2 − 2⎟ λ h ⎝ nf ni ⎠ Dividing through by c we obtain a theoretically derived expression in a form that is equivalent to the generalised Balmer equation: 1 − E1 ⎛ 1 1⎞ = ⎜ 2 − 2⎟ λ hc ⎝ nf ni ⎠



Describe how Bohr’s postulates led to the development of a mathematical model to account for the existence of the hydrogen spectrum:  1 1 1 = R  2 − 2 λ  n f ni 

PRACTICAL EXPERIENCES Activity 12.1

Activity Manual, Page 105

The expression –E1| hc corresponds to Rydberg’s constant and, when evaluated, agrees with the experimentally derived value. Bohr’s theory had successfully provided an explanation for some spectral phenomena and permitted the calculation of Rydberg’s constant. The theoretical derivation of Balmer’s equation was a major accomplishment and provided strong support for the Rutherford– Bohr atomic model of the hydrogen atom.

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Worked example Question An electron makes a transition between the energy levels –0.85 eV and –3.40 eV in a hydrogen atom. a Determine to which spectral series this photon would belong. b Calculate the wavelength of the emitted photon without using Balmer’s equation. c Predict if the spectral line associated with this transition is visible to the human eye. d Evaluate the principal quantum numbers (n) to which each energy level corresponds. e Substitute your values of ni and nf into the generalised Balmer equation and calculate the wavelength of the emitted photon. (This should be the same value as in part b.)

Solution a From Figure 12.4.3, the energy level –0.85 corresponds to n = 4 and energy level and –3.40 eV corresponds to n = 2. The n = 2 tell us that this transition belongs to the Balmer series. b The difference between these two energies corresponds to the energy of the emitted photon, E = 2.55 eV. Convert this to the SI unit joules (remember 1 eV = 1.602 × 10–19 J). E = 2.55 × 1.602 × 10–19 = 4.09 × 10–19 J



Now, using E = hf, we can calculate the frequency of the emitted photon with this energy: E 4.09 × 10−19 f= = = 6.17 × 1014 Hz h 6.626 × 10−34 c Using the relationship v = f λ where v  is c  the speed of light, we have λ = . f c 3.0 × 108 −7 λ= = = 4.86 × 10 m f 6.17 × 1014

c Light of this wavelength is in the visible range. d From Figure 12.4.3, the energy level –0.85 corresponds to n = 4 and the energy level –3.40 eV corresponds to n = 2. 1 e Using the generalised form of Balmer’s equation to calculate  : λ  1 1 1  1 1 = R  2 − 2  = 1.097 × 107  2 − 2 = 2 05 × 106 m–1 2 4  λ n n  f



i

Now take the inverse to determine the wavelength of the emitted photon for the transition between ni = 4 and nf = 2. λ = 4.86 × 10–7 m

Checkpoint 12.6 Explain the relationships between Bohr’s atomic model and observed line spectra.

238

quanta to quarks

12.7 Limitations of the Rutherford–Bohr model The Rutherford–Bohr model provided a simple visual model and accurately predicted line spectra for the hydrogen atom. It synthesised classical mechanics with the concept of quantisation, and addressed a number of experimental observations. Despite its successes, it was nevertheless a hybrid model and many experimental observations remained unresolved.

Discuss the limitations of the Bohr model of the hydrogen atom.

The spectra of larger atoms The observed spectral series of larger atoms appeared to present patterns that could be explained in a similar manner to the hydrogen atom. Despite the efforts of Bohr and his colleagues, they were unable to develop an arrangement of stationary states (orbits) that matched the experimental observations. Atoms larger than hydrogen have more than one electron, and these electrons were obviously interacting with each other in a complex manner. Bohr’s simple quantised planetary model explained only the hydrogen atom, the helium ion (He+) and the lithium ion (Li2+), which all have single electrons orbiting the nucleus. a

b

c

PRACTICAL EXPERIENCES Activity 12.2

Activity Manual, Page 110

Sr

Ba

Ca 400

Figure 12.7.1

450

500

550 600 Wavelength (nm)

650

700

Emission spectra for (a) stronium, (b) barium and (c) calcium

The relative intensity of spectral lines The spectra of the hydrogen atom and larger atoms all displayed three identifiable types of spectral lines based upon their width. These were categorised as sharp (s) lines, primary (p) lines and diffuse (d) lines. Also within these categories the Bohr’s postulates provided no intensity of the individual lines varied. explanation for these observations.

Figure 12.7.2

750

Emission spectrum for hydrogen showing the differences in relative widths

Measuring the Magnetic field of our Sun

A

stronomers use a spectroscope and the Zeeman effect to measure the strength of the magnetic field associated with the surface structures of the Sun. The splitting of the lines in the hydrogen spectrum is proportional to the magnetic field strength.

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From Rutherford to Bohr

The existence of fine and hyperfine spectral lines

Universal units for Length and Time

T

he plaque on the Pioneer space probe used the hyperfine transition of hydrogen, which is the most abundant element in the universe, to define a base unit for length and time.

The fine structure of closely spaced spectral lines that are 0.1–0.5 nm apart is a result of an additional property of the electron called ‘spin’, which was proposed in 1925 by Ralph Kronig (1904–1995). In 1881, using interferometry, Albert Michelson (1852–1931) observed that some even finer spectral lines (called hyperfine) existed; these were about 0.001 nm apart. The observation was not addressed until 1924, when Wolfgang Pauli (1900–1958) proposed the existence of a small nuclear magnetic moment Bohr’s model did not caused by a non-spherical atomic nucleus. provide any explanation for these observed phenomena. a

0.1 nm 0.1 nm

b

0.001 nm

Figure 12.7.3 0.001 nm Normal spectral line:

With no magnetic field applied—single spectral line observed.

‘Normal’ Zeeman effect:

Magnetic field applied— triple spectral line observed. Central line polarised parallel to applied magnetic field. Other two lines polarised perpendicular to applied field.

‘Anomalous’ Zeeman effect: When magnetic field applied— more spectral lines observed.

Figure 12.7.4 A single spectral line splits into more lines when the source is subjected to a strong magnetic field.

A comparison of (a) fine structure and (b) hyperfine structure

The Zeeman effect In 1862, Michael Faraday placed a sodium flame between the poles of a magnet to see if the bright sharp spectral lines were influenced by the magnetic field, but he observed no change. Some 30 years later (in the 1890s) Pieter Zeeman (1865–1943) repeated the experiment using both a more advanced spectroscope and a stronger magnetic field. Zeeman found that the spectral lines were indeed influenced and he observed that the previously single lines had each split into three. Classical physics and Bohr’s model could provide an explanation for these triplets. This was called the Zeeman effect. In 1897 Zeeman used an even stronger magnetic field and found that the triplet lines were also split. The additional splitting was also associated with the yet to be discovered spin property of electrons. Bohr’s atomic model provided some explanation for the ‘normal’ Zeeman effect, but none for the ‘anomalous’ Zeeman effect.

Checkpoint 12.7 Compare the origins and observed separations of the splitting in the ‘normal’ Zeeman effect and the hyperfine spectral lines.

240

PRACTICAL EXPERIENCES

quanta to quarks

CHAPTER 12

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 12.1: Hydrogen and the atom Connect a high-voltage source to a low-pressure hydrogen discharge tube, and observe the emitted light through a spectroscope. Identify the visible part of the hydrogen spectrum. Equipment: low-pressure hydrogen gas discharge tube, gas discharge tube high-voltage power supply, spectroscope. Discussion questions 1 List the wavelengths of the emission lines in the visible part of the hydrogen spectrum. 2 Identify the energy levels of the electron transition that creates these wavelengths. 3 Outline Bohr’s explanation of the Balmer series of emission lines.

Figure 12.8.1

Perform a first-hand investigation to observe the visible components of the hydrogen spectrum. Process and present diagrammatic information to illustrate Bohr’s explanation of the Balmer series. Solve problems and analyse information using:  1 1 1 = R  2 − 2 λ  n f ni 

Experiment set-up and the visible emission spectral lines of hydrogen

Activity 12.2: Problems with the Rutherford–Bohr model In this activity you will look at the difficulties with the Rutherford–Bohr model in explaining certain observations. Discussion questions 1 Identify the ways in which this model is an improvement over the previously accepted model of the atom. 2 Explain the limitations of this model.

Analyse secondary information to identify the difficulties with the Rutherford–Bohr model, including its inability to completely explain: • the spectra of larger atoms • the relative intensity of spectral lines • the existence of hyperfine spectral lines • the Zeeman effect.

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•

•

•

•

• •

•

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From Rutherford to Bohr

Chapter summary

Atomic theory had its origins in Greece more than 2000 years ago with ideas put forward by Democritus and Aristotle. JJ Thomson in 1904 proposed the ‘plum pudding’ model comprising a positive sphere in which the electrons were distributed like ‘plums in a pudding’. Ernest Rutherford in 1911 used the results of Geiger and Marsden’s scattering experiments to propose a ‘Rutherford’s planetary model’. Rutherford’s model comprised a massive, positively charged nucleus with a set of orbiting electrons (like planets orbiting the Sun), inferring that the atom is mostly empty space. Despite the success of Rutherford’s planetary atomic model in explaining the scattering of alpha particles, the model failed to explain: — what the nucleus is made of — how the orbits of the electrons are arranged around the nucleus — what keeps the negatively charged electrons from losing energy and spiralling into the positive nucleus. Planck in 1901 proposed a theory to model the spectrum of a black body and introduced the concept of quantisation E = nhf where n = 1, 2, 3, 4 … A spectroscope uses a prism or diffraction grating to reveal the constituent colours present in the light. There are two types of spectra: — emission spectra — absorption spectra Spectra are a ‘window’ into the hidden atomic structure. Each element has its own unique spectral ‘fingerprint’.

•



• •

•

•

Rydberg’s generalised Balmer equation allows you to calculate the wavelength of the photon emitted or absorbed for any transition of an electron between stationary states:  1 1 1 = R −   n 2 n 2 λ f i where λ is the wavelength of the spectral line R is Rydberg’s constant 1.097 × 107 m–1 nf is the final state ni is the initial state. Hydrogen has several spectral line series including the Lyman, Balmer, Paschen, Brackett and Pfund series. Bohr in 1913 described the revised planetary model of a hydrogen atom based upon the quantisation of energy and angular momentum of the electron. Bohr’s three postulates for his atomic model: 1 Electrons exist in stable orbits called stationary states. 2 Electrons absorb or emit specific quanta of energy when they transition between stationary states (orbits), described by the Planck–Einstein relation hf = Ei – Ef. 3 Angular momentum of electrons is quantised. The Rutherford–Bohr model, despite its successes, was nevertheless a hybrid model and many experimental observations still remained unresolved including: — the spectra of larger atoms — the relative intensity of spectral lines — the existence of hyperfine spectral lines — aspects of the Zeeman effect.

Review questions

quanta to quarks

Physically speaking The following table is all jumbled up. Copy the table into your workbook and correctly arrange the information. You may also add additional features associated with each atomic model. For each atomic model, draw and label a pictorial representation.

Scientist and date

Features of the atomic model

Democritus

• Planetary orbits of the electrons • Three postulates

c 400

b ce

Aristotle c 300

b ce

Dalton 1801

Thomson

Draw a pictorial representation of the model

A positive sphere with electrons embedded in it

• Planetary orbits of the electrons • Small positively charged nucleus

1904

Limit to how small you could divide matter; the smallest is called an atom

Rutherford

Electrons act like waves

1911

Bohr 1913

de Broglie 1924

There are four elements: earth, air, fire and water

• Elements only contain one type of atom • Different elements contain different atoms • Compounds contain more than one type of atom

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From Rutherford to Bohr

Reviewing

Solving Problems

1 Explain why flame tests of unknown samples are so

13 A photon has an energy of 6 eV. Calculate: a its energy in joules b the frequency of the photon c the wavelength of the photon.

useful to scientists.

2 Compare Max Planck’s and Albert Einstein’s ideas relating to ‘quanta’.

3 Outline Rutherford’s model of the atom. 4 Explain how emission line spectra are formed. 5 Identify the major events that led Niels Bohr to propose his model of the atom.

6 Define Bohr’s postulates associated with his 1913 model of the atom. Describe how these postulates led to the development of a mathematical model to account for the existence of the hydrogen spectrum. Describe how Bohr’s postulates led to the development of a mathematical model to account for the existence of the hydrogen spectrum:  1 1 1 = R  2 − 2 λ  n f ni 

7

Recall the relationship between energy levels in the Bohr model of an atom and the observed Balmer series spectra.

8

Discuss the limitations of the Bohr model of the hydrogen atom.

9

When excited atoms were placed in a strong magnetic field some previously observed single spectral lines split. Explain why the Rutherford–Bohr model could not completely explain the observed phenomena.

10 Marsden and Geiger in 1909 fired alpha particles at thin platinum and gold foils. a Recall what the then current model of the atom proposed by JJ Thomson theoretically predicted. b Recall the name of the scientist they were working with during these experiments. c Outline their experimental findings. d Discuss the implications these experiments had for the development of a new model of the atom. Discuss the structure of the Rutherford model of the atom, the existence of the nucleus and electron orbits.

11 Recall why the Rutherford–Bohr model could not explain spectra of atoms larger than hydrogen.

12 Compare the Zeeman effect and hyperfine spectral lines.

244

14 Using the formula En =

1

× E1 and given that the n2 energy of the first stationary state (orbit) in the hydrogen atom is –13.6 eV, calculate: a the energy of the fourth stationary state in the hydrogen atom b the energy difference between the fourth and first stationary states c the frequency of a photon emitted by an electron transitioning from stationary state n = 4 to n = 1 d the wavelength of a photon for this transition e the spectral series to which this photon belongs.

15 An electron makes a transition between the –3.40 eV and –13.6 eV energy levels of a hydrogen atom. a Calculate the wavelength of the emitted photon without using Balmer’s equation. b Predict if the spectral line associated with this transition is visible to the human eye. c Evaluate the principal quantum numbers (n) of each energy level. d Substitute your values of ni and nf into the generalised Balmer equation and calculate the wavelength of the emitted photon. (This should be the same value as part a.) e Identify the spectral series to which this photon would belong.

16 Using the Bohr model, construct diagrams of the following electron transitions and describe the role of the photon in these transitions: a an electron being excited from the ground state (n = 1) to the fourth excited state (n = 5) b an excited electron dropping from the n = 6 energy level to the n = 2 energy level c an excited electron dropping from the n = 4 to n = 3 and then back to the ground state (n = 1)

17 Using the information in the following table, construct an energy level diagram for n = 1 to n = 5 for the hydrogen atom.

Principal quantum number (n)

Energy (eV)

1 2 3 4 5

–13.6 –3.4 –1.51 –0.85 –0.54

quanta to quarks a Identify the electron transition between the energy levels in your diagram that would absorb the highest frequency photon. b Identify which transition of an electron between energy levels would emit the highest energy photon. c Demonstrate on your energy level diagram how the Balmer series of spectral emission lines is produced. d Identify the transition between energy levels in the Balmer series that would produce the longest wavelength photon. Solve problems and analyse information using:  1 1 1 = R  2 − 2 λ  n f ni 

18 Rydberg generalised Balmer’s equation for the hydrogen atom:

⎛ 1 1 1⎞ = R ⎜ 2 –+ 2 ⎟ λ ⎝ nf ni ⎠

19 A photon is emitted from a hydrogen atom with a wavelength of 410.12 nm. The electron associated with the transition that produced the photon is now in energy level nf = 2. a Using Balmer’s generalised equation, calculate the value of ni. b Construct a diagram showing the transition and identify all relevant features. c Identify the series associated with this transition.

20 The Lyman series for a hydrogen atom has the following energy levels: E1 = –13.6 eV, E2 = –3.40 eV, E3 = –1.51 eV, E4 = –0.85 eV, E5 = –0.54eV, E6 = –0.38 eV. a Construct a diagram using the Bohr model to represent the series. b Construct an energy level diagram for the Lyman series. c Calculate the energy of the photon emitted in the transition from n = 5 in the Lyman series. d Calculate the wavelength for this emission.

Re

iew

Q uesti o

n

s

v

a Recall what each term in the equation represents. b Calculate the wavelength of the emitted photon for a transition between n = 6 and n = 1. c Predict the values for nf and ni for the transition of the second-longest wavelength in the Balmer series. d Calculate the wavelength for the second-longest wavelength in the Balmer series.

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PHYSICS FOCUS Quanta to Quarks Timeline H1. Evaluates how major advances in scientific understanding and technology have changed the direction or nature of scientific thinking

H3. Assesses the impact of particular advances in physics on the development of technologies

Physics is not just about learning facts. Physics is a human journey of exploration seeking to explain the universe in which we live. Understanding the history of events and people will allow you to ‘stand upon the shoulders of giants’ and continue this quest. Often the ‘stories’ told in modern textbooks skip over the years of frustration and confusion that scientists faced and present only the milestones and important discoveries. When you study the history of scientific advancement, it can become quite confusing if you do not see the situation in the context of the period of history. For example, many of the experiments carried out from the late 1800s to the modern particle accelerators relied on vacuum pump technology. How did this technology frustrate the pioneers who worked with low-pressure gas experiments? Consider how the ability for the scientist to produce a strong magnetic field may have influenced the outcome of an experiment. For example in 1862, Michael Faraday placed a sodium flame between the poles of a magnet to see if the bright sharp spectral lines were influenced by the magnetic field; Faraday observed no change. Some 30 years later in 1896, Pieter Zeeman repeated the experiment using both a more advanced spectroscope and a stronger magnetic field, and found that the spectral lines were indeed influenced. Zeeman also observed that some previously single lines had each split into three.

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Figure 12.8.2

JJ Thomson in his Cavendish laboratory

Maybe you are now thinking that the interplay between technology, people, events and ideas might be quite interesting. Construct a timeline that includes people, experiments, advances in technology, models and theories for the period from 1885 and Balmer’s observations of the hydrogen spectrum through to 1913 when Niels Bohr proposed his atomic model. You can include pictures of people, models and experiments. Investigate and explore the relationships between events. You might construct your own or, as a class, build a huge timeline and display it in your laboratory. Some technologies you might research: • Advances in vacuum pump technology • Advances in spectroscope technology • Sources of electricity

Extension Some additional technologies you might research: • Methods of detecting the non-visible electromagnetic spectrum, including infra-red, ultraviolet light, X-rays and gamma rays • Types of magnets available • Electric and electronic circuitry As you progress further through this module extend and include more details on your timeline.

13

Birth of quantum mechanics A new paradigm Niels Bohr’s hybrid atomic model, which combined classical and quantum concepts, held centre stage from 1913 through to the early 1920s. From a historical perspective we must remember that World War I, 1914–1918, disrupted research laboratories throughout Europe. The 1920s saw the move away from classical ideas and culminated at the 1927 Solvay Conference in Copenhagen where modern day quantum mechanics was born. In the years leading up to this conference Louis de Broglie proposed that all matter has wave properties, Erwin Schrodinger developed a mathematical model for describing wave mechanics, Heisenberg revealed the statistical nature of quantum theory and formulated his most famous discovery—the Heisenberg uncertainty principle— and Pauli developed his exclusion principle.

13.1 The birth Newtonian classical physics was under attack throughout the 1920s and the outcome, which was based upon theoretical and experimental evidence, gave birth to quantum mechanics. The major debates came to a head at the 1927 Solvay Conference in Copenhagen where the old deterministic view of matter was cast aside in favour of a world that was ruled by probabilities. This outcome, however, was certainly not unanimously accepted and most notably Bohr and Einstein continued to debate the interpretations for more than 20 years.

Checkpoint 13.1 Outline how quantum mechanics was started.

quantum mechanics, de Broglie, Schrodinger, wave mechanics, diffraction, quantum numbers, wave function, Heisenberg, uncertainty principle, Pauli, exclusion principle

A Great Loss

H

arry (Henry) Moseley was one of the brilliant young scientists who worked with Rutherford. In about 1910, using X-ray diffraction of crystals, he discovered a systematic relationship between wavelength of characteristic X-rays and atomic number, which is now known as Moseley’s law. When World War I broke out, he enlisted and joined the British Royal Engineers. He was killed in action by a sniper shot in 1915 at Gallipoli. The loss of Moseley and other scientists during this war prompted British and other world governments to implement policy to not allow scientists to enlist for combat postings.

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Birth of quantum mechanics

Figure 13.1.1

Participants at the 1927 Solvay conference Back row: A Piccard, E Henriot, P Ehrenfest, E Herzen, T De Donder, E Schrödinger, JE Verschaffelt, W Pauli, W Heisenberg, RH Fowler, L Brillouin Middle row: P Debye, M Knudsen, WL Bragg, HA Kramers, P Dirac, A Compton, L de Broglie, M Born, N Bohr Front row: I Langmuir, M Planck, M Curie, HA Lorentz, A Einstein, P Langevin, CE Guye, CTR Wilson, OW Richardson

13.2 Louis de Broglie’s proposal By 1915, William Henry Bragg (1862–1942) had provided convincing evidence that X-rays have particle properties; his son William Lawrence Bragg (1890– 1971), an Australian, had developed an equation based upon the wave nature of X-rays that enabled a detailed analysis of X-ray diffraction patterns. In the early 1920s, Prince Louis de Broglie (pronounced ‘de broy’), with the knowledge that X-rays and light possessed both wave and particle properties, began to consider the wave–particle duality as a natural symmetry.

Matter waves In 1923, Louis de Broglie began with a supposition that was based upon the Planck–Einstein equation linking energy quanta to frequency. He equated Einstein’s special relativity energy–mass relationship E = mc2 with the Planck– Einstein equation E = hf: mc2 = hf He rearranged it to derive an expression for the momentum of a photon: mc =

hf c

Now mc is mass times velocity, hence the photon’s momentum is: p =

hf c

c c Using the relationship λ = in the form f = and substituting, f λ he obtained: h p= λ 248

quanta to quarks He had derived an expression for the momentum of a photon in terms of its wavelength. In 1924, de Broglie proposed the concept of matter waves by applying his belief in symmetry to propose that all ‘particles’ of energy should also possess an associated wavelength. At this time, particles of matter such as electrons, atoms and alpha particles were known to possess the properties of mass and charge, but there was no experimental evidence to indicate that these particles exhibited wave phenomena. De Broglie’s proposed matter waves could be calculated mathematically by using what is now known as the de Broglie equation: λ=

h h or λ = p mv

where λ is the wavelength, h is Planck’s constant, m is the mass of the particle, v is the velocity of the particle and p is the momentum of the particle. (Remember p = mv.) Louis de Broglie presented his hypothesis in his doctorial dissertation. Unfortunately the combination of the involvement of de Broglie and other French scientists in unfriendly debates with Niels Bohr and his supporters and the experimentally unsupported proposition that particles such as electrons possessed wave properties meant that the majority of scientific community did not take de Broglie’s ideas seriously.

Figure 13.2.1

Louis de Broglie

Solve problems and analyse information using: h λ= mv

Worked example Question a Calculate the wavelength of an electron moving with a velocity of 6.0 × 105 m s–1. b Compare this calculated wavelength with the wavelength of visible light (400–650 nm).

Solution

h and substituting in the following values: mv h = 6.63 × 10–34 J s

a Using de Broglie’s equation λ =

me = 9.11 × 10–31 kg



v = 6.0 × 105 m s–1



λ=



we obtain

6.63 × 10−34 J s 9.11 × 10

−31

5

kg × 6.0 × 10 m s

−1

= 1.21 × 10−9 m

Therefore the wavelength of the de Broglie matter wave is 1.21 × 10–9 m.

b The visible spectrum ranges from violet light, with the shortest wavelength (~400 nm), to red with a wavelength of 650 nm. For the purpose of this comparison we will use the average wavelength of these extremes 525 nm, which corresponds to a yellow/green colour.

First we need to convert 525 nm into metres (5.25 × 10–7 m). Now, comparing the ratio of the wavelength of the electron (1.21 × 10–9 m) with the average wavelength of visible light (5.25 × 10–7 m), we discover that the wavelength for an electron travelling at 6.0 × 105 m s–1 is approximately 430 times smaller than the wavelength of visible light. 249

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Birth of quantum mechanics

Always keep your eye on the physics

A

n important point to consider when you are calculating the de Broglie wavelength for macroscopic objects such as cars and balls is that these are not discrete fundamental particles. Rather they are a ‘bulk’ collection of such. Therefore, if you calculate the de Broglie wavelength for an object at near rest (very close to zero velocity) you would obtain a large wavelength, which of course is not what we observe. This is because you have not considered that the atoms and molecules making up the object are actually vibrating at high velocities due to their thermal energy. Also it is important to remember that, even at absolute zero (–273°C), all atomic particles including atoms and electrons will still have some motion and therefore cannot have zero velocity.

Describe the impact of de Broglie’s proposal that any kind of particle has both wave and particle properties.

Initially Einstein was the only high-profile physicist to support de Broglie’s hypothesis. Erwin Schrodinger, a loner, who was then based at the University of Zurich, was intrigued and took up the idea proposed by de Broglie. He became the architect of wave mechanics, a fundamental component of quantum mechanics.

Checkpoint 13.2 1 2 3

Outline the significance of the work of the Braggs in Louis de Broglie developing his theorical research. Determine the wavelength of an electron moving at 3.0 × 105 m s–1. Calculate the momentum of a proton that has a wavelength of 1.01 × 10–9 m.

13.3 Diffraction Define diffraction and identify that interference occurs between waves that have been diffracted.

250

A defining property of waves is their ability to ‘flare’ or seemingly bend around corners. Waves in water, sound waves and light can all exhibit this property called diffraction. Let’s first look at water waves. If you generate a set of plane waves in a shallow trough of water using a ruler, you will see the waves propagate down the length of the trough. Now, if you place an object such as an edge, a block, a gap or a narrow opening in the path of the waves, you will observe a strange behaviour. Instead of ‘casting a shadow’, each wavefront bends around the object and enters the shadow region. Let’s now observe the behaviour of light as it passes an edge, through a narrow slit and a small aperture (Figure 13.3.1). Again you can see that the light does not cast a sharp shadow, instead it forms a series of bright and dark lines. These patterns are known as diffraction patterns.

quanta to quarks a

b

Figure 13.3.1

c

Light passing (a) around a razor blade, (b) through a slit and (c) through a small aperture

You can’t escape them

S

ometimes when you are sitting back relaxing or looking upwards into a clear blue sky, you might occasionally see tiny fuzzy, out-of-focus ‘hairs’ or ‘blobs’. These are called floaters and they are, literally, that. They are little bits of jelly broken off the vitreous humour (the clear gel that fills the space between the lens and the retina of the eye) floating just in front of your retina. The fuzziness is caused by the diffraction of light around the edges of these floaters.

Try This! Observing a diffraction pattern Simply take two pens or pencils (you can even use your fingers). Place them very close together, almost touching, to form a narrow slit. Now hold them up in front of one eye (about 3–4 cm in front works well) now close the other eye. Look through the slit at a bright light source such a fluorescent light. Vary the width of the slit and observe carefully! What do you see?

Checkpoint 13.3 1 2

Define diffraction. Sketch a diffraction pattern that would result as light moves around a hair.

13.4 Confirming de Broglie’s hypothesis Clinton Davisson (1881–1958) and Lester Germer (1896–1971), using electron scattering, reported the experimental discovery of electron waves in 1927. Davisson and Germer’s discovery was preceded by an accident in the laboratory in which an explosion shattered the evacuated glass vessel holding the nickel target. The introduction of air oxidised the surface of the nickel target, which then had to be degassed by heating it to a high temperature. This process allowed a number of large nickel crystals to form. Experiments with this target produced a new set of data in which sharply defined currents of electrons were present. Davisson and Germer quickly realised that these new results were a result of the recrystallisation, and they constructed new nickel targets of single crystals.

Describe the confirmation of de Broglie’s proposal by Davisson and Germer.

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In the family

G

eorge P Thomson, the co-discoverer of the wave properties of electrons, was the son of JJ Thomson the discoverer of the electron.

Up until this time Davisson and Germer had been unaware of de Broglie’s hypothesis, but during a meeting at Oxford in 1926 Davisson heard of de Broglie’s proposed matter waves. He immediately realised that the new results he and Germer had observed were similar to those of X-ray diffraction patterns. With this insight, Davisson and Germer quickly identified that the scattered electrons were indeed producing diffraction patterns. At the same time, George Thomson (1892–1975) conducted a different experiment in which he fired a beam of electrons through a thin gold foil and observed diffraction rings. Therefore, in 1927, two different research teams using different experimental techniques had verified Louis de Broglie’s hypothesis concerning the wave nature of electrons. a

b I

V = 54 V

electron gun 50° 0

power supply

15° 30° 45° 60° 75° 90°

electron detector

c

electron beam (in vacuum)

θ

incident waves in phase λ 50°

nickel crystal

Figure 13.4.1

scattered waves in phase

GP Thomson

crystal surface

d

Figure 13.4.2

Davisson and Germer’s apparatus and results. With the single crystal nickel target, Davisson and Germer detected several peaks in scattered electrons.

Checkpoint 13.4 1 2

Outline how matter waves were first observed. Discuss what other evidence has been observed to support Louis de Broglie’s hypothesis.

13.5 Electron orbits revisited PRACTICAL EXPERIENCES Activity 13.1

Activity Manual, Page 112

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Bohr’s first postulate stated that an electron can exist in any of several circular orbits with no emission of radiation. However, he provided no supporting reasoning to explain why the orbiting electron would not radiate away its energy, as predicted by Maxwell’s classical electromagnetic theory, and simply spiral into the nucleus. When de Broglie formulated his concept of matter waves, he envisioned that electron orbits were standing waves. Each orbit (stationary state) was occupied by an integral number of wavelengths that fitted around the circumference of the orbit. The circumference of a circular orbit is 2πR and hence an integral number of wavelengths: nλ = 2πR

quanta to quarks where n is 1, 2, 3, 4, 5 …, λ is the wavelength and R is the radius of the orbit. To visualise these standing waves we first imagine a standing wave on a slinky spring or rope, obeying the condition nλ where n can have the values n = 1, 2, 3 etc. When you shake a rope or a slinky spring you can produce standing wave patterns that contain whole or integral numbers of wavelengths. Now we can take these standing waves and wrap them end-to-end to form a circular standing wave that corresponds to the circular Bohr orbit. These values of n are called the principal quantum numbers and they tell you the number of wavelengths fitting into the Bohr orbit. Remember that the standing waves in Figure 13.5.1 represent the matter wave nature of the electron. Often students imagine the electrons travelling in wavy orbital paths around the nucleus—this is not the correct interpretation.

Explain the stability of the electron orbits in the Bohr atom using de Broglie’s hypothesis. λ n=1 λ

Try This! Standing waves Using a slinky spring or a rope, generate standing waves by having one person hold one end firmly while the other person shakes their end. Once you get a stationary wave you will notice that if you stop shaking, the standing wave is quite stable and continues to vibrate. On a rope or slinky you can generate standing waves other than those shown in Figure 13.5.1. Why are these other standing wave patterns not suitable for forming circular standing waves? (Hint: Consider interference.)

n=2

λ λ λ

λ

Figure 13.5.1

n=3

λ

n=4

Comparison of standing waves wrapped in a circle and on a rope. For a standing wave to be produced in a circle, a whole number of matter waves must fit into the circumference.

Checkpoint 13.5 1 2

Outline how de Broglie’s matter waves help to explain the stable orbits of electrons. Calculate the radius of the smallest orbit of an electron with wavelength 1.21 × 10–9 m.

13.6 Further developments of atomic theory 1924–1930 Quantum mechanics, the modern version of quantum theory, was developed between 1925 and 1930. During this period, physicists moved from the hybrid mixture of classical and quantum concepts to a fully mathematical model, to describe the behaviour of matter. The strangeness of wave–particle duality and the many associated paradoxes were sidelined as a new physics was born.

Erwin Schrodinger (1887–1961) Schrodinger’s interests were wide and spanned most of modern physics including statistical mechanics, X-ray diffraction, relativity and field theory. After World War I, Schrodinger moved between several academic posts and in 1921 he settled for 6 years at the University of Zurich. Here Schrodinger, encouraged by Einstein, 253

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Birth of quantum mechanics

developed wave mechanics by synthesising the works of de Broglie, Planck, Einstein, ideas from Hamilton and the fundamental differential equations from optics. In 1926 Schrodinger published the equation known by all physicists as ‘Schrodinger’s equation’: −

h2 d2ψ + V ( x )ψ = E ψ 2m dx 2

When solved, the equation provides a wave function. The wave function contains all the measurable information about a particle. If you square this function you obtain a probability density, which allows you to predict the likelihood of finding a particle. This very quickly successfully addressed a large variety of atomic and molecular problems. The price paid for this new way of looking at matter was that the classical and deterministic view of the atomic world was totally replaced by a system in which you could only calculate the probability of finding a particle in a certain place and time.

Werner Heisenberg (1901–1976) Figure 13.6.1

Erwin Schrodinger

A famous cat and inspiring a new career path to DNA

E

veryone has heard of Schrodinger’s cat—the one that is both alive and dead—well that is until you open the box and see. But a little less known fact is that in 1944 Schrodinger published a little book entitled What is Life?, which looked at the new field of molecular biology. Francis Crick the codiscoverer of the DNA double helix said that this book influenced him to change his career from physics to molecular biology.

254

Heisenberg met Bohr in 1922, he was just 20 years old and working toward his PhD. At the end of a lecture Heisenberg raised an objection to what Bohr had been discussing. Bohr was so impressed that he invited Heisenberg on a walk and they discussed the difficulties of quantum theory for more than 3 hours. Bohr liked physical models of atoms, but Heisenberg thought it was nonsense to talk about electron orbits when it was obvious that you could not observe the electrons or their orbital paths. Heisenberg went on to discover the first complete version of quantum mechanics. The new theory that was developed in parallel with Schrodinger had taken a completely different path. Heisenberg’s theory had abandoned any attempt to create a visualisation of the atom, ignored the ideas of waves and particles, and in itself was purely mathematical. Heisenberg took de Broglie’s proposal and Schrodinger’s wave equation and was able to ascertain that there was a limit to how precisely you would measure pairs of quantities. The pair he identified was that of position and momentum. In 1927 Heisenberg announced his uncertainty principle, which is expressed mathematically as: h h Δx Δp ≥ and ΔE Δt ≥ 2π 2π where ∆x is the uncertainty in the position of a particle, ∆p is the uncertainty in the momentum of a particle and h is Planck’s constant. You can gain some insight into understanding the uncertainty principle by considering the following example. If you want to know the position of a particle, you will need to use a high-energy photon that has a small wavelength as a high-resolution probe to locate your particle. Now consider that the photon ‘bounces’ nicely off the particle and you detect the photon and calculate the position of your particle. You have located the particle very precisely, but the particle is now recoiling and you will have very little information about its momentum. An even more bizarre implication of Heisenberg’s uncertainty principle (pointed out by Einstein) is that particles (i.e. energy) can appear out of the nothingness of space for a very brief period of time, and then disappear.

quanta to quarks These particles are called virtual particles and we will discuss some aspects of them in Chapter 15 ‘The particle zoo’. In summary, Heisenberg’s contributions to atomic theory include: 1 the development of his matrix mechanics, which, like Schrodinger’s wave equation, allowed the nature of the atom and especially spectra to be explained 2 the uncertainty principle, which set limits on the precision of measurements.

Wolfgang Pauli (1900–1958) Pauli and Heisenberg were great friends and studied as research students under Arnold Sommerfield at the University of Munich. Pauli’s introduction to quantum theory was as a student listening to Sommerfield’s lectures. Sommerfield had extended the Rutherford–Bohr model to include elliptical orbitals, and the Bohr–Sommerfield theory attempted to describe the hydrogen molecule. During the decade after 1913, physicists focused on the role of quantum numbers. A major outstanding question was ‘How many quantum numbers were required to account for the observed chemical and physical behaviour of atoms?’ Pauli responded by presenting an ad hoc solution to address this puzzle. He found that by assigning four quantum numbers to each electron in an atom, combined with a set of rules, he was able to produce a system that explained a number of outstanding issues including the structure of the periodic table. Several prominent scientists including Heisenberg, Bohr, Compton and Pauli had considered that a fourth quantum number related to electron spin might be required to describe the behaviour of electrons in an atom. The more experienced physicists hesitated, but two Dutch graduate students George Uhlenbeck and Samuel Gouldsmit published and described the fourth quantum number as spin on the basis of the ideas presented in a paper Pauli had published on his exclusion principle. In summary, Pauli made several important contributions to atomic theory: 1 He proposed that each electron in an atom would be described by four quantum numbers. 2 He proposed, through his exclusion principle, that no two electrons in an atom could have a set of four identical quantum numbers. 3 His exclusion principle and rules provided a system to explain the arrangement and number of electrons in each atomic orbital, thus providing an explanation for the structure of the period table. 4 He proposed that a neutrino was also emitted in beta decay, which provided an explanation for the spectrum of beta-particle energies observed.

Checkpoint 13.6 1 2 3

State what Schrodinger’s equation allows you to find. Outline Heisenberg’s contributions to quantum mechanics. Recall how Pauli’s work explained outstanding problems in the theories.

Figure 13.6.2

Werner Heisenberg

Figure 13.6.3

Wolfgang Pauli

Gather, process, analyse and present information and use available evidence to assess the contributions made by Heisenberg and Pauli to the development of atomic theory.

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PRACTICAL EXPERIENCES CHAPTER 13 This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 13.1: Atomic theory Gather, process, analyse and present information and use available evidence to assess the contributions made by Heisenberg and Pauli to the development of atomic theory.

Research and present a 5 minute speech assessing the contributions of Heisenberg and Pauli to the development of atomic theory. Discussion questions 1 Outline the contributions of Heisenberg and Pauli to atomic theory. 2 Assess how these contributions have affected our understanding of the atom.

Chapter summary • •

•

• •

256

Louis de Broglie proposed that all matter has wave properties. These are called matter waves. The wavelength of a de Broglie matter wave can be determined by using de Broglie’s equation: h h λ= or λ = p mv where λ is the wavelength h is Planck’s constant m is the mass of the particle v is the velocity of the particle p is the momentum of the particle. Diffraction is a property of waves and it is characterised by the waves bending around corners or spreading out after passing through a narrow slit. Diffraction patterns can only be produced by waves. In 1927 two different research teams using different experimental techniques had verified Louis de Broglie’s hypothesis concerning the wave nature of electrons: – Clinton Davisson and Lester Germer, using electron scattering – George Thomson firing a beam of electrons through a thin gold foil and observing diffraction rings.

•

•

•

Louis de Broglie explained the stability of the electron orbits in the Bohr atom by stating that each orbit (stationary state) was occupied by an integral number of wavelengths that fitted around the circumference of the orbit. Heisenberg’s contributions to atomic theory include: 1 the development of his matrix mechanics 2 the uncertainty principle, which set limits on the precision of measurements. Pauli’s contributions to atomic theory include: 1 proposing that each electron in an atom would be described by four quantum numbers 2 proposing through his exclusion principle that no two electrons in an atom could have a set of four identical quantum numbers 3 his exclusion principle and rules provided an explanation for the structure of the periodic table 4 proposing that a neutrino was also emitted in beta decay, which provided an explanation for the spectrum of beta-particle energies observed.

Review questions

quanta to quarks

Physically speaking Copy and complete the following text. ______________ atomic model published in ______________ was a mix of ______________ and ______________ concepts. In ______________ Louis de Broglie proposed that particles had both ______________ and ______________ properties. ______________ developed the idea of ______________ waves, which, when applied to the situation of an electron ______________ in Bohr’s atomic model, provided

a mechanism to explain the stability of ______________. The concept of a ______________ wave in which the ______________ of the orbit was equal to the h de Broglie ______________, calculated using λ = .

mv

1 E1 where E1 = –13.6 eV n2 to calculate the ______________ of the electron in a given orbit (n). You can then rearrange the relation E = hf to calculated the frequency. Substituting the frequency into λ = c/f  you can calculate the ______________ associated with the electron for the orbit.

For the hydrogen atom you can use the equation E n =

You calculate the wavelength for n = 1 and the first orbit (n = 1) has a circumference equal to one ______________. You recalculate the wavelength for n = 2 and the second orbit (n = 2) has a ______________ equal to ______________ wavelengths. In 1927 ______________ and ______________ fired electrons at a ______________ target and observed ______________ phenomena that verified the ______________– ______________ duality of electrons.

Reviewing

1 2 3 4 5

Define diffraction. Describe how interference can be achieved. Recall instances in which diffraction occurs. Outline an experiment you could conduct to observe interference. In terms of the Bohr atomic model, explain how matter waves were used to account for the stability of electron stationary states.

6 Describe the experiment performed by Davisson and Germer in 1927 and outline its significance.

7 Louis de Broglie proposed that any kind of particle has both wave and particle properties. Recall the response given by the scientific community.

8 Recall and list the contributions made by Heisenberg to the development

Define diffraction and identify that interference occurs between waves that have been diffracted. Explain the stability of the electron orbits in the Bohr atom using de Broglie’s hypothesis. Describe the confirmation of de Broglie’s proposal by Davisson and Germer.

of atomic theory.

9 Recall and list the contributions made by Pauli to the development of atomic theory.

10 If the speed of an electron is increased, predict what would happen to its de Broglie wavelength.

11 In terms of de Broglie’s hypothesis, discuss the implications for locating an electron in the ground state.

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13

Birth of quantum mechanics

Solving Problems Solve problems and analyse information using: h λ= mv

12 Louis de Broglie equation is λ = h .

mv a Recall what each term represents. b Calculate the wavelength of an electron in a television tube with a velocity of 3.0 × 107 m s–1.

13 Consider an electron of wavelength of 1.0 m. a Calculate its velocity. b Is this wavelength possible?

14 Using de Broglie’s equation, calculate the wavelength of: a an electron travelling at 6.0 × 105 m s–1 b a proton travelling at 6.0 × 105 m s–1.

15 Calculate the velocity of a proton with a wavelength of 1.38 × 10–12 m. 16 a Calculate the wavelength of:

i a person (80 kg) walking at 2 m s–1 ii a soccer ball (0.43 kg) kicked at a speed of 5 m s–1 iii a car (1000 kg) travelling at 80 km h–1. b Assess the validity of your calculations against real-world observation.

17 Consider an electron at rest. a Calculate the wavelength of the electron. b Is this situation possible?

18 A car of mass 850 kg is travelling very slowly at 7.0 × 10–37 m s–1.

Re

258

iew

Q uesti o

n

s

v

a Calculate the wavelength associated with the car. b Will you see the wave properties of the car?

quanta to quarks

PHYSICS FOCUS

a

b

The Microscope: How small can we see? H3. Assesses the impact of particular advances in physics on the development of technologies Describe the impact of de Broglie’s proposal that any kind of particle has both wave and particle properties.

Figure 13.7.1  (a) Light microscope, (b) electron microscope

The maximum magnification of microscopes in most schools is usually about 400–600 times. More expensive research microscopes can achieve magnifications of up 2000 times. You would think that if you made better lenses and optics you could increase your magnification to whatever you wanted. Yet this is not the case and the reason for this limit is related to the wavelength of the light that is illuminating the object you are examining. You can think of the wavelength of the light as your probe, and any details smaller than your probe would be difficult to see (or resolve). A useful analogy is to consider that you are blindfolded and asked to examine an unknown object using only the palms of your hands instead of your fingers. Your fingers are smaller and therefore able to detect finer details. The same is true for the wavelength you use to look at an object through a microscope. Now consider using something that has a shorter wavelength than light to illuminate the object. You might consider using X-rays or gamma rays, which certainly have shorter wavelengths. Unfortunately you cannot easily build an optical system that focuses these forms of radiation, and if you could the radiation tends to have such high energies that it would just pass straight through the object. Of course various types of X-ray machines use these attributes to image bones and airport luggage. Now consider using a beam of electrons. They have an associated de Broglie wavelength that can be many times smaller than the wavelength of visible light.

You can easily focus and manipulate the beam using electrostatic and electromagnetic lenses. You can also easily ‘see’ the electrons using detectors ranging from a simple fluorescent zinc sulfide screen to specialised complex semiconductor arrays. 1 Discuss how the wavelength of the radiation that illuminates the object affects the detail you can see with a microscope. 2 Determine the size of the smallest feature you could visualise using a light microscope. 3 Research to find the maximum magnification possible using an electron microscope. 4 Research and investigate the different types of electron microscopes, including: a transmission electron microscope b scanning electron microscope c scanning transmission electron microscope d reflection electron microscope.

Extension 5 Outline the different ways in which specimens are prepared if they are to be examined with an electron microscope. 6 Discuss why the way in which most specimens need to be prepared differs from they way they are prepared when we use an optical light microscope. 7 Investigate the advantages and disadvantages of using an electron microscope.

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14 neutron, beta decay, atomic mass, nucleons, strong nuclear force, atomic number, nuclides, transmutation, alpha decay, decay series, atomic mass unit, fission, nuclear reactor, controlled nuclear reaction, atomic pile, critical mass, uncontrolled nuclear chain reaction, neutron scattering

20th century alchemists Base metals to gold The idea of changing base metals into gold has been an age-old quest of alchemists. Their attempts were not successful because all the types of reactions the alchemists performed were merely chemical reactions. To achieve their dream, they needed to change the identity of the atom by changing the number of protons in the nucleus. In this chapter we will trace the history of nuclear physics, the ‘modern alchemy’, from the early 1930s through to the construction of the first man-made nuclear reactor.

14.1 Discovery of the neutron

Figure 14.1.1

260

James Chadwick

Rutherford proposed the idea of the neutron in 1920. At the time he considered the neutron to be a system comprising a proton and an electron tightly bound together. This system provided an explanation for the ejection of electrons from the nucleus during beta decay. James Chadwick (1891–1974) set to work to find the neutron. Chadwick set up two main experiments in which he fired alpha particles at a beryllium target. He allowed the unknown radiation generated in the first experiment to pass through paraffin blocks and in the second experiment to pass through nitrogen gas. He then applied the conservation laws of momentum and energy, by setting up simultaneous equations that involved the known masses and velocity measurements of the alpha particle and ejected nuclei from both experiments. Chadwick solved these equations in order to determine the likely mass of the ‘unknown radiation’. A month after commencing these experiments, Chadwick submitted for publication a paper entitled ‘Possible existence of a neutron’ in which he reported the neutron to be just slightly more massive than a proton.

quanta to quarks

Many late nights

Table 14.1.1  The properties of nucleons

C

P Snow, a research student at the Cavendish Laboratory, recalled that Chadwick worked night and day for 3 weeks in his quest to identify the neutron.

Property

Proton

Neutron

Charge Mass

+1.6 × 10–19 C 1.673 × 10–27 kg

Neutral 1.675 × 10–27 kg

With the discovery of the neutron, the nucleus of the atom was considered to consist of protons and neutrons; the number of protons equalling the number of electrons in the neutral atom. The number of neutrons could vary to account for the observed differences in the atomic mass of the atom. The proton and the neutron are referred to as nucleons, when they are in a nucleus. Be foil

α particles

Figure 14.1.2

Be foil

paraffin

?

protons (hydrogen nuclei)

N2 gas

Discuss the importance of conservation laws to Chadwick’s discovery of the neutron. Define the components of the nucleus (protons and neutrons) as nucleons and contrast their properties.

nitrogen nuclei α particles

?

Chadwick’s two main experiments

Checkpoint 14.1 1 2 3

Describe Rutherford’s vision of the neutron. Outline how Chadwick determined the mass of the unknown radiation from his experiment. Describe the structure of the atom prior to and after the discovery of the neutron.

14.2 The need for the strong force The new model of the nucleus posed a new problem. What forces held the protons and neutrons together to form a stable nucleus? The new model of the nucleus posed a new problem. What forces hold the protons and neutrons together to form a stable nucleus? Is it gravity? Assuming the size of the nucleus is roughly ~1 × 10–15 m, the gravitational force acting on a proton in the nucleus of a helium-4 atom due to the other proton and 2 neutrons is approximately 5.6 × 10–34 N (can you calculate it?) The force of Coulomb (electrostatic) repulsion between the two protons in the nucleus is approximately 200 N, which is ~1035 times larger! (See Physics Phile ‘Repulsive protons’.)

Evaluate the relative contributions of electrostatic and gravitational forces between nucleons.

Repulsive Protons

T

he electrostatic force between charged particles follows Coulomb’s law, an almost identical law to Newton’s gravity, except that it’s much stronger and it can be repulsive. Let’s calculate the repulsive force between two protons in a helium nucleus: qq Felectrostatic = k 1 22 d (1.6 × 10−19 C)(1.6 ×10−19 C) = 9.0 ×109 (1.0 × 10−15 )2 ≈ 230 N

where k is the electrostatic constant, q1 and q2 are the charges of the protons and d is the distance between them.

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20th century alchemists

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Clearly, the force of gravity was too weak to overcome the huge electrostatic repulsive force produced by the interaction between protons—a third fundamental force of nature was required to explain the stability of the nucleus. Experiments over the period from 1930 to 1950 showed that this new strong nuclear force possessed the following properties. 1 The strong nuclear force does not depend on the charge, therefore all nucleons (protons and neutrons) bind together with the same force. This is supported by evidence that protons and neutrons are equally likely to be ejected from a nucleus in a collision. 2 The strong nuclear force acts over short distances of about 1 × 10–15 m, the diameter of a nucleus, and within this range the force is much stronger than the electrostatic forces. The evidence supporting this is that otherwise the nucleus would have a tendency to attract more protons and neutrons. 3 The strong nuclear force between the nucleons acts only between adjacent nucleons. The evidence supporting this is based upon the observed stability of the nucleus.

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repulsion

repulsion

Figure 14.2.1 shows the forces between two nucleons, and from the graph you can see that the strong nuclear force is at a maximum at a distance of approximately 1.3 × 10–15 m. If the nucleons are less than 0.5 × 10–15 m apart, a repulsive force is present. If we consider the resultant force acting on a proton or an alpha particle approaching a nucleus, we see that the Coulomb repulsion increases and then sharply decreases once the particle is within the range of the strong nuclear force (see Figure 14.2.2). Coulomb repulsion

3

Figure 14.2.1

4

Separation of nucleons × 10–15 m

Nuclear force versus separation between nucleons

Force

2

attraction

1

attraction

Force

+

range of the nuclear force

Distance from nucleus

nuclear attraction

Figure 14.2.2

Force acting on a positive charge as it approaches the nucleus

Checkpoint 14.2 1 2

What is the difference in the magnitude of the electrostatic and gravitational forces between protons in the nucleus? List the properties of the strong nuclear force.

Account for the need for the strong nuclear force and describe its properties.

262

14.3 Atoms and isotopes The number of protons in a nucleus is referred to as the atomic number (Z). The total number of protons and neutrons is called the atomic mass number (A).

quanta to quarks The number of neutrons (N) can be calculated by subtracting the atomic number (Z) from the mass number (A). N=A–Z The nuclear structure of an element is represented by ZA X where X is the element symbol. Atoms of each element have the same number of protons in their nuclei, but the number of neutrons can vary, and these atoms are called isotopes. There are 91 naturally occurring elements and we have identified more than 2500 different nuclei (or nuclides) associated with these. Of these isotopes, 90% are unstable and often quickly decay into other nuclei.

Checkpoint 14.3 Copy and complete the following table to identify the number nucleons and electrons in elements. Element

Mass number

Atomic number

Protons

Neutrons

Electrons

C He F Xe

14.4 Transmutation Artificial transmutation The process of changing one element into another is called transmutation. Transmutation occurs naturally in stars through the process of nuclear fusion and here on Earth via natural radioactivity by which certain elements decay spontaneously. In 1919, Rutherford fired alpha particles into nitrogen gas and detected a highly energetic particle that he identified to be a proton. This was actually the first artificially induced transmutation—the alpha particle had collided with the nitrogen nucleus to produce an oxygen nucleus and a highly energetic proton. Rutherford initially detected the production of very high energy protons and, after further experiments in which he used a cloud chamber, he identified that this was not just a simple collision. Rather his calculations indicated that the nitrogen nucleus had absorbed the alpha particle and then ejected a high-energy proton. Rutherford then showed that the new nucleus was oxygen. He had changed nitrogen-14 into oxygen-17, as prescribed by this equation: 4 2

He +

14 7

Define the term ‘transmutation’.

N → 178 O + 11 H

Natural transmutation Today we know that nuclei with more than 83 protons (Z > 83) or atomic mass numbers greater than 209 (A > 209) are unstable and decay. In these atoms, the repulsive electrical forces between the protons overcome the strong nuclear force. We also know that instability due to odd number pairing of protons and neutrons The two most common natural decay processes can occur in smaller nuclei. are the emission of alpha and beta particles.

Describe nuclear transmutations due to natural radioactivity.

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20th century alchemists

N 238U

146

4.47 × 109 y

145

234Th

144

24.10 d

143 142 141 140 138 1600 y

137 136 134

131

214Pb 27 min

129

Tl

19.9 min 214Po

α decay β decay

1.637 × 10–4 s

1.30 min

127

214Bi

19.9 min

130210 128

218Po

3.10 min

210Pb 22.3 y 210Bi 5.01 d

126 125

210Po

138.38 d

206Pb 124 80 81 82 83 84 85 86 87 88 89 90 91 92 93 Z Hg Tl Pb Bi Po At Rn Fr Ra Ac Th Pa U Np

Figure 14.4.1

Alpha-particle emission usually occurs with large unstable nuclei. An alpha particle is identical to a helium nucleus 42 He . After the nucleus has emitted the alpha particle, the atomic mass A decreases by four, corresponding to the loss of two protons and two neutrons, the atomic number Z decreases by two, corresponding to the loss of two protons, and the number of neutrons N decreases by two, corresponding to the loss of two neutrons. The general equation for this process is: A A−4 4 → AZ −−42Y + α    or more formally   Z X → Z − 2Y + 2 He X represents the nuclei of the parent element and Y represents those of the daughter element. A common naturally occurring alpha decay is that of uranium-238 decaying to thorium-234: A ZX

226Ra

222Rn

3.82 d

135

132

230Th

7.54 × 104 y

139

133

234Pa

70 s 234U 5 2.46 × 10 y

Alpha decay

This Segrè chart is a plot of number of neutrons (N  ) against the atomic number (Z  ) for the uranium-238 decay series.

238 92 U

→

234 90Th

238 92 U

+ α    or more formally  

→

234 90Th

+ 42 He

Beta decay There are three different types of beta-decay processes: beta-minus (β–), beta-plus (β+) and electron capture.

Beta (minus) decay The beta-minus particle is identical to an electron. Beta-minus decay occurs when the ratio of neutrons to protons is too high, and involves the transformation of a neutron into a proton, an electron and an antineutrino νe . The transformation process is represented as: n → p + β− + ν e

The generic equation for this process is: A ZX

The Unseen

I

n many parts of the world the bricks, cement and soil contain significant quantities of isotopes from the uranium-238 series. Radium-226 decays to radon-222—a colourless, odourless, inert radioactive gas that is heavier than air and often settles in poorly ventilated cellars and basements. Now consider breathing in a radioactive gas! It decays by ejecting a high-energy alpha particle that can easily damage lung cells. But that is not the worst of it! A radioactive nucleus of polonium-218 remains; it is not a gas and is very likely to lodge itself in your lung tissue and continue to decay down the 238U series. So avoid hiding in the cellar or basement!

264

→

A Z +1Y

+ β − + νe    or more formally  

A ZX

→

A Z +1Y

+

0 −1 e

+ 00 νe

The atomic mass Z remains the same, as the total number of nucleons is unchanged. The atomic number A increases by one and the number of neutrons N decreases by one, corresponding to the transformation of a neutron into a proton. A commonly occurring beta-minus decay is thorium-234 to protactinium-234: 234 90Th

→

234 91 Pa

+ β − + νe   or again formally 

234 90Th

→

234 91 Pa

+

0 −1 e

+ 00 νe

Beta (plus) decay The beta-plus particle is identical to an anti-electron commonly called a positron. The process of beta-plus emission involves the transformation of a proton into a neutron, positron and a neutrino. The transformation process is represented as p → n + β + + νe . The general equation for this process is: A ZX

→

A Z −1Y

+ β + + νe

or more formally

A ZX

→

A Z −1Y

+

0 +1 e

+ νe

A commonly occurring beta-plus decay is the transformation of neon-19 to flourine-19: 19 10 Ne

→ 199 F + β + + νe or again formally

19 10 Ne

→ 199 F +

0 +1 e

+ 00 νe

quanta to quarks A

Electron capture

238

This is the process in which an electron from an inner shell is captured by the nucleus, resulting in the conversion of a neutron to a proton, for example:

234

+ e − → 73 Li + νe

The three types of beta decay are mediated by the fourth force of nature called the weak nuclear force, which was proposed by Enrico Fermi in 1934 to explain beta decay. As Rutherford and Soddy’s 1903 paper stated, often a series of decays occurs. For example, in nature the decay series for urainium-238 is common. Figure 14.4.1, a Segrè chart, and Figure 14.4.2, a plot of atomic mass against atomic number, both show the decay series for uranium-238 ( 238 92 U ), which undergoes eight alpha decays and six beta decays, terminating at the stable isotope ). The original nucleus for each transition is called lead-206 ( 206 82 Pb the parent and the resultant nucleus is called the daughter.

Gamma radiation After the emission of an alpha or beta particle, the daughter nucleus is sometimes left in an excited state. The change in energy from this excited energy state of the nucleus to the ground state of the nucleus will result in the emission of a very high energy photon called a gamma ray and is represented by the Greek letter γ. Gamma radiation does not change the atomic number (Z) or the atomic mass (A) and therefore it is not an example of transmutation. A general equation for gamma emission can be written as:

24 d 6.7 h

2.5 x

α decay

105

y

7.5 x 104 y 226Ra

226

1600 y 222Rn

222

3.8 d

218

3.1 min

214

218At

218Po

214Pb 27 min 20 min

3.1 min 1.6 s

218Rn

1.6 s 0.04 s

214Bi 214 Po 20 min

1.6 x 10–4 s 210Tl 210 Pb 210Bi 210Po 210 1.3 22 y 5 d min 5 d 138 d 206Tl 206Pb

206 4.2 81min82 Tl Pb

83 Bi

84 Po

Figure 14.4.2

85 At

86 Rn

87 Fr

88 Ra

89 Ac

90 Th

91 Pa

92 Z U

Plot of atomic mass (A  ) against the atomic number (Z   ) for the uranium-238 decay series

PRACTICAL EXPERIENCES Activity 14.1

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Checkpoint 14.4 Define a transmutation and outline how Rutherford made this happen. Identify how Rutherford showed that the high-energy protons were not the result of a collision. List the properties of an atom that will naturally transmute. List the types of decay that can occur to help stabilise a nucleus and the properties each possess.

14.5 The neutrino In alpha-particle decay, the energies of the ejected alpha particles have well-defined values of kinetic energy, whereas in beta-particle decay a broad spectrum of kinetic energies for the beta particles is observed. Many scientists including Otto Hahn (1879–1968), Lise Meitner (1878–1968) and Chadwick had all carefully studied beta decay. By 1930 experiments clearly showed that the spectra of the kinetic energies of ejected beta particles when graphed was a smooth and continuous curve ranging from just above zero to a maximum

Relative number of β particles

1 2 3 4

234U

230Th

230

A ZX

→ +γ The * denotes the nucleus is in an excited state.

234Pa

234Th

β decay

e

A * ZX

4.5 x 109 y

I

17 4 Be

238U

9 8 7 6 5 4 3 2 1 0

1 2 3 4 5 6 7 8 9 10 11 12 Kinetic energy of the β-particles (MeV)

Figure 14.5.1

Energy distribution for beta particles 265

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20th century alchemists

Discuss Pauli’s suggestion of the existence of the neutrino and relate it to the need to account for the energy distribution of electrons emitted in β-decay.

value that was dictated by the parent nuclei. Refer to Figure 14.5.1 for a typical energy distribution for beta particles. In 1930, Wolfgang Pauli (1900–1958), in a letter addressed to Lise Meitner and Hans Geiger, proposed that a new particle was also emitted during the beta-decay process. Pauli proposed that this new particle had no charge and would only very weakly interact with matter. This newly proposed particle would allow the distribution of energies to be explained. The energy released during a decay could be shared between the beta particle and the neutrino in any ratio; thus, the spectrum of energies observed could be explained. Enrico Fermi formally proposed a theory for beta decay whereby a neutron in the nucleus of an atom was transformed into a proton, an electron and an antineutrino. His theory also proposed a fourth force of nature, the weak nuclear force (commonly referred to as the ‘weak force’). The scientific community accepted the theory, even though Reines and Cowan did not experimentally detect the neutrino until 1953.

The elusive neutrino

Checkpoint 14.5

he thermonuclear reactions in the core of the Sun create an environment that produces streams of neutrinos. As you stand at midday with the Sun overhead, approximately 1013 neutrinos per second pass through your body. At midnight when the Sun is on the other side of the Earth, amazingly still about 1013 neutrinos per second pass through your body. They have passed straight through the Earth! From this you can see how weakly neutrinos interact with matter, and shows the extent of the task physicists had to surmount to detect these elusive particles.

Explain what led to the need for the neutrino.

T

Are You Nuclear Free?

D

id you know that one very widespread application of nuclear physics is the smoke detector? The detector commonly used in houses has a small amount of americium-241, which emits alpha particles, which ionise the air between two charged electrode plates. A small current normally flows between the plates; but if smoke enters the detector, it disrupts the normal current and the alarm sounds.

14.6 Was Einstein right? Using a mass spectrometer the masses of electrons, protons and nuclei (as ions) can be determined by examining their radius of curvature as they move through a magnetic field. The common unit used to measure these extremely small masses on an atomic scale is the atomic mass unit (amu), which is used rather than the kilogram (kg), the SI unit for mass. Using this unit the neutral carbon-12 atom has a mass of exactly 12 amu. It is easy to convert between atomic mass units (amu) and kilograms (kg). 1 amu = 1.6605 × 10–27 kg

Worked example Question a Convert 4.05 amu to kilograms. b Convert 5.023  × 10–27 kg to amu.

Solution a 4.05 × 1.6605 × 10–27 = 6.73 × 10–27 kg b

5.023 × 10−27 1.6605 × 10−27

= 3.025 amu

Einstein was right! Using his energy–mass equivalence E = mc2 allows us to also convert an energy value into an equivalent mass or, vica versa, a mass into an energy. The common unit of energy in atomic physics is the MeV (megaelectron volts 1 × 106 eV). Converting between amu and MeV is also easy: 1 amu = 931.5 MeV. 266

quanta to quarks Worked example Question Calculate the equivalent energy contained in 0.0015 amu.

Solution 0.0015 × 931.5 = 1.4 MeV

Mass defect The total mass of a stable nucleus is less than the total sum of the masses of the protons and neutrons it contains. How can this be? Einstein linked mass to energy in his equation E = mc 2. In our case, when a nucleus forms, some energy is radiated away. This loss of energy reduces the mass of the nucleus. Conversely, when the unstable nucleus of a large atom splits into two or more fragments, the combined mass of the daughter nuclei (and possible neutrons) is less than the mass of the parent nucleus. This loss of mass is caused by some of the original parent mass being transformed into the kinetic energy of the fragments and also possibly gamma radiation. The difference in mass between the total original mass and the total final mass is called the mass defect.

Explain the concept of a mass defect using Einstein’s equivalence between mass and energy.

Solve problems and analyse information to calculate the mass defect and energy released in natural transmutation and fission reactions.

Worked example Question Calculate the mass defect for a helium-4 atom ( 42 He), given the rest mass of a helium atom is 4.002602 amu.

Solution A helium-4 atom consists of a nucleus (2 protons and 2 neutrons) and 2 electrons. Calculate the total mass of these constituent particles: • • • •

Mass of protons = 2 × 1.007276 amu Mass of neutrons = 2 × 1.008665 amu Mass of electrons = 2 × 0.000549 amu The total mass of the constituent particles is 4.032980 amu.

Calculate the difference in masses by subtracting the total mass of the constituent particles from the actual mass of the atom: 4.032980 amu – 4.002602 amu = 0.030378 amu Therefore the mass defect for the helium-4 atom is 0.030378 amu.

Figure 14.6.1

The mass of constituent parts is greater than the mass of the whole nucleus.

Checkpoint 14.6 1 2 3 4

State what amu stands for and give reasons for its use. Convert 3.1 amu to kilograms. Calculate the number of MeV from 0.3 amu. Define mass defect.

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20th century alchemists

14.7 Binding energy Another useful way of interpreting this mass defect in atoms and nuclei is to look We can use Einstein’s equivalence of mass and at it in terms of energy. energy to convert this mass defect into what is referred to as the binding energy of the nucleus. The binding energy tells you how much energy you would need to separate the nucleus of the atom back into separate protons and neutrons.

Worked example Question Calculate the binding energy for the helium-4 ( 42 He) nucleus.

Solution In our previous calculation we determined the mass defect for the helium-4 atom ( 42 He) to be 0.030378 amu. Using Einstein’s equivalence between mass and energy we simply convert the mass defect into the units of energy by using the relationship 1 amu = 931.5 MeV. The binding energy for helium-4 nucleus ( 42 He) = 0.030378 × 931.5 = 28.3 MeV.

To investigate the stability of various nuclei, the binding energy (Eb) per nucleon can be calculated by dividing the binding energy by the total number of nucleons (protons and neutrons) in the nucleus. E binding energy Binding energy per nucleon = = b atomic mass number A

Worked example Question Calculate the binding energy per nucleon for helium-4 nuclei ( 42 He).

Solution In our previous calculation we determined the binding energy for the helium-4 atom ( 42 He) to be 28.3 MeV. We know that the atomic mass number for helium-4 is 4. E binding energy Therefore: Binding energy per nucleon = = b atomic mass number A 28.3 MeV = 7.075 MeV 4 We can see from Figure 14.7.1 that elements with atomic mass numbers between 40 and 80 have nuclei that are tightly bound. The binding energy of elements with atomic mass number greater than 80 is slightly less, and the binding energy of elements with atomic mass number less than 40 decreases sharply. You will notice that 42 He has quite a high binding energy per nucleon, which makes it very stable and explains why alpha particles 42 He rather than single protons are ejected from nuclei in alpha decay. The graph also shows that if you fuse light nuclei together you would increase the binding energy per nucleon, thus energy would be released. This process, called fusion, is the nuclear reaction that provides the energy source for stars. You can see that if the nucleus of a heavy atoms splits, the binding energy per nucleon will increase. This process is called fission, and is the energy source used in nuclear reactors. =

268

Binding energy per nucleon Eb/A (MeV/nucleon)

quanta to quarks 9

20 16O Ne

8

12C

7

4He

Ca Fe

Kr Hg

greatest stability

fission

U

6 5 4 3 fusion

2 1 0 0

Figure 14.7.1

50

100 150 Mass number (A)

200

Binding energy is greatest for elements with atomic mass numbers between 40 and 80.

Checkpoint 14.7 1 Define binding energy. 2 Referring to the graph (Figure 14.7.1), explain how binding energy can help you understand when energy is released or absorbed.

14.8 Nuclear fission Between 1934 and 1938, Enrico Fermi and his fellow researchers fired neutrons at a variety of target elements and produced many new unstable radioactive nuclei. In the majority of cases some nuclei in the target would absorb a neutron and then emit a beta particle. This beta-decay process transforms the neutron into a proton, an electron and an antineutrino, thus increasing the atomic number by one. Fermi predicted that he should be able to produce transuranic elements (elements beyond uranium) with atomic numbers greater than 92 (Z > 92). Fermi bombarded uranium with neutrons and tested the properties of the new radioactive elements. He was sure he had produced element number 93. Otto Hahn (1879–1968), Lise Meitner (1878–1968), Otto Frisch (1904– 1979), Fritz Strassman (1902–1980), Irene Joliot-Curie and Pavle Savitch (1909– 1994) all repeated Fermi’s experiments in order to identify the new isotopes. With Hitler’s invasion of Austria in 1938, Lise Meitner (an Austrian Jew) emigrated to the safety of Sweden. Hahn and Strassman continued the work and, a few months later, Hahn wrote to Meitner outlining the final results of their research, which indicated that smaller nuclei were present. Further research supported these results, confirming that the decay products of the bombardment of uranium were not transuranic elements but rather unstable isotopes that included barium (Z = 56), radium (Z = 88) and lanthanum (Z = 57), all of which have nuclei with atomic numbers much less than that of uranium (Z = 92).

Describe Fermi’s initial experimental observation of nuclear fission.

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Checkpoint 14.8 1 Identify what a transuranic particle is. 2 Outline the experiment performed by Fermi which resulted in nuclear fission and not the production of transnuranic elements.

Earth’s Own Fission Nuclear Reactor

L

ong ago, actually about 1.7 billion years ago in what is now Gabon, Africa, there was a large natural uranium deposit. At that time in geological history there was more urainium-235 present in ore bodies; approximately 3% compared to today’s average 0.7%. The scene was set and nature provided the final ingredient: a flowing aquifer provided a water source that could act as a natural moderator. In the 1970s this ore body was mined and the ratio of 235 U to 238U was found to be significantly lower than normal, revealing the telltale fingerprint of this ancient natural nuclear reactor.

Describe Fermi’s demonstration of a controlled nuclear chain reaction in 1942.

270

14.9 Chain reactions Leo Szilard (1898–1964) realised that if the fission fragments included two or more neutrons, these neutrons could cause other uranium nuclei to split, resulting in a self-sustaining chain reaction. Further experiments by American and French scientists showed that indeed the fission of a uranium nucleus releases between two and four energetic neutrons. There are many dozens of possible decay pathways, for example: 238 92

U+n→

239 92

U→

90 36

Kr + 146 Ba + 3n 56

The scientists working on this research realised that this chain reaction if controlled could be used as a useful power source, but if it was allowed to proceed uncontrolled, it could produce a huge explosion. In August 1939, war with Germany seemed imminent and the American-based scientists felt that America should investigate the possibility of building a bomb. Leo Szilard, Eugene Wigner and Edward Teller drafted a letter that was signed by Einstein and sent to the US President Franklin D Roosevelt. It was known that 238 92 U would not support a chain reaction because the neutrons released during the fission process did not have the right energy to cause other 238 92 U nuclei to split. However, the rarer naturally occurring isotope 235 acted differently and was known to fission when struck by a slow neutron. 92 U In 1935 Enrico Fermi and his team set about designing and carrying out an experiment to see if a controlled nuclear reaction was possible. In 1942 Fermi’s team designed and commenced building a fission reactor in a squash court at Chicago University. Some 40 000 graphite bricks, weighing 350 tonnes, were used as a moderator to slow down the high-energy neutrons. They also provided the reactor with a structural component that gave the reactor its name, ‘the pile’. Natural uranium in the form of uranium oxide, which contains about 0.7% 235 , provided the fuel for this reactor, and a set of control rods made from 92 U sheets of cadmium nailed to sticks of timber acted as ‘neutron sponges’. These could be withdrawn or inserted into the ‘pile’ to control the number of neutrons and thus the rate of the fission reactions. The atomic pile took 6 weeks to construct, and in early December 1942 the control rods were withdrawn 15 cm at a time. With each withdrawal the neutron counters would climb and level off. At 2:00 pm, Fermi announced that, based upon his calculations, the next withdrawal would result in a self-sustaining reaction. As predicted, the neutron intensity increased more and more rapidly. 1 After 4 2 minutes Fermi raised his hand and announced: ‘The pile has gone critical’. The control rods where pushed back in and the reactor shut down.

quanta to quarks

Figure 14.9.1

Fermi’s atomic pile

Controlled and uncontrolled nuclear chain reactions A chain reaction requires: 1 fuel that is capable of fission; referred to as fissile material 2 an amount of fuel that produces sufficient neutrons to cause new fission; this is referred to as the critical mass 3 neutrons with an energy that allows them to be absorbed by other fuel nuclei. Slow (or ‘thermal’) neutrons are often best, as the de Broglie wavelength is larger at low speeds. This larger wavelength increases the likelihood of a neutron interacting with nearby nuclei and hence the likelihood of it being absorbed.

Compare requirements for controlled and uncontrolled nuclear chain reactions.

The conditions for a controlled nuclear chain reaction are such that the available neutrons, which cause the fission, are regulated. The control mechanisms in a nuclear reactor include the use of neutron-absorbing materials within the physical structure and the manipulation of control rods. In an uncontrolled nuclear chain reaction, such as in a nuclear reactor meltdown or in a fission atomic bomb, the production of neutrons goes unchecked and the fission reactions increase at an accelerating rate. This process releases an enormous amount of energy in a very short period of time and results in a significant explosion.

Checkpoint 14.9 1 Explain how a chain reaction can be sustained in nuclear fission. 2 Outline Fermi’s experiment to show nuclear fission. 3 Give reasons why artificial nuclear fission is a desired process. 271

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14.10 Neutron scattering

Describe how neutron scattering is used as a probe by referring to the properties of neutrons.

The discovery of the neutron and the further development of nuclear reactors provided scientists and industry with a powerful analysis tool. Chadwick’s work had revealed that neutrons were subatomic particles with a mass approximately the same as that of the proton, and no net electric charge. From de Broglie’s work we also know that neutrons will exhibit wave properties and their de Broglie wavelength will vary, depending on their velocity. Interestingly, although neutrons have no net charge, it was discovered they do have the property of spin; they therefore possess a magnetic property and act like little subatomic magnets. These properties make the neutron an excellent tool with which to probe aspects of atomic structure. There are four main applications of neutron scattering: • Spacing: The de Broglie wavelength associated with slow (thermal) neutrons allows investigations to determine the structure and spacing between atoms in solids and liquids. • Motion: The energy of slow (thermal) neutrons is similar to the energies associated with the movements of atoms in solids and liquids, allowing researchers to investigate atomic vibrations and the forces between atoms. • Magnetic structure: The neutron acts like a subatomic magnet, which allows researchers to examine at the atomic level the structures of semiconductors and magnetic materials. • Inner structure: X-rays and electrons are scattered by atomic electrons, but neutrons are scattered by atomic nuclei and, unlike alpha particles, are not influenced by electrostatic forces. This allows neutrons to penetrate dense materials to depths of several centimetres, therefore making it possible for researchers to study material deep inside large pieces of equipment (such as aircraft engines), and inside containment vessels that have varying conditions of pressure, temperature and environment. The first neutron-scattering experiments were carried out by Ernest Wollan and Clifford Shull in 1945 using another graphite pile reactor built at Oak Ridge. Today neutron diffraction is used in combination with X-rays to investigate the structures and properties of a wide range of materials. The Bragg Institute at the Australian Nuclear Science and Technology Organisation (ANSTO) OPAL reactor facility, located just south of Sydney, currently houses eight neutron-beam instruments. The neutron-scattering and X-ray techniques are used to solve complex research and industrial problems across a wide range of fields including engineering, pharmaceuticals, mining, plastics and biology.

Checkpoint 14.10 1 Outline how neutrons can be used to probe atomic structure. 2 List applications of neutron scattering.

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PRACTICAL EXPERIENCES

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CHAPTER 14

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 14.1: Detecting radiation Using different pieces of detection apparatus, determine the type of radiation and the penetration ability of the radiation being emitted. Equipment list: alpha, beta and gamma radiation sources, aluminium foil of varying thicknesses, paper, 2 mm thick lead sheets, Geiger–Müller tube, spark counter, cloud chamber, dry ice, methylated spirits, ruler. Discussion questions 1 List the apparatus best suited to detect each form of radiation. 2 In order of penetrating ability, list each radiation. 3 Identify the properties that make each radiation identifiable.

Perform a first-hand investigation or gather secondary information to observe radiation emitted from a nucleus using the Wilson Cloud Chamber or similar detection device.

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14 • •

• •

•

•

•

• •

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Chapter summary

Rutherford proposed the idea of the ‘neutron’ in 1920. In 1934 Chadwick fired alpha particles at a beryllium target and allowed the unknown radiation generated to pass through paraffin blocks and nitrogen gas. Chadwick then applied the conservation laws of momentum and energy to determine the mass of the neutron. The proton and the neutron are referred to as nucleons when they are in a nucleus. A third fundamental force called the strong force was required to explain the stability of the nucleus, as the force of gravity was too weak to overcome the huge electrostatic repulsive force produced by the interaction between protons. The strong nuclear force has the following properties: – It does not depend on the charge, therefore all nucleons (protons and neutrons) bind together with the same force. – It acts over a short distance, in the order of the size of a nucleus, and the force within this range is much stronger than the electrostatic forces. – It acts only between adjacent nucleons. An element’s nuclear structure is represented by ZA X where X is the element symbol. – The number of protons in a nucleus is referred to as the atomic number (Z). – The total number of protons and neutrons is called the atomic mass number (A). – The number of neutrons (N) can be calculated by subtracting the atomic number (Z) from the mass number (A). Isotopes are atoms of an element that have the same number of protons but a different number of neutrons in their nuclei. The process of changing one element into another is called transmutation. The first verified artificially induced transmutation was reported in 1919 by Rutherford when he fired alpha particles into nitrogen gas.

• •

• •

•

• •

•

•

•

•

•

•

The two most common natural decay transmutation processes are the emission of alpha and beta particles. Alpha decay usually occurs with large unstable nuclei and an alpha particle is identical to a helium nucleus 4 2 He . There are three different types of beta-decay processes: beta-minus (β–), beta-plus (β+) and electron capture. In 1930 Pauli proposed the existence of the neutrino to account for the energy distribution of electrons emitted in beta decay. The total mass of a stable nucleus is less than the total sum of the masses of the protons and neutrons it contains. The difference in mass between the total original mass and the total final mass is called the mass defect. The binding energy tells you how much energy you would need to separate the nucleus of the atom back into separate protons and neutrons. Fermi’s initial nuclear fission experiments involved firing neutrons at a variety of target elements and producing many new unstable radioactive nuclei. In 1942 Fermi’s team designed and built a fission reactor (an atomic pile) in a squash court at Chicago University. A chain reaction requires: 1 fuel that is capable of fission 2 a ‘critical mass’ of fuel, which produces sufficient neutrons to cause new fission 3 neutrons with an energy that allows them to be absorbed by other fuel nuclei. A controlled nuclear chain reaction requires the number of available neutrons that cause the fission to be regulated. An uncontrolled nuclear chain reaction occurs when the production of neutrons goes unchecked and the fission reactions increase at an accelerating rate. Neutron-scattering techniques are used to solve complex research and industrial problems across a wide range of fields including engineering, pharmaceuticals, mining, plastics and biology.

Review questions

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Physically speaking Pretend you are Charles Wilson, the inventor of the cloud chamber, shown in Figure 14.11.1. Name the components of your new invention. Prepare a short scientific report on your new invention for presentation to your class.

A

B

a Discuss the features of the apparatus; you

C

could also include a description of what happens when you place a magnet on top of the apparatus.

E

b Complete your report, outlining the importance of it to reveal the nature of matter.

F D

Figure 14.11.1 Wilson cloud chamber

Reviewing

7 Copy and complete the following table. Process

1 Name the scientist who predicted the existence of

Fission

the neutron.

Fusion

2 Copy and complete the following table for the

Alpha emission

properties of the nucleons of an atom.

PROPERTIES

Neutron

Beta emission

Proton

Symbol

Describe Fermi’s demonstration of a controlled nuclear chain reaction in 1942.

Charge Mass (kg)

Compare requirements for controlled and uncontrolled nuclear chain reactions.

3 Define the term transmutation. Discuss the importance of conservation laws to Chadwick’s discovery of the neutron.

4 Recall how the conservation laws used by Chadwick

8 Outline Fermi’s 1942 demonstration of a controlled nuclear reaction.

9 Describe the requirements for: a a controlled chain reaction b an uncontrolled nuclear reaction.

allowed him to identify the neutron.

5 Assess why the invention and development of the Wilson cloud chamber was important in identifying transmutation.

a brief description of each process



6 Describe the properties of the strong force.

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10 Copy and complete the following table which outlines the events associated with the discovery and identification of the neutron.

Scientist(s) 1920 1930 1932

Contribution

Firing alpha particles into nitrogen gas

Observed transmutation and also predicted the existence of the neutron Identified the production of a weak but penetrating ‘radiation’ Published a paper describing …

Bothe and Becker Joliot Curie and ______________

1934

Experiment

______________

Fired alpha particles into a beryllium target and allowed the ‘radiation’ to pass through a paraffin block Fired ______________ into a beryllium target and allowed the radiation to pass through a ______________ and ______________

Used the conservation laws on his measurement observations and identified that the ‘radiation’ was most probably composed of ______________

11 Referring to Figure 14.11.2, describe in words the

repulsion

force acting on a proton or an alpha particle that is fired directly toward the nucleus.

Force

Coulomb repulsion

range of the nuclear force

distance from nucleus

N

KE of beta particles

Figure 14.11.3

attraction

13 Describe the following processes associated with nuclear attraction repulsion in the nucleus

Figure 14.11.2 Discuss Pauli’s suggestion of the existence of the neutrino and relate it to the need to account for the energy distribution of electrons emitted in β-decay.

12 This question relates to the graph in Figure 14.11.3. a Explain why the distribution of kinetic energies for the beta particles was so baffling for the scientists. b Name and recall the properties of the new particle suggested by Pauli. c Account for the energy distribution of the electrons emitted in beta decay.

276

nuclear transmutations due to natural radioactivity. Include any other particles that may be present in the process and write word equations for each process. a Alpha emission (α-decay) b Beta emission (β-decay) c Emission of gamma radiation (γ) d Another naturally occurring transmutation

Solving problems 14 Consider a helium nucleus where the distance

between the protons is 1 × 10–15 m and given that qq mm Felectrostatic = k 1 2 and Fgravitational = G 1 2 d2 d2 a Calculate the electrostatic force between the two protons. b Calculate the gravitational force between the two protons. c Account for the need of the strong force.

quanta to quarks Solve problems and analyse information to calculate the mass defect and energy released in natural transmutation and fission reactions.

15 Calculate the mass defect for a lithium-6 atom ( 63Li),

given the rest mass of a lithium-6 atom is 6.015122 amu and the following rest masses: proton = 1.007276 amu neutron = 1.008665 amu electron = 0.000549 amu

16 Helium ( 42He) has a binding energy of 7.1 MeV

per nucleon. Given that the mass of an electron is 0.000549 amu, a proton is 1.0073 amu and a neutron is 1.0087 amu, calculate: a the total binding energy in the 42He nucleus in MeV b the mass defect of the 42He nucleus in amu c the mass of a 42He atom in amu.

Binding energy per nucleon (MeV )

9 8 7

16O

4He

56Fe

208Pb

12C

235U

9Be

6

6Li

5 4

2.9 MeV/nucleon

3 3H 3He 2

1.15 MeV/nucleon

2H 1 1H

0

0

50

100

150

200

250

Atomic mass (A)

Figure 14.11.4

18 Consider this fission reaction of the uranium-235 nucleus: 235 1 131 92U + 0 n → 53I +

? + 3 01n

a Identify the missing product in the above equation. b From the information in Figure 14.11.4, estimate the binding energies of the original uranium-235 and the two daughter nuclei.

19 Identify the missing components in the following reactions: a

1 65 66 0n + 29Cu → 29?

b

22 ? 11Na → 10?

+ +01e + ν

20 Tritium and deuterium are isotopes of hydrogen and the symbols T and D are often used when writing nuclear reactions.



Consider the following fusion reaction: 3 T + 2D → 4He + 1n + γ 1

1

2

0

Given that the mass of deuterium is 2.014 amu, tritium is 3.016 amu and helium is 4.003 amu, calculate the energy of the γ-ray. iew

Q uesti o

n

s

v

energy versus mass number. a Define the term binding energy. b Construct an argument based on the information in Figure 14.11.4 why alpha particles rather than just protons are ejected from some radioactive nuclei. c Estimate the binding energy of iron.

Re

17 Figure 14.11.4 shows a graph of average binding

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PHYSICS FOCUS Probing Atomic Structure In 1934 Ernest Rutherford’s prediction of neutrons was proved by his associate James Chadwick, work that gained Chadwick the Nobel Prize in Physics in 1935. More recently, Bertram Brockhouse and Clifford Shull shared the Nobel Prize for Physics in 1994 for pioneering contributions to the development of neutron-scattering techniques for studies of condensed matter. This is the modern application of neutron scattering—determining the arrangement and motion (structure and dynamics) of atoms and molecules, both of which can determine the physical properties of the material they form. Today, these methods are practised at neutronscattering centres such as the Bragg Institute at the Australian Nuclear Science and Technology Organisation (ANSTO) located at Lucas Heights, south of Sydney. Neutrons are used to study atomic positions, motions and magnetic properties of materials. Neutron scattering provides unique information about a material that cannot be obtained by other methods.

H3. Assesses the impact of particular advances in physics on the development of technologies Define the components of the nucleus (protons and neutrons) as nucleons and contrast their properties.

1 Discuss why neutrons are so useful in probing the structure of materials. 2 Discuss the role of neutrons in fission reactors and outline how they could be used as a neutron source. 3 You are a physicist at ANSTO, describe how you would construct apparatus to provide your researchers with a beam of neutrons.

Extension 4 Investigate and report on the ways in which you could detect scattered neutrons. 5 Neutron scattering is very sensitive to different isotopes of an element (atoms that have nuclei with the same number of protons but different numbers of neutrons). Outline some reasons why this would be the case.

Figure 14.11.5  Dr Vanessa Peterson is changing a sample in a high-temperature furnace near Echidna, the highresolution neutron powder diffractometer at ANSTO. In this experiment, the structure of the material was determined, as it changed with temperature.

278

The particle zoo The atomic age The unleashing of the power stored in the nucleus of an atom dramatically shaped the world we live in today. The detonation of two atomic bombs dropped on Japan in 1945, and the post-World War II threat of atomic weapon proliferation, leading to the Cold War between the Soviet Union and the West, changed the lives of millions of people. On the scientific front, the development of particle accelerators allowed the three constituents of matter known in the 1930s to grow to several hundred by the mid-1970s. The secrets of the atom were being uncovered and the actual fabric of our universe was being unveiled. The giant accelerators were recreating the processes that occurred during the first few fractions of a second during the birth of time and space—the Big Bang.

15 Manhattan Project, isotope, Los Alamos, fusion, fission, fuel rods, core, moderator, control rods, coolant, radiation shielding, radioisotopes, half-life, particle accelerator, cyclotron, positron, radiopharmaceuticals, radiotherapy, irradiated, cosmic rays, muons, synchrotron, linear accelerators, quarks, hadrons, bosons, pions, Standard Model, mesons, baryons, leptons

15.1 The Manhattan Project The United States of America’s top-secret nuclear bomb research project, code named the Manhattan Project, ushered in the atomic age. The project comprised many sub-projects including Fermi’s ‘atomic pile’ in Chicago. Another reactor was built near Richland in Washington to produce plutonium, and a gaseous diffusion plant in Oak Ridge Tennessee was constructed to extract the from uranium ore. At the famous Los Alamos facility in fissile isotope 235 92 U New Mexico, the theoretical and experimental work for the design, development and testing of the atomic bomb was carried out. At Los Alamos, research into the construction of two types of atomic bombs was carried out. A uranium-235 bomb named ‘Little Boy’ was never tested prior to being dropped on Hiroshima on 6 August 1945. A more complex plutonium-239 bomb, which was tested at the Trinity site in the New Mexico desert, resulted in the construction of the bomb named ‘Fat Man’, which was dropped on Nagasaki on 10 August 1945. The driving fear that fuelled the Manhattan Project centred on the possibility that Germany was building its own bomb. After World War II had ended, it was found that Hitler’s military had invested the majority of its scientific research and development into conventional weapons and rocketry. They had access to a heavy water plant in Norway, but very little progress had been made toward the construction of an atomic bomb.

PRACTICAL EXPERIENCES Activity 15.1

Activity Manual, Page 119

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The Manhattan Project: its legacy

Gather, process and analyse information to assess the significance of the Manhattan Project to society.

The Manhattan Project left a legacy that shaped the 20th century in ways that few could have imagined. The project led to the post-World War II nuclear arms race between the Soviet Union and the West. After the test detonation of the Soviet Union’s first bomb in 1949, relations between the Soviet Union and the West deteriorated. This tension became known as the Cold War and continued until the early 1990s; it ended with the dissolution of the Soviet Union. The United States of America, Britain and the Soviet Union developed fusion bomb technology, known as the thermonuclear bomb, the H-bomb or the hydrogen bomb. By the mid-1960s France and China had tested atomic weapons, and today Israel, India, Pakistan and North Korea also possess atomic weapons. Despite the Nuclear Non-proliferation Treaty, which pursued a reduction in weapon stockpiles, today there is a worldwide arsenal of approximately 20 000 nuclear weapons. In tandem with the military developments, civil projects using nuclear energy have also proliferated. Today there are 440 nuclear power plants worldwide, which supply 15% of the world’s electricity demand, and there are more than 200 research reactors. The use of reactors for nuclear medicine, neutron scattering and other industry-based applications highlights the peaceful endeavours pursued by scientists and engineers today.

Checkpoint 15.1 1 2 3 4

Outline the parts and purposes of the Manhattan project. Justify the project being ‘fuelled’ by the race to beat Germany to building the bomb. What was the Cold War? List developments that have come about due to the knowledge gathered from this project (other than weapons).

15.2 Nuclear fission reactors Nuclear reactors have a variety of applications in modern society. They are used as an energy source to generate electricity for power grids and to power naval vessels. Research laboratories and industry use reactors as a neutron source to probe the structure and properties of materials, and irradiate materials to make a range of industrial, agricultural and pharmaceutical isotopes.

Components of a typical nuclear fission reactor

Explain the basic principles of a fission reactor.

280

Fission reactors are carefully designed to control the rate of fission (splitting) of large nuclei to ensure their safe operation. The fission process releases energy, which manifests itself as heat and high-energy gamma radiation. For the purpose of this course we will limit our discussion to uranium fuel nuclear fission reactors, which comprise six key components: fuel rods, core, moderator, control rods, coolant and radiation shielding. • Fuel rods: These are tubes filled with pellets of uranium oxide. The fuel rods are located in the centre of the reactor known as the core. • Core: The core of the reactor houses the fuel rods, control rods, a coolant system and a moderator material (which sometimes also acts as a coolant).

quanta to quarks pressuriser (prevents water boiling)

containment building primary coolant circuit (high pressure water)

boiler

pump

water

condenser

turbines and electricity generator to grid

control rod

Figure 15.2.1 sea water to condense steam

fuel rod

water acts as coolant and moderator in reactor core

• Moderators: These materials slow down neutrons, improving the chance of the neutron being captured by a nucleus. Commonly used moderators include ordinary water, heavy water and graphite. The neutrons are slowed by colliding inelastically with the moderator nuclei. For water, the neutron has about 20 collisions, which reduce its kinetic energy from about 2 MeV to 1 eV. • Control rods: These are usually made from boron or cadmium and are located between the fuel rods. The control rods absorb neutrons and are adjusted so that the chain reaction proceeds at a constant rate. A constant rate is achieved when, for the average fission, each ejected neutron initiates one new fission. If the reactor needs to be shut down for service or in an emergency, the control rods are inserted fully into the core. These capture the neutrons and the chain reaction ceases. • Coolant: The reactor generates heat and the coolant transfers this heat away from the core. In some reactors normal or heavy water is used as a coolant and also serves as a moderator. In nuclear electrical power stations, the heat is used to make steam, which powers a turbine and generator. • Radiation shielding: Reactor cores emit large quantities of gamma radiation and neutrons. Lead and graphite are used to absorb and reflect radiation, which protects the containment walls of the reactor vessel. The walls themselves are made from high density concrete. These radiation shields are designed to protect people and the environment and prolong the working life of the reactor facility.

A schematic diagram of a pressurised water reactor. A typical 1000 MW power plant consumes about 6 000 000 tonnes of black coal each year, or about 25 tonnes of enriched uranium that has been obtained from around 75 000 tonnes of ore.

Nuclear energy of fuel

Heat energy of coolant

Kinetic energy of steam and turbine

Electric energy

Light, heat, sound, etc.

Figure 15.2.2

The energy stored in the nuclear fuel undergoes a number of transformations.

The nuclear reaction resulting from the absorption of a slow (or thermal) neutron by uranium-235 has many possible pathways for the decay process to proceed. Following are three examples: 235 92

U+n→

235 92 235 92

236 92

U→

U+n→

236 92

U+n→

236 92

140 55

93 Cs + 37 Rb + 3n

U→

147 57

U→

90 36

La + 87 Br + 2n 35

Kr + 143 Ba + 3n 56 281

15

The particle zoo

Checkpoint 15.2 1 2

Define fission. Construct a table that lists the six key elements in a nuclear reactor and outline their purpose.

15.3 Radioisotopes Radioisotopes are increasingly being used in medicine, scientific research and industry. They provide a unique tool that is often cheaper and more effective than alternative techniques and processes. Radioisotopes are atoms that have an unstable ratio of protons to neutrons and will decay via alpha or beta decay to attain a more stable configuration. Some may also emit gamma radiation. Each radioisotope has a specific half-life—the time it takes half of the radioisotope sample to decay. For example, if a sample contains 12 grams of a radioisotope with a half-life of 5 hours, in 5 hours’ time half of the radioisotope will have decayed and 6 grams will remain. Five hours later there will be 3 grams of the original radioisotope present in the sample, and so on. Radioisotopes occur naturally or can be produced by changing the ratio of nucleons in the nuclei of a target material in a nuclear reactor or a particle accelerator. A nuclear fission reactor provides a source of neutrons that can be used to irradiate a target material. A particle accelerator such as a cyclotron can accelerate protons or ions and fire them into a target material. Both processes are required to produce the range of radioisotopes used in medical and industrybased applications. In Australia, the Australian Nuclear Science and Technology Organisation (ANSTO) facilities in Sydney, produces a range of radioisotopes. ANSTO operates Australia’s only nuclear research reactor, OPAL, at Lucas Heights and the National Medical Cyclotron at The Royal Prince Alfred Hospital. It is necessary to have both the research reactor and the cyclotron, as each produces different types of radioisotopes. Pharmaceutical companies, research centres and hospitals operate other medical and industrial cyclotrons.

Radioactive tracers Often radioisotopes are used in medical and industrial applications to track the movement, flow or absorption of materials. The radioisotope can be incorporated into a molecule or introduced into the system. These isotopes then act as tracers and allow the position, flow or absorption to be mapped. In complex biological systems it is possible to map the pathway of specific chemicals being transported and used by cell tissue. For example, glucose is metabolised by brain tissue and so the radioisotope carbon-11 can be placed into the glucose molecule to act as a tracer in medical imaging. The carbon-11 via beta-decay is a positron emitter, and the radioactive labelling of the glucose molecules can be tracked and imaged using positron emission tomography (PET). An industrial application is the use of a radioactive tracer in mapping the dispersion of a sewerage ocean outfall.

282

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Medical radioisotopes Nuclear medicine has been used routinely since the 1970s. Radioisotopes that decay quickly are said to have short half-lives, and are often used in diagnostic procedures. Radioisotopes with longer half-lives are often used therapeutically to target diseased organs and tumours.

Describe some medical and industrial applications of radioisotopes.

Radiopharmaceuticals Radiopharmaceuticals can be classified as diagnostic or therapeutic. The overarching process on which this pharmacology is based is the specific uptake and absorption of chemicals by organs and specific body tissues. This provides an ideal mechanism that enables radioisotopes to be incorporated into or attached to molecules, which are then tracked or used to target specific tissues. Diagnostic radiopharmaceuticals are used to assess the functioning of organs including the lungs, heart, liver and brain; identify bone fractures not visible in X-rays; and assess the flow of fluids such as blood. The decay of the diagnostic pharmaceuticals can be monitored by detectors that may comprise a simple device such as a Geiger counter or a complex array of detectors that convert the information into an image. Diagnostic radiopharmaceuticals subject the body to a very low radiation dose, usually comparable to a routine diagnostic X-ray. Therapeutic radioisotopes generally contain radioisotopes with longer half-lives and once absorbed by the specific tissue or organ will deliver a target dose of radiation.

Radiotherapy In radiotherapy the radiation emitted from a radioactive source is directed at an area of diseased tissue. This procedure is referred to as teletherapy when the source is located outside the body. In brachytherapy a radioactive implant is used to target the specific tissue.

Positron emission tomography (PET) This imaging technique also relies on the principle that specific molecules are absorbed by specific organs or tissues. In this imaging procedure the radioisotope used will decay and produce a positron via beta decay. The positron (an antimatter electron) will collide with an electron and the two will annihilate, producing two gamma rays. These are detected and triangulated with other events by a computer, and the location and activity of the targeted organ or tissue is imaged. (See Chapter 19 ‘Imaging with gamma rays’ for more details.)

Industrial radioisotopes Radioisotopes are used widely in industry across applications as diverse as checking the structural integrity of bridges, determining wear in engine components, examining welds in gas lines, imaging internal structures in jet aircraft engines, and thickness control. Radioisotopes commonly used in measuring the thickness of materials include iridium-190 and cobalt-60. The most common (often unrecognised) application of a radioactive isotopes is smoke detection, and americium-241 is the radioisotope most commonly used in home-based detector systems.

Figure 15.3.1

Brachytherapy for prostate cancer can be administered using ‘seeds’, small radioactive rods implanted directly into the tumour.

283

15

The particle zoo Eyes

phosphorus-32

Thyroid

technetium-99m iodine 123 iodine-131 iodine-132 selenium-75

Brain

iodine-131 mercury-197 technetium-99m oxygen-15 carbon-11

Blood

Heart

iron-59

rubidium-81m thallium-201 cesium-137

Lungs

xenon-127 xenon-133 nitrogen-13 oxygen-15 carbon-11 indium-113m technetium-99m

Spleen

chromium-51 rubidium-81 technetium-99m

Kidney

technetium-99m mercury-197 iodine-131

Liver

technetium-99m iodine-131 gold-198

Pancreas

selenium-75

Bladder

gold-198

Placenta

iodine-123 iodine-131 carbon-11

Knee

yttrium-90 rhenium-186

Prostate

gold-198 indium-111

Bone

strontium-85 strontium-87 fluorine-18 iron-52 phosphorus-32 technetium-99m

phosphorus-32 arsenic-74 indium-133m iron-18

Lymph

gold-198

Figure 15.3.2

Commonly used radioisotopes and the organs in which they act

Agricultural radioisotopes

PRACTICAL EXPERIENCES Activity 15.2

Activity Manual, Page 122

Radioactive isotopes are used in the agricultural industry as tracers in plants to explore chemical and biological processes. As in medical applications, specific chemicals are taken up and used by specific parts of plants. Radioisotope tracers are attached or incorporated into molecules, which are then taken up by plants. For example, the radioisotopes phosphorous-32 or nitrogen-15 can be added to fertilisers and the uptake of the fertiliser measured by a sensitive Geiger counter. Similarly, toxic heavy metal compounds such as those of mercury or cadmium can be introduced into a test soil. The uptake of these tracer radioisotopes can be used to investigate potential problems associated with livestock feed or food for human consumption when planted in areas containing these toxins.

Food irradiation In many countries certain foods are irradiated in order to increase the shelf life and make some foods safer to eat. The exposure of the food to gamma radiation targets disease-causing bacteria or those that cause spoilage. A common, widely used radioisotope is cobalt-60. The gamma rays passing through the food possess enough energy to destroy many bacteria without Identify data sources, and gather, changing the texture or flavour of the food. The food never comes into contact process, and analyse information with the radioactive source or other forms of radiation and therefore is not at risk to describe the use of: of itself becoming radioactive. Examples of irradiated foods found worldwide • a named isotope in medicine include potatoes and onions in which sprouting is reduced, grains, meats • a named isotope in agriculture • a named isotope in engineering. including poultry and some fish, many spices and dried herbs, and fresh fruits. 284

quanta to quarks

PHYSICS FEATURE Hydrogen as an energy carrier

R

ecent scientific interest in hydrogen as an alternative energy carrier to carbon-based fossil fuels such as petrol has generated questions that can be answered using neutrons generated by a fission reactor, and neutron-scattering techniques. One of the problems with using hydrogen as an energy carrier is the difficulty in storing such a small molecule. Before we can study new hydrogen-storage materials, we need to know where the hydrogen is in the material and how it is interacting with the storage material. At Australia’s OPAL research reactor, neutron diffractometers such as ANSTO’s high-resolution neutron powder diffractometer (Echidna) can be used to ‘see’ hydrogen. The protium (1H) in the material is replaced with deuterium (2H), and the neutron scattering by 2H that is obtained in a neutrondiffraction experiment allows the position of the deuterium in the material to be established. For example, this technique was used to determine nine positions on the inner surface (referred to as adsorption sites for molecular deuterium) within the porous material Cu3(1,3,5benzenetricarboxylate)2, which adsorbs hydrogen. The neutron scattering arising from the deuterium molecules in this material is shown in yellow in Figure 15.3.3.

Figure 15.3.3

Neutron scattering was used to identify the hydrogen sites in the porous material Cu3(1,3,5-benzenetricarboxylate)2.

Neutron spectrometers give information about the dynamics and motions of atoms and molecules. Neutron spectroscopy takes advantage of the exceptionally strong scattering from hydrogen 1H, which dominates the spectrum obtained. The scattering from 1H gives information about the dynamics (motions) of the hydrogen and can be used to understand the types of interactions that the hydrogen has with the host material. In the case of Cu3(1,3,5-benzenetricarboxylate)2, some of the nine sites for molecular hydrogen have a stronger adsorption enthalpy with the material than others.

Checkpoint 15.3 1 2 3 4 5 6 7 8 9

Define the term radioisotope. Explain what half-life is. Outline how radioisotopes can be made. Recall what is a radioactive tracer and what it is used for. Explain the significance of using different half-lives in medical radioisotopes. Outline what radiopharmaceuticals are used for. Compare and contrast radiotherapy and brachytherapy. Recall one use of radioisotopes in industry and one in agriculture. Why can radiation be used to sterilise food safely? 285

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15.4 Particle accelerators Identify ways by which physicists continue to develop their understanding of matter, using accelerators as a probe to investigate the structure of matter.

Prior to the development of particle accelerators, scientists had observed showers of particles caused by cosmic rays colliding with the Earth’s atmosphere. By the late 1940s, Carl Anderson (1905–1991) had identified positrons and muons created by cosmic rays. Cosmic rays bombard the Earth continuously, and they provided many challenges to scientists, who had to use balloons to lift equipment high into the atmosphere to carry out many experiments. This was not very convenient and what was required was the development of a more controllable ground-based laboratory. The subsequent development of a variety of particle accelerators has provided scientists with the tools to probe and investigate the structure of matter.

Basic accelerator design and principles All particle accelerators have three basic components: • a source of charged particles (ions or elementary particles) • a tube or chamber that is highly evacuated so that the particles can travel without colliding with air molecules • a mechanism to accelerate and control the trajectory of the particles. There are a number of common designs for particle accelerators including electrostatic, linear, cyclotron, betatron, synchrotron and storage-ring colliders for particle accelerators. Modern accelerator facilities often use a combination of these; for example, a Van de Graaff generator is often used to initially accelerate and inject a particle beam into a larger linear or synchrotron accelerator.

Electrostatic accelerators There are two common electrostatic accelerators, which were developed in the 1930s: the Cockroft–Watson accelerator and the Van de Graaff accelerator. You maybe familiar with the school version Van de Graaff generator (200 kV). The early accelerators were scaled-up versions capable of accelerating particles across potentials of 1.5 MV. The Science Museum in Boston, Massachusetts, now houses the (physically) largest air-insulated Van de Graaff ever built. At the top it has two joined domes, one housing the top of the belt mechanism and the other housing the top of the original accelerator tube. The museum now uses the generator to produce bolts of artificial lightning. Modern Van de Graaff accelerators are smaller and use a tandem system that can accelerate ions across a potential of 20 MV.

Linear accelerators Linear accelerators comprise a very long straight evacuated tube that contains a set of hollow metal drift cylinders (Figure 15.4.1). Each alternate cylinder is electrically connected to an alternating power supply, which provides an alternating potential difference. An ion is injected into the tube and is initially attracted toward the first cylinder. Once the ion is in the cylinder, the electric field drops to zero and the potential polarity is reversed, so that when the ion emerges from the cylinder it is repelled from the first cylinder and attracted to the second. When the ion enters the second cylinder, it enters a region of zero electric field and drifts at a constant velocity through the cylinder. The polarity again reverses and when the ion emerges it is repelled by the second cylinder and 286

quanta to quarks attracted to the third. This process is repeated for the entire journey of the ion down the tube. The alternating potential remains at a constant frequency and, therefore, as the ions increase in velocity, the lengths of the hollow metal drift cylinders are increased. The Stanford Linear Accelerator Center (SLAC) houses the most well-known and longest linear accelerator (Figure 15.4.2). It is 3.2 km long.

Cyclotron In 1932 Ernest Lawrence (1901–1958) and Stanley Livingston (1905–1986) developed a small, compact accelerator that electrostatically accelerated ions or elementary particles across a small gap between two semicircular D-shaped hollow metal cavities called ‘dees’ (see Figure 15.4.3). Once accelerated across this gap, the particle is turned through 180º in a near semicircular path by a constant uniform strong magnetic field. The particle is again accelerated across the gap and again turned through 180º. The two dees are connected to a constant high-frequency alternating power source, which provides the electric field that accelerates the particle across the gap. The polarity across the gap is reversed as the particles are being turned. The electric field inside the semicircular hollow metal dee is zero, and therefore does not affect the path of the particle. As the velocity of the particle increases, the radius of its orbit also increases, resulting in an outward spiral trajectory. One of the dees has an exit point and a beam of particles is produced. A design limit is imposed upon the cyclotron because, if you attempt to accelerate particles beyond energies of 20 MeV, the increase in relativistic mass causes the particles to become out of phase with the constant high-frequency source. At the ANSTO-operated National Medical Cyclotron at The Royal Prince Alfred Hospital, Sydney, the particle beam is used to produce radioisotopes.

source

+

Figure 15.4.1

–

+

–

Structure of linear accelerator components

electron injector damping rings

positron electron

b

c

D1 +

transverse magnetic field

bending magnets

focusing magnets

collision detector

– E +

– D2

high frequency alternating voltage

D1

E

B

d

3 km positron source

Figure 15.4.2 a

+

D2

The Stanford linear accelerator accelerates alternate bunches of positrons (e+) and electrons (e–) to 50 GeV. Magnets separate the particles and bend them in arcs to produce head-on e+– e– collisions. The damping rings and focusing magnets help bunch and focus the beam.

ion or electron source

accelerated particle beam

Figure 15.4.3

(a) Schematic diagram of a cyclotron. (b) Particles are accelerated across the gap by an electric field, so that each semicircular spiral has a larger radius. (c) By the time the particle reaches the gap again, the voltage has reversed and the particle is accelerated. These diagrams show the acceleration of positive ions.

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Van de Graaff accelerator

deflecting magnets

Synchrotron target

high frequency alternating voltage

injection point deflecting magnets

Figure 15.4.4

A Van de Graaff accelerator injects particles into the synchrotron at high speeds. The magnets force the particles into a curved path. Each time the particle passes the high-frequency alternating voltage, it is accelerated.

+ proton source

– antiproton source

Figure 15.4.5

main ring

Using the accelerated particle beams

magnets are coaxial with rings

Tevatron ring

collision detector

The Fermilab accelerator accelerates protons in one ring and antiprotons in the other. The beams are made to collide with a total energy of ~2000 GeV. Superconducting magnets force the particles to stay in the rings.

Switzerland

Geneve LHC–B Point 8

The modern accelerators that strive to produce the highest energy collisions are based upon the synchrotron design (see Figure 15.4.4). Charged particles are injected into a circular evacuated tube that is shaped into a ring. The ring has a series of magnets whose fields can be adjusted to maintain a circular trajectory for the accelerated particles. The ring also has one or more locations where an adjustable high-frequency (radio frequency) alternating power source provides a region in which the particle is accelerated. Both the alternating power source and the magnets are adjusted to increase the speed of the particles and maintain a circular orbit inside the evacuated tube. Modern synchrotrons use superconducting magnets to maintain the trajectories of the accelerated charged particles. The synchrotron at the Fermilab Tevatron has a circumference of 6.28 km (see Figure 15.4.5) and the newly commissioned Large Hadron Collider (LHC) at CERN has a circumference of 27 km (see Figure 15.4.6). The LHC has 1232 superconducting magnets and will accelerate protons up to 10 TeV and collide them head-on, travelling in opposite directions. The high-energy particles can either be fired at a stationery target or collided with other high-energy particles travelling in the opposite direction. Both collisions are governed by the laws of conservation of energy and momentum. Fixed target collisions can involve a wide range of target materials, and they produce secondary beams of particles that can be used in a variety of experiments to probe the nature of matter. In the late 1970s scientists at CERN developed a way to collide protons and antiprotons—a collision that has nearly zero overall momentum. This allowed the total energy of the collision to be available to produce the new varieties of particles that these types of experiments were uncovering.

France

CERN Point 1ATLAS

ALICE Point 2

Point 5 CMS

TI 8 LHC-B

SPS

TI 2

ATLAS ALICE

CMS

Figure 15.4.6 288

Overall view of the LHC experiments

C

/LH

LEP

Figure 15.4.7

An engineer checks one of the magnets in the tunnel of the LHC.

quanta to quarks

tracking device

particle electromagnetic hadronic identification detector calorimeter calorimeter

photons

muon chamber hadron calorimeter E-M calorimeter tracking

electrons or positrons muons

e+

e–

pions or protons neutrons

Figure 15.4.8

The diagram shows typical interactions with sub-detectors for some common particles.

Figure 15.4.9

Schematic design for a typical particle detector

Particle detectors Particle detectors are designed to record and provide a way of ‘seeing’ the explosion of particles that can be generated as a result of the collision. Modern particle detectors are made up of a series of layers of sub-detectors, each specialising in identifying particular types of particles or specific properties (see Figure 15.4.8). These sub-detectors can be classified into three main types: • Tracking devices detect and reveal the trajectory of a particle. Often a magnetic field is used to bend the trajectory of charged particles into a curve. The curvature provides physicists with information about the momentum of the particle. Early detectors such as the cloud chamber and the bubble chamber provided a direct visual record of the paths of particles. Modern tracking devices produce electrical signals that are processed by computers to reconstruct the path of particles. The outer layer of many detectors contain a muon chamber that detect these weakly interacting particles, which can travel through many metres of dense material. • Calorimeters used in detectors are of two types: the electromagnetic calorimeter, which is positioned close to the collision, and the hadronic calorimeter. These sub-detectors are designed to stop, absorb and measure the energy of a particle. The electromagnetic calorimeter measures the energy of lighter particles such as electrons and photons. Hadronic calorimeters absorb the energy of heavier particles, which contain quarks (these types of particles are called hadrons, and include pions, protons and neutrons). The interactions inside both types of calorimeters often cause successive cascades or ‘showers’ of new particles and with each interaction or collision the momentum and energy is shared between more particles, until eventually all are slowed and stopped. To accurately measure the total energy requires the detector to stop nearly all particles, and thus explains the huge overall size of the detectors used. Calorimeters can stop most known particles except muons and neutrinos. • Particle identification detectors identify the type of particle by various techniques. Two commonly used methods to detect the nature of charged particles are Cherenkov radiation, produced when charged particles travel faster than the normal speed of light of that medium, and transition radiation, produced when a charged particle crosses the boundary between certain materials. 289

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PHYSICS FEATURE The world’s largest microscope

H

ow do we study the fundamental building blocks of matter, and the forces that determine how they interact with each other, at the very tiniest of distances? Ironically, we do it by building the world’s largest ever particle accelerator, which acts a bit like a very sophisticated microscope. A 100 years or so ago, Ernest Rutherford showed us that we could learn about what was inside an atom by bombarding it with an energetic ‘probe’, in this case an alpha particle, and observing what happens to it. This principle has served us well, and by applying it Figure 15.4.10 Jason Lee at the site of the giant ATLAS detector, which is in its final stages of construction and testing physicists have burrowed down into the nuclei of atoms to find protons and neutrons, and into the These detectors ‘track’ the emerging charged particles, protons and neutrons to find quarks and gluons. bending them in magnetic fields to work out their charge Many other particles, some of which can only be and momentum from their trajectories. Further out from made in accelerators, have turned up along the the collision point, calorimetry is employed to absorb way. The ones which we view, at least today, as particles such as photons and electrons and measure truly fundamental have been identified, and two of their energy. the forces of nature, the electromagnetic and the Mounting scientific experiments on this scale takes weak, have been unified. the combined efforts of thousands of physicists from all On the outskirts of Geneva, in a 27 km circumference over the world, including Australia. In Figure 15.5.10, tunnel 50–100 m underground, the Large Hadron postgraduate student Jason Lee from the University of Collider (LHC) has the task of colliding two beams of Sydney can be seen standing in front of the giant ATLAS protons at very high energies, in order to allow us to take detector in its final stages of construction. ATLAS will the next step along our road to understanding. Using scour the products of the proton collisions at its centre Einstein’s famous principle relating energy and mass, to search for things such as the ‘Higgs field’, which is the two colliding protons are destroyed and some of believed to be involved in giving mass to the other the energy they carry is transformed into the mass of fundamental particles; yet-to-be discovered new particles. We can understand why the protons ‘supersymmetric’ particles that are predicted by some need to have large energy and momentum by recalling new theories; and possibly even hints of extra de Broglie’s idea that particles can be assigned a dimensions. Jason completed high school in Sydney and wavelength that is inversely proportional to their an undergraduate degree at Oxford University before momentum. The higher the momentum and the shorter returning to Sydney to do his postgraduate studies. Like the wavelength, the shorter the distances we can study. other Australian students in experimental high-energy Many of the particles produced in the collisions live for physics, he spends some of his time at CERN, where he only the tiniest fraction of a second, and their existence can interact with physicists from all over the world and must be inferred from the products of the collision gain experience of the LHC project first hand. For his (i.e. their decay). To observe all of the products of the postgraduate studies he is investigating how well ATLAS collisions in detail, very complex detectors the size of can identify electrons and their antimatter partners, several-storey buildings have been constructed.

290

quanta to quarks

the positrons, and trying to improve the ability of the detector to pick them up. Electrons and positrons turn up in the expected decay products of many of the new particles being hunted for. An example of what a Higgs particle decay might look like in the ATLAS detector can be found in Figure 15.4.11. In this example, the products of the decaying Higgs particle travel upwards in the picture and a spray of other particles, known as a ‘jet’, travels downwards (the yellow cone of particles that is absorbed

in the calorimeters, ending in blue and orange ‘blobs’). The Higgs particle decays to two other short-lived particles known as Z bosons, with one Z boson decaying to an electron and a positron (shown in blue travelling upwards) and the other to a muon and antimuon (shown in red). The other particles in the picture (in yellow) are the debris from the two protons that collided. The human shown in the figure gives an indication of scale. She would not be standing there if the LHC was operating!

Figure 15.4.11 ATLAS detector schematic with a simulated detection of a Higgs boson

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Checkpoint 15.4 1 2 3 4

List the three basic parts of an accelerator. Identify each of the types of accelerators and their uses. List the types of collisions that can be used to identify information about particles. Create a table that identifies the types of detectors that can be used and specify what they will show.

15.5 The Standard Model Discuss the key features and components of the Standard Model of matter, including quarks and leptons.

292

In the early 1930s the list of identified sub-nuclear particles comprised the electron, the proton and the neutron; the positron and the neutrino had been proposed by Pauli. It was not long before the muon and the positron were identified in cloud chambers while examining cosmic showers, and by the 1960s a hundred other particles had been identified. The physicists were confronted with this ever-increasing ‘zoo’ of particles, as the energy of the collisions increased. They had no overall unifying theory to explain the behaviour or adequately classify these particles. The quest to uncover which of these particles were truly fundamental became a primary area of research. A few of the major discoveries that led to the development of the Standard Model of matter are discussed here. In 1964 Murray Gell-Mann (1929–) and George Zweig (1937–) proposed the existence of quarks to explain the properties of a family of particles called hadrons. The hadrons are further subdivided into two types: mesons, those containing two quarks (comprising a quark and antiquark pair), and baryons those containing three quarks. In 1967 the electro–weak theory, unifying the electromagnetic and weak nuclear force, was proposed by the collaboration of Steven Weinberg (1933–) and Sheldon Glashow (1932–), and independently by Abdus Salam (1926–1996). In 1969 the first evidence of quarks was reported by Jerome Friedman (1930–), Richard Taylor (1929–) and Henry Kendall (1926–1999). Then, during the period of 1970–1973, the Standard Model was formulated. All the components of the model except the Higgs boson were verified over the next 25 year period. Today the Standard Model of Matter provides a mechanism to classify and explain the nature of matter. It includes three of the four fundamental The model has three main forces but excludes gravity (Figure 15.5.1). components: quarks, leptons and force carriers. The experiments at Fermilab and the new CERN LHC to search for the Higgs boson will either confirm the Standard Model’s validity or send the experimentalists and theorists back to formulate a new or revised model. Either way, the major challenge facing particle physicists is the need to develop a grand unified theory that will unite all four of the fundamental forces.

quanta to quarks

Quarks There are six varieties (commonly called flavours) of quarks and in all there are 12 types of quarks. The first generation of quarks are named up and down, the second generation are charm and strange, and the third are top and bottom. Each quark is considered to be a fundamental point-like particle that carries the properties of mass, charge and colour. The electric charge of a quark is fractional and is either –1/3e or +2/3e, and they possess a ‘colour charge’ that can be either red, blue or green. Quarks also have antimatter counterparts that possess the same magnitude of electric charge with the opposite sign, and colours that are antired, antiblue and antigreen. Quarks with different colour charge attract each other, as do quark–antiquark pairs of the same colour–anticolour (e.g. red and antired). The colour charge is associated with the strong force and is constantly changing as they interact with gluons. Quarks do not exist as bare entities, and only form composite particles called hadrons. Within this classification, quarks can either pair with an antiquark to form a meson, or three quarks can combine to form a baryon. All hadrons have a total electric charge of 1e, and the total colour charge will sum to white (achieved by the quarks in mesons having a colour and an anticolour, or in baryons by the combination of all the quarks having different colours). atom

neutron

electron

quarks

matter

proton

nucleus Matter particles

These particles transmit the four fundamental forces of nature although gravitons have so far not been discovered

1st FAMILY

Force particles

QUARKS

Electron neutrino Particle with no electric charge, and possibly no mass; billions fly through your body every second

Up Has an electric charge of plus two-thirds; protons contain two, neutrons contain one

Down Has an electric charge of minus one-third; protons contain one, neutrons contain two

2nd FAMILY

These particles existed just after the Big Bang. Now they are found only in cosmic rays and accelerators

LEPTONS Electron Responsible for electricity and chemical reactions: it has a charge of –1 Muon A heavier relative of the electron; it lives for twomillionths of a second

Muon neutrino Created along with muons when some particles decay

Charm A heavier relative of the up; found in 1974

Strange A heavier relative of the down; found in 1964

Tau Heavier still; it is extremely unstable. It was discovered in 1975

Tau neutrino Recently discovered

Top Heavier still

Bottom Heavier still; measuring bottom quarks is an important test of electroweak theory

Photons Particles that make up light; they carry the electromagnetic force

Intermediate vector bosons Carriers of the weak force

3rd FAMILY

All ordinary particles belong to this group

Gluons Carriers of the strong force between quarks

Felt by: quarks The explosive release of nuclear energy is the result of the strong force

Figure 15.5.1

W– Felt by: quarks and charged leptons

Electricity, magnetism and chemistry are all the results of electromagnetic force

Z0

Gravitons Carriers of gravity

W+ Felt by: quarks and leptons

Some forms of radioactivity are the result of the weak force

Felt by: all particles with mass All the weight we experience is the result of gravitational force

A tabular presentation of the three main components of the Standard Model and the three generations of particles classified as quarks, as well as leptons and a description of the gauge bosons that mediate the four forces of nature 293

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Table 15.5.1  Properties of quarks Symbol

Charge

Table 15.5.2  Properties of hadrons Mass

Colour charge

Up

u

+2/3e

Red, blue or green

Down

d

–1/3e

Red, blue or green

Charm

c

+2/3e

Red, blue or green

Strange

s

–1/3e

Red, blue or green

Top

t

+2/3e

Red, blue or green

Bottom

b

–1/3e

Red, blue or green

Hadron

Example

Meson (2 quarks)

Positive pion

π+

Neutral kaon Proton Neutron Sigma plus

0

Baryon (3 quarks)

Symbol

K p n ∑+

Constituent quarks

u d¯ ds¯

uud udd uus

Note: ud ¯ is an up–antidown pair.

Quark road rules

W

hen remembering what quarks are in protons and neutrons, think of overtaking lane signs. The passing lane is a proton—two up quarks and one down quark. No overtaking is a neutron—one up quark and two down quarks. Passing

No passing

proton uud

neutron udd

Leptons There are six varieties and 12 types of leptons. The most commonly known is the electron. All leptons appear to be fundamental point-like particles. The electron, the heavier muon and the massive tau all have an electric charge of –1e and their antimatter counterparts have a charge of +1e. Each of these leptons has an associated neutrino that is named accordingly; for example, the electron (e) has an electron neutrino (νe ).

Checkpoint 15.5 1 2

294

Draw a chart that shows the classes and subclasses of particles, starting with hadrons. List the properties of quarks, hadrons and leptons.

PRACTICAL EXPERIENCES

quanta to quarks

CHAPTER 15

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 15.1: The Manhattan Project Research the Manhattan Project and the events that surrounded it. Assess the impact it has had on society then and now. Discussion questions 1 Outline the discoveries that were made during this project. 2 Describe the effect of the outcome of the project on the war. 3 Assess the significance to society today.

Gather, process and analyse information to assess the significance of the Manhattan Project to society.

Activity 15.2: Isotopes in society Research a series of isotopes that have been developed for medicine, agriculture and engineering purposes. List their advantages over prior methods. Discussion questions 1 Name an isotope that is used in: a medicine   b agriculture   c engineering. 2 Describe the purpose of one of these isotopes and how it is used. 3 Explain how the use of that isotope has replaced previous technology.

Identify data sources, and gather, process, and analyse information to describe the use of: • a named isotope in medicine • a named isotope in agriculture • a named isotope in engineering.

295

15 • • •

•

•

•

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The particle zoo

Chapter summary

The Manhattan Project was the code name for the USA’s top-secret nuclear bomb research project. Nuclear fission releases energy that manifests itself as heat and high-energy gamma radiation. Fission reactors are carefully designed to control the rate of fission (splitting) of large nuclei to ensure their safe operation. Uranium-fuelled nuclear fission reactors, comprise six key components: – Fuel rods are tubes filled with pellets of uranium oxide and are located in the centre of the reactor known as the core. – The core of the reactor houses the fuel rods, the control rods, a coolant system and a moderator material (which sometimes also acts as a coolant). – Moderators slow down neutrons and so improve the chance of the neutrons being captured by a nucleus. Commonly used moderators include ordinary water, heavy water and graphite. – Control rods absorb neutrons and are adjusted so that the chain reaction proceeds at a constant rate. The rods are usually made from boron or cadmium and are located between the fuel rods. If the reactor needs to be shut down for service or in an emergency, the control rods are fully inserted into the core. – The coolant transfers the heat generated by the reactor away from the core. In nuclear power stations the heat is used to make steam, which powers a turbine and generator. – Radiation shielding protects people and the environment and prolongs the working life of the reactor facility. Lead and graphite are used to absorb and reflect the large quantities of gamma radiation and neutrons emitted by the core. Radioisotopes are atoms that have an unstable ratio of protons to neutrons. They will decay via alpha or beta decay to attain a more stable configuration; some may also emit gamma radiation. Radioisotopes occur naturally, or they can be produced by changing the ratio of nucleons in the nuclei of a target material in a nuclear reactor or a particle accelerator.

•

•

•

• •

•

•

•

•

Radioisotopes have a specific half-life, which describes the time it takes half of the radioisotope sample to decay. Radiopharmaceuticals can be classified as diagnostic or therapeutic. – Diagnostic radiopharmaceuticals are used to assess the functioning of organs including the lungs, heart, liver and brain; identify bone fractures not visible in X-rays; and assess the flow of fluids such as blood. – Therapeutic radioisotopes generally contain radioisotopes with longer half-lives, and once absorbed by the specific tissue or organ will deliver a target dose of radiation. Radioisotopes have applications as diverse as checking the structural integrity of bridges, determining wear in engine components, examining welds, imaging internal structures in jet aircraft engines, and thickness control. Particle accelerators provide scientists with the tools to probe and investigate the structure of matter. Particle accelerators have three basic components: – a source of charged particles (ions or elementary particles) – a tube or chamber that is highly evacuated so that the particles can travel without colliding with air molecules – a mechanism to accelerate and control the trajectory of the particles. Common particle accelerator designs include electrostatic, linear, cyclotron, betatron, synchrotron and storage ring colliders. Particle detectors are designed to record and provide a visualisation of the ‘explosion’ of particles that can be generated as a result of the collision. The four fundamental forces in nature are gravity, the electromagnetic force, the strong nuclear force and the weak nuclear force. The Standard Model includes three of the four fundamental forces; it excludes gravity. The model has three main components: quarks, leptons and bosons (the force carriers).

Review questions

quanta to quarks

Physically speaking Very quarky 1 3

4

2

5

6 7 8 9

10

11

12 13

14

15

16 17

18

19

Across 1 Discovered the neutron

Down 2 Type of accelerator

6 Slows things down

3 Family name for the carrier of all the forces

9 Made up from any three quarks

4 The anti one often accompanies the beta particle

10 Atom ‘smasher’

5 A project

12 Planck has one

7 You will find them in a nucleus

15 Same number of protons

8 An electron is one

17 A type of scattering

11 Lived 1871–1937

18 Keeps things cool

12 The mass you require to start a chain reaction

19 Type of reactor

13 A constant R 14 They are the reason for the strong force 16 Fermi’s first name

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Reviewing

10 Discuss the purpose of accelerating particles and colliding them with a target (or their antiparticle) in terms of: a the de Broglie wavelength b Einstein’s equation E = mc2 c probing matter d producing new types of particles.

1 Recall three of Enrico Fermi’s contributions to physics. 2 Define the term critical mass. Explain the basic principles of a fission reactor.

3 Describe the roles of the following components in a fission nuclear reactor. a fuel b control rods c moderator d coolant e radiation shields

4 Explain using a schematic sketch the basic processes involved in a nuclear-powered electricity station.

5 Name two materials that could be used in a fission reactor as: a a moderator b control rods c shielding.

6 Define the term radioisotope. Describe some medical and industrial applications of radioisotopes.

7 Name and describe the use of a radioisotope used in: a medicine b agriculture c engineering. Describe how neutron scattering is used as a probe by referring to the properties of neutrons.

8 Neutron scattering is used to probe structures of many materials. Explain how each of the following properties is useful. a the wave nature of the neutron b the magnetic moment of the neutron c strong interaction of the neutron with nuclei d the range of energies the neutron can possess

9 Draw, label and describe the main features of the

Re

iew

298

Q uesti o

n

s

v

following particle accelerators. a Van de Graaff b linear accelerator c cyclotron d synchrotron

11 Identify the ways by which physicists continue to develop their understanding of matter using accelerators. Discuss the key features and components of the Standard Model of matter, including quarks and leptons.

12 13 14 15

Name the six quarks. Name three leptons. Compare fermions and bosons. Recall the constituents of these particles. a meson b baryon c proton d neutron e pion

16 Complete the following table. Types of boson

Role in Standard Model

Photon Strong force Graviton

17 Describe the purpose of the Manhattan Project. 18 Describe the long-term ramifications of the Manhattan Project on society.

19 The Standard Model classifies the constituents of matter into three families. a Name the three families. b Describe the properties of each family.

20 Explain why the top quark was the last of the six quarks to be found.

21 Recall why quarks are never found as individual free particles.

quanta to quarks

PHYSICS FOCUS Linking the very big with the very small The very big and the very small are inextricably linked as cosmology and particle physics enter a new era of research. Many of the recent developments in cosmology have depended on advances in high-energy physics. 1 Explain why it is necessary to accelerate particles to extremely high velocities. 2 Describe how high-energy accelerators such as the Large Hadron Collider can test aspects of theories such as the Big Bang.

H1. Evaluates how major advances in scientific understanding and technology have changed the direction or nature of scientific thinking Identify ways by which physicists continue to develop their understanding of matter, using accelerators as a probe to investigate the structure of matter

Extension 3 Research how high-energy physics was and is still carried out using cosmic rays.

299

4

The review contains questions that address the key concepts developed in this module and will assist you to prepare for the HSC Physics examination. Please note that the questions on the HSC examination that address the option modules are different in structure and format from those for the core modules. Past exam papers can be found on the Board of Studies NSW website.

Multiple choice

4

A nuclear chain reaction requires certain conditions. Which of the following statements provides the best description? A Fuel that is capable of fission B An amount of fuel that produces sufficient neutrons to cause new fission C Neutrons with an energy that allows them to be absorbed by other fuel nuclei. D All of the above

5

Today the Standard Model of Matter provides a mechanism to classify and explain the nature of matter. Which of the following statements is false in relation to the model? A Quarks are the subatomic particles that make up all matter. B The model includes three of the four fundamental forces but excludes gravity. C The model has three main components: quarks, leptons and force carriers. D Hadrons are subdivided into two types: mesons and baryons.

(1 mark each) 1 Which of the following lists only contains people who provided science with atomic models in the period up until 1913? A Bohr, Democritus, Heisenberg, Schrodinger B Aristotle, Bohr, Pauli, Rutherford C Bohr, Dalton, Rutherford, Thomson D Bohr, Einstein, Rutherford, Schrodinger

2

Which of the following scientists provided an explanation for the spectrum of beta-particle energies observed, and proposed that each electron in an atom could be described by four quantum numbers and a system that provided an explanation for the structure of the period table? A Erwin Schrodinger B Werner Heisenberg C Wolfgang Pauli D Louis de Broglie

3

In 1924, de Broglie proposed the concept of matter waves. Which of the following statements is false in relation to de Broglie’s ideas at this time? A All particles of matter would have an associated wavelength. B An electron in orbit around the nucleus would follow a ‘wavey’ path. C The majority of the scientific community did not take de Broglie’s ideas seriously. D There was no experimental evidence to support de Broglie’s ideas.

300

quanta to quarks

Short response 1

4

The diagram below shows the kinetic energy distribution of electrons for the beta decay of bismuth-214.

In 1913 Niels Bohr published a paper ‘On the Constitution of Atoms and Molecules’, which described the planetary model of a hydrogen atom. a In terms of the Rutherford–Bohr model, present a diagrammatic representation of how ONE of the Balmer series spectral lines is produced.  (2 marks) b Calculate the wavelength of the electromagnetic radiation emitted from a hydrogen atom when an electron drops from the third shell to the ground state.  (3 marks) c Outline the ONE spectral observation that the Rutherford–Bohr model could not completely explain.  (1 mark)

Account for the energy distribution of the electrons

2

State the two laws of physics used by James Chadwick when he discovered the neutron.  (1 mark)

5

3

The Standard Model has the components shown in the table.

Quark

Up

Charge

Mass GeV/c2

Down

Strange

Charm

Bottom

N

KE

emitted in beta decay.  (3 marks)

Top

2 + e 3

1 − e 3

1 − e 3

2 + e 3

1 − e 3

2 + e 3

0.005

0.01

0.2

1.5

4.7

180

a Identify ONE way in which physicists have developed their understanding of matter using accelerators.  (2 marks) b Compare neutrons and protons in terms of their constituents.  (2 marks)

Naturally occurring polonium-218 emits an alpha particle to produce a new element X. This new element emits a beta particle to become bismuth-214. Po → X + α + γ a Identify the new element X.  (1 mark) b Describe, using an equation, the decay of element X to produce bismuth-214.  (2 marks) c Calculate the number of protons and neutrons in element X.  (1 mark)

Extended response 6

You have gathered, processed and analysed information to assess the significance of the Manhattan Project. a Describe how you ensured that the information you gathered was reliable.  (2 marks) b Assess the impact the work carried out at Los Alamos has had on society and the environment.  (5 marks)

301

5 Context

Figure 16.0.1

This image shows the pattern of brain tissue loss in methamphetamine users, relative to healthy adults, mapped using MRI scans.

medical physics The human body is a complex machine and many things can go wrong with it. However, unlike a machine, we cannot do a total shutdown to locate and correct a problem. Most of the functions of our body need to continue working when we are looking for a problem and when we are trying to fix it. This is where physics comes to the aid of modern medicine by providing the tools to investigate and, in some cases, repair problems in the living body. We ‘see’ using light, and optical-fibre technology offers a minimally invasive way of looking inside the body to actually see what may have gone wrong. Listening is also an important way for doctors to find evidence of problems but, remarkably, we can now use sound to look deep within the body without ever breaching the surface. Although they present some potential damage to human tissues, X-rays can also give us vital information of ‘structural’ problems below the surface. For more than a century, X-rays offered the only ‘non-invasive’ way to see inside the body. Now, when combined with enhanced computer technologies, X-rays provide powerful tools for creating images of the structures within the body. Modern physics also offers magnetic resonance imaging (MRI) as a powerful method to investigate both the form and function of individual tissues. Carefully chosen radioisotopes offer similar possibilities, because they can be used to identify where atoms are accumulating within our bodies. Other radioisotopes provide a therapeutic tool to help combat certain problems. None of these techniques will reveal everything, but physics has provided doctors with a range of diagnostic and therapeutic tools that earlier generations could never have imagined.

Areas of greatest loss emotion, reward (limbic system) memory (hippocampus) 0% Loss

302

3% Loss

5% Loss

Figure 16.0.2

A drawing of the human brain by Vesalius

Inquiry Activity What makes you tick? The ancient Egyptians knew that the heart was centre of blood supply in the human body. It’s pretty obvious when you look inside the body. However, just how it worked remained a mystery, because looking inside dead bodies was not permitted in many societies. Great advances were made by Andreas Versalius (1514–1564) in the 16th century precisely because dissection was permitted in Italy in some circumstances. In 1543 he published De Humani Corporis Fabrica (On the Fabric of the Human Body), which marks the foundation of the scientific study of human anatomy. The function of many of the organs of the human body has been poorly understood until recently—none more so than the brain. However, modern imaging techniques such as PET and MRI are slowly revealing the operation of the brain by uncovering what parts are involved in different actions or affected by different diseases. In this exercise you should seek out reliable sources of information on the web to investigate the difference between modern structural and functional imaging of the brain. 1 Define what is meant by the structural and functional imaging. 2 How would you classify the images in Figures 16.0.1 and 16.0.2—are they structural or functional? 3 What imaging techniques allow structural differences to be seen in the brain? 4 What diseases might be revealed by structural images of the brain? 5 What imaging techniques allow functional differences to be seen in the brain? 6 Outline one study of the brain that might be conducted using functional images. 7 How reliable are your sources? Why do you expect them to be reliable?

303

16

Imaging with ultrasound See like a bat

transducer, sonogram, Doppler ultrasound, piezoelectric transducer, piezoelectric effect, convex array transducer, sector scan, linear array transducers, resolution, acoustic impedance, impedance matching, acoustic coupling, A-mode scan, B-mode scan, M-mode scan, 2D real-time scan, 3D ultrasound, 4D ultrasound, bone density, dual X-ray absorptiometry, Doppler ultrasound, colour Doppler imaging, Doppler effect, echocardiography

Bats, dolphins and some birds have evolved the ability to produce and sense sound waves that are reflected from objects in their surroundings, enabling them to hunt prey and to navigate in the dark. The principle used by these animals has been applied to radar (using radio waves) and sonar (using sound waves). Since the 1960s, the principles of sonar have also been applied for medical purposes. Ultrasound imaging is a medical procedure used to produce pictures of the inside of the body by using sound waves with a frequency much higher than that audible to humans.

16.1 What is ultrasound? low high low high pressure pressure pressure pressure

compression

Air pressure

high

Distance low

Figure 16.1.1

304

atmospheric pressure

rarefaction

A longitudinal wave with compression and rarefaction pressure variations graphed versus distance

Sound is a vibration in an elastic, mechanical medium that can be solid, liquid or gas. A sound wave is made up of periodic pressure variations in the medium through which it travels. It propagates as a longitudinal or pressure wave of alternating compressions and rarefactions (see Figure 16.1.1 and in2 Physics @ Preliminary section 5.5). The speed of sound is different in different media, as shown in Table 16.1.1. As well as speed, you should recall three other important properties of sound: frequency (the number of vibrations per second), wavelength (the distance between adjacent compressions) and amplitude (the maximum displacement of the particles in the medium from their equilibrium position). The greater the amplitude of a sound wave, the more energy the wave carries. Humans can hear sounds with frequencies between about 20 Hz and 20 kHz. Ultrasound is any sound that has a frequency greater than this upper limit. Other mammals such as bats and dolphins use

medical physics ultrasound waves with frequencies as high as 125 kHz for navigating and sonar visualisation. Ultrasound used for medical purposes has frequencies that range from 500 kHz to 30 MHz, but for most imaging applications, the ultrasound used is in the range from 3.5 MHz to about 10 MHz. The exception is ultrasound used for imaging blood vessels in the body. Tiny probes on the end of a catheter (tube) are inserted into the blood vessels and operate at frequencies up to 30 MHz. The choice of ultrasound frequency that is most suitable for producing a medical image is a compromise between conflicting criteria. Higher frequency ultrasound produces images with a better resolution of detail, but they have poorer penetration through tissues. Imaging depth into tissue is limited by attenuation (absorption) of the ultrasound waves, and this increases as the frequency is increased.

Checkpoint 16.1 1 2 3

Table 16.1.1  Velocity of sound in air and ultrasound in different human tissues

Velocity of sound (m s–1)

Material Air Fat Water Brain Liver Kidney Blood Muscle Lens of eye Skull bone Average soft tissue

330 1450 1480 1540 1550 1560 1570 1580 1620 4080 1540

Identify the differences between ultrasound and sound in normal hearing range.

Explain how sound is made by moving air particles. Outline the difference between sound and ultrasound. List the advantages of low frequency and high frequency ultrasound.

16.2 Principles of ultrasound imaging In ultrasound imaging, ultrasound waves are emitted and detected by a device called a transducer. A transducer is any device that converts energy from one form to another. For example, a loudspeaker is a transducer that converts electrical energy to sound energy; a microphone does the reverse. The ultrasound transducer combines the functions of a speaker and a microphone. It can transmit and detect ultrasound. The ultrasound transducer emits pulses of ultrasound for only ~1% of the time and detects for ~99% of the time. The transducer detects ultrasound waves that are reflected from boundaries between different tissues in the body. The amount of reflection depends on differences in the properties of the two tissues at the boundary. The intensity of the reflected wave returned to the transducer is determined by: 1 the difference in the acoustic properties of the tissues at the boundary— the greater the difference, the more energy is reflected 2 the characteristics of the intervening tissue—some tissues absorb more energy than others 3 the angle at which the ultrasound is reflected from the boundary between two tissues—the production of an image is easiest when the reflection occurs perpendicular to the tissue boundary.

PRACTICAL EXPERIENCES Activity 16.1

Activity Manual, Page 127

A strong reflection from a tissue boundary in the body produces a sharply defined image. A poor image will be produced if the reflection is weak or if tissues in the body absorb the energy before it returns to the transducer. 305

16

Seeing with ultrasound

To produce an ultrasound image or sonogram, a sonographer applies a gel to the area of the patient’s skin overlying the region to be imaged and the transducer is moved across the skin. For example, to obtain images of the heart, the transducer is placed in contact with the chest. A computer processes information from the reflected waves received by the ultrasound transducer to produce real-time images, which are displayed on a screen (Figures 16.2.1 and 16.2.2). The patient does not feel any discomfort. Because the images are produced while the procedure is being done, the sonographer can manipulate the transducer to obtain the best images for the doctor to examine. A variety of display different types of images can be produced, depending on the way the data were collected and how they were processed (Figures 16.2.3 and 16.2.4). disk Ultrasound imaging is one of the safest and least storage keyboard/ CPU expensive medical imaging technologies. It is useful for cursor producing images of soft tissues and organs. Two major printer advantages of ultrasound imaging are that it is non-invasive Transducer and it does not involve use of ionising radiation (as in pulse controls transducer X-rays) and it is therefore very safe to patients. More • frequency • duration invasive procedures carry the risk of infection, and ionising • scan mode sound backing radiation damages DNA. Ultrasound imaging is commonly used to produce PZT crystals images of a developing foetus during pregnancy, where its use can reveal foetal development and movement. Ultrasound imaging is also used to produce images of the kidneys, liver, pancreas and heart. Doppler ultrasound imaging is a special type of imaging used to measure blood flow rate to diagnose heart disease and blocked blood vessels. Figure 16.2.1 Principles of using ultrasound as a diagnostic tool

Figure 16.2.2

306

An ultrasound examination of a pregnant woman. Note the transducer in the technician’s hand and in contact with the patient’s abdomen.

Figure 16.2.3

Colour ultrasound of a kidney

medical physics

PHYSICS FEATURE

3. Applications and uses of physics

Obstetrics

P

robably the most well-known use of medical ultrasound imaging is in obstetrics. It began in 1966, but now most women have ultrasound imaging at some time during their pregnancy. Among the things that can be determined using ultrasound are the number, position and size of babies, if a baby’s internal organs are growing normally and how much fluid is around the baby. Measuring the head size is used to help determine the age of the foetus. Figure 16.2.4 illustrates ultrasound imaging of a foetus. Some doctors have expressed possible concerns about the safety of using ultrasound in making threedimensional images (movies) of foetuses simply so parents can have images of the baby before it is born. Although the risk is low, they argue that if the procedure is not necessary for a medical diagnosis, it should be avoided.

4. Implications of physics for society and the environment

Figure 16.2.4

Ultrasound of a foetus

Ultrasound can be classified as low or high intensity. For sound, intensity corresponds to the loudness of a sound, and it is a measure of the amplitude and the energy of a wave. Low-intensity ultrasound has a minimal effect on tissues as it passes through them. High-intensity ultrasound heats the tissues it passes through. The intensity of ultrasound used to produce images is kept as low as possible to minimise the risk to the patient. Low-intensity ultrasound is considered safe enough to produce images during pregnancy and to produce images of vital organs such as the heart. The use of ultrasound to produce images is referred to as a diagnostic application. High-intensity ‘therapeutic ultrasound’, on the other hand, affects the medium and can be used to heat injured muscles or to disintegrate kidney stones.

Limitations of ultrasound imaging High-frequency sound waves cannot effectively penetrate bone or air, so ultrasound cannot be used to produce images of parts of the body if there is bone or air between the ultrasound transducer and the body part of interest. As a result, ultrasound cannot image the adult brain and parts of the body obscured by gas in the intestines, and produces poor quality images of the lungs. Ultrasound can pass through parts of the skeleton of a foetus before the bone calcifies and hardens.

307

16

Seeing with ultrasound

Table 16.2.1  Advantages and disadvantages of ultrasound Advantages

Disadvantages

It is non-invasive and does not require surgical procedures.

Many reflections occur within the body and therefore good imaging is operator dependent. An ultrasound image is harder to interpret than an X-ray image.

Patients can be examined without sedation, and relatively quickly and conveniently. Since sound is low energy and doesn’t ionise the tissue, it does not damage DNA, cells and tissues. It is relatively cheap compared with other scanning technologies.

It is difficult to produce clear images in obese patients because of attenuation and reflection from fat. The presence of air and bone pose problems because their acoustic impedances are so different from those of soft tissue.

Checkpoint 16.2 1 2 3 4

Explain the purpose of the transducer in ultrasound imaging. Define what is meant by ‘intensity’ of ultrasound imaging. Explain how high frequency sound waves can make images of internal organs. State the difference in the effects of low- and high-intensity ultrasound.

16.3 Piezoelectric transducers Describe the piezoelectric effect and the effect of using an alternating potential difference with a piezoelectric crystal.

force + voltage

– voltage force

Figure 16.3.1

Apply force to a piezoelectric crystal and it will produce a voltage; apply a voltage and the crystal will expand (or contract) slightly.

Ultrasound scanners use a piezoelelectric transducer made from a crystal that responds to an oscillating voltage placed across it by oscillating slightly in size. The oscillating crystal produces a sound wave. It also works in reverse, converting the sound energy into electrical energy. The process is called the piezoelectric effect, and was discovered in 1880 by Pierre Curie (1859–1906) and his brother Jacques (1856–1941). A ceramic called lead–zirconate–titanate (PZT) is commonly used to make piezoelectric transducers. This material responds to relatively small voltages or forces; that is, it has high electromechanical efficiency. The properties of PZT can be altered by modifying the ratio of zirconium to titanium, or by adding small amounts of other materials such as lanthanum. Figure 16.3.2 shows how PZT rods are embedded in plastic, which is then used to construct different types of transducers for ultrasound machines. In ultrasound imaging, a convex array transducer is used to produce a sector scan. The scanning surface of these transducers is curved outwards a

b PZT rods

epoxy resin matrix

Figure 16.3.2 308

(a) PZT rods are embedded into plastic to construct transducers of various shapes, such as (b) a convex array transducer.

medical physics (convex), resulting in a divergent beam (Figure 16.3.3). This is different from earlier linear array transducers, which had parallel beams. Convex array transducers use as many as 512 piezoelectric elements to display a wedge or pie-shaped image. The size of the area of skin through which the ultrasound enters the body (the acoustic window) depends on the curvature of the transducer surface. There is generally a narrow window at the skin surface, while still providing a wide angle deeper beneath the skin (Figure 16.3.4). The narrow field of view close to the transducer makes structures near the transducer difficult to view. The smallest size of objects that can be imaged (resolution) using ultrasound depends on the wavelength (or frequency). All waves have the property of being ‘diffracted’ by objects; that is, their paths are bent around the object. After passing around the object, the paths of the waves recombine and undergo interference (see in2 Physics @ Preliminary section 7.4), which tends to ‘smudge’ the reflection of the object. If the size of the object is approximately one wavelength or less, the smudging becomes so significant that the individual object is not visible, limiting the ability to resolve fine detail. This ‘diffraction limit’ applies to all waves. In ultrasound, this limits resolution both in the direction along the beam (axial) and across the beam (lateral) and is summarised in the Table 16.3.1. Low frequency waves are more effective in penetrating human tissue. This results in a dilemma for ultrasound imaging. In order to produce a clear image of an object deep in the body, low frequency waves should be chosen over high frequency waves. However, to obtain a clearer image of smaller structures, a high frequency needs to be used. Clearly the final choice is a compromise.

data processing computer convex array

organs sector scan

Figure 16.3.3

monitor

A convex array transducer

Table 16.3.1 Frequency and resolution for

Figure 16.3.4

linear array transducers with parallel beams

MHz 3.0 4.0 5.0 7.5 10.0

Axial resolution 1.1 0.8 0.6 0.4 0.3

mm mm mm mm mm

lateral resolution 2.8 1.5 1.2 1.0 1.0

mm mm mm mm mm

The head of a foetus in the womb. Note the small acoustic window— the area through which the ultrasound enters the body at the top of the image, which is the abdomen of the mother.

Checkpoint 16.3 1 2 3 4

Define the term transducer. Outline the energy transformations that are happening in an ultrasound transducer. Define the piezoelectric effect and describe how it is useful in ultrasound imaging. Recall the advantages of convex sector systems. 309

16

Seeing with ultrasound

16.4 Acoustic impedance

Define acoustic impedance: Z = ρυ and identify that different materials have different acoustic impedances. Solve problems and analyse information using: Z = ρυ

The ability of a medium to transmit sound varies between materials. Even for a The acoustic single material, it also varies with the frequency of the sound. impedance describes how readily a specific sound frequency will pass through a material. It is defined as the product of the density ρ (in units of kg m–3) of the material and the acoustic velocity u (in m s–1) in the material: Z therefore has units of kg m

Z = ρυ s , which is called a rayl.

–2 –1

Worked example Question Bone has a density of 2 × 103 kg m–3. The speed of sound in bone is 4080 m s–1. Calculate the acoustic impedance of bone. ρ = 2 × 103 kg m–3 υ = 4080 m s–1

Solution Z = ρυ = 2 × 103 kg m–3 × 4080 m s–1 Z = 8.16 × 106 rayl

Acoustic impedance differs for different tissues in the body because of differences in their densities and the speed of ultrasound in the tissue. Acoustic impedance can be used to predict the proportion of the energy of an ultrasound wave that is reflected from a boundary between two tissues.

Reflection of ultrasound at tissue boundaries Define the ratio of reflected to initial intensity as: 2

Ir Z 2 − Z1  = Io Z + Z  2  2 1 Identify that the greater the difference in acoustic impedance between two materials the greater the reflected proportion of the incident pulse.

When ultrasound waves meet a boundary between tissues with different acoustic impedances, some of the ultrasound energy (intensity) will be reflected and some will be transmitted. Conservation of energy requires that the sum of the reflected and transmitted energies must be equal to the original energy incident on the boundary. Note that for any reflection to occur, the acoustic impedances of the two media Z1 and Z2 must be different. A large difference in impedance results in a large reflection. The incident ultrasound intensity can be represented by Io, with the reflected intensity Ir and the transmitted intensity It, as illustrated in Figure 16.4.1. At a tissue boundary, the ratio of the reflected intensity of ultrasound to the original intensity of the ultrasound is equal to the ratio of the square of the difference of the acoustic impedances to the square of the sum of the acoustic impedances, as represented by the equation: 2 I r  Z2 − Z1  = Io  Z + Z  2  2 1 IO Ir Z1 = ρ1υ1

310

It = IO – Ir

Z2 = ρ2υ2

Figure 16.4.1

Reflection occurs at a boundary between two materials with different acoustic impedance.

medical physics Table 16.4.1  Acoustic impedance of ultrasound in human tissue Density (kg m–3)

Ultrasound velocity (m s–1)

1000

1540

1.54 × 106

920

1450

1.33 × 106

Lens of eye

1100

1620

1.78 × 106

White matter (brain)

1040

1541

1.60 × 106

Grey matter (brain)

1040

1541

1.60 × 106

Muscle

1040

1585

1.65 × 106

Kidney

1040

1561

1.62 × 106

Blood

1060

1570

1.66 × 106

Liver

1065

1549

1.65 × 106

Bone (other than skull)

1810

4080

7.38 × 106

Skull bone

1910

4080

7.79 × 106

Substance Water Fat

Impedance (rayl)

Identify that different materials have different acoustic impedances.

Worked examples Question Calculate the percentage of ultrasound intensity reflected at a boundary between fat and muscle. Fat:

Z1 = 1.334 × 106 rayl

Muscle:

Z2 = 1.648 × 106 rayl

Solution

2

I r Z 2 − Z 1  [(1.648 × 106 ) − (1.334 × 106 )]2 = = 0.011 = Io Z + Z  2 [(1.648 × 106 ) + (1.334 × 106 )]2  2 1 Hence only 0.011 of the incident intensity is reflected. This is 1.1% of the incoming intensity.

Solve problems and analyse information using: 2

Ir Z 2 − Z1  = Io Z + Z  2  2 1

Question Calculate the percentage of ultrasound intensity reflected at a boundary between air and muscle. Air:

Z1 = 0.0004 × 106 rayl

Muscle:

Z2 = 1.648 × 106 rayl

Solution

( (

) ( ) (

6 6 2 I r Z 2 − Z 1  1.648 × 10 − 0.0004 × 10 = = I o Z + Z  2  1.648 × 10 6 + 0.0004 × 10 6  2 1 

) 2 = 0.999 ) 2

Hence 99.9% of the incoming intensity is reflected. Since intensity is proportional to energy, we can say ‘99.9% of the energy’ is reflected.

Bone has a much higher acoustic impedance than other tissues, so most of the ultrasound energy that encounters bone is reflected. For this reason, ultrasound cannot image the internal structure of bone or a brain inside an adult skull. Some ultrasound penetrates the skull of a foetus, because the bone has not yet calcified— it is cartilage, similar to the tissue that gives ears their semi-rigid nature. If ultrasound travelling from air or other gas strikes a soft-tissue boundary of any type, then most of the ultrasound energy is reflected. For this reason, when an ultrasound is being carried out, a gel is placed on the skin and the ultrasound probe is moved over this gel. The most important function of the gel is to exclude any air from the region between the transducer and the skin. The gel has an acoustic

Describe how the principles of acoustic impedance and reflection and refraction are applied to ultrasound.

311

16

Seeing with ultrasound

Solve problems and analyse information to calculate the acoustic impedance of a range of materials, including bone, muscle, soft tissue, fat, blood and air and explain the types of tissues that ultrasound can be used to examine.

impedance similar to that of human tissue. The term used to describe this Impedance process is impedance matching or acoustic coupling. matching between two media results in most of the energy being transmitted through the interface, with almost no reflection. A further complication arises at the boundary between two different tissues, because ultrasound is also refracted at a boundary (see section 18.1 and in2 Physics @ Preliminary section 8.3). This bending of the sound path can complicate the analysis of the data.

Checkpoint 16.4 1 2

Describe the conditions necessary for reflection of ultrasound waves to occur at a boundary. Calculate the percentage of ultrasound energy that reflects from a muscle and bone boundary. (Use values in Table 16.4.1 to help.) Explain why it is necessary to use gel on the transducer in order to get clear images.

3

16.5 Types of scans To perform an ultrasound scan, a transducer is placed against the patient’s skin, directly over the region to be imaged. The transducer sends a very brief pulse of ultrasound into the tissue. The pulse travels into the body as a beam, similar to a searchlight. Interfaces along the way reflect some of the ultrasound energy back to the transducer. The transducer converts the energy of the reflected echo into electrical signals, which are sent into amplifiers and signal-processing components of the ultrasound machine. The exact delay between when the transducer first emits the ultrasound pulse and when it picks up an echo allows the machine to calculate how far the reflecting interface is from the transducer, given the known speed of ultrasound in the tissue. Ultrasound can be used in different ways, called modes, which produce different data and images.

Describe the situations in which A scans, B scans and phase and sector scans would be used and the reasons for the use of each.

A-mode scans 1200 high-frequency signal absolute value of signal Hibert envelope of signal

Amplitude

800 400 0 –400 –800 –1200

2

3

4

5

Time of flight (µs)

Figure 16.5.1

312

An A-mode scan signal

6

7

When a single-element transducer is used and the transducer is stationary, only a single dimension can be recorded. That dimension is the distance of the echoing interface from the transducer. A typical A-mode scan (or A scan) is shown in Figure 16.5.1. Returning echoes produce a vertical displacement of the signal display. The amount of displacement is proportional to the amplitude. This was the first use of ultrasound as it is relatively simple and requires little signal processing to display the data. New ultrasound devices use computers to analyse the various time values. Typically the A scan is used to examine midline structures in the brain, solid or cystic structures, foreign bodies in the eye, abnormal fluid accumulation around the heart or lungs, or to guide biopsy and amniocentesis needles.

medical physics

The most common type of medical ultrasound is called a B-mode scan In a B-mode scan the returning (or brightness-mode scan). echoes are displayed on the screen as dots, rather than as a signal plotted versus time, as in A-mode scans. The intensity or brightness of the dot is proportional to the amplitude of the reflected pulse (Figure 16.5.2). In the simplest scanning mode, a single-element transducer can be moved around to provide different lines probing through the body. Like the beam of a searchlight, it picks out structures along the line of the beam. An image can be built up from these scans (Figure 16.5.3). probe placenta legs

foetal skull

probe

spine probe

organ

to scan display

vertebra

pulse echo

transducer A scan display

Echo strength

B-mode scans

Time

B scan display

Figure 16.5.2

The timing of a return echo indicates depth in the body. The intensity of reflection may be indicated as a signal versus time in A-mode scans or as a bright spot in B-mode scans.

ultrasound image synthesised from scans

Figure 16.5.3

B-mode scan with a moving transducer can be used to reconstruct a two-dimensional ultrasound image of a foetus.

M-mode scans A single-ultrasound beam can be used to produce an M-mode image in which movement of an individual structure such as a heart valve can be displayed versus time. An M-mode scan uses a high-sampling frequency (up to 1000 pulses per second) to capture the rapid motion. This display is essentially a series of B-mode scans lined up side-by-side in time (Figure 16.5.4).

2D real-time scans Using a convex-array transducer with many piezoelectric elements, beams of ultrasound can be launched in different directions into the body. Each beam produces a B-mode scan. Once all the echoes from the first beam have been picked up, the transducer sends a second Figure 16.5.4 An M-mode scan through the left ventricle of the heart allows pulse along a slightly different beam direction measurement of the ventricle size (green markers) as the heart beats. into the tissue. Echoes are again picked up and sent for processing to produce an image, and another pulse is launched in another direction, and so on. Like the beam of a searchlight swept across the night sky, the pulsed beam from an ultrasound transducer is swept through the body, mapping out reflecting surfaces and forming a 2D real-time scan. The echo’s image is two-dimensional and shows detail along 313

16

Seeing with ultrasound

one slice through the body, originating at the transducer. Beams are swept and ultrasound images are formed very rapidly in ‘real-time’ images. As the operator holds the transducer in contact with the skin, the image appears live on a video monitor. Real-time images are usually made with a sector scanner, so the images will be pie shaped. The B-mode scan is continuously updated and displayed in real time, with the image changing as the tissues move in the field of the ultrasound beam. The movement of the object being scanned can be monitored on the image. Interpreting ultrasound images requires considerable skill and knowledge, as the images in Figures 16.5.5 and 16.5.6 demonstrate. An untrained observer would have difficulty making sense of the image without explanation. Figure 16.5.5

This sector scan image of a gall bladder reveals a strong (bright) echo from an abnormally thick wall.

Where to from here? Complex ultrasound transducers are used for producing real-time 3D ultrasound images. These simultaneously produce sector scans in many planes (Figure 16.5.7). Combined with tilting or moving the transducer, this yields ultrasound data of body structures from many angles. High-speed digital signal processing permits the large amount of gathered data to be assembled into a 3D image (Figure 16.5.8).

a

Figure 16.5.6

This image of a kidney reveals a large, dark cyst that shows no internal echoes because of its homogeneous structure.

b

Figure 16.5.7

Examples of scanning patterns produced by modern transducers. Each small blue triangle indicates a sector scan (a) moved in an angle or (b) rotated relative to the transducer.

Ultrasound using this 3D imaging technique is sometimes used to produce a movie. This is usually called 4D ultrasound. There is a section of the ultrasound industry that promotes 4D ultrasound for non-medical reasons—trying to persuade parents that it is fun to have such a record of the child before it is born.

Checkpoint 16.5 1 2 Figure 16.5.8

314

Example of a 3D ultrasound image of a foetus

Summarise the process of (a) A-mode scanning, (b) B-mode scanning and (c) M-mode scanning. Outline examples of when each mode of scanning would be used.

medical physics

16.6 Ultrasound at work Bone-density measurements Osteoporosis literally means ‘porous bones’. Bones that once were strong typically become fragile as a person gets older, due to loss of calcium. The problem is common in postmenopausal women, but a significant number of men also suffer from a loss of bone density. A person may not realise they have osteoporosis until they suffer a vertebral fracture when doing ordinary activities such as lifting a bag of groceries, or break a hip in a fall. To determine if a person has osteoporosis, it is necessary to measure bone density (or bone mineral density). Bone-density testing can measure the amount of bone in different parts of the skeleton, and can predict the risk of future fractures and monitor changes in bone-mineral density due to medical conditions or therapy. There are several ways to measure bone density, most of which involve the use of X-ray radiation. X-ray based methods usually take a bone-density measurement of the hip, spine, forearm or heel by passing two X-ray beams of different energy through the bone. The technique is called dual X-ray absorptiometry (DXA or DEXA). Alternatively, ultrasound can be used to estimate the bone density of the heel, as the heel bone contains a high percentage of the kind of bone most affected by osteoporosis. During an ultrasound examination, two soft rubber pads are placed in contact with either side of the heel. Transducers within these pads send and The speed and absorption receive ultrasound waves through the heel bone. of ultrasound vary with bone density. The test takes about a minute and is performed in a seated position (Figure 16.6.1). No injections or invasive procedures are necessary and the test results are processed immediately.

What are the relative advantages of ultrasound imaging and DXA? • Ultrasound systems are smaller and less expensive than DXA systems. • Ultrasound requires no exposure of the body to ionising radiation (although X-ray levels are very low with DXA—lower than in most other X-ray procedures). • Studies show that ultrasound mainly measures the bone mass while DXA is better able to predict bone strength. • Reproducibility of results using ultrasound is not quite as good as with DXA. • Relative risk for hip fracture is predicted as well by ultrasound of the heel as by DXA of the hip.

Blood flow and the Doppler effect Doppler ultrasound scanning is used to image any structure or tissue that pulsates or moves. It is based on the principle that whenever the reflector surface moves with respect to the transducer, there is a shift in the frequency of the ultrasound received by the transducer compared with the frequency it emitted (the Doppler effect; see in2Physics @ Preliminary section 13.8). This concept is illustrated for blood flow in Figure 16.6.2, while Figure 16.6.3 illustrates a case in which Doppler imaging reveals abnormal blood flow in a heart. Colour Doppler imaging is colour-coded, using redder colours to show flow or movement towards the transducer and bluer colours to show flow away from it. (Interestingly, this is the opposite of the choice an astronomer would make in presenting Doppler images.)

Identify data sources, gather, process and analyse information to describe how ultrasound is used to measure bone density.

Figure 16.6.1

A machine used to measure the bone density of your heel

PRACTICAL EXPERIENCES Activity 16.2

Activity Manual, Page 128

Describe the Doppler effect with respect to sound and how it is used in ultrasonics to obtain flow characteristics of blood flow through the heart.

315

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Seeing with ultrasound

Outline some cardiac problems that can be detected through the use of the Doppler effect.

The Doppler shift effect has been used for a long time in foetal heart rate detectors. Further developments in Doppler ultrasound technology in recent years have resulted in its increased use in obstetrics to assess and monitor the well-being of a foetus.

PRACTICAL EXPERIENCES Activity 16.3

B

A

C

D

Activity Manual, Page 130

skin beam direction

Doppler again and again

A

T

he Doppler effect (see in2 Physics @ Preliminary section 13.8) is one of the most versatile techniques for measuring speed. As well as its use in Doppler ultrasound, it is used by astronomers to measure the motion of stars and by police radars to detect speeding drivers.

Figure 16.6.3

Colour Doppler ultrasound of a newborn’s heart, revealing blood flow (green) through a hole in the septum (grey) between the ventricles

vessel

flow B

C

Sonogram

Figure 16.6.2

Different transducers (A–D) in this system ‘see’ blood flowing in different directions—with a component towards or away from the transducer. This produces a Doppler effect that can be used to calculate a velocity at that point.

Echocardiography Echocardiography is the name given to ultrasound imaging of the heart. It is used to investigate heart-valve function, ventricular function, congenital heart diseases (including holes in the septum separating the left and right halves of the heart), cardiac tumours and obstructions in cardiac blood vessels. Pulsed ultrasound from a transducer is used and a continuous recording is made of the echoes received from the various parts of the heart, showing the motion of the various parts. A standard transthoracic echocardiogram can be conducted through the thorax (chest wall). Higher resolution imaging can be achieved by a trans-oesophageal echocardiogram in which a transducer on a specialised probe is passed down the patient’s oesophagus, thereby getting much closer to the heart. Blood-flow characteristics in the foetal blood vessels can also be examined using Doppler ultrasound. Diminished flow, particularly in the diastolic phase of a pulse cycle (as the heart muscle relaxes) is associated with problems with the circulation of blood in the foetus. Colour Doppler can also be used to assess the success of coronary bypass surgery, and is particularly useful in the diagnosis and assessment of congenital heart abnormalities such as atrial or ventricular septal defects (a hole between the atria or the ventricles; see Figure 16.6.3).

Checkpoint 16.6 1 Discuss why the heel is an ideal place to use ultrasound to determine bone density. 2 Explain how the Doppler effect can help diagnose the well-being of a foetal heart. 3 Outline how echocardiography works. 316

D

PRACTICAL EXPERIENCES

medical physics

CHAPTER 16

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 16.1: Ultrasound images Collect some images made using ultrasound and interpret information from the image. Discussion questions 1 Interpret the grey scale that accompanies most ultrasound images. 2 Determine what parts of the body are best imaged with ultrasound. 3 List things that can be determined from ultrasound images.

Activity 16.2: Doppler ultrasound Use several data sources to find and observe a Doppler ultrasound image that shows the flow of blood through the heart. Discussion questions 1 Define the Doppler effect. 2 Explain how different directions in the video image are shown. 3 Outline what can be determined from these images.

Activity 16.3: Bone density Find out how bone density measurements are made using ultrasound and present the information in a 5 minute talk to the class. Discussion questions 1 Define bone density measurements and discuss why they are needed. 2 Outline how the measurements are taken. 3 Explain why ultrasound is a good method for obtaining these measurements.

Gather secondary information to observe at least two ultrasound images of body organs.

Identify data sources and gather information to observe the flow of blood through the heart from a Doppler ultrasound video image.

Identify data sources, gather, process and analyse information to describe how ultrasound is used to measure bone density.

317

16 • • •

•

•

•

• •

• •

Seeing with ultrasound

Chapter summary

Ultrasound can travel through the body but is reflected from interfaces between different tissues. For most medical imaging applications, ultrasound in the range from 3.5 MHz to about 10 MHz is used. The reflected sound can be detected at the skin surface, allowing imaging of structures deep within our bodies in a non-invasive way with little stress to the patient. Higher frequency ultrasound produces images with a better resolution, but has poorer penetration through tissues. Lower frequency ultrasound has better penetration through tissues, but produces images with a lower resolution. The intensity of the reflected wave is determined by the difference in the acoustic properties of the tissues and the absorption and angle of the intervening tissue. Ultrasound waves are emitted and detected by a transducer. A piezoelectric transducer uses a crystal that responds to an oscillating voltage by oscillating slightly in size and responds to the pressure of a sound wave by producing an oscillating voltage. A convex-array transducer uses a diverging beam to produce a sector scan. Acoustic impedance describes how readily a specific sound frequency will pass through a material. Acoustic impedance: Z = ρυ

•

When ultrasound waves meet a boundary between tissues with different acoustic impedances, some of the ultrasound energy will be reflected and some will be transmitted. The greater the difference in acoustic impedance, the greater the intensity of the reflected signal. 2 I r  Z2 − Z1  = Io  Z + Z  2  2 1

•

Impedance matching (acoustic coupling) between two media results in most of the energy being transmitted through the interface, with almost no reflection. An A-mode scan shows returning echoes as a vertical displacement, proportional to the amplitude of the reflected signal. A B-mode scan shows the returning echoes as dots with intensity proportional to the returning echo amplitude. These dots are used to construct a 2D real-time scan. A movie made using 3D ultrasound images is called 4D ultrasound. Bone density scans use the fact that the speed and absorption of ultrasound vary with bone density. Doppler ultrasound scanning uses the Doppler effect to image any structure or tissue that moves. Colour Doppler imaging is colour-coded to show flow or movement towards or away from the transducer. Echocardiography is ultrasound imaging of the heart.

•

•

• • • • •

Review questions Physically Speaking Below is a list of topics that have been discussed throughout this chapter. Create a visual summary of the concepts in this chapter by constructing a mind map linking the terms. Add diagrams where useful.

318

Ultrasound

Transducer

Doppler effect

Frequency

Piezoelectric

Acoustic impedance

Reflection

Linear array

Impedance matching

Bone-density measurement

Echocardiography

A-mode scan

B-mode scan

Sector scan

Blood flow

medical physics

Reviewing

11 Describe the effect of increasing the difference in acoustic impedance at a tissue boundary on the proportion of the incident ultrasound energy that is reflected at the boundary.

1 Compare quantitatively the frequencies of medical ultrasound with sound in the normal hearing range of humans.

12 Describe the conditions that must be satisfied for reflection to take place at a tissue boundary in the body.

2 Contrast medical ultrasound and the sound that humans hear in terms of the media in which they mainly occur.

3 Figure 16.7.1 is a colour ultrasound image. Identify the organ that is imaged here and outline the main benefit of this type of image.

13 Explain why ultrasound imaging is not used for investigating the function and structure of the brain.

14 Outline how refraction complicates the process of obtaining clear images of organs in the ultrasound process.

15 Discuss current issues associated with the use of obstetric ultrasound imaging.

16 Describe the Doppler effect and outline how this effect may be observed using sound waves.

17 Explain how ultrasound can be used to obtain information about the flow of blood through the heart.

18 Describe how ultrasound is used to measure bone density.

19 Compare the nature and use of A-mode and B-mode ultrasound scans.

20 Explain why ultrasound is not used to scan the lungs. 21 Assess the statement ‘Tissues having the same Figure 16.7.1

density have the same acoustic impedance’.

Example of an ultrasound image

4 Outline the piezoelectric effect and its place in ultrasound imaging.

5 Assess the advantages of ultrasonic examination of a pregnant woman against the dangers associated with an ultrasound examination.

6 Define acoustic impedance in words (without using equations).

7 Identify the units for the quantities represented in the equation Z = ρυ that defines acoustic impedance.

8 Account for the fact that different materials have different acoustic impedances.

Solving Problems 22 Bone has a density of 2 × 103 kg m–3. The speed of

sound in bone is 4080 m s–1. Calculate the acoustic impedance of bone.

23 Calculate the density of kidney tissue, given that the

acoustic impedance is 1.623 × 106 rayl and that the ultrasound velocity in kidney tissue is 1561 m s–1.

24

Calculate the percentage of the ultrasound energy that would be reflected from an interface between blood and brain tissue.

9 The acoustic impedance of two tissues in the body is the same, and yet one has a greater density than the other. Explain how this is possible.

10 Identify the quantities represented by each symbol in the equation



2

I r Z 2 − Z 1  = Io Z + Z  2  2 1 Re

iew

Q uesti o

n

s

v

State the units for each quantity.

319

17 X-rays, cathode, radiographs, tomography, computed axial tomography (CAT), characteristic X-rays, Bremsstrahlung, anode, radiographer, filter, collimator, grid, fluorescent, contrast agents

Imaging with X-rays The invisible reveals the hidden Wilhelm Röntgen (1845–1923) was studying cathode rays in a Crookes tube wrapped in cardboard, when he noticed a faint green glow from the fluorescent screen painted with barium platinocyanide—the rays emitted from the tube were passing through the cardboard. He found the rays could also pass through other papers and books. This phenomenon intrigued Röntgen and he dedicated the next few months to systematically investigating the properties of these rays. He named this new type of radiation X-rays.   On 22 December 1895, Röntgen took the first X-ray, an image of the hand of his wife Anna. After this, many scientists and companies began investigating X-rays. Thomas Edison began a rigorous investigation into fluorescent materials. He developed the fluoroscope using calcium tungstate, which became the standard for medical X-ray examinations. Today, with the development of semiconductor detectors and imageprocessing software, X-rays are used in computed axial tomography (CAT scans) to provide medical doctors with high-resolution 3D images of skeletal, vascular and soft tissue.

17.1 Overview and history: types of X-ray images X-rays are a form of electromagnetic radiation that lies between ultraviolet and gamma radiation on the electromagnetic spectrum. X-rays are able to pass through our body. In medicine, they are used for diagnosis of medical problems and injuries, and for therapeutic purposes such as treatment of certain tumours. From 1896, many applications of X-rays were immediately recognised and generated a wide range of interest. Companies began researching and developing a range of specialised versions of Crookes tubes to produce X-rays. However, Crookes tubes are classified as cold cathode tubes and were often unreliable, as they needed to contain a small quantity of gas to operate. In 1913 William D Coolidge (1873–1975) invented the Coolidge X-ray tube, which permitted a stable and controllable continuous production of X-rays. Coolidge drew upon the thermionic diode (vacuum tube), which used a hot cathode (a filament) that permitted current to flow in a high vacuum. Most modern X-ray tubes are variations on Coolidge’s basic design.

320

medical physics A range of film-based and ‘fluoroscope’ devices were developed and refined, providing the familiar two-dimensional X-ray radiographs. In 1930, Alessandro Vallebona (1899–1987) proposed the idea of tomography in which a single slice of the body could be recorded on radiographic film. In the 1970s the development of affordable computers capable of processing image data, and the parallel development of semiconductors and image intensifiers allowed a new generation of X-ray devices to be developed. The technology was now available to re-engineer Vallebona’s tomography. Allan McLeod Cormack (1924–1998) developed the theory and Godfrey Hounsfield (1919–2004) independently built a prototype of the new generation of X-ray tomography scanner. This became known as computed tomography (CT) or computed axial tomography (also called CAT). Today, with the development of more sensitive detectors and advances in image processing software, CAT scans can provide two- and three-dimensional images with resolutions of 1 cm. The combination of traditional radiographs and CAT scans provides the medical profession with important therapeutic and imaging diagnostic tools.

Checkpoint 17.1 1 2

Describe the properties of X-rays. Construct a timeline that shows the development of technology in X-ray devices. Figure 17.1.1

17.2 The X-ray tube X-rays are emitted when high-energy electrons strike a target. There are two processes by which X-rays can be produced: characteristic X-rays and Bremsstrahlung. The efficiency of both processes in converting the kinetic energy of the electrons into X-rays is often less than 2%; the remainder of the energy is converted into heat. The Coolidge X-ray tube developed in 1913 has remained the basis for the majority of today’s diagnostic devices. Several improvements have been made to increase efficiency, control and cooling of the anode target. Figure 17.2.1 shows a schematic representation of a modern rotating-anode diagnostic X-ray tube. rotating anode

Wilhelm Röntgen’s first X-ray taken on 22 December 1895. It is the hand of his wife Anna Berthe Röntgen and you can see her wedding ring.

Describe how X-rays are currently produced.

stator of induction motor

glass envelope bearings rotor/anode support

cathode block rotor filament

focusing cup electrons

target exit window

Figure 17.2.1

Schematic representation of a rotating-anode diagnostic X-ray tube 321

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Imaging with X-rays

The tube is highly evacuated and generally constructed from glass. It contains an anode and a cathode. A potential difference of between 20 000 and 300 000 volts is applied between the cathode and the anode. The cathode has a small filament (similar to an incandescent lamp filament), which is heated to a dull red glow by a small current. The hot filament readily releases electrons and the high potential difference then accelerates these electrons from the cathode towards the target anode. The cup shape of the cathode behind the filament focuses the electron beam towards a point on the target anode. When the electrons hit the angled tungsten anode, X-rays are emitted through the exit window at the side of the X-ray tube. Because of the inefficiency of the process, a huge amount of thermal energy is produced. To ensure the tungsten anode does not melt, it is often rotated to more evenly distribute the energy. The anode is often mounted on a copper heat-sink that may also have an additional cooling system attached. Shielding, to reduce the emission of X-rays in unwanted directions, surrounds the tube itself.

Checkpoint 17.2 1 2 3

Name the two types of X-rays. Discuss the energy conversion of electrons as they strike a surface to produce X-rays. Describe the structure of a Coolidge tube.

17.3 Types of X-rays Compare the differences between ‘soft’ and ‘hard’ X-rays.

X-rays are often categorised as ‘hard’ and ‘soft’ X-rays, descriptive terms that indicate the relative penetrating ability of the X-ray beam. ‘Soft’ X-rays do not penetrate through body tissue and are absorbed easily. These absorbed X-rays are not useful for producing radiograph images and they pose an increased risk to patients, of genetic mutations and cancers. X-ray machines have aluminium filters that attenuate these lower energy ‘soft’ X-rays from the X-ray beam. ‘Hard’ X-rays have a higher energy and greater penetration into and through tissue, thus producing a sharper image. Changing the potential difference between the cathode and anode can vary the distribution of ‘hard’ and ‘soft’ X-rays produced by the X-ray tube. Increasing the voltage to the tube will produce more ‘hard’ X-rays. The radiographer can adjust the voltage, to obtain the best quality image of the particular organ, tissue, vessel or bone under examination. Table 17.3.1  Properties of X-rays

322

‘Hard’ X-rays

• • • • •

Higher potentials are applied to the X-ray tube Produced by the impact of high-energy electrons onto the anode Shorter wavelength ~0.01 nm Higher penetration (the X-ray photons have more energy) Produce higher resolution images

‘Soft’ X-rays

• • • • •

Lower potentials are applied to the X-ray tube Produced by the impact of lower energy electrons onto the anode Longer wavelength ~1 nm Lower penetration (the X-ray photons have less energy) Produce lower quality images

medical physics

Characteristic X-rays

Kα

ejected

e–i

2p

1s 2s – e

electron e–

e–

e

Kβ

2p

1s 2s

characteristic X-ray

tic

is ter

c ara ch ay X-r 2 posible paths Au to fill the gap ger ele ctr o – n e 2p

e– 1s 2s

e–

0.02 0.04

0.06 0.08 0.10

0.12

Wavelength (nm)

Figure 17.3.1

X-ray wavelength 10

Characteristic X-rays show up as distinct sharp peaks within the X-ray spectrum.

X-ray continuum radiation (Bremsstrahlung) 50 kV

X-rays can be produced by firing high-speed electrons at a metal target. These electrons can eject electrons from the inner shells of the atoms of the target. Vacancies will be quickly filled by electrons dropping down from higher levels, emitting X-rays with specific defined wavelengths. Occasionally an emitted characteristic X-ray’s photon is re-absorbed by another electron, which is then ejected from the atom (an ‘Auger electron’).

8

Relative intensity

Figure 17.3.2

X-rays from a molybdenum target at 35 kV

1

e–

a gap remains

characteristic X-rays

2

Brehmsstrahlung continuum

X-ray intensity

incident electron

3 Relative intensity

The accelerated electrons can interact with and knock out an electron from the inner electron shell of the anode. If an outer shell electron drops down and fills the vacant position, an X-ray photon of a specific energy and wavelength, peculiar to the elements of the target will be produced. In the total X-ray spectrum, the characteristic X-rays appear as sharp well-defined peaks. Figure 17.3.1 shows the characteristic peaks for X-rays produced by a molybdenum target anode bombarded by electrons accelerated through a potential of 35 kV. The details of the process are shown in Figure 17.3.2.

Bremsstrahlung The word Bremsstrahlung means ‘braking radiation’ in German. It is used to describe the radiation emitted when electrons are decelerated or ‘braked’, when fired into a metal target. When an electron rapidly changes its velocity, the lost kinetic energy is converted into photons of electromagnetic radiation. When the energy of the incident electrons is high enough, the radiation emitted will be X-rays. Decelerating electrons from the beam interact and produce a continuous distribution of radiation. Figure 17.3.3 shows a set of curves for a tungsten target with electron beams of four different energies. The details of the Bremsstrahlung process are shown in Figure 17.3.4.

40 kV 6

4

30 kV

2 20 kV 0.02

0.04

0.06

0.08

0.10

Wavelength (nm)

Figure 17.3.3

e–i incident electron

Bremsstrahlung (braking radiation) is characterised by a continuous distribution of wavelengths. The curves in this graph are based on the 1918 paper of Clayton Ulrey (1884–1963).

incident electron e–i

hard X-ray

e–i

Figure 17.3.4

X-ray intensity

soft X-ray

Wavelength (nm)

e–i

The processes associated with high-energy and low-energy Bremsstrahlung 323

17

Imaging with X-rays

Checkpoint 17.3 1 2 3 4

Distinguish between hard and soft X-rays. Create a table to list the properties of hard and soft X-rays. Explain how characteristic X-rays are produced. Explain how Bremsstrahlung X-rays are produced.

17.4 Production of X-ray images The key components of a medical X-ray machine are an X-ray tube, a filter, a collimator, a platform to situate the patient, a grid and a detector. Figure 17.4.1 shows a typical X-ray machine. The filter is designed to remove ‘soft’ X-rays, which pose a risk to the patient and are not useful for imaging purposes. The collimator provides a mask to shape the output X-ray beam to minimise unwanted X-ray exposure to the patient. The grid situated just in front of the detector absorbs scattered secondary X-rays that would make the image fuzzy.

Conventional radiographs In traditional radiographs, fluorescent screens absorb the energy in the X-ray beam that has penetrated the patient. This energy is converted into visible light that has same information as the original X-ray beam. This light is then used to expose the X-ray film. The more efficient this light conversion, the less the patient needs to be exposed to X-ray radiation. In most systems, the film is sandwiched between two fluorescent screens in a cassette so that the film emulsion is exposed from both sides. In modern X-ray machines, film is replaced by arrays of electronic detectors. Conventional X-ray systems are used to image a variety of patient conditions including suspected skeletal bone fractures (Figure 17.4.2), cancer in the breast (Figure 17.4.3) and lung conditions.

a

Figure 17.4.1 324

A basic diagnostic X-ray machine

Figure 17.4.2

b

(a) An incomplete (greenstick) fracture of the radius and ulna of the lower arm and (b) a simple fracture of the ulna

medical physics

Imaging vascular and hollow structures

a

b

X-ray imaging does not show vascular (veins and arteries) or hollow soft tissue structures such as the digestive tract To provide radiologists with a in great detail. mechanism to reveal these structures, non-toxic contrast agents containing barium or iodine, which absorb X-rays, are used. These contrast compounds can be ingested, or injected into an artery or vein. These procedures were first carried out between 1906 and 1912 and allowed blood vessels, the stomach, digestive tract, the gall bladder and bile ducts to be seen in situ for the first time. Today the process is very similar. An initial X-ray is taken, then contrast material containing iodine is injected into the bloodstream and a second X-ray taken. These two images are then digitally subtracted, leaving only Figure 17.4.3 This breast X-ray, called a mammogram, shows the image of the blood vessels (Figure 17.4.4). (a) a healthy breast and (b) a breast with a tumour (arrowed). Another once common procedure is called a barium meal X-ray (Figure 17.4.5). This procedure involves the patient drinking a suspension of barium sulfate and then a series of radiographs are taken of the oesophagus, stomach and duodenum. Barium meal tests are declining with the increasing use of endoscopy (section 18.2), which allows the doctor to directly visually inspect the oesophagus, stomach and duodenum.

Figure 17.4.4

Renal artery angiogram showing blood vessels, a kidney and the spine

Figure 17.4.5

Barium swallow X-ray of the large intestine of a patient with cancer of the sigmoid colon

Checkpoint 17.4 1 2 3

List the components of an X-ray machine and their functions. List the possible uses of conventional X-rays. Outline how it is possible to image hollow structures within the body. 325

17

Imaging with X-rays

17.5 X-ray detector technology In the early 1900s a head X-ray would require the patient to be exposed to 10 minutes of radiation. Today a similar higher resolution image would expose the patient to one-fiftieth of the radiation used 100 years ago. Technological advances over the last 100 years have improved the quality of X-ray images and dramatically reduced the X-ray exposure to patients. Originally radiographic images were produced on glass photographic plates. These were replaced by film cassettes with scintillating screens which intensified the images. In 1955, the X-ray image intensifier was developed and X-ray images were displayed on a television monitor. The image intensifier allowed doctors to image blood vessels and the heart in real time. New digital technology and semiconductor detectors (silicon or germanium doped with lithium) began to be developed in the 1970s. Today, with the advent of large semiconductor array detectors, it has become possible to collect digital data that can be processed into high resolution images. These images can be enhanced and processed, transmitted for remote diagnosis and easily stored and retrieved. Over the next 15 years most conventional X-ray systems will be upgraded to all digital technology.

Checkpoint 17.5 1 2

Discuss how image intensifiers have changed what doctors can do in diagnosis. List the benefits of digital technology.

17.6 Production of CAT X-ray images A computed axial tomography (CAT) or computed tomography (CT) scanner is a diagnostic device that uses an X-ray tube that is rotated around the patient. A detector collects the X-rays and digitises the information. A series of linear scans is processed by a computer and a series of two-dimensional slices or a three-dimensional rotatable image can be constructed. The development of CAT scan technology was enabled by the development of microcomputers and detector technology that could digitise data. In medical circles, the term CT is more commonly used than CAT, but they are interchangeable. Figure 17.6.1

A 64-slice CAT scanner

Explain how a computed axial tomography (CAT) scan is produced.

326

Basic design There are several designs of CAT scanners in use. One commonly used scanner has the patient placed in a fixed position on a bed. The X-ray tube and detector array are attached to a movable C-shaped gantry that is rotated 180º around the patient and then progresses forward and records the next set of scans. Another more modern system has a moveable bed is that moved slowly through a fixed toroidal structure (a donut) that houses the X-ray tube and detector. As the bed progresses, the X-ray tube and detectors continually rotate around the toroidal structure, scanning the patient in a spiral motion.

medical physics

The scanning process In older style devices (Figure 17.6.2a), a linear scan of data is taken at a large number of positions, producing the digitised data for that slice through the patient’s body. The source and detector are then rotated to produce a new scan. Modern devices speed up this process and direct a narrow fan-beam through the patient. The intensity of the emerging transmitted radiation is recorded by a line of adjacent detectors (Figure 17.6.2b and c). Fast CAT scans can now also be performed with a cone-shaped X-ray beam and an area-array detector panel. a

X-ray tube

45 n

5

sc a

13 detectors

detector array

c fan beam

direction of X-ray tube motion

patient

Figure 17.6.2

n 180°

X-ray source

collimator

X-ray source

odfrey Hounsfield’s original CT scan in 1972 took several hours to acquire a single slice of image data and then more than a day to reconstruct the data into a single image. Today a CAT scan can construct an image in less than 1 second.

scan 90

0°

b

G

a sc

detector

A Long Wait

X-ray source

ring of stationary detectors

The history of CAT: (a) A pencil-beam X-ray source and a detector scan linearly across the patient from a series of angles. (b) A fan-beam X-ray source and an arc of detectors rotate around the patient. (c) Only the fan-beam X-ray source rotates; the detector ring is stationary.

Constructing the two-dimensional slice image To form an image, we need to calculate the radiation that is absorbed at each small volume element (voxel) pixel within the scanned structure. These scans are three-dimensional versions of two-dimensional picture elements (pixels) that make up a picture from your digital camera. With this information we can construct a digital image. To assist in the explanation of the process used in CAT scans, we will use the simplified example shown in Figure 17.6.3. In this example, two linear scans of a hollow cross have been made (Figure 17.6.3a). One linear scan has been made from the left and the other from the top. The intensities for each beam have been measured by the detectors opposite the X-ray sources. The computer then translates the information from the detectors into the total amount of the incident X-ray radiation that was absorbed as it passed through the structure. These values have been recorded down the right-hand side and along the bottom edge of the diagram (Figure 17.6.3a). This is the same information as obtained in a normal radiograph. Now to obtain specific information about each voxel, we can sum the vertical and horizontal absorption values and record the value in each corresponding voxel (Figure 17.6.3b). A grey scale can then be overlaid, with the value ‘10’ being white, ‘0’ being black and the other values appropriate shades of grey. Figure 17.6.3c shows this grey-scale overlay.

a

X-rays

X-rays

5 4 2 4 5

b

10 9 7 9 10

9 8 6 8 9

7 6 4 6 7

9 8 6 8 9

5 4 2 4 5

10 9 7 9 10

c

d

Figure 17.6.3

Mechanism for constructing the 2D image slice

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The image generated shows some sort of cross structure but it is fuzzy. Now if we reassign the grey scale and assign light grey to all voxel values greater than 6 and a dark grey to all voxels with values of 6 or less, we obtain a clearer image of the original structure (Figure 17.6.3d). This is the basic process of a CAT scan. For a real CAT scan in which there are complex sets of absorption calculations, a lot of data processing is required to recreate a high-resolution image from the data within a second. Figure 17.6.4 shows a single two-dimensional slice from a scan. Because the data contains three-dimensional information, the imaging software can combine a series of slices to produce a three-dimensional image. Computer imaging software can also remove distracting tissues from the scan images and generate an image that shows the specific organ, tissue or structure under examination. Figure 17.6.5 shows a CAT angiogram of the blood vessels in the brain, with the surrounding tissue removed. Common applications for CAT scans include: • identifying trauma injuries to the lungs, heart, spleen, kidneys and liver • planning for and assessing the results of surgery • planning radiation treatments for tumours • detecting osteoporosis by measuring bone density.

Figure 17.6.4

A CAT scan of the lungs with the window level set to demonstrate the vessels and airways. This is used to look for diseases such as pneumonia or lung cancer.

Checkpoint 17.6 1 2 3 4

328

Outline how CAT images are produced. Describe how a two-dimensional scan is produced. Outline the benefits of a CAT scan. Outline the uses of a CAT scan.

Figure 17.6.5

An intracranial angiogram showing the blood vessels of the brain

medical physics

17.7 Benefits of CAT scans over conventional radiographs and ultrasound Conventional radiographs provide an image of all the structures present, superimposed on each other. For example a chest X-ray of the lungs will show all the ribs. The ribs therefore obscure details associated with the lungs. A CAT scan allows the ribs to be removed from the image and provides the doctor with a full image of the lungs. When the head is imaged, CAT scans provide a detailed image that is not as affected by the skull bone as a radiographic image. A CAT scan of the head, for example, can distinguish between the white matter, grey matter and spinal fluid. This enhanced discrimination allows the doctor to more easily identify vascular problems, tumours and subtle abnormalities. CAT scans contain the information to produce a three-dimensional image, allowing many angles of a particular area of interest to be viewed by the doctor. Software can be used to remove distracting tissues or structures. Modern ultrasound equipment can also produce three-dimensional images (see section 16.5). Ultrasound can provide an initial diagnosis but cannot clearly identify the level of tissue damage or sites of internal bleeding. It is unable to penetrate bone or gas, and therefore tissues beyond these regions cannot be imaged. For this reason, ultrasound cannot be used to examine the brain, which is surrounded by the skull, or tissue masked by gas pockets in the abdominal region. A CAT scan can image all these regions, with resolution that is superior to that of ultrasound imaging.

Describe circumstances where a CAT scan would be a superior diagnostic tool compared to either X-rays or ultrasound.

PRACTICAL EXPERIENCES Activity 17.1

Activity Manual, Page 133

Checkpoint 17.7 1 2

Identify the uses of CAT for which a radiograph can not be used. List problems that CAT scans can overcome that ultrasound imaging can not.

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PRACTICAL EXPERIENCES CHAPTER 17 This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Gather information to observe at least one image of a fracture on an X-ray film and X-ray images of other body parts. Gather secondary information to observe a CAT scan image and compare the information provided by CAT scans to that provided by an X-ray image for the same body part.

330

Activity 17.1: CAT versus X-rayS Find a CAT scan and an X-ray radiograph of the same part of the body and compare the information that can be determined from each. Discussion questions 1 Outline characteristics that allow you to determine that an image is a CAT scan. 2 Outline characteristics that allow you to determine that an image is an X-ray radiograph. 3 Identify applications that would make a CAT scan more appropriate than an X-ray radiograph. 4 Identify applications that would make an X-ray radiograph more appropriate than a CAT scan.

Chapter summary •

•

•

• • • •

•

•

X-rays are a form of electromagnetic radiation that lies between ultraviolet and gamma radiation on the electromagnetic spectrum. X-rays are used for the diagnosis of medical problems and injuries, and for therapeutic purposes such as the treatment of certain tumours. The modern X-ray tube is based upon the 1913 design by William Coolidge, which used a hot cathode and permitted a stable and controllable continuous production of X-rays. The CAT scan was independently conceived by Allan McLeod Cormack and Godfrey Hounsfield. X-rays are emitted when high-energy electrons strike a target. There are two processes by which X-rays can be produced: characteristic X-rays and Bremsstrahlung. X-rays are often categorised by the terms ‘hard’ X-rays and ‘soft’ X-rays, which are descriptive terms indicating the relative penetrating ability of the X-ray beam. The basic components of an X-ray machine are an X-ray tube, a filter, a collimator, a platform to situate the patient, a grid and a detector. Contrast compounds can be ingested or injected to assist X-ray imaging of vascular (veins and arteries) or hollow soft tissue structures.

•

•

•

• •

•

medical physics

Benefits of digital technology to all X-ray systems include: – lower X-ray doses to patients – images that can be enhanced and processed, transmitted for remote diagnosis and easily stored and retrieved. A computed axial tomography (CAT) or computed tomography (CT) scanner is a diagnostic device that uses an X-ray tube that is rotated around the patient. A detector collects the X-rays and digitises the information. The development of CAT scan technology was enabled by the development of microcomputers and detector technology that could digitise data. CAT scanners use powerful image-processing software to recreate images from the scan data collected. Common applications for CAT scans include: – identifying trauma injuries to the lungs, heart, spleen, kidneys and liver – planning for and assessing the results of surgery – planning radiation treatments for tumours – detection of osteoporosis by measuring bone density. CAT scans generally provide higher resolution images across a greater range of body tissues than either radiographs or ultrasound.

Review questions Physically speaking This is a list of topics that have been discussed throughout this chapter. Create a visual summary of the concepts in this chapter by constructing a mind map linking the terms. Add diagrams where useful.

X-ray

Hard

Soft

Characteristic

Bremsstrahlung

Penetration ability

Resolution

Diagnostic

Benefits

2D scan

Image intensifier

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Imaging with X-rays

Reviewing 1 Outline two ways in which X-rays are produced in an X-ray tube.

2 Describe the differences between characteristic X-rays and Bremsstrahlung.

3 Using a chest X-ray as an example, recount how an X-ray image of the body is produced.

4 Compare the wavelengths of ‘hard’ and ‘soft’ X-rays. 5 A student wrote the statement: ‘Soft X-rays could also be called ultraviolet rays’. Assess this statement.

6 Assess the impact on society of doctors being able to examine images such as Figure 17.4.2, showing part of a patient’s arm.

7 For more than 100 years, X-ray images have been used to examine one type of tissue in the body more than any other. Identify this tissue and outline why X-rays are so successful in producing images of this type of tissue.

8 Figure 17.8.1 shows the chest area of two patients. Identify the types of X-ray images shown and compare the information provided by each image.

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n

s

v

Figure 17.8.1

9 Explain why it is sometimes necessary for radiographers to take two radiographs of the chest area, one from the front and one from the side to provide the same information that can be seen in a single CAT scan of the chest.

10 Compare the effectiveness of CAT scans and radiographs to resolve soft tissue in the body of a patient.

11 Compare the advantages of the 2D and 3D images. 12 Compare the passage of the X-rays through a patient given a conventional chest X-ray (radiograph) and a CAT scan of the chest region.

13 Explain how a CAT scan is produced. 14 Outline the main technological development that took place in the 1970s that made CAT scans possible.

15 Assess the impact of particular advances in physics on the development of CAT scans.

Imaging with  light Look and see Optical fibres have had a great impact on society, although the fibres themselves are usually hidden from view. One of those impacts is in a doctor’s endoscope, a device to peer inside the human body without major surgery. Optical fibres allow light to be directed onto an area of interest and then lenses relay an image back to the doctor. Combined with tiny surgical instruments, these devices allow ‘minimally invasive’ surgery to reduce the trauma of surgery, having a radical effect on the treatment and recovery of patients with problems ranging from a torn knee cartilage to leaky heart valves.

18 endoscope, endoscopy, total internal reflection, critical angle, optical fibres, core, cladding, coherent fibre bundle, non-coherent fibre bundle, biopsy

18.1 Endoscopy An endoscope is a medical device used to shine visible light into a patient’s body and relay an image out, to allow organs, tissues and cavities to be seen. Light is directed to the area of interest through a flexible bundle of optical fibres. Lenses are used to focus an image of the area of interest onto a separate bundle of optical fibres that transfer the image out of the body. The image is usually processed electronically to display it on a screen in real time or to record still or moving images for later use. The process of using an endoscope to examine a patient is called endoscopy. The tube-like part of the endoscope, which enters the body, is typically less than a centimetre in diameter. Depending on the purpose of the endoscopy, the tube is inserted via a natural opening of the body, or through a small cut made in the skin. Miniature surgical instruments can be attached to the part that enters the body and these are controlled from outside the body by a doctor using control wires that pass through another narrow tube that is also a part of the endoscope. Tissue samples can be removed from the patient and surgical procedures can be carried out. The main advantages of endoscopy are that it is minimally invasive and provides real-time images in true colour. The operator can manipulate the endoscope to obtain the best view of the area of interest inside the patient.

Figure 18.1.1

A gastric ulcer seen through an endoscope

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Imaging with light

PHYSICS FEATURE

3. Applications and uses of physics

Optical fibres

Explain how an endoscope works in relation to total internal reflection.

Y

ou will recall learning about the properties and behaviour of light (see in2 Physics @ Preliminary Chapter 8). The path of a light ray meeting a boundary between two transparent materials changes if there is a difference in the refractive indices of the two media. This process is called refraction. Total internal reflection may occur at a boundary if the light is travelling from a medium with a higher refractive index ni to one having a lower refractive index nr and the angle of incidence exceeds the critical angle (θc) given by: sin θc =

optical fibre is transmitted along the optical fibre by total internal reflection, as illustrated in Figure 18.1.4. Most optical fibres used for communications purposes transmit light in the near infra-red region of the electromagnetic spectrum at wavelengths of 800–1600 nm. The glass used in optical fibres has maximum transparency at these wavelengths. (Remember that the wavelength of visible light is between about 400 and 700 nm.) Fibres used in medical endoscopy must be able to transmit light in the visible region of the spectrum. Because of the short distances involved (usually less than about 2 m), the fibres do not need to have to have the same transparency as fibres used for communication technology, nevertheless endoscopic fibres are made from high quality, homogeneous, high transparency glass. Homogeneity is an important property of the fibre because any irregularities would distort and degrade the image.

ni nr

Total internal reflection (Figure 18.1.2) is the basic principle underlying the operation of relatively large diameter optical fibres. Optical fibres are usually made of glass with two layers—the light-carrying, higher index inner core and the surrounding lower index cladding (Figure 18.1.3). Light entering the core at one end of the normal

normal

normal

normal

r

air

nr

air

water

ni

water

i

r

air

ni

water

i

a

b

Figure 18.1.2

nr

90°

i

nr

air

nr

ni

water

ni

c

i

c

c

r

i

d

Light moving from high to low refractive index materials may be totally internally reflected.

jacket

core

core cladding cladding

Figure 18.1.3 334

Figure 18.1.4 A cross-section of an optical fibre

Light is transmitted along an optical fibre by total internal reflection.

medical physics The main parts of an endoscope are shown in Figure 18.1.5. The endoscopic tube that is inserted into the patient typically contains the following parts: • a bundle of optical fibres to transmit light to the point of observation • a coherent bundle of optical fibres to carry the image of the tissue to the observer • a system of lenses to focus an image of the tissues under examination onto the optical fibre bundle. At the viewing end of the optical fibre bundle, more lenses allow direct viewing of the image by eye, or connect to a camera that feeds a video screen or computer • suction tube to remove blood and other loose, obscuring tissue material from the area under inspection • an inlet and an outlet to permit the area under observation to be flushed with clear saline (salt water) solution to increase visibility • control lines to manipulate the tube inside the patient. These vary in complexity, depending on the task undertaken. A simple gastroscopy may not require complex controls. Endoscopes used for surgical procedures may require fine control over the position of the lens at the tissue end • miniature remote-controlled surgical instruments may be present, ranging from simple suction tools to more elaborate surgical tools used in operations.

power source video out etc.

The endoscope is a thin, flexible fibreoptic ‘telescope’.

light to see inside Various devices can be passed down side channels. These can be manipulated by the doctor to take specimens etc.

doctor looks down endoscpe

oesophagus

stomach

duodenum

Figure 18.1.5

Worked example Question Calculate the critical angle for a glass optical fibre with a core refractive index of 1.48 and a cladding refractive index of 1.46.

endoscope passed down oesophagus into stomach

Gastroscopy is the examination of the upper gut (oesophagus, stomach and duodenum) using a flexible fibre optic endoscope.

PRACTICAL EXPERIENCES Activity 18.1

Activity Manual, Page 136

Solution The critical angle will be a value of θi such that the angle of refraction is 90°. ni = 1.48, nr = 1.46, θr = 90° Snell’s law states:

ni sin θi = nr sin θr

Rearrange this to make θi the subject: Substitute values:

 n sin θr  θi = sin−1  r   ni   1.46 × sin 90  = 80.5° θi = sin−1   1.48 

Coherent and non-coherent bundles of fibres If all of the fibres in an optical fibre bundle are parallel along the full length of the bundle, so that there is a uniform one-to-one correspondence between the positions of the fibres at one end of the bundle and the positions of the opposite ends of each fibre at the other end of the bundle, then the bundle 335

18

Imaging with light

is said to be a coherent fibre bundle. The light from an object projected onto the ends of the fibre bundle by a simple lens will travel along the fibres and emerge as a corresponding image of the object at the other end of the bundle (Figure 18.1.6). This is the principle of the medical endoscope. In a non-coherent fibre bundle, one or more of the fibres swap positions relative to each other at opposite ends of the fibre. Typically, the arrangement of fibres along the bundle is random, so that although the light travels along each fibre, no clear image is produced at the exit end (Figure 18.1.7). A non-coherent fibre bundle is adequate to simply transmit light to the point where observations are being made with the endoscope. White light is used so that the doctor can observe the tissues in true colour.

Discuss differences between the role of coherent and noncoherent bundles of fibres in an endoscope.

object

object

no image

image 1 2 3 4 5

1 2 3 4 5

Figure 18.1.6

1 2 3 4 5

The arrangement of fibres in a coherent bundle allows an image to be transmitted by the bundle.

Figure 18.1.7

2 4 5 1 3

A non-coherent bundle does not transmit a sensible image.

Checkpoint 18.1 1 2 3 4

Outline how an optical fibre allows you to see something inside the body. Define total internal refraction. Explain why infra-red light is used in communication applications but visible light is used in medical applications. Identify the difference between coherent and non-coherent bundles of fibres.

18.2 Medical uses of endoscopes Endoscopes are used to visually examine the inside of a patient’s body. Being able to see tissues allows doctors to diagnose diseases such as ulcers and tumours, and also to determine the nature and extent of injuries such as damaged cartilage and ligaments in joints. Endoscopes used for specialised purposes have different names. An arthroscope is used to examine joints, a bronchoscope to view the lungs, a laparoscope to view female reproductive organs and an otoscope is used to view the ear. A disadvantage of endoscopic examination is that it only allows the surface of tissues to be viewed, and so its use is limited to problems that cause visible changes to the surfaces of tissues. It can, however, be linked with ultrasound imaging to produce better results than ultrasound scans from outside the body. 336

medical physics A risk in using an endoscope is that the part of the endoscope inside the body can tear tissues while it is being moved about. Endoscopic examination of the bowel presents a particular risk, because the bowel contains bacteria, which, if they enter the bloodstream, can produce a fatal infection. However, the risk in using an endoscope is much less than the risk encountered if the abdomen had to be opened up in conventional surgery. The fact that a patient usually has to be sedated or anaesthetised presents another minor risk to the patient, although this type of risk is the lesser problem when weighed against the alternative of an undiagnosed or untreated problem. Endoscopes that have been modified with surgical instruments can be used to remove tissue samples for testing. This process is called a biopsy, and this is one of the most common endoscopic procedures. A common example of a biopsy is the removal of polyps or other growths from the intestine for further examination and testing. Endoscopy reduces this risk because the incisions and amount of cutting is minimised. Figure 18.2.1 shows a biopsy being performed within the abdomen. The sample of tissue cut off can then be placed or sucked into a tube attached to the endoscope and withdrawn from the body. Minimally invasive surgery is conducted using optical fibre instruments that are often an integral part of the endoscope. Surgery that is commonly carried out with the aid of an endoscope includes removal of the gall bladder and the prostate, and repairs to damaged tissues in joints. A common joint operation is the repair of the anterior cruciate ligament in the knee (Figure 18.2.2). This part of the anatomy is frequently torn in sports such as netball and football, which involve vigorous twisting forces on the knees. People once condemned to months off the sporting field by knee injuries are now returning to their sport within weeks, because of endoscope-aided surgery.

Explain how an endoscope is used in: • observing internal organs • obtaining tissue samples of internal organs for further testing.

Figure 18.2.1

A biopsy from within the abdomen

Figure 18.2.2

External view of surgery to repair an anterior cruciate ligament

Developments in endoscopy

A

n endoscopic capsule (Figure 18.2.3) is an endoscope with no optical fibres! This small capsule can be swallowed by the patient and contains a wireless camera that can pass through the intestinal system and report via video link what is observed. This will improve endoscopic observation of the digestive tract.

optical dome lens holder illuminating LEDs lens battery

antenna

Figure 18.2.3

An endoscopic capsule

Checkpoint 18.2 1 2

List the advantages and disadvantages of endoscopy. Outline how a biopsy is done. 337

18

Imaging with light

PRACTICAL EXPERIENCES CHAPTER 18

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Perform a first-hand investigation to demonstrate the transfer of light by optical fibres. Gather secondary information to observe internal organs from images produced by an endoscope.

Activity 18.1: Optical fibres Many light shops sell products known generally as ‘optical fibre lights’, which consists of numerous fibre optic tubes through which coloured light is passed via some colour-changing mechanism. Obtain one of these tubes and use a LED or laser light to shine light through the tube. Change the direction of the illumination by moving the end of the tube in different directions. Equipment: optical fibre (e.g. from an optical fibre lamp), light source (LED), power supply. Discussion questions 1 Identify what optical fibres are made of and explain how something so brittle can be made so flexible. 2 Explain how light is transferred down an optical fibre. 3 Explain how optical fibres are used in an endoscope to transfer images from inside the body.

Figure 18.3.1

338

An optical fibre lamp

Chapter summary • • • • •

An endoscope is an optical instrument that allows real-time observation of internal organs. Light is transmitted through optical fibres by total internal reflection. Cheaper non-coherent bundles of fibres carry the light into the body. More expensive coherent bundles carry the image out of the body. Advantages of endoscopy – The tissues and organs are seen in real colour. – Imaging is in real time, enabling the doctor to respond to what is seen.

•

medical physics

– It allows minimally invasive tissue sampling and minor surgery that is safer and cheaper, with quicker recovery than open surgery. – The process uses non-ionising radiation, namely light, an advantage over X-rays. Disadvantages of endoscopy – It is more time consuming than ultrasound and X-rays. – It presents minor risks to the patient, especially if an anaesthetic is required. Operations on the bowel involve a risk of infection. – Only the surface of tissues is visible.

Review questions Physically Speaking

5 Explain how an endoscope is used to obtain tissue

Use some of the chapter key words to complete the following paragraph.

6 Recall an investigation that you carried out to

The medical technique known as ________________ allows minimally invasive procedures such as ________________ to be performed via a body opening or a small incision.

samples from the stomach of a patient. demonstrate the transfer of light by optical fibres.

7 Assess the advances in medical techniques as the result of the use of endoscopes.

8 Explain why endoscopic surgery is often referred to as ‘keyhole surgery’.

The ________________ allows a surgeon to view internal tissues through the ________________ while it is illuminated via the ________________. An ________________ works because light is

Solving problems

confined to the ________________ enclosed by the

9 Determine the angle of refraction of light that passes

________________ of a fibre by ________________.

Reviewing 1 Explain the importance of total internal reflection to the operation of an endoscope.

2 Compare the structure of coherent and non-coherent fibre bundles.

3 Compare the function of coherent and non-coherent

from water (n = 1.33) to glass (n = 1.48) at an incident angle of 30º.

10 Determine the critical angle for a material with refractive index of 1.4 that is immersed in: a glass (n = 1.48) b water (n = 1.33)

11 A critical angle of 48.75° is measured at the boundary between air (n = 1) and another medium. Calculate the refractive index of the medium. Can you identify the probable medium?

fibre bundles in an endoscope.

4 Explain how an endoscope is used to observe

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internal organs.

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19 radioactive decay, radiation, radioactive, radioisotopes, nucleons, atomic number, mass number, isotopes, alpha decay, alpha particle, beta decay, beta particle, positron decay, antiparticle, positron emission tomography (PET), gamma decay, half-life, radiopharmaceuticals, nuclear reactor, cyclotron, gamma camera, bone scan, collimator, scintillator, single-photon emission computed tomography (SPECT)

Imaging with gamma rays Radioactivity can be good! Images made using ultrasound, X-rays and visible light can show anatomical structures rather well. They all involve sending various forms of energy into the body. A rather different approach is to introduce radioactive elements into a person’s body and study the radiation that emerges. Images can be made of the bones as well as soft tissues including the brain, heart, liver and thyroid. Rather remarkably, this approach allows the production of images from outside the body that show how a person’s organs are functioning. Here we consider two of these minimally invasive but powerful diagnostic tools: bone scans using radioactive tracers and positron emission tomography.

19.1 Isotopes and radioactive decay Outline properties of radioactive isotopes and their half-lives that are used to obtain scans of organs.

340

For the 30 or so lightest elements, the number of protons is roughly the same as the number of neutrons in the nucleus in most of their naturally occurring isotopes. These isotopes are stable. However, many elements have isotopes whose nuclei have too few or too many neutrons. These isotopes are unstable and undergo radioactive decay in which they change and emit radiation. The type of radiation that is emitted depends on the nature of the decay (see in2 Physics @ Preliminary section 15.5). There are 82 elements that have at least one stable isotope. The stability depends on the ratio of protons to neutrons. As the atomic number increases, the ratio of neutrons to protons needed for stability also increases. Many elements have naturally occurring unstable isotopes. These are called radioactive isotopes or radioisotopes. The nucleus of a radioisotope (the parent nucleus) usually transforms itself into another nucleus (the daughter nucleus) by emitting particles and energy. It will decay repeatedly until it forms a daughter nucleus that is stable.

medical physics

increasing distance

Isotopes

A

maximum shielding

reducing time of exposure

Figure 19.1.1

Reducing the danger from radiation involves increasing distance, maximising shielding and reducing the time of exposure.

Alpha decay Some unstable nuclei decay by emitting a particle that contains two protons and two neutrons in a process known as alpha decay. The remaining nucleus has a mass number that is reduced by 4 and an atomic number that is reduced by 2. This particle emitted from the nucleus is called an alpha particle (α-particle). Alpha particles are helium nuclei ( 42 He ) and they rapidly become helium atoms, as they gain electrons from the surroundings. Such reactions are the source of most of the helium on Earth. For example, radioactive uranium-238 undergoes alpha decay to produce thorium-234. The daughter nucleus has 2 protons less than the parent nucleus and so it is a different element. In a nuclear reaction, both mass number and charge are conserved, and the decay process can be described by an equation: 238 92 U

→

234 90Th

n atomic nucleus consists of nucleons—   protons and neutrons. The number of protons in the nucleus is called the atomic number, while the total number of nucleons is called the mass number. Atoms of the same element with different numbers of neutrons are called isotopes of that element. Many isotopes occur naturally, but some are made artificially.   In section 15.4 of in2 Physics @ Preliminary we represented this information in a compact form. For example, an important isotope of fluorine is: Mass number 18 Atomic number 9

It is called fluorine-18, with 18 being the mass number. Other isotopes that are important in medicine include carbon-14, iodine-131, phosphorus-31 and technetium-99. Hydrogen is the only element that has special names for its three isotopes: hydrogen, deuterium and tritium.

electron proton

Beta decay When a radioactive nucleus undergoes beta decay, a neutron changes into a proton, releasing a high-energy electron in the process. The electron is ejected from the nucleus with such a high velocity that it totally escapes the atom. An electron (represented as e– or −10e) emitted from the nucleus in this way is called a beta particle (β-particle). An electron has only 1/1836

1 1H hydrogen

neutron 2 neutrons

+ 42 He

or illustrated by a diagram (see Figure 19.1.3). An alpha particle can only travel a few centimetres in air before it loses its kinetic energy and gains electrons to become a helium atom. In living tissue the range is about 50 µm, about half the width of a human hair. Due to their relatively large mass, alpha particles carry a lot of energy and have a high ability to ionise the surrounding medium, making them very dangerous to living cells. They are not used very much in medicine.

F

proton electron

Figure 19.1.2

proton 3 1H tritium

electron 2 1H deuterium

Isotopes of hydrogen—each with one proton and one accompanying electron

daughter nucleus Th-234 parent nucleus U-238

Figure 19.1.3

4 2He

alpha particle (helium nucleus)

Uranium-238 undergoes alpha decay

341

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Imaging with gamma rays

14 7N 14 6C

daughter nucleus

parent nucleus

Figure 19.1.4

beta particle

of the mass of the proton, and its mass number is 0. Its negative charge gives it an atomic number of –1. Beta decay increases the atomic number of the nucleus by one but the mass number does not change. There is one less neutron but one more proton and hence the total number of nucleons does not change. Carbon-14 decays by beta decay to nitrogen-14. The process can be described using an equation or a diagram (Figure 19.1.4).

Carbon-14 decays to nitrogen-14 and an electron.

14 6C

→ 147 N +

0 −1 e

The two products are a stable nitrogen nucleus and a beta particle (electron). Beta particles have a range that depends on their energy. In air they travel several metres, but they travel only a few millimetres in human tissues.

Beta decay: positron production

Identify that during decay of specific radioactive nuclei positrons are given off.

Some nuclei of radioactive elements are unstable because they have too many These become protons in the nucleus, relative to the number of neutrons. more stable by a form of beta decay called positron decay. A positron 0 (represented as e+ or β+ or sometimes +1 e ) is the antiparticle of the electron. It has the same mass as an electron and the same magnitude charge as the electron, but it is positively charged. In positron decay, a proton in the nucleus decays to a neutron and a positron. The atomic number of the decaying nucleus decreases by one but the mass number remains the same. The positron is ejected with high kinetic energy from the nucleus. The most important positron-emitting isotope is fluorine-18, an artificially produced isotope. Its decay equation is: 18 18 0 9 F → 8 O + +1 e

Discuss the interaction of electrons and positrons resulting in the production of gamma rays.

Carbon-11 and oxygen-15 are also artificially produced positron emitters. When a positron and an electron collide, they mutually annihilate (see Physics Phile ‘Evil twins’ p 73). Their total mass is converted into energy, producing two identical gamma (g) ray photons with a total energy consistent with Einstein’s famous equation E = mc2 where m is the total mass of the two particles (the mass of each electron is 9.1 × 10–31 kg) (see section 3.4). The energy of each gamma ray is E = 9.1 × 10–31 × (3 × 108)2 = 8.2 × 10–14 J = 511 keV The two gamma rays emerge in opposite directions to conserve momentum (Figure 19.1.5). In positron emission tomography (PET) (section 19.5) it is these gamma rays, not the positrons, that are detected and used to produce medical images. Both kinds of beta decay are always accompanied by the creation of neutrinos, which are almost undetectable and not relevant to medical applications.

positron emitter e+

Gamma ray emitters gamma ray photon

Figure 19.1.5

342

e– + e

gamma ray photon

A positron and an electron annihilate to produce two gamma rays.

The most widely used medical isotope is technetium-99. It is a beta emitter that decays to ruthenium-99. The higher energy form of technetium-99, called metastable technetium-99 or technetium-99m, is also unstable and becomes more stable by emitting electromagnetic radiation. At the energy involved, this The gamma ray carries away no charge or mass radiation is a gamma ray. and so the nucleus remains technetium-99. It is the gamma decay of technetium-99m that is the medically important decay.

medical physics Gamma decay also frequently accompanies alpha or beta decay. Cobalt-60 is a well-known beta and gamma emitter. Gamma rays have no charge and so can pass easily through matter. To achieve the 95% absorption of the gamma energy from a cobalt-60 source, the rays must pass through 60 mm of lead, 100 mm of iron, or 330 mm of concrete.

alpha beta X-ray (medical) gamma ray neutrons

concrete human hand

Figure 19.1.6

aluminium

thin lead

thicker lead

The penetrating ability of different forms of radiation

Checkpoint 19.1 1 2 3

Describe what happens to an unstable nucleus. Identify two naturally occurring isotopes of the same element. Compare an electron with a positron.

Atom-sized energy units

P

articles or gamma rays produced in radioactive decay carry away energy. The SI unit for energy is the joule, but that’s a lot of energy for a tiny particle or even a single gamma ray. A more ‘atomsized’ unit of energy is the electron volt (eV)—the energy acquired by an electron in accelerating through a potential difference (voltage) of one volt. More generally, the energy acquired by a particle of charge q is given by: E = qV So the energy acquired by the electron is: E = (1.6 × 10–19 C)(1 V) = 1.6 × 10–19 J So 1.6 × 10–19 J = 1 eV   As with other units, a large amount of energy can be expressed as keV (kilo-electron volts) and MeV (mega-electron volts) and so on.

19.2 Half-life

Amount (grams)

The time it takes for half the mass of the parent isotope to decay into Outline properties of daughter nuclei is defined as the half-life of the isotope. After one half-life, only radioactive isotopes and their 50% of the original parent isotope remains; 50% of that remaining amount half-lives that are used to obtain scans of organs. decays after another half-life, leaving just 25% of the original parent isotope and so on. The half-life of a radioactive isotope can be deduced from a graph showing the mass of the remaining 2000 Strontium-90 decay radioactive atoms of the element plotted against time. In 1800 Figure 19.2.1, the time taken for 2000 g of strontium-90 1600 to be reduced to 1000 g is 28.1 years, the half-life of 1400 1200 strontium-90. The daughter isotope is yttrium-90, which 1000 rapidly decays to zirconium-90, which is stable. 800 This mathematical model, called exponential decay, is 600 applicable to all forms of radioactive decay. The rate of 400 200 radioactive decay of an isotope is not affected by changes 0 in physical conditions such as temperature or pressure. 0.0 28.1 56.2 84.3 112.4 140.5 168.6 196.7 224.8 252.9 281.0 309.1 Time (years) The decay rate is unchanged by any chemical reactions (or compounds in which the radioactive isotope may be Figure 19.2.1 The mass of strontium-90 remaining versus time, incorporated). Every radioisotope has its own half-life. from an original sample of 2000 g 343

19

Imaging with gamma rays

Checkpoint 19.2 1 2

Define half-life. Outline how you would find the half-life of an isotope from a graph showing its mass in a sample versus time.

19.3 Radiopharmaceuticals: targeting tissues and organs

Outline properties of radioactive isotopes and their half-lives that are used to obtain scans of organs.

Radioisotopes are used to produce functional images of the body. They are used to examine blood flow to the brain, to assess functioning of the liver, lungs, heart or kidneys, to assess bone damage, and to confirm other diagnostic procedures. In contrast to the imaging techniques already discussed—ultrasound images, X-ray images and CAT scans—the use of radioisotopes can show how the body is functioning, rather than simply showing detailed images of tissues and organs in the body. Diseases such as cancer alter chemical processes in the body and images produced using radioisotopes can reveal these changes. Radioactive chemicals used in medicine are called radiopharmaceuticals.

Choose your element

Describe how radioactive isotopes may be metabolised by the body to bind or accumulate in the target organ.

344

Different isotopes of the same element have identical chemical properties. This is important in medical applications because radioactive isotopes can be substituted for non-radioactive atoms normally used by the body. Radiopharmacologists are able to attach various radioisotopes to biologically active substances and introduce them into the body. The radioisotope is then incorporated into the normal biological processes. Radiopharmaceuticals are chosen so that when they enter the body they will circulate around the body and be absorbed by the organ of interest. The radiopharmaceutical may be a substance that is specifically used by a particular organ or it may be part of a molecule that is used by the organ to be imaged. Radiopharmaceuticals are prepared by replacing one of the atoms in the molecules of that substance with a radioactive atom. This process is sometimes referred to as ‘tagging’ or ‘labelling’. When the radiopharmaceutical is placed in the body, it accumulates in the target organ and so the radiation that is emitted from that organ during the imaging process will be greater than the amount emitted from other organs and tissues. The choice of a radioisotope used for medical imaging is based on the following criteria. It must: • produce gamma rays (directly or indirectly) since only gamma radiation is likely to leave the body • have a half-life that is long enough for the molecule to enter the metabolic processes, yet short enough to minimise the radiation dose to the patient • be taken up rapidly into the desired tissue—this is achieved by incorporating the radioisotope into an appropriate compound that is metabolised by the target tissue in the body • be rapidly excreted from the body. Appropriately chosen compounds are broken down and excreted as part of the normal body chemistry.

medical physics Radiopharmaceuticals are used in very small quantities for diagnostic work. Just enough is administered to obtain the required information before the radiopharmaceutical decays, therefore minimising cell damage from the radiation. The radiation dose received is similar to that from diagnostic X-rays.

Where to get your radioisotopes Radioisotopes are produced in two main ways: in a nuclear reactor or in a cyclotron particle accelerator. In a nuclear reactor, fission (splitting) of heavy nuclei produces large numbers of neutrons. The target element to be converted into a radioisotope is placed in the path of these neutrons and the nuclei absorb one or more neutrons, producing an unstable isotope with an excess of neutrons. These typically decay via beta decay. Different isotopes can be produced in a cyclotron. In a cyclotron, protons are accelerated in a vacuum and fired into the nucleus of a target atom, to create isotopes that have an excess of protons. These isotopes typically decay via positron emission. In Australia, most radioisotopes for medical purposes are prepared at the Australian Nuclear Science and Technology Organisation (ANSTO) OPAL nuclear reactor at Lucas Heights in Sydney and at the National Medical Cyclotron at the Royal Prince Alfred Hospital in Sydney (see Physics Focus p 353). The ANSTO reactor and the cyclotron make various radiopharmaceuticals, including some of those listed in Tables 19.3.1 and 19.3.2. Nuclear reactors and cyclotrons are expensive to build and operate. The cost of these machines and the increasing demand for radioisotopes has added significantly to health costs in Australia and this has an impact on society through increased medical insurance costs. However, it is not practical to import many medical radioisotopes from overseas because of the short half-lives and the difficulty of transporting them safely.

Figure 19.3.1

The National Medical Cyclotron Facility at Royal Prince Alfred Hospital

Table 19.3.1  Examples of radioisotopes produced in a nuclear reactor Isotope

half-life

Cobalt-60

5.3 years

Iodine-131

8 days

Phosphorus-32

14.3 days

Molybdenum-99

2.75 days

Technetium-99m

6 hours

Emission β, γ β, γ β

γ

Uses External beam radiotherapy Cancer diagnosis and imaging of the thyroid gland Treatment of excess red blood cells ‘Parent’ isotope used in a generator to produce technetium-99m, the most widely used isotope in nuclear medicine To image the skeleton and heart muscle in particular, but also for brain, thyroid, lungs, liver, spleen, kidney, gall bladder, bone marrow, salivary and lacrimal glands, heart blood pool, infection and numerous specialised medical studies

Table 19.3.2  Examples of radioisotopes produced in a cyclotron Isotope

Half-life

Carbon-11

20.3 minutes

Emission Uses β+

Nitrogen-13

10 minutes

Oxygen-15

2.03 minutes

Fluorine-18

109.8 minutes

Iodine-124

4.5 days

β+ β+ β+ β+

In PET for studying brain physiology and pathology, for investigating epilepsy; have a useful role in cardiology In PET scans to tag ammonia and trace protein metabolism In water for PET scan on blood flow in muscle tissue For investigating tumours in the breast, prostate, liver and brain To investigate cancer, notably in the thyroid gland

345

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PHYSICS FEATURE

3. Applications and uses of physics

Technetium-99m

T

he radioisotope technetium-99m (Tc-99m) is used in more than half of all nuclear medicine procedures. It is an isotope of the artificially produced element technetium and is almost ideal for use in nuclear medicine scanning. Technetium-99m has a half-life of 6 hours and when it decays it emits a single gamma ray with energy of 140 keV. It does not emit any alpha or beta particles—an advantage since these cause biological damage. A further advantage of Tc-99m is that it has several electron valence states, so it has versatile chemical properties and can be used to produce a wide variety of pharmaceuticals. A technetium generator, consisting of a lead vessel enclosing a glass tube containing the radioisotope molybdenum-99, is supplied to hospitals by ANSTO. Molybdenum-99 (Mo-99) has a half-life of 66 hours and it decays to Tc-99m by normal beta decay. The

Tc-99m is washed out of the lead or glass pot as required by passing a saline solution through the glass tube, a process called elution. After about 2 weeks the generator is returned for recharging. The two nuclear reactions involved in Tc-99m production can be represented as:

99 m 99 42Mo → 43 Tc



99 m 43 Tc

→

+ −01e half-life 2.75 days

99 43Tc +

γ

half-life 6.01 hours

Technetium-99m has a half-life of 6 hours, which is long enough to examine metabolic processes, yet short enough to minimise the radiation dose to the patient. It is usually injected into the patient’s body and eliminated from the body via the kidneys. Technetium-99m allows the diagnosis of many conditions in the brain, lungs, heart, bones and other organs, avoiding the use of surgery.

Checkpoint 19.3 1 2 3 4

Explain the process of tagging. Discuss why only small amounts of radioisotopes are used in patients. Outline how radioisotopes are made. Explain why we do not just import all radioisotopes from overseas.

digital signal processing unit photomultipliers NaI (Tl) detector lead collimator

gamma photons

radioactive tracer in body

Figure 19.4.1

346

Principle of a gamma camera

19.4 The gamma camera A gamma camera (Figure 19.4.1) detects gamma rays emitted by a radiopharmaceutical in the patient’s body. The camera reveals the distribution of radioactive material in a patient and this distribution is determined by the uptake of the radiopharmaceutical, which is dependent on the functions—normal or otherwise—of the body. In a bone scan, for example, technetium-99m methylene diphosphonate (Tc99m-MDP) is usually injected into the patient. The radioactive phosphate travels through the blood and is metabolised and accumulates in the bones. The patient lies under a gamma ray camera. The gamma rays coming from the radioactive tracer in the body travel out of the body in all

medical physics directions. Unlike visible light, gamma rays cannot be focused to form an image. Instead a lead collimator with parallel holes passing from the bottom to the top is used to allow only those gamma rays directed through the holes to reach the detector itself. The detector is a scintillator, made of a crystal of sodium iodide containing traces of thallium. The thallium impurity causes the crystal to scintillate (i.e. to emit a small flash of light) when a gamma ray enters the crystal. The flash is very faint, so sensitive devices called photomultipliers are used to detect the flash and amplify its effects, converting the result into an electrical signal. A computer processes the signals to produce an image that may be viewed directly on a screen, printed or stored electronically, since the information is all digitally encoded. Regions of bone having a high metabolic activity appear darker on the bone scan because more of the radiopharmaceutical is absorbed (Figure 19.4.2). Conditions that may cause increased metabolic activity include cancer cells multiplying in the bone, healing of a fracture or damage from arthritis. A gamma camera is also used in single-emission computed tomography (SPECT). The camera acquires two-dimensional images from multiple directions. These are then assembled into a three-dimensional image in a computer, in a similar way to other three-dimensional tomographic imaging techniques.

PRACTICAL EXPERIENCES Activity 19.1

Activity Manual, Page 140

Checkpoint 19.4 1 2

Outline how a gamma camera creates an image of the body using gamma radiation. Explain how the ‘grey scale’ in the bone scan (Figure 19.4.2) can be interpreted. Figure 19.4.2

19.5 Positron emission tomography Positron emission tomography (PET) is a ‘functional’ imaging technology that allows physicians to assess chemical changes related to how the body is functioning. In this way, PET is like a bone scan. PET imaging can analyse sugar metabolism, blood flow, oxygen use and a long list of other vital physiological processes. It is often used to investigate brain function and diseases affecting the brain, such as epilepsy, schizophrenia and Parkinson’s disease, as well as investigating which parts of the brain are active during specific activities. PET imaging uses positron-emitting radiopharmaceuticals to obtain images. Positron-emitting isotopes exist for carbon, oxygen, nitrogen, fluorine and others, which allows these radioisotopes to be substituted into many naturally occurring substances used by the body (e.g. proteins, water and sugars—especially glucose). An increased level of metabolic activity in a part of the body associated with the chosen radiopharmaceutical causes a higher radiopharmaceutical concentration in that part and hence the gamma rays from those parts are more intense than those from parts in which there is less metabolic activity.

This bone scan shows numerous hotspots along the spine, pelvis, shoulders and ribs that indicate that in this patient the breast cancer has spread to the bone.

347

19

Imaging with gamma rays

seeing words

hearing words

Figure 19.5.1

This PET image contrasts the visual and auditory stimulation of the brain.

For example, if glucose labelled with fluorine-18, a positron emitter, is given to a patient who is then asked to perform a task, the regions in the brain in which the isotope concentrates most are the regions that are actively involved in that behaviour. Figure 19.5.1 shows PET images that reveal the different active parts of the brain of a person reading (seeing) written words or hearing the same words spoken by another person. The radiopharmaceuticals are usually injected into a patient, although oxygen-15 can be inhaled and absorbed into the body through the lungs. The radioactive atoms decay, emitting positrons that travel about 2 mm before encountering an electron. Mutual annihilation occurs and two gamma rays are emitted in opposite directions. In a PET scanner these gamma rays trigger pairs of gamma-ray detectors that are arranged in a ring around the patient (Figure 19.5.2). When two detectors directly opposite each other record photons within a few nanoseconds of each other (called a coincidence pair), the data are stored for processing to produce the image. Non-coincident hits on a single detector are ignored because they do not provide useful information about the location of the annihilation that produced them. Signals reaching opposite detectors are analysed by a computer, which determines the position of the source of the annihilation event using the intersections of the gamma ray trajectories. Regions that emit the most gamma-ray photons produce a stronger signal than regions that emit fewer photons. A large number of measurements from the ring of detectors surrounding the patient is required to compute the distribution of the radioisotope in the body. PET images don’t show anatomical structures well, so they are usually shown superimposed onto other medical images such as CAT scans, to clarify the location of the PET signal (Figure 19.5.3).

a

b detectors

coincidence unit

Figure 19.5.2

348

valid event

(a) A PET scanner showing coincidence detection of gamma rays by opposite detectors and (b) an example of a PET scanner

medical physics PET is a valuable diagnostic tool because the images that it produces show differences in chemical activity in different parts of the body. These differences may result from normal processes or they may be the result of diseases such as cancer or other functional abnormalities in the body. The great diagnostic benefit of PET results from the fact that it is possible to incorporate positron emitters into a wide variety of radiopharmaceuticals which can be produced to target specific organs or chemical processes in the body. PET has provided new knowledge, in particular about the functions of the normal brain, as well as of diseases that affect the brain. This knowledge has also greatly modified brain surgery.

Figure 19.5.3

PRACTICAL EXPERIENCES Activity 19.2

Activity Manual, Page 143

Describe how positron emission tomography (PET) technique is used for diagnosis.

In this false colour PET scan (centre) two regions of high metabolic activity typical of cancerous tumours are indicated by the bright white hot-spots (in the spine and the prostate). This image is combined with a CAT image (left) to produce the fused image (right) that accurately indicates their location.

Checkpoint 19.5 1 2 3

Explain what features in the body PET scans can be used to identify that X-rays and ultrasound can not. Outline how the radioisotopes for a PET scan can be introduced into the body. Explain how coincidence pairs in a PET scanner allow an image to be constructed.

349

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Imaging with gamma rays

PRACTICAL EXPERIENCES CHAPTER 19 This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 19.1: Bone scans Perform an investigation to compare a bone scan with an X-ray image.

A bone scan is performed to obtain a functional image of the bones and so can be used to detect abnormal metabolism in the bones, which may be an indication of cancer or other abnormality. Because cancer mostly involves a higher than normal rate of cell division (thus producing a tumour), chemicals involved in metabolic processes in bone tend to accumulate in higher concentrations in cancerous tissue. This produces areas of concentration of gamma emission, indicating a tumour. Compare the data obtained from the image of a bone scan with that provided by an X-ray image. Discussion questions 1 Identify the best part of the body for each of these diagnostic tools to image. 2 Compare and contrast the two images in terms of the information they provide.

Figure 19.6.1

Comparison of an X-ray and bone scan of a hand

Gather and process secondary information to compare a scanned image of at least one healthy body part or organ with a scanned image of its diseased counterpart. a

Figure 19.6.2

350

b

Activity 19.2: Healthy or diseased? Typical images of healthy bone and cancerous bone are shown. The tumours show up as hot-spots. Use the template in the activity manual to research and compare images of healthy and diseased parts of the body. Discussion questions 1 Examine Figure 19.4.2. There is a hot-spot that is not cancerous near the left elbow. Explain. 2 In the normal scan (Figure 19.6.2a), the lower pelvis has a region of high intensity. Why is this? (Hint: It may be soft tissue, not bone. Looking at Figure 19.6.2b might help you with this question.) 3 State the differences that can be observed by comparing an image of a healthy part of the body with that of a diseased part of the body.

Bones scans of (a) a healthy person and (b) a patient with a tumour in the skeleton

Chapter summary •

• • • •

•

The number of protons in a nucleus is given by the atomic number, while the total number of nucleons is given by the mass number. Atoms of the same element with different numbers of neutrons are called isotopes of that element. Many elements have naturally occurring unstable radioisotopes. In alpha decay an unstable nucleus decays by emitting an alpha particle (α-particle). In beta decay, a neutron changes into a proton and a high-energy electron that is emitted as a beta particle (β-particle). In positron decay, a positron—the antiparticle of the electron—is emitted.

•

• •

• • •

medical physics

When a positron and an electron collide, their total mass is converted into energy in the form of two gamma-ray photons. In gamma decay a gamma ray (g) is emitted from a radioactive isotope. The time it takes for half the mass of a radioactive parent isotope to decay into its daughter nuclei is the half-life of the isotope. Artificial radioisotopes are produced in two main ways: in a nuclear reactor or in a cyclotron. A gamma camera detects gamma rays emitted by a radiopharmaceutical in the patient’s body. PET imaging uses positron-emitting radiopharmaceuticals to obtain images using gamma rays emitted from electron–positron annihilation.

Review questions Physically Speaking

Reviewing

Below is a list of topics that have been discussed throughout this chapter. Create a visual summary of the concepts in this chapter by constructing a mind map linking the terms. Add diagrams where useful.

1 Recall how the bone scan produced by a radioisotope

Radioactive decay

compares with that from a conventional X-ray.

2 Analyse the relationship between the half-life of a radiopharmaceutical and its potential use in the human body.

3 Explain how it is possible to emit an electron from the Radiation

Radioisotope

Nucleon

nucleus when the electron is not a nucleon.

4 Assess the statement that ‘Positrons are radioactive particles produced when a proton decays’.

5 Discuss the impact that the production and use of Neutron

Proton

Isotope

Alpha decay

radioisotopes has on society.

6 Describe how isotopes such as Tc-99m and F-18 can be used to target specific organs to be imaged.

7 Use the data in Table 19.6.1 to answer the questions: Beta decay

Gamma decay

Antimatter

PET

Half-life

Bone scan

Positron decay

Scintillator

a Which radioactive isotope would most likely be used in a bone scan? Justify your choice. b Propose two reasons why cesium-137 would not be a suitable isotope to use in medical imaging.

Table 19.6.1  Properties of some radioisotopes Radioactive source

Radiation emitted

Half-life

C-11 Tc-99m TI-201 I-131 Cs-137 U-238

b+, g g g b, g a a

20.30 minutes 6.02 hours 3.05 days 8.04 days 30.17 years 4.47 × 109 years

351

19

Imaging with gamma rays

8 Figure 19.6.3 shows a bone scan and an X-ray image.

14 Describe the interaction between electrons and

Identify which is which and justify your choice. a

b

positrons.

15 Explain how a PET scan is produced and compare the image with that obtained from a CAT scan.

16 Identify the key characteristic of a PET image that gives it a diagnostic advantage over a CAT image.

Solving problems 17

Figure 19.6.3

Two different images showing the chest region of the human body

9 Compare the processes used to produce a bone scan image and an X-ray image.

10 Compare the radiation used to produce a bone scan image and an X-ray image.

11 Compare the resolution of a bone scan with that of an X-ray.

N

α

146 234Th

144

dependent on the design of the lead collimator used in the gamma camera.

13 Explain why tumours are visible in the patient in Figure 19.6.4b.

234U

230Th

140

226Ra

138 222Rn

136

218Po

134

214Pb

130

210TI

128

218At 214Bi 214Po

210Pb

126

210Bi

210Po

206Pb

124 122 78

b

238U 234Pa

142

132

12 Explain why the resolution achieved in a bone scan is

a

Complete the decay series of U-238 shown in Figure 19.6.5 by including the decay product—an α, β or γ particle—in each step, as shown for the α-emission seen in the first step.

80

82

Figure 19.6.5

84

86

88

90

94 Z

92

The decay series of U-238

18 Figure 19.6.6 shows the percentage of a radioactive isotope that remains as a function of time. a Estimate the half-life of this isotope. b Would the isotope represented by this decay curve be suitable for medical imaging? Justify your answer.

50.0%

25.0% 12.5%

Re

iew

352

Q uesti o

A bone scan of (a) a healthy patient and (b) a patient with tumours

6.3% 0

2

n

s

v

Figure 19.6.4

Figure 19.6.6

4

6

8 Years

10

Decay curve of a radioisotope

12

3.1% 14

16

medical physics

PHYSICS FOCUS

4. Implications of physics for society and the environment

ANSTO Radiopharmaceutical production Lucas Heights production facility The Lucas Heights Facility in Sydney, operated by ANSTO, manufactures radiopharmaceuticals such as molybdenum-99 generators for technetium-99m and iodine-131, together with other associated products including chromium-51, thallium and gallium. The facilities are ideal for simultaneous production of large quantities of different isotopes used for diagnosis and therapy. An average of 3500 medical isotopes and 2500 packages are dispatched per month. The packaging, transport and delivery of all products comply with strict national and international regulations.

Molybdenum plant ANSTO Health is in the process of commissioning a new plant that will be used to manufacture molybdenum-99. ANSTO is currently importing this product from overseas to meet domestic needs; however, this is costly (as a heavy airfreight product) and shipment is not always reliable, which emphasises the importance of local Australian production. Molybdenum-99 is used as a raw material for 80% of nuclear medicine procedures performed around the world. Molybdenum-99 is formed by the fission of uranium-235 that is itself formed by irradiation of low enriched uranium (LEU) in ANSTO’s nuclear reactor. The molybdenum-99 is then used as a generator for technetium-99m, which can then be formulated into a plethora of radiopharaceuticals. The new plant is located behind shielding because of radiation, and has been carefully designed to protect workers and the external environment during and after processing.

5. Current issues, research and developments in physics

National Medical Cyclotron Facility (NMC) The NMC was established in 1991 and produces specialised isotopes for SPECT (single photon emission computed tomography) imaging. The facility houses a 30 MeV cyclotron, the largest in Australia. Located at Sydney’s Royal Prince Alfred Hospital, Camperdown, the cyclotron accelerates protons to the required energy and then fires them at different targets to generate the radioisotope required. The radioactive isotope is then recovered and purified for dispensing into the finished dosage at Lucas Heights. The products produced at the cyclotron facility are gallium-67, thallium-201, iodine-123 and iodine-123. 1 Discuss why the correct packaging and transport of radiopharmaceuticals is important. 2 Discuss why it is important to have facilities to produce radiopharmaceuticals in Australia. 3 Explain why molybdenum-99 is an especially important radioisotope.

Extension 4 The OPAL nuclear reactor opened in 2007, replacing the earlier HIFAR reactor. Some critics of the nuclear reactor program claim a new reactor was unnecessary.

Investigate this issue and discuss your conclusions in terms of: a the need for radiopharmaceuticals b the safety of a nuclear reactor.

353

20 spin, magnetic moment, parallel, antiparallel, precession, Larmor frequency, radio frequency (RF), magnetic resonance imaging (MRI), resonance, relaxation, longitudinal relaxation time constant, transverse relaxation time constant, RF transceiver coils, gradient coils, functional MRI,

Imaging with radio waves Hydrogen calling Magnetic resonance imaging (MRI) uses radio waves emitted by hydrogen atoms to produce high-resolution images of the body. Magnetic resonance has been used as a tool for studying atomic structure since 1946, but it was not until 1973 that medical imaging was suggested, and another 15 years before useful images were obtained. Like CAT scans, MRI can produce tomographic images (slices) of the body, enabling detailed two- and three-dimensional images to be constructed from the data. Magnetic resonance imaging was initially used to produce images showing structural detail; however, recent developments have led to functional MRI, which has greatly advanced knowledge of how the brain and other organs work.

20.1 Spin and magnetism Identify that protons and neutrons in the nucleus have properties of spin and describe how net spin is obtained.

spin vector

protons

spin vector

354

opposite spins resulting in zero net spin

Particles such as protons, neutrons and electrons have a property called spin, which can be visualised as a rotation (spinning) of the particles. However, spin is a quantum-mechanical property of the particles and the classical view of a spinning ball is a useful ‘model’, but it is not reality. Protons are not simply tiny spinning charged balls, but some of their properties are similar. For protons, neutrons and electrons the spin property can only have one of two possible directions: up or down (sometimes called ‘spin up’ and ‘spin down’). If the total number of protons and neutrons in a nucleus is even, then their spins align in pairs in opposite directions so that the net spin of each pair and the whole nucleus is zero (Figure 20.1.1). If there is an odd number of nucleons, then the nucleus has a net spin, since there must be one unpaired nucleon. Nuclei having a net spin include hydrogen, phosphorus-31, fluorine-19, nitrogen-15 and carbon-13.

Figure 20.1.1

Using our model of spin, we can visualise two particles of opposite spin resulting in zero net spin.

medical physics Spin produces a magnetic effect called a magnetic moment or Identify that the nuclei of magnetic dipole. You can think of this as being similar to the magnetic field certain atoms and molecules produced by moving charge (a current) around the wire of a solenoid (see Figure behave as small magnets. 4.1.10). Therefore, charged particles such as electrons and protons behave like tiny magnets. Even uncharged neutrons have a magnetic moment, because they contain fundamental charged particles called quarks (see section 15.5). Nuclei with an odd number of nucleons will therefore have a net spin and always have an unpaired magnetic moment and so behave like tiny magnets. When placed in a strong magnetic field these nuclei behave a little like a compass needle in the A compass needle always points north, but the Earth’s magnetic field. Explain that the behaviour of nuclei magnetic moments may assume one of two possible alignments—they can nuclei with a net spin, particularly align parallel or antiparallel to the magnetic field. External magnetic fields or hydrogen, is related to the magnetic field they produce. the magnetic field of electromagnetic radiation may affect the nuclei because of this magnetic moment. north Hydrogen is the most important element in the process of magnetic resonance imaging because of magnetic the magnetic properties of the nucleus and the fact moment that it is present in water molecules, proteins, fats and carbohydrates—in fact, in most of the molecules in the body. Since it normally has just a single proton forming the nucleus, a hydrogen nucleus will clearly have a magnetic moment. The magnetic properties of hydrogen are not normally evident, because the magnetic moments of the south nuclei are randomly aligned, resulting in a zero net magnetic effect (Figure 20.1.2). Figure 20.1.2 The net spin of hydrogen results in a magnetic moment that is often not apparent in large groups of hydrogen nuclei.

Checkpoint 20.1 1 2 3 4

Outline what is meant by the ‘spin’ of a particle. Describe the situations in which an atomic nucleus has zero net spin. Define the term magnetic moment. Describe the two possible alignments of the magnetic moment of a proton.

20.2 Hydrogen in a magnetic field When placed in a very strong magnetic field, the proton will tend to align with the applied field, because of the interaction between the external field and the magnetic properties of the proton. In a medical context, the strength of the magnetic field used is typically between 0.5 and 5 T, although fields used in research applications may be more than 10 T. The ‘parallel’ and ‘antiparallel’ alignments to the field are not strictly parallel to the field. They are actually The parallel configuration has a angled as illustrated in Figure 20.2.2. slightly lower energy than the antiparallel configuration; just how much lower depends on the strength of the magnetic field. In a 1 T field, the difference is just 0.18 µeV (0.18 × 10–6 eV or 2.8 × 10–26 J). As a lower energy state, it is

Describe the changes that occur in the orientation of the magnetic axis of nuclei before and after the application of a strong magnetic field.

355

20

Imaging with radio waves

no external field

therefore slightly preferred—but only about 0.0003% more protons favour the parallel configuration. Injecting the energy difference that corresponds to a photon with a radio frequency of 42.6 MHz can cause the proton to flip from the parallel to the antiparallel configuration. This same energy will be released if it flips back. When the applied external field is removed, the protons become randomly orientated again because of collisions occurring between the randomly moving atoms.

Precession external magnetic field

parallel

antiparallel

Figure 20.2.1

Randomly orientated hydrogen nuclei align to an externally applied magnetic field.

Define precessing and relate the frequency of the precessing, (i.e. Larmor frequency), to the composition of the nuclei and the strength of the applied external magnetic field.

When you have a magnetic moment angled to the direction of a magnetic field, the field will exert a force and hence a twist (a torque) on the magnetic moment. This will cause the direction of the magnetic moment to rotate about the magnetic field direction. This is called precession. It is similar to the precession of a toy spinning top caused by the force of gravity. In our model of a spinning proton, the direction of the magnetic moment corresponds to the imagined spin axis of the proton, which is analogous to the spin axis of the top (Figure 20.2.2). The frequency of precession of external magnetic field a magnetic moment such as a proton is called the Larmor frequency. It depends on the magnetic properties moment of the proton (µp) and is proportional to precession the magnitude of the external field (B). It is given by: ⎛ 2μ p B ⎞ f Larmor = ⎜⎜ ⎟⎟ ⎝ h ⎠ where h is Planck’s constant. For a proton in a 1 T magnetic field, the Larmor frequency is 42.6 MHz, corresponding to radio frequency (RF) electromagnetic waves.

Try this!

Figure 20.2.2

Precession of a hydrogen nuclei around the applied magnetic field

precession

A spinning top When you try to push over a child’s spinning top, it doesn’t fall over. Try it! Instead, the top’s axis of rotation itself starts to rotate around the vertical axis—it ‘precesses’ (at least until frictional forces cause it to tip too far and it hits the ground). The force of gravity produces a twisting effect (a torque) on the top. If the top weren’t spinning, this would cause it to simply fall over. Precession is the motion that results when the top is spinning. The axis of rotation of the top sweeps out a conical motion.

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rotation

force of gravity

Figure 20.2.3

Precession of a spinning top

medical physics

Checkpoint 20.2 1 2 3

Explain what happens to a proton in an external magnetic field. Using the photon energy given by E = hf, demonstrate that the frequency f required to cause a proton to flip its alignment in a 1 T magnetic field is 42.6 MHz. Explain how the frequency required to flip a proton will change if the applied magnetic field is doubled.

20.3 Tuning in to hydrogen Excitation

PRACTICAL EXPERIENCES Activity 20.1

Activity Manual, Page 146

In magnetic resonance imaging (MRI) a person is placed in a strong Discuss the effect of subjecting magnetic field and a pulse of RF electromagnetic radiation is sent into their precessing nuclei to pulses of body. The frequency is tuned to match the Larmor frequency of protons within radio waves. the field. The RF pulses have two effects on the protons in the person’s body. First, they cause a ‘spin flip’ to occur. A few of the protons aligned parallel to the external field absorb the RF energy because it matches the energy necessary to flip from the parallel to the slightly higher energy antiparallel configuration. Before the RF pulse there was a small imbalance in numbers, with slightly more protons in the parallel configuration. The net magnetic a b effect of the alignment of protons along the magnetic field direction (usually called the longitudinal S S magnetisation) was not zero. The RF pulse results in the number of parallel and antiparallel protons becoming more nearly equal, causing the net magnetic effect in this Z direction to become zero (Figure 20.3.1). This direction X is along the head-to-toe axis of the person in the MRI Y machine. N N The second effect of the RF pulses is to cause the protons to precess in step (‘in phase’) with each other. Figure 20.3.1 (a) Parallel (blue) and antiparallel (red) protons are aligned This can be compared to soldiers all marching in step. within a strong magnetic field. (b) RF radiation causes some The protons precessing in phase with each other create protons to jump to the antiparallel configuration. a net magnetic effect, Bxy, in the plane perpendicular a b to the field (Figure 20.3.2). This net magnetic effect S S rotates at the Larmor frequency. The RF wave is used to transmit energy into the tissues to be imaged. The energy of the RF electromagnetic waves is absorbed effectively by protons Bxy in hydrogen, because the frequency of the radiation and the precession frequency of the protons are the same. Said another way, the energy of the RF photons matches N N the energy difference between the parallel and antiparallel Figure 20.3.2 (a) The protons precess in phase in response to the RF pulse. configurations. This energy exchange in which properties (b) The net magnetic field is in the xy plane. are matched is called resonance.

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Imaging with radio waves

Resonance

1st and 2nd strings

R

esonance occurs between two oscillating objects if they have the same natural frequency and if there is some means by which energy can be transferred from one object to the other. The everyday experience of pushing a child on a swing demonstrates the principle. To make the child swing higher, you must push at the same frequency as the child is swinging—push every time they get to the top of the back swing. If you push at the wrong time, you will just interfere with the swing (and hurt yourself!).   Resonance can also be observed in a guitar when pressing down on the fifth fret of the first string and plucking the second string. If the guitar is tuned correctly, the first string will also vibrate. This occurs because under these conditions the two strings have the same natural frequency and the energy can be transferred from one string to the other via the body of the guitar.

nut 1st fret

5th fret

Figure 20.3.3

Resonance can be observed in a guitar.

Longitudinal magnetisation

Relaxation 100%

Time

T1

Figure 20.3.4

The relationship between the longitudinal relaxation time constant T1 and the increasing longitudinal magnetisation

Table 20.3.1  Relaxation times for various tissues in a 1 T field Tissue Brain grey matter Brain white matter Cerebrospinal fluid Fat Muscle Blood Water

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T1 (ms) 520 390 2000 180 600 800 2500

T2 (ms) 100 90 300 90 40 180 2500

After the RF pulse, the protons return to their original state and, as they do so, they re-emit the energy they absorbed from the radio wave. They do this over a period varying from about 0.01 to 0.1 s. The emitted energy is again an RF wave that is detected using an antenna. The signals emitted by the protons are used to create the MR image. The return of the proton to the less excited state after absorbing the RF energy is called relaxation. The emission of the energy from the hydrogen nuclei returning to the lower energy state is described by two time constants, T1 and T2, associated with two different electromagnetic processes (Table 20.3.1). The longitudinal relaxation time constant T1 is a measure of the time taken for protons to return to their normal ratio of parallel and antiparallel configurations relative to the field. In this process, the longitudinal magnetisation returns to its non-zero strength. The signal that is detected as a result of this process increases in strength from zero as the number of protons aligned parallel to the external field approaches its equilibrium state (Figure 20.3.4). The time constant T1 is defined as the time taken for the signal from the protons aligned in this state to reach a particular value (about 63% of the final equilibrium value). T1 is dependent on the interaction between the precessing protons and the other atoms in the material. It is sometimes called the spin–lattice interaction, because the process was first observed in crystals, which have a lattice structure.

medical physics

100%

Transverse magnetisation

The transverse relaxation time constant T2 is a measure of the time taken for the magnetic component perpendicular (transverse) to the external field to return to zero. This happens as the precessing protons become out of phase with each other (in the analogy, this is like the soldiers getting out of step). As a result of the loss of phase between the precessing protons, the transverse magnetisation, Bxy , decreases exponentially to zero (see Figure 20.3.5). The time constant T2 is dependent on the interaction between the precessing protons and is sometimes referred to as the spin–spin relaxation time.

Figure 20.3.5

Checkpoint 20.3 1 2 3 4

Outline the sequence of events during an MRI scan. Explain the two consequences of applying correctly tuned RF pulses to the hydrogen in the body. Outline the general concept of resonance. Define the relaxation times T1 and T2.

Bxy

T2

Bxy

Time

The relationship between the transverse relaxation time constant T2 and the decreasing transverse magnetisation. The arrows in the circles represent the components of the magnetic properties of the protons becoming progressively more random, causing the transverse magnetisation, and the RF signal associated with it, to decrease to zero.

Explain that the amplitude of the signal given out when precessing nuclei relax is related to the number of nuclei present.

20.4 It depends on how and where you look Each type of tissue in the body contains different amounts water and therefore different amounts of hydrogen. Hydrogen is also present in carbohydrates, proteins and fats, and the amounts of these also vary The amplitude of the radio in different tissues. signal emitted as the relaxation processes occur increases as the number of protons in the tissue increases. An MRI can be created which shows these differences in proton density and thereby highlights the difference between different tissues. Each tissue has unique values for T1, and T2. Magnetic resonance images can also be obtained to show specific tissues more clearly using the differences in the values of T1 and T2 that are characteristic of each tissue type. If the contrast in an image results mainly from differences in the T1 values, the images are called ‘T1 weighted’. Images dominated by T2 differences are called ‘T2 weighted’. Figure 20.4.1 illustrates the difference in contrast between T1 and T2 weighting.

Bxy

a

Figure 20.4.1

b

Contrast differences in (a) T1 weighted and (b) T2 weighted images centred on the knee of a 9-year-old boy. The arrows indicate the lesion responsible for his knee pain.

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Explain that large differences would occur in the relaxation time between tissue containing hydrogen-bound water molecules and tissues containing other molecules.

Table 20.4.1  Water content of various tissues Tissue

Water content (%)

Brain grey matter Brain white matter Heart Blood Bone

70.6 84.3 80 93 12.2

The difference in relaxation times T1 and T2 is significantly greater for tissues As a result, MRI is very containing relatively large amounts of water. sensitive to variations in the water content of tissues (see Table 20.4.1) and this is a significant factor in MRI’s ability to produce high-resolution, high-contrast images (see Figure 20.4.2). Because tumours are characterised by rapid cell division and high growth rates, they typically have a higher percentage of water than similar non-cancerous tissue, and can thus be clearly imaged using MRI. Haemoglobin molecules in red blood cells also provide a clearly identifiable resonance signal and so MRI can be used to compare the blood content of different tissues. This is often greater in cancerous tissue, again because of the high growth rates. MRI contrast agents are unique in radiology because it is not the chemical of the contrast agent that is detected, but rather the effect that the chemical has on the protons in nearby hydrogen atoms. Gadolinium (bonded to a carrier molecule) is a contrast agent that, when injected into the bloodstream, shortens the relaxation time T1 of protons near it, making the tissues appear brighter in a T1-weighted image. Because gadolinium normally stays in blood vessels, it has the effect of making these blood vessels and and areas in the body where blood leakage is occurring appear brighter. Another powerful tool used to enhance image contrast is to vary the time intervals between successive repetitions of a sequence of applied RF signals. This changes the weighting of the relaxation times and can improve the contrast in the image between different tissues.

Checkpoint 20.4 1 2 3 4

Figure 20.4.2

MRI scans of the brain show more contrast and detail than conventional X-ray or CAT scans because MRI is more sensitive to the differences in water content of grey matter and white matter.

PRACTICAL EXPERIENCES Activity 20.2

Activity Manual, Page 152

360

Account for the different MRI signal amplitudes observed from different tissues in the body. Explain how different parts of the body can be highlighted in an MRI image. Outline why tumours are so clear on scans. Explain why gadolinium is injected during an MRI scan.

20.5 The MRI scanner There are approximately 200 MRI scanners in Australia. All major hospitals and many private imaging facilities in Australia have one, although they cost more than $2 million to purchase and about $1 million per year to operate and staff. Part of this cost is for liquid helium to cool the superconducting electromagnets. These produce the strong magnetic fields—up to 5 T—required for MRI. Compare this with the field of about 50 × 10–6 T at the Earth’s surface. The patient is placed in the MRI scanner tunnel and the current in the superconducting coils is turned on to produce the strong magnetic field that will align the proton spins. These coils are shown in Figure 20.5.1, but not in Figure 20.5.2. When the patient lies in the MRI scanner, the strong magnetic field is

medical physics RF transceiver

Z

Z

Y coil

Z coil patient patient table

gradient coils

X coil

Y

superconducting magnet coils scanner

Y

RF transceiver X

Figure 20.5.1

Cutaway view of an MRI scanner

Figure 20.5.2

parallel to a line running in the head-to-toe direction, which is normally referred to as the longitudinal axis or the z direction. The pulse of radio waves from the RF transceiver coils is transmitted through the patient’s body. This flips the magnetic axes of some of the protons, resulting in a zero magnetic field from the protons in the z direction. It also causes the protons to precess in phase with each other. As some of the protons flip back and return to precession that is out of phase, the protons in each volume element (voxel) emit radio waves with an intensity proportional to the number of hydrogen atoms in the voxel. Gradient coils produce a small systematic variation in the net magnetic field throughout the patient’s body. The Z coils (Figure 20.5.2) produce variation in the z direction, which defines a slice through the body, represented by the blue region in Figure 20.5.3. The X and Y coils change the net magnetic field in their respective directions so that its strength is different in each voxel. As a result of these variations in the imposed field, the protons precess at slightly different frequencies. These frequencies allow the positions of the protons emitting RF signals to be determined. Hundreds of pulse cycles are used to obtain proton density data from each voxel in the plane of interest. The RF signals emitted by the protons provide data that is processed by a computer to produce an image for each slice through the body, and hence to create a 3D image that can be manipulated and viewed from any angle.

Checkpoint 20.5 Explain why the cost of running MRI scanners is so high.

X

patient

Gradient in an MRI scanner

Z

Y

X

Figure 20.5.3

Image slices and voxels in an MRI scan

PRACTICAL EXPERIENCES Activity 20.3

Activity Manual, Page 155

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20.6 Applications of MRI

Figure 20.6.1

One of the first highly successful applications of MRI was its use in diagnosing multiple sclerosis, an incurable disease that affects the nervous system. MRI clearly shows the scar-like plaque resulting from this disease, marking the destruction of the insulating myelin sheaths around the axons of nerve cells (Figure 20.6.1), something other imaging technologies are unable to do. Other examples of MRI imaging are shown in Figures 20.6.2 and 20.6.3. MRI has changed scientific thinking about the progress of diseases such as cancer and multiple sclerosis, because these diseases can now be detected earlier and different treatments can then be used. MRI also allows the progress of a disease and its treatment to be monitored more safely, increasing the chances of survival. False-colour MRI of the brain of a person with multiple sclerosis shows regions of myelin destruction (black–orange).

Advantages of MRI • MRI is non-invasive and has no known side effects, although a few patients experience anxiety in the confined, noisy space of the MRI scanner. • MRI provides excellent soft-tissue imaging, providing better contrast than CAT or conventional X-rays and much better resolution than ultrasound. • Image contrast can be altered by changing the T1 and T2 weightings. • There are few cases of adverse reaction to MRI contrast agents.

Disadvantages of MRI

Figure 20.6.2

MRI shows blood flow in the arteries of the brain as white, relative to stationary brain tissue.

• Patients with heart pacemakers or metal parts in their bodies cannot be imaged because of the strong magnetic fields. • Scans may take up to an hour to complete. • MRI scanners are very expensive and the running costs are high. • MRI scanners are very sensitive and must be screened from outside radio and magnetic interference. • The strong magnetic field produced can interfere with nearby electronic equipment and is dangerous in the presence of iron tools or furniture. Most current MRI images show structure within the body. An important area of technical development is the production of functional MRI images that reveal differences in chemical processes happening in the body. Other developments include the use of even stronger magnetic fields, allowing better resolution and contrast, and ultimately imaging using elements other than hydrogen. The possibility of future ‘high’ temperature superconductors, requiring only liquid nitrogen cooling, could impact significantly on MRI technology by removing the need for expensive, non-renewable liquid helium. Research and development has already produced portable MRI scanners for specialised applications. These machines do not require the use of superconducting magnets, greatly reducing the size and cost of the machine.

Checkpoint 20.6 1 Figure 20.6.3

362

MRI shows cancer (colour coded orange) that has spread from the lungs to the brain.

2

Describe what is seen in an MRI scan in a patient that has multiple sclerosis. Create a table to contrast the advantages and disadvantages of MRI and CT scans.

PRACTICAL EXPERIENCES

medical physics

CHAPTER 20

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 20.1: magnetic resonance images Look at the MRI scans and list their characteristics. Use images of healthy and damaged tissue and identify what characteristics allow you to distinguish between the two. Discussion questions 1 List characteristics of an MRI scan. 2 Outline how damaged tissue can be identified on an MRI scan. 3 Give reasons why MRIs are used to identify areas of high blood flow. 4 Construct a table that lists the parts of an MRI, identifying their functions.

Activity 20.2: A comparison of scanning techniques Research each of the diagnostic methods and compare the advantages and disadvantages of each technique. Prepare a summary of your findings. Discussion questions 1 Identify the most appropriate application for each of the methods listed. 2 List advantages for each of the diagnostic methods listed. 3 Identify limiting factors for each method.

Activity 20.3: Medicine and physics Research how medical applications of physics have affected society and write a report to assess these impacts. Discussion questions 1 Recall some medical applications of physics. 2 Identify the areas of society these applications have affected.

Perform an investigation to observe images from magnetic resonance image (MRI) scans, including a comparison of healthy and damaged tissue. Identify data sources, gather, process and present information using available evidence to explain why MRI scans can be used to: • detect cancerous tissues • identify areas of high blood flow • distinguish between grey and white matter in the brain. Gather and process secondary information to identify the function of the electromagnet, radio frequency oscillator, radio receiver and computer in the MRI equipment.

Identify data sources, gather and process information to compare the advantages and disadvantages of X-rays, CAT scans, PET scans and MRI scans.

Gather, analyse information and use available evidence to assess the impact of medical applications of physics on society.

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20 •

• • • •

•

•

•



Imaging with radio waves

Chapter summary

Particles have a property called spin, which can be visualised as a rotation (spinning) of the particles, although this is not reality. Protons, neutrons and electrons can only have two spin states: ‘spin up’ and ‘spin down’. If there is an odd number of nucleons in a nucleus, then the nucleus has a net spin. Spin produces a magnetic effect called a magnetic moment or magnetic dipole. Nuclear magnetic moments may align parallel or antiparallel to a magnetic field. The parallel configuration has a slightly lower energy and is therefore slightly preferred. Injecting energy equal to the energy difference between the spin states can cause the proton to flip from the parallel to the antiparallel configuration. This same energy will be released if the proton flips back. A magnetic moment angled to a magnetic field direction experiences a force and hence a twist (a torque) that causes precession of the direction of the magnetic moment. The frequency of precession of a magnetic moment is called the Larmor frequency given by: ⎛ 2μ p B ⎞ ⎟⎟ f Larmor = ⎜⎜ ⎝ h ⎠

•

• •

•

• • • •

Magnetic resonance imaging (MRI) involves sending a pulse of radio frequency (RF) radiation into a person’s body while located in a strong magnetic field. The frequency is tuned to match the Larmor frequency of protons within the field. The return of the proton to the less excited state after absorbing the RF energy is called relaxation. The longitudinal relaxation time constant (T1) is a measure of the time taken for protons to return to their normal ratio of parallel to antiparallel configurations relative to the field. The transverse relaxation time constant (T2) is a measure of the time taken for the magnetic component perpendicular to the external field to return to zero. MRI is very sensitive to variations in the water content of tissues. MRI contrast agents affect the protons in nearby hydrogen atoms. Gradient coils produce small variations in the net magnetic field throughout the patient’s body. Functional MRI images reveal differences in chemical processes happening in the body. 

where B = magnetic field and h = Planck’s constant.

Review questions Physically Speaking Complete the passage below by filling in the missing words from the list provided. antiparallel, field direction, energy, flip, higher energy, intensity, relaxation time constants, magnetic resonance imaging, quantum mechanical, parallel, protons, model, radio, randomly, reality, rotation, magnetic field, spin

MRI stands for _____________. This diagnostic tool uses a property of atoms called _____________ which can be visualised as a _____________ of the particles. However, spin is a _____________ property of the particles and the classical view of a spinning ball is a useful _____________, but is not _____________. Single _____________ (hydrogen nuclei)

generally all have their spins oriented _____________. When a _____________ is applied, they all line up _____________ or _____________ to the _____________. _____________ waves

transmitted through the patient’s body can cause protons to momentarily _____________ to the _____________ antiparallel state. As they return to their original orientation, they re-emit the _____________. The _____________ of the emission tells us how many protons there are in that area of the body. The _____________ tell us about the environment of the protons in

that area of the body.

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medical physics

Reviewing 1 Explain why fluorine-19 has a nuclear magnetic moment, but nitrogen 14 does not.

2 Identify the property of tissues that permits images like those in Figure 20.7.1 to be produced using an MRI machine.

3 Describe the effect on the energy of protons in an MRI scanner that allows an image to be produced.

4 Outline why cancerous tissues usually have a greater amount of blood flowing to them than normal tissue of the same type.

5 Describe the behaviour of a proton in a strong magnetic field.

6 Describe the process of precession in a mechanical system such as a child’s toy top.

7 Describe the process of precession involving protons in a strong magnetic field.

8 Summarise the conditions required for radio waves to cause protons precessing in a strong magnetic field to undergo a ‘spin flip’.

9 Justify the need for superconducting electromagnets in an MRI scanner.

10 Describe the function of the radio frequency transceiver coils in an MRI scanner.

MRI machine.

13 Describe what is meant by the term relaxation time. 14 Compare the advantages and disadvantages of MRI and CAT scans.

15 Briefly assess the impact of MRI scanners on medicine.

16 Assess the impact of superconductivity on the development of MRI.

17 Identify possible future directions of physics research that would contribute to improvements in MRI.

Solving problems 18 When a 1 T field is applied to a portion of the body containing 2 million protons, estimate how many extra protons favour the parallel configuration over the antiparallel configuration.

19 Calculate the magnetic moment of a proton, given that the Larmor frequency for a proton in a 1 T magnetic field is 42.6 MHz.

20 Calculate the Larmor frequency of an electron, given that the magnetic moment for an electron is –9284.764 × 10–27 J T–1.

iew

Q uesti o

n

s

v

MRI images show a normal brain (left) and the brain of a patient with a tumor.

Re

Figure 20.7.1

11 Explain the role of the computer in an MRI machine. 12 Describe the purpose of the gradient coils in an

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5

The review contains questions that address the key concepts developed in this module and will assist you to prepare for the HSC Physics examination. Please note that the questions on the HSC examination that address the option modules are different in structure and format from those for the core modules. Past exam papers can be found on the Board of Studies NSW website.

Multiple choice

4

(1 mark each) 1 Two differences between the normal hearing range of sound and ultrasound are: A B C D

Using the data in the table below, determine which answer is closest to the percentage of ultrasound intensity that is reflected from a boundary between kidney and fat.

Substance

Normal sound

Ultrasound

Not easily scattered by human tissue, 20 Hz to 20 kHz Shorter wavelength, audible 20 Hz to 20 kHz, longer wavelength Above 20 kHz, audible

Easily scattered by human tissue, above 20 kHz Longer wavelength, inaudible Above 20 kHz, shorter wavelength 20 Hz to 20 kHz, inaudible

2

An advantage of a CAT scan over an X-ray as a diagnostic tool would be: A seeing broken ribs more clearly. B the images contain three-dimensional information that allows for greater discrimination of structure. C the ability to carry out a biopsy. D lower cost.

3

Figure 20.8.1 shows two images of the same patient’s brain, using two different medical imaging techniques. The two techniques were: A Left: Ultrasound; Right: MRI B Left: Ultrasound; Right: CAT C Left: CAT; Right: MRI D Left: MRI; Right: CAT

Fat Kidney

A 0.29% B 1%

5

Ultrasound velocity (m s–1)

Impedance (rayl)

920

1450

1.33 × 106

1040

1561

1.62 × 106

C 12% D 18%

In medical imaging, the term relaxation time refers to: A how long it takes for an MRI machine to recharge. B the time for proton spin to return to equilibrium. C the minimum time between scans. D the maximum time before a radioisotope decays to untraceable amounts.

Figure 20.8.1

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Density (kg m–3)

medical physics

Short response

Extended response

6

The ultrasound image in Figure 20.8.2 shows an organ in a child that is not usually imaged in adults. Identify the organ and describe the features of the image that make it recognisable as an ultrasound image.  (4 marks)

10

7

Explain how an endoscope can be used to obtain tissue samples from the bladder of a patient.  (2 marks)

8

For each of the radioactive isotopes in the table below explain why it would or would not be a good choice for use in a bone scan.  (5 marks)

9

Radioactive source

Radiation emitted

Half-life

C-11

b+, g

20.30 minutes

Tc-99m

g

6.02 hours

I-131

b, g

8.04 days

Cs-137

a

30.17 years

U-238

a

4.47 × 109 years

Assess the impact of MRI on diagnosis of disease, in comparison to other medical imaging techniques.  (5 marks)

1

The nucleus of 1H has magnetic properties that enable it to be used for MRI. 4 a Explain why 2He is not used.  (2 marks) b Identify the primary factor that determines the 1 amplitude of the signal produced by 1H nuclei after an RF pulse at the resonant frequency has been applied to a part of a person’s body.  (2 marks )

Figure 20.8.2

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6 CONTEXT

Astrophysics

‘I render infinite thanks to God, for being so kind as to make me alone the first observer of marvels kept hidden in obscurity for all previous centuries.’ Galileo Galilei in Siderius Nuncius (Sidereal Messenger)

Figure 21.0.1 The 2-degree field ‘top end’ of the Anglo-Australian telescope (AAT) uses optical fibres to direct light from up to 400 astronomical targets to spectrographs.

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The Italian scientist Galileo Galilei (1564–1642) was not quite right in his claim to be the first to see the universe revealed by the newly invented telescope. That accolade may belong to Englishman Thomas Harriot (1560–1621). However, it was certainly Galileo’s short record of his observations in Siderius Nuncius (Sidereal Messenger), published in March 1610, that announced a revolution in astronomy to the educated world in Europe. Observations with telescopes revealed previously unknown objects, unimagined by thousands of years of naked-eye observers. Nonetheless, astronomy remained largely a ‘collector’s science’ for another two centuries. Observers used telescopes of increasing size to discover objects and note their positions and details, but knew little of their true nature. It was not until the early years of the 19th century that physicists discovered lines in the spectrum of sunlight, and later realised that they were the fingerprints of the elements making up the Sun and the stars. Although Newton’s law of universal gravitation explained motion in the sky, it was the understanding of spectral lines that really put physics into astronomy. Spectroscopy remains at the heart of modern astrophysics. In this module we will explore some aspects of both gravity and spectroscopy in astrophysics, leading to our current understanding of the life history of stars.

Figure 21.0.2 The stars of the Southern Cross

(right) and the ‘pointers’ α (alpha) and β (beta) Centauri (left)

INQUIRY ACTIVITY Learn about a star A lot of astronomical data gathered by astronomers is now available on-line in various catalogues. One important catalogue of stars was produced by the Hipparcos satellite in the early 1990s. It observed the positions and basic observable properties of more than 100 000 stars with high precision and more than 2.5 million stars to lower precision. In this exercise, you will use the Hipparcos catalogue to find out about one of the bright stars of the Southern Cross (the constellation Crux) or a neighbour in the constellation of Centaurus. 1 Find a map of Crux and Centaurus, perhaps in a star atlas or a computer program that shows the stars (e.g. Google Sky or WikiSky). You’ll need to orient the map to match Figure 21.0.2 above. 2 Identify the brightest stars of Crux and Centaurus seen in Figure 21.0.2. 3 Select one of these stars and look it up in the Hipparcos online catalogue, which can be accessed on the companion website at Pearson Places . To do this, enter the name (e.g. Alpha Crucis) into the ‘Target Name’ field and then click on ‘Submit Query’. The database should respond with a single line of data about your star.

4 Can you interpret the data? Use the cursor to select each column heading to understand what each value means. 5 Record the following important values (and the units) from the catalogue for each star: • Trigonometric parallax and the standard error in this value • V magnitude • Colour index B – V At the end of this Module, you should have a better understanding of many of these data values.

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21

Eyes on the sky Who needs a telescope?

telescope, magnification, refracting telescope, field of view, focal length, reflecting telescope, radio telescope, sensitivity, angular resolution, diffraction, Airy disc, active optics, seeing, scintillation, adaptive optics, interferometer, interferometry

Looking at the sky with the naked eye involves an optical system with a maximum diameter of about 7 mm—the size of the pupil of your dark-adapted eye. The brightness of the image and the ability to resolve fine detail are both set by that size. Using a telescope increases the diameter of the light-collecting ‘aperture’ to perhaps 60 mm in a small backyard telescope, or 10 m in the largest optical telescopes currently in use. The aperture is the most important property of any telescope operating at any wavelength because it governs the ability to ‘see’ faint objects and resolve fine detail.

21.1 The first telescopes The magnifying ability of lenses has been known since ancient times, but it seems that the combination of two lenses to form a practical optical telescope was first achieved by spectacle makers in the Netherlands in 1608. By mid-1609, Galileo Galilei (1564–1642), at the University of Padua, had built his first telescopes with magnification of 3×. He quickly progressed to instruments with increasing magnification and, for just a few months, his superior instruments gave him an unchallenged ability to observe the sky (Figure 21.1.1). This turning point in the history of science prompted the United Nations to declare 2009, the 400th anniversary of Galileo’s initial observations, as the International Year of Astronomy.

Figure 21.1.1 Two of Galileo’s original telescopes, mounted for display. The longer telescope had an aperture of 26 mm, a magnification of 14× and was about 1 m long. 370

astrophysics

Galileo’s influence was immense because, in March 1610, he quickly published a short record of his initial observations in Sidereus Nuncius (Sidereal Messenger). Among his major discoveries was the presence of mountains and valleys creating shadows on the Moon’s surface, which changed as the lighting of the surface changed. This rugged nature of the surface of the Moon contradicted Aristotle’s concept of heavenly perfection (Figure 21.1.2).

What is magnification?

T

he magnification of an optical telescope describes how much bigger an object appears in the telescope compared to the nakedeye view. Technically, it is the ratio of the angular size of the object with and without the telescope. A pair of 7 × 50 binoculars, for example, has an angular magnification of 7 times (7×) and 50 mm diameter main lenses. The view through the binoculars is as if the object was only 1/7 th as far away.

Figure 21.1.2  One of Galileo’s drawings of the Moon compared with a photograph

He also saw four ‘stars’ moving back and forth relative to Jupiter in a pattern he recognised as orbital motion around Jupiter. The observation of these ‘Galilean’ satellites almost led to the discovery of the planet Neptune, more than two centuries before its actually discovery. Galileo’s instruments were refracting telescopes that used a planoconvex lens (Figure 21.1.3) as the ‘objective lens’. The properties of the glass and the shape of this lens bend the light and bring it to a focus (Figure 21.1.4). The smaller ‘eyepiece lens’ then relays the light to your eye, which creates a new image on the retina at the back of the eye—and you ‘see’ the object! Galileo’s telescopes used a plano-concave lens (Figure 21.1.3) as the eyepiece. This results in an image that is upright; but the telescope then has a very narrow field of view—the area of sky you can ‘see’ at any moment—a little like looking down a narrow pipe at the object. In 1611, Johannes Kepler (1571–1630) pointed out that a plano-convex eyepiece lens would also work (Figure 21.1.4), although the image was inverted. Christoph Scheiner (1573–1650) popularised this form of ‘Keplerian’ refractor by pointing out that the field of view was larger than in the ‘Galilean’ telescope (as illustrated in Figure 21.1.5). With the field-of-view problem reduced in the Keplerian (or ‘astronomical’) refractor, a race began for higher magnification. This was most easily achieved by increasing the focal length of the objective lens—the distance between the lens and the image it forms (see section 21.3 for more on calculating magnification). By the 1670s, long telescopes achieved magnifications of more than 100×, but at the cost of enormous size and resulting difficulty in mounting and operation.

Discuss Galileo’s use of the telescope to identify features of the Moon.

plano-convex

plano-concave

Figure 21.1.3 Early telescopes used simple lenses that were flat on one side and either concave or convex on the other. Modern single lenses are often curved on both sides. 371

21

Eyes on the sky

a

Keplerian telescope objective lens

focus to eye

starlight eyepiece

focal length b

Galilean telescope objective lens to eye

starlight

eyepiece

focal length

Figure 21.1.4 (a) Keplerian and (b) Galilean telescopes using an objective lens

a

of the same focal length for easy comparison b

Figure 21.1.5 These images, shown through (a) Keplerian and (b) Galilean telescopes at the same magnification, have very different fields-of-view.

Figure 21.1.6 A replica of Newton’s original reflecting telescope

Another advantage of a long focal length objective lens was to minimise distortions inherent in the lens design. One of these was chromatic (colour) aberration caused by the fact that different colours of light come to a focus at slightly different distances from the lens. This causes a coloured halo around a white star image. It was Isaac Newton (1642–1727) who realised that this was caused by the properties of the glass itself. Combining different types of glass offered the possibility of partially cancelling the colour effect of one glass with the effect of the other. It was not until 1733 that such an ‘achromatic’ objective lens was first made. Today, any good-quality small telescope or set of binoculars has an achromatic objective lens and an eyepiece also composed of several lenses. Despite these developments, and the construction in the late 19th century of some large refractors with apertures up to 1 m, the dominance of the refracting telescope in astronomy is long past. Again it was Newton who made the breakthrough by constructing a reflecting telescope in 1668, using a mirror rather than an objective lens to collect and focus the light (Figure 21.1.6). It offered better images, without a coloured halo, in a telescope that was much smaller than any comparable refractor. Newton’s telescope used a polished metal primary mirror to focus the light, and a smaller, flat secondary mirror to divert the light out the side of the telescope tube to the eyepiece. In 1672, a variation of the Newtonian telescope was developed by Laurent Cassegrain (~1629–1693), who used a small hyperbolic secondary mirror to reflect light back through a hole in the primary mirror (Figure 21.1.7). By the mid-1700s reflecting telescopes with up to 15 cm

Newtonian telescope to eye flat secondary mirror

Cassegrain telescope hyperbolic secondary mirror

eyepiece

prime focus

parabolic primary mirror

starlight

eyepiece

prime focus

to eye

starlight

focal length

focal length

Figure 21.1.7 Newtonian and Cassegrain telescopes using a primary mirror of the same focal length for easy comparison. In large telescopes, if no secondary mirror is in place, the prime focus position can be used. 372

parabolic primary mirror

astrophysics apertures were available, and in 1789 William Herschel (1738–1822) built a reflector with a 124 cm aperture! This was the first ‘giant’ reflector—the forerunner of the 8 and 10 m aperture reflecting telescopes of today.

Checkpoint 21.1 1 2 3

Identify what sets the light-gathering ability and resolving power of an optical system. Outline key astronomical observations made by Galileo. Describe chromatic aberration and its cause.

21.2 Looking up For more than 200 years from the time of Galileo, all telescopes were optical telescopes designed to focus visible light onto the retina of the human eye. From about 1840, the use of photography revolutionised astronomy by allowing images of the sky to be permanently recorded. For the next century telescopes grew in size but remained fundamentally unchanged. Discuss why some wavebands In 1933, Karl Jansky (1905–1950), a physicist working for Bell Telephone can be more easily detected Laboratories in New Jersey, realised that part of the background ‘hiss’ heard in from space. Discuss the problems associated radio communication was coming from the sky. Janksy’s discovery was followed with ground-based astronomy up in 1937 by Grote Reber (1911–2002), who built a reflecting telescope in his in terms of resolution and backyard in Chicago (Figure 21.2.1). This was a radio telescope, with a mirror absorption of radiation and made from sheet metal and a diameter of 9.5 m—almost four times the size of atmospheric distortion. the largest optical reflecting telescope of the era. The basic optical principles of Reber’s radio telescope were exactly the same as those of an optical reflecting telescope, with the exception that the radio ‘light’ was detected by a radio receiver rather than by eye or photography. Radio astronomy really began to develop after World War II, spurred by technical developments during the war This was possible because a wide range of radio years. wavelengths, as well as visible light, penetrate the Earth’s atmosphere without significant absorption and can therefore be observed by telescopes on the ground. Electromagnetic (EM) radiation of most other wavelengths is absorbed by the atmosphere before it reaches the ground (Figure 21.2.2). At wavelengths longer than about 10 m, the Earth’s ionosphere (see in2 Physics @ Preliminary section 8.4) blocks radio energy from space. The radio ‘window’ opens at metre wavelengths and remains open down to millimetre wavelengths for telescopes on the ground. However, observing at these shortest radio wavelengths requires a high, dry location such as the site of the ALMA mm–sub-mm radio telescope, under construction in the Atacama Desert in Chile. Figure 21.2.1 Grote Reber’s radio telescope

373

Antarctic astronomy

I

t may surprise you to learn that the highest and driest sites on Earth are in Antarctica. The altitude in the centre of the Antarctic continent and the extreme cold of the atmosphere make it a tempting site for an infra-red observatory. Australian astronomers have studied the conditions for several years and a design study for a 2.4 m aperture optical/IR telescope called PILOT has been conducted. However, a large Antarctic telescope is still in the future.

Far infra-red (FIR) light at wavelengths of about 100 µm (0.1 mm) is absorbed by water vapour in the atmosphere. However, this blocking becomes patchy as wavelength decreases, with windows of less absorption at certain infra-red (IR) wavelengths between 1 and 10 µm. As a result, some infra-red astronomy can be done from mountaintop observatories or even aircraft. However, observing across the full IR range requires a telescope above the atmosphere—a role most recently and spectacularly filled by the 0.85 m aperture Spitzer Space Telescope. At wavelengths shorter than IR (i.e. below 1 micron), the atmospheric window opens again for the narrow optical band between about 350 and 750 nm. It then quickly closes in the ultraviolet (UV) and the shorter wavelength, higher energy X-ray and γ-ray wavebands. All of these are now observed from space.

Absorption by atmosphere

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Eyes on the sky

Galex

Fermi

Spitzer

Hubble

100%

50%

0%

Parkes Gemini 0.1 nm

1 nm

10 nm

100 nm

1 μm

10

μm

100 μm

1 mm

1 cm

10 cm

1m

10 m 100 m 1 km

Wavelength

Figure 21.2.2 The Earth’s atmosphere stops light from many parts of the electromagnetic spectrum reaching the ground.

Checkpoint 21.2 1 2

Discuss how and why radio astronomy started. Construct a table listing the various wavebands of the electromagnetic spectrum and describe their ability to penetrate the Earth’s atmosphere.

21.3 The telescopic view Most telescopes operate on the same basic principles, no matter what their operating wavelength. They use some type of mirror to collect the ‘light’ (EM radiation) and focus it onto an appropriate detector—often using the same prime focus or Cassegrain configurations illustrated in Figure 21.1.7. The combination of telescope and detector will set the field of view.

Magnification Perhaps surprisingly, the concept of magnification (sometimes called ‘power’) is not too important in most astronomical telescopes. It is easily calculated if the focal length of the objective lens or mirror, fo, and the focal length of the eyepiece, fe, are known. The magnification m is then given by: m= 374

fo fe

astrophysics Worked example Question A small refracting telescope has an objective lens diameter D of 70 mm and focal length fo of 700 mm. If this telescope has a typical ‘low power’ eyepiece focal length fe of 25 mm, what would be its magnification?

Solution m =

fo fe

=

700 = 28 (i.e. 28×) 25

Sensitivity The sensitivity of a telescope system describes its ability to ‘see’ faint objects. Sensitivity is sometimes called ‘light-gathering power’. It depends on how much light the telescope collects and how much of that light is delivered to the detector. The primary factor controlling light-gathering power is the diameter D of the telescope’s objective lens or mirror. Most telescopes have circular mirrors and so the collecting area is proportional to D2. A 10 m diameter telescope therefore collects 100 times the light of a 1 m diameter telescope. This is the primary reason for building bigger telescopes, although it is often not quite that simple, as illustrated by the example of the Anglo-Australian Telescope (AAT).

Define the terms ‘resolution’ and ‘sensitivity’ of telescopes.

Worked example Question What is the collecting area of the 3.9 m AAT, and how does it compare with that of a 70 mm diameter refracting telescope?

Solution The refracting telescope has D = 70 mm, so its collecting area can be calculated using: Collecting area = p

2 D

2

The AAT has D = 3.9 m; however, it also has an obstruction caused by the secondary mirror and its housing, as seen in Figure 21.3.1. In the f/8 Cassegrain configuration, this obstruction is about 20% of the collecting area. So the ratio of the collecting areas of the two telescopes is: 3900 2 p × 0.8 2 Collecting area AAT = 2 Collecting area 70 mm 70 p 2

 

 

≈ 2500 The AAT has 2500 times the light-gathering power of the 70 mm telescope.

Once the light has entered the telescope, it must be delivered to the detector and this may involve a series of lenses and mirrors. Each of these may reflect, absorb or scatter a little of the light and reduce the amount that reaches the detector. In the AAT there are just two mirrors between the sky and a detector placed at the Cassegrain focus, yet 10% of the light is lost.

Figure 21.3.1 The 3.9 m Anglo-Australian Telescope, showing the black 2-degree field (2dF) secondary mirror structure at the top of the telescope 375

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Eyes on the sky

Apart from the telescope itself, the overall sensitivity also depends on several other factors: • Atmospheric transmission—For telescopes on the Earth’s surface, the proportion of the light that penetrates the atmosphere depends on wavelength, as illustrated in Figure 21.2.2. • Detector efficiency—Often much less than 100% of the light striking the detector is actually recorded. • Observing time—Unlike the human eye, most detectors record more light if they are left exposed for longer times. • Background—The sky is not truly dark because of natural sources and man-made pollution at optical and radio wavelengths.

Angular resolution

a

Angular resolution is often simply called resolution. Adding ‘angular’ reinforces the idea that we are concerned with an ability to discern (resolve) detail separated In a practical sense, angular resolution by very small angles in the sky. describes whether we can see two closely spaced objects—or do they look like one? The limit to resolution is set by the wave-like properties of light. Whenever light encounters an obstacle, such as the aperture of a telescope, it doesn’t cast a sharp shadow. Instead diffraction caused by the edges of the obstacle alters the For a telescope with a circular objective path of the light near the edge. lens or mirror this creates a pattern of light known as an Airy disc (Figure 21.3.2) at the focus—not an image of a star. For a telescope of diameter D using light of wavelength λ, the angular size R of this pattern is given by the equation:

b

R (in arc seconds) = 1.22 RD

R2D

Figure 21.3.2 (a) A view of an Airy disc, and (b) the same view through an aperture with twice the diameter

 Dλ  × 206 265

(Recall the arc second as a unit of angle from in2 Physics @ Preliminary section 13.4.) The angular size of this pattern depends only on the wavelength of the light and the diameter of the telescope. For most telescopes, R is much bigger than the apparent size of even the largest star as seen from Earth. If two stars are close together, their Airy discs may overlap so that it is impossible to see that there are two stars. They are ‘unresolved’. The stars can theoretically be ‘resolved’ and seen as two stars (Figure 21.3.3) if they are separated by at least R (in arc seconds) = 1.22

 Dλ  × 206 265

Resolution also affects images of larger objects such as the Moon or planets, since each point of the object acts like a distant star and is blurred into an Airy disc (Figure 21.3.3). Figure 21.3.3 A magnified view of an Airy

a

b

pattern from an unresolved pair of stars (left), and a resolved pair (right)

Figure 21.3.4 Simulated views of Saturn, showing the effect of resolution using (a) a 10 cm diameter telescope and (b) a 30 cm diameter telescope 376

astrophysics Notice that the angular resolution of a telescope is expected to be ‘better’ (R is smaller and the image is sharper) for a big telescope. It is also better if the wavelength is small (e.g. much better for visible light than for radio waves). This is illustrated by the theoretical resolution values for 10 m diameter telescopes listed in Table 21.3.1. For comparison, the size of the Sun or Moon in the sky is around 1800 arc seconds, while a $1 Australian coin seen from a distance of 100 km would appear just 0.05 arc seconds across. Table 21.3.1  Theoretical resolution of 10 m diameter telescopes at different wavelengths Resolution R  for D = 10 m

Band

Typical wavelength

Ultraviolet (UV)

100 nm

0.0025

Optical

500 nm

0.013

Near infra-red (IR)

2000 nm (2 µm)

0.050

Radio (millimetre)

1 mm

25

Radio (centimetre)

21 cm

5300

(arc seconds)

PRACTICAL EXPERIENCES Activity 21.1

Activity Manual, Page 158

Identify data sources, plan, choose equipment or resources for, and perform an investigation to demonstrate why it is desirable for telescopes to have a large diameter objective lens or mirror in terms of both sensitivity and resolution.

TRY THIS!

Checkpoint 21.3

The man in the Moon

1

When you look at the Moon, do you see the face of the ‘man in the Moon’? We’re all limited by the angular resolution of our eyes. If the pupil of your eye is open to, say, 5 mm, the theoretical resolution is about 1/100 the size of the Moon in the sky, and not even quite that in practice. The result is that you can only see the broad details of the dark ‘mare’ and bright lunar highlands. Your imagination does the rest—and often in not the same way as someone else!

2 3 4

Calculate the magnification of a telescope that has an objective lens with a focal length of 500 mm and an eyepiece with a focal length of 12.5 mm. Define sensitivity of a telescope system. Define angular resolution. Describe an Airy disc.

21.4 Sharpening the image In practice, the theoretical resolution of a telescope may not actually be achieved if the imaging performance of the telescope is not good enough. How good is good enough? As a ‘rule of thumb’, the surface of a lens or mirror must be within one-eighth of a wavelength of its correct shape if it is going to produce sharp images. At optical wavelengths of about 500 nm, this means the surface must be smooth and the correct shape to within about 60 nm—about one-thousandth the thickness of human hair! This extraordinary precision is routinely achieved with optical polishing techniques. At radio wavelengths of 21 cm, one-eighth of a wavelength is 2.6 cm and this is easily achieved with metal plates and even wire mesh attached to a metal frame.

Active optics These tight tolerances may be impossible to achieve if the mirrors are so large that they bend under their own weight. Alternatively, the telescope itself may bend as it points in different directions. In the past, both effects could largely be overcome by building massive telescopes. The dish of the Parkes radio telescope (Figure 21.4.1) is 64 m across and weighs 300 tonnes, but also has a clever design to allow for flexure of the structure.

Figure 21.3.5 The face of the man in the Moon—in the southern hemisphere 377

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Outline methods by which the resolution and/or sensitivity of ground-based systems can be improved, including: • adaptive optics • active optics.

This approach has been abandoned in recent optical telescopes in favour of much lighter mirrors and structures that are controlled by an active optics system that keeps the telescope at optimum performance. For example, the mirrors of the 8.1 m diameter Gemini telescopes (Figure 21.4.2) are only 20 cm thick—much too thin to hold their shape. Instead, a system of 180 computer-controlled actuators push and pull on the back of the mirror to keep it in shape. The distortions are measured and corrections made about once a minute.

Figure 21.4.1 The 64 m Parkes radio telescope

Figure 21.4.2 One of the 8 m Gemini telescopes, in which Australia owns a small share

Seeing Discuss the problems associated with ground-based astronomy in terms of resolution and absorption of radiation and atmospheric distortion.

PRACTICAL EXPERIENCES Activity 21.2

Activity Manual, Page 161

378

For all wavelength ranges, coping with the slowly changing distortions in the telescope itself is an issue. In contrast, the much faster effects of ‘seeing’ are problems restricted to ground-based optical and infra-red telescopes. Seeing refers to the effects on images of tiny, rapidly changing temperature variations in the Earth’s atmosphere. These effects distort the path of the visible and, to a smaller extent, the infra-red light passing through the atmosphere so that even a perfect optical/IR telescope will not produce a sharp image. Images in a large telescope are distorted into a constantly moving and changing ‘speckle pattern’ that averages over time into a blurred ‘seeing disc’ (Figure 21.4.3). The same effects produce the familiar scintillation, or twinkling, of stars. Scintillation describes rapid changes in the brightness of a star, while seeing refers to blurring of a star image. The turbulence that produces seeing is usually concentrated near the ground and in discrete layers in the atmosphere. It depends on local weather conditions and geography. The best locations for good seeing are typically tall mountains that project above the relatively smooth stream of air flowing across open oceans. Most of the world’s largest optical or IR telescopes are therefore sited on islands,

astrophysics

a

b

c

1”

1”

1”

Figure 21.4.3 Simulated images of a binary star seen through a large telescope in ‘good’ seeing—(a) without seeing effects, (b) the ‘speckle’ pattern seen in a 10 ms exposure time and (c) the ‘seeing disc’ seen in longer exposure times. The arrow indicates 1 arc second.

or mountains on the western edges of continents. Siding Spring Observatory in New South Wales, site of the Anglo-Australian Telescope, is relatively poorly sited in comparison. The long-wavelength radio waves are unaffected by atmospheric turbulence and so a mountaintop site is unnecessary for most radio telescopes. Protection from artificial sources of radio waves is essential, and so many radio telescopes are in very remote sites, away from natural guide star population centres.

science target laser guide star created in sodium layer at 90 km

Adaptive optics Obviously, the ultimate way to beat the seeing effects is to place a telescope in space, above the atmospheric turbulence. The 2.4 m diameter Hubble Space Telescope (HST) has spectacularly demonstrated the power of observing from space since its launch in 1990. However, it is incredibly expensive and the largest optical/IR telescopes are on mountain tops, not in space. For at least 40 years, astronomers and engineers have been working towards ‘beating’ the seeing effects for ground-based telescopes by using adaptive optics. Adaptive optics is essentially active optics, working much faster—correcting hundreds of times a second to overcome the rapidly changing seeing effects. An adaptive optics system (Figure 21.4.4) works by using a wavefront sensor to measure the image distortion of a target star. A tilting mirror is used to stop the image moving and a deformable mirror is used to sharpen the image. The result is a much sharper image, although still not perfect (Figure 21.4.6). The target star may not be the real target of scientific interest, but it must be very close and bright enough for the adaptive optics system to work. A single system can measure and correct only a tiny patch of sky, since the blurring is different just a few arc seconds away. If there is not a bright enough adaptive optics target star close to the scientific target, then astronomers can provide one! A laser guide star is created by firing a high-power laser up from the telescope, causing the atmosphere to glow at about 90 km altitude. To the adaptive optics system it looks like a star that can be moved to where it is needed.

layers of turbulence in the atmosphere laser projection system telescope

sodium laser

distorted wavefronts producing a moving blurred image

control signals for deformable mirror deformable mirror

tilting mirror

control signals for tilting mirror

smoothed wavefronts producing a sharp image

real-time control computer

wavefront sensor

moving, blurred image of guide star

observing instrument

sharp image of science target

Figure 21.4.4 Layout of an adaptive optics system, including the science target, a natural guide star and a laser guide star 379

21

Eyes on the sky

Outline methods by which the resolution and/or sensitivity of ground-based systems can be improved, including: • adaptive optics • interferometry • active optics.

All large optical/IR telescopes now have some sort of adaptive optics system. The next generation of even larger telescopes will rely on multiple adaptive optics systems and laser guide stars.

Seeing is not believing!

T

he water in a swimming pool provides a very clear illustration of how light is affected by the medium through which it passes. In Figure 21.4.5 the surface of the water bends the light paths, concentrating the light in places and generally distorting the view of the bottom of the pool. A similar effect makes stars twinkle. Another example is the ‘mirage’ seen above water or a road on a hot day, due to the effects of temperature variations in the air.

Figure 21.4.6 The blurred image (left) of the star IW Tauri reveals a pair of stars separated by 0.3 arc second (right) when imaged using adaptive optics on the 5 m Hale Telescope.

Checkpoint 21.4 1 2 3 4 5

Identify the factors that limit the resolution of an optical telescope. Explain the physical problems that make active correction necessary. Describe what is meant by the term seeing. Outline the need for adaptive optics. Distinguish the differences between active and adaptive optics.

21.5 Interferometry

Figure 21.4.5 A distorted image of the bottom of a swimming pool, caused by ripples on the surface of the water

The goal of adaptive optics is to achieve the resolution limit set by the diameter of the telescope and the wavelength of the light. Given the long wavelengths of radio waves, it is impractical to build a single large radio telescope to achieve The the resolution possible with even a very small optical telescope. solution is to use two or more radio telescopes and link them together to form an interferometer. Provided the effective path to the detector via each telescope is the same to within a few wavelengths, interference will be seen. The interference pattern contains information about the image of the astronomical object. The ability of an interferometer to resolve fine details is governed by the equation: R (in arc seconds) = 1.22

 Dλ  × 206 265

where D is now the distance between any two telescopes, not their individual diameters. The resolution of the interferometer is about the same as having a large telescope of diameter D—but only in the direction along the line between the telescopes. At right angles to that direction the resolution is the same as using only one telescope. If the interferometer is formed from an array of several telescopes, then each pair provides good resolution in one direction. With enough telescopes, the array 380

astrophysics can produce good resolution in many different directions—enough to build up an image of the target. Also, as the Earth rotates, these directions move relative to the target and the array can ‘synthesise’ an even better image. An interferometer produces the resolution of a single large telescope, but with much less sensitivity, because only a small proportion of the light is collected by the array of telescopes (Figure 21.5.1). Interferometry is fundamental to the operation of almost all large radio telescopes. The main interferometric array of radio telescopes in Australia is the Australia Telescope Compact Array (ATCA), located near Narrabri, NSW (Figure 21.5.2). It has six 22 m dishes spread over 6 km in a pattern that can be varied. However, it is sometimes used with other dishes spread around Australia or the world to form a much larger array. Interferometry is also now being applied in optical/IR astronomy. The Sydney University Stellar single telescope mirror of the Interferometer (Figure 21.5.3), also same resolution located at Narrabri, is one of the pioneering instruments in this field. The much shorter optical wavelengths make the tolerances much tighter on the optical components and their positions, and seeing effects are once again a major problem. individual telescopes

Figure 21.5.2 Some of the dishes of the Australia Telescope in a compact configuration

2 3

G

enerally when you combine two beams of light from a source you just get more light. If you split the light and then recombine those two beams you may get something different. Provided the two beams travel the same distance, they ‘interfere’ with each other in a way that depends on the path of the beams and the characteristics of the light source. An optical device used to do this is called an interferometer (see section 3.2).

Figure 21.5.1 The resolution of a large telescope can be achieved using an array of small telescopes. Each white line joins an interferometric pair of telescopes.

Figure 21.5.3 The Sydney University Stellar Interferometer

Checkpoint 21.5 1

What is interferometry?

has an array of fixed stations along a 640 m baseline. Interference of the light occurs in the central building (upper centre of the image).

Outline how greater resolution is achieved in interferometry than with single telescopes. Calculate the maximum theoretical resolution of the Australia Telescope Compact Array that has six 22 m dishes spread over 6 km. Explain why the resolution of an optical interferometer differs from that of a radio interferometer. 381

21

Eyes on the sky

21.6 Future telescopes The forefront ground-based optical/IR telescopes in the world today are 8 m and 10 m in diameter. Planning is now underway for the next generation of Extremely Large Telescopes (ELTs) with apertures up to 42 m! Two or three of these may be operational after 2016. Smaller telescopes still have a role of course, but even 4 m telescopes are only truly competitive if, like the AAT, they have excellent instrumentation. Also by about 2016, radio astronomers hope to be commissioning their own new instrument—the Square Kilometre Array (SKA). This will actually be an interferometric array of radio telescopes spaced over 3000 km, with a collecting area of one square kilometre. By then, the replacement for the 2.4 m diameter Hubble Space Telescope will be operational. The James Webb Space Telescope (JWST) will be a 6.5 m diameter IR telescope. Other smaller space telescopes will continue to observe the sky at other wavelengths inaccessible from the ground. Observations at many wavelengths are essential to understanding most astronomical objects (Figure 21.6.1). a

b

c

d

Figure 21.6.1 (a) A composite image of the nearby active galaxy Centaurus-A, with (b) an X-ray image and (c) a radio image revealing much more than is seen in (d) the optical image they overlay.

Checkpoint 21.6 Contrast the maximum theoretical resolution of a 42 m diameter ELT, the JWST and the SKA. (You will need to assume some appropriate wavelength values.) 382

PRACTICAL EXPERIENCES

astrophysics

Chapter 21

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For more detailed instructions and advice, use the in2 Physics @ HSC Activity Manual.

Activity 21.1: Mountains on the moon Use an image of the Moon to make the same type of measurements that Galileo used to estimate the size of the mountains on the Moon. Discussion questions 1 State what assumptions are made in estimating the size of the mountains. 2 Explain how advances in technology in astrophysics have allowed for more accurate methods to determine the height of lunar mountains.

Figure 21.7.1 A portion of the Moon’s surface around the crater Plato

Activity 21.2: A better telescope Measure the amount of light collected by lenses of differing diameter. Also determine the size of the lens that can best separate (resolve) two close objects. Discussion questions 1 Determine the relationship between diameter of a lens and its lightgathering ability. 2 Explain how the diameter of the lens is related to resolution and explain how this relationship was determined.

Identify data sources, plan, choose equipment or resources for, and perform an investigation to demonstrate why it is desirable for telescopes to have a large diameter objective lens or mirror in terms of both sensitivity and resolution.

383

21 •

• • • •

•

•

• •

Eyes on the sky

Chapter summary

The magnification of an optical telescope describes how much bigger an object appears in the telescope compared with the naked-eye view. Refracting telescopes use an objective lens to collect and focus light. Reflecting telescopes use a mirror to collect and focus light. ‘Field of view’ describes the area of sky an optical instrument can ‘see’ at any moment. The focal length of a lens or mirror is the distance between the lens or mirror and the image it forms of a distant object. The basic optical principles of most telescopes at all wavelengths are the same as for an optical reflecting telescope. Visible light and a wide range of radio wavelengths penetrate the atmosphere without significant absorption and can therefore be observed by telescopes on the ground. Most other wavelengths are absorbed by the atmosphere. Sensitivity (or ‘light-gathering power’) of a telescope system describes its ability to ‘see’ faint objects. Diffraction in a telescope with a circular objective lens or mirror creates a pattern of light known as an Airy disc at the focus—not an image of a star.

•

•

•

•

•

•

•

Angular resolution describes whether we can discern two closely spaced objects as separate (resolved) or not (unresolved). The surface of a lens or mirror must be within about one-eighth of a wavelength of its correct shape if it is going to produce sharp images. Active optics systems control telescope mirrors and structures to keep the telescope at optimum performance. ‘Seeing’ describes the blurring of images caused by tiny, rapidly changing temperature variations in the Earth’s atmosphere. Scintillation, or twinkling, describes rapid changes in the brightness of a star caused by tiny, rapidly changing temperature variations in the Earth’s atmosphere. Adaptive optics measures the image distortion of a target star by using a wavefront sensor, and then corrects it hundreds of times a second, to overcome the rapidly changing seeing effects. Two or more telescopes linked appropriately form an interferometer, provided the effective path to the detector via each telescope is the same to within a few wavelengths. The interference pattern contains information about the image of the astronomical object.

Review questions

Physically speaking

Copy and complete the following table with the knowledge that you have gained after studying this chapter.

Concept Telescope

Angular resolution

Sensitivity

Active optics

Adaptive optics

Interferometry

384

Definition of concept

Diagram to illustrate the concept

astrophysics

Reviewing 1 Recall how the introduction of technology has influenced the discoveries made in astronomy, with particular reference to Galileo and the telescope.

2 Describe some of the observations Galileo made with his telescope. 3 Name some telescopes that are used to view the universe using wavebands other than visible light, identifying the wavebands they use as well as where they are located.

4 Define the following terms and describe how they relate to a telescope: • collecting area • sensitivity • magnification

5 A small value for angular resolution is considered good. Explain what it means and how it can be achieved in a telescope.

6 Discuss the implications of ‘seeing’ in optical astronomy. 7 ‘Twinkle twinkle little star’ is a line from a popular nursery rhyme. Explain what physical phenomenon is being described in this phrase.

8 Outline how active optics works. 9 Compare and contrast active and adaptive optics. 10 Adaptive optics systems are complex because what they are attempting is difficult to achieve. Outline some of the factors that make adaptive optics difficult.

11 Justify the use of interferometry to increase resolution. 12 Assess the importance of telescopes observing at different wavelengths in gathering information about the universe.

Solving problems 13 Compare the sensitivity of a telescope with a diameter of 10 cm with one of 20 cm.

14 The nearby spiral galaxy M31 in the constellation of Andromeda is approximately 2.5 million light-years from the Sun. What is the size (in light-years) of the smallest features in M31 that can theoretically be resolved by a 10 m telescope at the visible light wavelength of 550 nm?

15 What is the size (in light-years) of the smallest features in M31 that can

Re

iew

Q uesti o

n

s

v

theoretically be resolved by the 6 km long baseline of the Australia Telescope Compact Array operating at the radio wavelength of 21 cm?

385

21

Eyes on the sky

PHYSICS FOCUS Beautiful death-star could threaten Earth The Earth may be in the firing line when one of the sky’s most beautiful objects explodes, according to University of Sydney astronomer Peter Tuthill. Dr Tuthill discovered the elegant rotating pinwheel system, named WR104, in the constellation Sagittarius. It includes a highly unstable star known as a Wolf-Rayet, widely regarded by astronomers as a ticking bomb—the last stop in a star’s life before a cataclysmic supernova explosion.

5. Current issues, research and developments in physics

‘When it finally explodes as a supernova, it could emit an intense beam of gamma rays coming our way’, says Dr Tuthill. New images of WR104 taken with the Keck Telescope in Hawaii by Dr Tuthill in 2008 show a glowing plume of hot dust and gas flung out into a whirling spiral as the two stars at the centre of the system orbit one another every 8 months.

Figure 21.7.2 Spiral pattern around the massive star WR104, as seen using masked aperture interferometry

386

astrophysics

But something odd about the images caught the attention of Dr Tuthill and his team: ‘Viewed from Earth, the rotating tail appears to be laid out on the sky in an almost perfect spiral. It could only appear like that if we are looking nearly exactly down on the axis of the binary system.’ Dr Tuthill and his team worry this box-seat view might put us in the firing line when the system finally explodes. ‘Sometimes, supernovae like the one that will one day destroy WR104, focus their energy into a narrow beam of very destructive gamma-ray radiation along the axis of the system. If such a ‘gamma-ray burst’ happens, we really do not want Earth to be in the way,’ warns Dr Tuthill. At only 8000 light-years distance, WR104 is just down the road in galactic terms, only one-third of the way to the centre of our Milky Way Galaxy. ‘Earlier research has suggested that a gamma-ray burst— if we are unfortunate enough to be caught in the beam—could be harmful to life on Earth out to these distances. Scientists have speculated that, eons ago, a gamma-ray burst from a distant star could explain mass extinctions seen in the fossil record,’ he said. ‘I used to appreciate this spiral just for its beautiful form, but now I can’t help a twinge of feeling that it is uncannily like looking down a rifle barrel.’ But Dr Tuthill is not panicking just yet. ‘There are still plenty of uncertainties: the beam could pass harmlessly to the side if we are not exactly on the axis, and nobody is even sure if stars like WR104 are capable of producing a fully fledged gamma-ray burst in the first place.’ ‘We probably have hundreds of thousands of years before it blows, so we have plenty of time to come up with some answers.’

1 What do the colours in the image (Figure 21.7.2) represent? 2 Do you believe that every detail you see in this image is real? Why or why not? 3 This image is ~0.5 arc seconds across. How does the resolution in this image compare with one you might expect from one of the Keck telescopes?

Extension 4 Peter Tuthill used a technique called ‘masked aperture interferometry’ to obtain this image. It uses a single telescope instead of two or more. How does interferometry work on a single telescope? 5 How does the resolution of this technique compare with that possible using an interferometer such as the Sydney University Stellar Interferometer or the Australia Telescope Compact Array? 6 How much of a risk to Earth do you think is posed by the star WR104?

387

22 astrometry, parallax, astronomical unit, light-year, parsec, spectroscopy, spectrum, Fraunhofer lines, continuous spectrum, emission line spectrum, absorption line spectrum, black body, black body curve, Planck curve, spectroscope, prism, spectrograph, diffraction grating, spectral classes, brightness, luminosity, Doppler effect, photometry, magnitudes, apparent magnitude, absolute magnitude, distance modulus, colour index, spectroscopic parallax

Define the terms parallax, parsec, light-year. Explain how trigonometric parallax can be used to determine the distance to stars. more distant stars photo taken now Earth now r

Sun

Earth 6 months from now

Figure 22.1.1

388

p

p d

relatively nearby star

photo taken 6 months from now

The parallax motion of a nearby star, showing the parallax angle p

Measuring the stars Fingerprinting the stars The invention of the telescope led to a revolution in astronomy. The Moon and planets were found to be unique worlds, faint stars were revealed and new mysteries such as ‘spiral nebulae’ were discovered. However, astronomers still couldn’t answer the most basic questions: How far away are the stars? What are they? The answer to the first question came in 1838 with the first successful measurement of the parallax of a star. The clue to the second question was already in place by then with Fraunhofer’s observation of spectral lines in light from the Sun. The understanding of these lines as fingerprints of the stars is the basis of modern astrophysics.

22.1 How far? Prior to the invention of the telescope, astronomy was the science of the positions and motions of the stars and planets. We now call this astrometry. The telescope made these measurements much more precise, but one of the prizes of astrometry remained elusive—a parallax measurement of the distance of a star. Stellar parallax (see in2 Physics @ Preliminary Figure 13.3.2) is the apparent change in position of a nearby star relative to more distant stars. Over the period of a year, any nearby star will trace out a tiny ellipse in the sky, mimicking the Earth’s orbital motion (Figure 22.1.1). The closer the star, the larger the ellipse will be. The parallax angle p measures the angular length of half the major axis of the ellipse. Parallax, as a surveying concept, was well known to ancient astronomers. It offers an unambiguous distance measurement based on simple geometry. Unfortunately, even the largest stellar parallax motion is less than 1 arc second— about the size a star appears because of seeing effects (see section 21.4). The first successful measurement was made by Friedrich Bessel (1784–1846) in 1838, using visual observation of the star 61 Cygni. Later in the 19th century photography replaced visual observations.

astrophysics The distance d of the star and the parallax angle p are simply related by: r tan p = d Using the small angle approximation (sin p ≈ tan p ≈ p) since p is so small, and measuring p in arc seconds, r in astronomical units (AU) and d in parsecs (pc) results in an even simpler form for this equation: p=

1 d

Worked example Question Proxima Centauri is the nearest star to Earth. It has a parallax p = 0.76887 arc seconds. What is its distance in parsecs, light-years and astronomical units?

Solution 1 1 d= = = 1.3006 pc = 4.2421 ly = 26 827 AU = 4.0133 × 1016 m p 0.76887

Ground-based measurements of parallax are hindered by seeing effects, yet the uncertainty in the measured angles is, remarkably, only about 0.01 arc seconds. Realistically, the limit of ground-based parallax measurements with an acceptable uncertainty (of, say, 30%) is about 0.03 arc seconds, corresponding to a distance of 30 pc. Between 1989 and 1993 Hipparcos (high precision parallax collecting satellite) measured the parallax of 118 000 stars to a precision of 0.001 arc seconds and more than 1 million stars to lower precision. This represents a factor of 10 increase in the distance range of parallax measurements to, say, 300 pc. However, compared with the distance of about 8000 pc to the centre of the Milky Way galaxy, the stars with accurate Hipparcos distances are still very local! In about 2012 the astrometric boundaries should begin to expand even further, with the launch of the Gaia spacecraft. It will survey 1 billion stars with a typical precision of 0.00002 arc seconds, reaching out to the centre of the galaxy.

Checkpoint 22.1 1 2

Explain exactly what the parallax angle p measures. Outline the reasons for using satellites to measure parallax angles.

Units of distance

P

arallax measurement is based on the size of the Earth’s orbit, so a useful unit of distance is the astronomical unit (AU). This is approximately the average distance between the Earth and the Sun.   1 AU ≈ 1.4960 × 1011 m The most familiar distance unit in astronomy is the light-year (ly). This is the distance that light covers in 1 year, travelling at 2.998 × 108 m s–1. 1 ly ≈ 63 241 AU ≈ 9.4605 × 1015 m.   The parsec (pc) is an alternative unit that represents the distance of a hypothetical star that has a parallax angle of 1 arc second. 1 pc ≈ 206 265 AU ≈ 3.2616 ly ≈ 3.0857 × 1016 m

Discuss the limitations of trigonometric parallax measurements.

PRACTICAL EXPERIENCES Activity 22.1

Activity Manual, Page 167

PRACTICAL EXPERIENCES Activity 22.2

Activity Manual, Page 171

22.2 Light is the key In his book The Positive Philosophy, the 19th century French philosopher August Compte (1798–1857) wrote about our knowledge of the stars and planets. ... We see how we may determine their forms, their distances, their bulk, and their motions, but we can never know anything of their chemical or mineralogical structure ... 389

22

Measuring the stars

The distances in astronomy are an enormous barrier to learning about the stars and planets, but Compte was wrong in thinking we would never know about their composition. Astronomical spectroscopy is the study of the light from objects in the universe to reveal their composition and physical characteristics. The first step had been made long before when Isaac Newton (1643–1727) realised that ‘white’ light is composed of component colours, revealed when the light is passed through a glass prism. In 1802, William Wollaston (1766–1828) looked more closely and noticed dark lines in the rainbow of colours that form the spectrum of the Sun. In 1814 Joseph von Fraunhofer (1787–1826) used a spectroscope to rediscover the lines and eventually catalogued over 570 of these Fraunhofer lines (Figure 22.2.1). By 1860, Gustav Kirchhoff (1824–1887) and Robert Bunsen (1811–1899) and others discovered that chemical elements produced bright spectral lines when heated and They some of these lines matched the dark Fraunhofer lines. deduced that the Fraunhofer lines were produced by absorption of light Figure 22.2.1 Fraunhofer’s spectral lines superimposed by these elements in the Sun, providing the key to understanding the on the rainbow spectrum of the Sun composition of the stars. Kirchhoff ’s empirical laws of spectrum analysis describe how to produce three of the key types of spectra we observe in astronomy: Account for the production of • A hot, dense gas produces a continuous spectrum. emission and absorption • A hot, low-density gas produces an emission line spectrum. spectra and compare these • A continuous spectrum source viewed through a cool, low-density gas produces with a continuous black body spectrum. an absorption line spectrum. Describe the technology needed to measure astronomical spectra.

PRACTICAL EXPERIENCES Activity 22.3

Activity Manual, Page 174

400

500

600

Wavelength (nm)

Figure 22.2.2

A continuous spectrum

Continuous spectra When you heat an object such as a steel bar it begins to glow red and orange and then perhaps white before melting. A very low resolution spectrograph reveals that at any of these temperatures the bar is emitting a continuous range of wavelengths—a continuous spectrum (Figure 22.2.2). Any relatively dense material—a solid, liquid or high density gas—will behave in the same way. They approximate a black body (see section 9.2), which is the idealised example of a hot object. A black body emits light at all wavelengths, with a distribution described by a 700 black body curve (or a Planck curve) that depends only on temperature (Figure 22.2.6), not composition. In astronomy, the interior of a star is very close to a black body source. The wavelength of peak emission of a black body curve λmax is given by Wien’s law (see in2 Physics @ Preliminary section 14.2): 2.9 × 10−3 λmax = T where T is the temperature on the kelvin scale. The change in λ max with temperature alters the balance of the visible colours and leads to a change in the overall colour of the object. Cool objects (T  4000 K) look red but emit most of their light in the infra-red. Peak emission from a hot object (T  10 000 K) is in the ultraviolet (UV) and the object looks blue-white to the eye. The total energy emitted per second per unit area of the surface is represented by the area under the black body curve. This power increases very rapidly with temperature according to the Stefan–Boltzmann law:

(

)

P = 5.67 × 10−8 T 4 390

astrophysics

PHYSICS FEATURE

light from telescope

Spectroscopes and spectrographs

slit

A

grating focusing mirror

camera

Figure 22.2.3

A basic spectrograph

a

b Intensity

spectroscope is a device that disperses light into its component colours to be viewed by eye. Early spectroscopes used a prism to refract the various colours to different angles, according to their wavelengths. Replacing the eye with a camera creates a spectrograph, the key instrument in astronomical spectroscopy. The spectral resolution of a spectrograph defines how much detail can be seen in the spectrum, in the same way that angular resolution is used for images. Optical spectral resolution ranges from low (~0.5 nm) to ultra-high (~0.001 nm). Anything beyond the lowest resolutions requires greater dispersion of the light than is possible with a prism. This can be achieved using a diffraction grating, often made by ruling very fine lines onto an optical surface. Interference between light beams diffracted by the lines results in a spectrum of the light striking the surface. Most spectrographs use a slit to restrict the light beam. The light is dispersed in the direction perpendicular to the slit (Figure 22.2.3). This direction measures wavelength. Depending on the optical set-up, distance along the slit may measure distance across the target object.

collimating mirror

measure intensity along a slice through the spectrum Hγ

Hβ Hα 500

600

700

Wavelength (nm)

Figure 22.2.4

A spectrum as (a) coloured light and (b) an intensity profile

The resulting spectrum of a single point such as a star can be shown as a coloured band (Figure 22.2.4a), but it is usually presented as an intensity profile versus wavelength (Figure 22.2.4b). Spectrographs are the key instruments on all optical telescopes. At the Anglo-Australian Telescope, AAOmega (Figure 22.2.5) is a modern spectrograph, fed by optical fibres bringing light from multiple targets in the image.

5. Current issues, research and developments in physics

Figure 22.2.1 Fraunhofer’s spectral lines superimposed rainbow Rob spectrum the Sun the grating feeding Figure 22.2.5 on the Astronomer Sharpofexamines a camera (the assembly in the right half of the image) in the AAOmega spectrograph.

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Measuring the stars

Worked examples PRACTICAL EXPERIENCES Activity 22.4

Activity Manual, Page 178

Question In Figure 21.0.2, the two bright stars on the left are α and β Centauri—the pointers to the Southern Cross. a Centauri is actually a double star, and the brighter star has a surface temperature of ~5800 K. b Centauri is a triple system, with the brightest star having a surface temperature of ~23 000 K. Which star is which in the image?

Solution

Intensity

Using Wien’s law for the two stars: α Centauri 2.9 × 10 −3 λ max = 5800 ≈ 500 nm This is in the middle of the visible band, so we expect a yellow overall colour.

12 000 K 7500 K 6000 K 4500 K 3000 K

0

500

1000

1500

2000

Wavelength (nm)

Figure 22.2.6

Black body curves for different temperatures show that the peak moves to shorter wavelengths with increasing temperature and the curves move up, indicating greater energy output at all wavelengths.

β Centauri 2.9 × 10 −3 λ max = 23 000 ≈ 130 nm This is in the ultraviolet, so we expect a blue-white overall colour. Clearly the left star in the image is α Centauri and the right star is β Centauri.

Question What is the relative power output of the two stars?

Solution Using the Stefan–Boltzmann law for the two stars: α Centauri P = (5.67 × 10−8 Wm−2 K −4 )(5800)4 = 6.4 × 107 Wm−2 β Centauri P = (5.67 × 10−8 Wm−2 K −4 )(23 000)4 = 1.6 × 1010 Wm−2 So, each square metre of the surface of the hotter star β Centauri emits nearly 250 times more energy per second than the surface of α Centauri. The hotter star is also about 8 times larger and so has 64 times the surface area of the cooler star. Thus the luminosity of the hotter star is about 16 000 times the intensity of the cooler one.

Emission line spectra 447.1

471.3 492.1 501.5

587.5

667.8

Wavelength (nm)

Figure 22.2.7

392

An emission spectrum

When you inject energy into a low-density gas, the spectrum is very different. It shows bright emission lines at wavelengths characteristic of the elements present in the gas (Figure 22.2.7). Familiar examples include the spectra from gas discharge lamps and neon signs. Low-density gas clouds in interstellar space also glow with an emission line spectrum.

astrophysics

The origin of emission lines lies in the structure of the atom (see Chapter 12 ‘From Rutherford to Bohr’). When an atom absorbs energy, an electron will make a transition to a higher energy state, provided the absorbed energy exactly matches the energy of the transition. It will often then quickly make one or more downward transitions back to lower energy states, emitting energy. This process is described as the absorption or emission of photons. In making a downward transition from energy E2 to energy E1, an atom emits a photon with energy E related to its wavelength λ (or frequency f ) by the relation: E = E 2 − E1 =

hλ = hf c

where Planck’s constant h = 6.626 × 10–34 J s and c = 3.00 × 108 m s–1 is the speed of light. These photon energies (and hence wavelengths or frequencies) correspond to the observed emission lines. The pattern of emission lines is a fingerprint for each element because it is matched to the energy levels within each atom of that element.

Absorption line spectra The spectrum of the Sun and stars is a continuous spectrum, but crossed by dark The origin of these absorption lines lines, not bright ones (Figure 22.2.8). lies in the relatively cool gas overlying the hotter, denser gas deep in the star. The denser gas produces a continuous spectrum. Atoms in the overlying gas absorb light from the continuous spectrum, but only at the wavelengths matching differences in their energy levels. They then re-emit the light, but in all directions, not just outwards. The net effect is less light in the outward direction at those wavelengths, creating dark lines in the spectrum. The spectral fingerprint of a gas is the same, whether seen in absorption or emission. In the Sun and stars, it is the outer layer of the star whose composition is imprinted on the spectrum as absorption lines (Figure 22.2.9). The spectra of other astronomical objects depends on their physical conditions of temperature, density and composition. Table 22.2.1 gives some examples.

black body source (e.g. deeper layers of star)

continuous spectrum

Where does black body light come from?

I

f light is emitted by atoms at very precise wavelengths, how is a continuous spectrum produced?   The answer lies in the density of the material emitting the light. If the density is low, the atoms are relatively far apart and they emit light at the discrete wavelengths expected. However, if the density is high, the atoms are closer together, and they influence one another and alter the energy levels of the atoms. The effect is to blur the lines into a continuous distribution.

Wavelength (nm)

Figure 22.2.8

An absorption spectrum

Describe how spectra can provide information on surface temperature, rotational and translational velocity, density and chemical composition of stars. Identify the general types of spectra produced by stars, emission nebulae, galaxies and quasars.

absorption spectrum

diffuse gas (e.g. outer layers of a star or a nebula) emission spectrum

Figure 22.2.9

All three types of spectra are produced by the Sun: continuous spectrum from deeper layers, absorption lines from the photosphere and emission lines from the chromosphere and corona. 393

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Measuring the stars

Table 22.2.1  Spectral characteristics of some astronomical objects Description

Spectrum

Emission nebulae

Regions of gas (mostly hydrogen and helium) that glow due to intense UV light from embedded young hot stars

Dominated by strong emission lines characteristic of the gas composition

Example 1 0.8 Relative flux

Object

0.6 0.4 0.2

450

500

550

600

650

λ (nm)

Collections of billions of stars, gas and dust; light output is generally dominated by the mix of stars of various types

Absorption spectra, often a mix of bright blue-white types with many more numerous but fainter yellow stars

0.3

Relative flux

Normal galaxies

0.2

0.1

0

400

500

600

700

800

λ (nm)

A type of galaxy with an active nucleus dominating the total energy output

A continuous spectrum with emission lines (that may be variable) suggesting fast-moving gas clouds

0.8

0.6 Relative flux

Quasars

Hγ

Hα

Hβ + [OIII]

0.4

0.2 400

500

600

700

800

λ (nm)

Checkpoint 22.2 1 2 3

Describe the Fraunhofer lines. Describe how emission lines are produced. Describe the relationship between the absorption lines and the emission lines of an element.

22.3 The stellar alphabet In 1885 astronomers at Harvard College Observatory began to compile a photographic catalogue of spectra that became the Henry Draper Catalogue of Stellar Spectra. When the final extension was published in 1949, almost 360 000 stars had been classified. 394

astrophysics

Spectral classes The Harvard classification scheme developed during the project remains in use today. It started as groups of spectra with similar spectral lines, given letters from A to N, with O, P and Q added for some unusual stars. Re-ordering and simplification of the sequence led to the current, seemingly random sequence The major spectral classes are: of letters.

Brightness and Luminosity

R

ecall the definitions of brightness and luminosity from in2 Physics @ Preliminary section 15.1. The brightness of a star is a measure of the energy received in a certain time per unit of collecting area (or power per unit area) W m–2. Its luminosity is the total power output of the star in watts. A star will look brighter when seen through a telescope. It obviously isn’t really more luminous; the telescope has a larger collecting area than your eye alone, making the star appear brighter.

OBAFGKM (We can remember this odd sequence using various mnemonics, the best known being ‘Oh Be A Fine Girl [Guy] Kiss Me’). With additions of other classes, the sequence can be represented as: W–OBAFGKM–LT The black body curves underlying the shape of the spectra clearly imply a sequence in surface temperature from hot O stars to cool M stars. It was not until the 1920s that the understanding of electron energy levels in atoms progressed enough to understand how surface temperature also controlled the appearance of the spectral lines. Composition differences between stars are relatively small and are not the reason why spectra vary so much. A sample of spectra from various spectral classes is shown in Figure 22.3.1. Some of the key features of each spectral class are described in Table 22.3.1, emphasising that the colour of a star, its spectral class and surface temperature are all closely related. Each spectral class has been divided into 10 sub-classes. The Sun, for example, is a yellow G2 star with a surface temperature of ~5770 K, placing it at the hot end of spectral class G.

Luminosity classes When plotting a Hertzsprung–Russell (HR) diagram (see in2 Physics @ Preliminary section 15.3), we can use the spectral class on the horizontal axis, since it is closely related to the surface temperature (Figure 22.3.2).

Describe the key features of stellar spectra and describe how these are used to classify stars.

Table 22.3.1  Characteristics of spectral classes of stars Spectral class W

Effective temperature (K) >50 000

Colour Blue

Strength of hydrogen lines Weak

Other spectral features He, C, N emission lines +

% of main sequence stars Extremely rare

O

31 000–50 000

Blue

Weak

Ionised He lines, strong uv continuum

0.00003

B

10 000–31 000

Blue-white

Medium

Neutral He lines

0.1

A

7500–10 000

White

Strong

Ionised metal lines

0.6

F

6000–7500

White-yellow

Medium

Weak ionised Ca+

3

G

5300–6000

Yellow

Weak

Ionised Ca+, metal lines

8

+

K

3800–5300

Orange

Very weak

Ca , Fe, strong molecules, CH, CN

12

M

2100–3800

Red

Very weak

Molecular bands, e.g. TiO, neutral metals

76

L

1200–2100

Red

Negligible

Neutral metals, metal hydrides

Brown dwarf numbers uncertain

T

m2), then the balance point is closer to star 1. 407

23

Stellar companions and variables orbit of star 2

star1 m1

r1

focus 2

focus 2

centre of mass

r2

orbit of star 1

star2 m2

Figure 23.1.1

Try this!

Worked example

Make your own binary system

Question

Some of the principles of a binary star system are easily demonstrated with some balls and a rod. Cut a hole through each ball so that it can be pushed onto the thin wooden or metal rod but not slide off easily. Rest the rod on your finger and slide it along until it is balanced horizontally. Tie a string to the balance point (the centre of mass) and suspend your ‘binary system’ by the string. Gently rotate the binary by giving one of the balls a push. Try a different ball and see how the centre of mass changes.

m1

m2

r1

Figure 23.1.2

r2

The centre of mass of the binary system is the balance point.

Explain the importance of binary stars in determining stellar masses.

A planet orbiting the Sun is a similar situation to a binary star, but with a larger ratio between the two masses. Where is the centre of mass between the Sun and Jupiter?

Solution mSun = 1.99 × 1030 kg mJupiter = 1.90 × 1027 kg ≈ 0.001 mSun Use the equation: rSun =

(

mJupiter m Sun + mJupiter

)

r

Even without calculating exactly, we can see that, since Jupiter has only ~one-thousandth the mass of the Sun, rSun is only about one-thousandth of r. Knowing the size of the Sun (see in2 Physics @ Preliminary section 16.1) leads us to the conclusion that the balance point between the two is very close to the surface of the Sun. For smaller planets such as the Earth, the balance point is within the Sun. So, although we usually say that the planets orbit the Sun, this isn’t quite true.

The apparent size and shape of the orbit is interesting, but the real importance of binary stars is the possibility of determining their masses. Mass cannot be determined from spectroscopic information alone. We need to study the gravitational effect on another object, and a binary star system provides a laboratory in which to do this. We can see how this works if we assume the orbit is circular and use equations from section 2.2. The centripetal force needed to maintain circular motion of star 1 around the centre of mass is provided by gravitational force between the two stars: Fgravity = Fcentripetal Gm1m2

408

The centre of mass of a binary system is the balance point between the two stars.

r2

m v2 = 11 r1

astrophysics 2π r1 We can include the period T of the orbit of star 1 by using v1 = . T This gives: 2 Gm2 4π r1 = r2 T2

Now we can insert the result for r1 derived earlier: Gm2 4π 2 = 2 r2 T

Rearranging this produces:



 

⎡ mm22 ⎤ ⎢ m + m ⎥ rr ⎢⎣ (m1 1+ m22) ⎥⎦

m1 + m2 =

or

4π 2r 3 GT 2

r 3 G (m1 + m2 ) = T2 4π 2

This is a form of Kepler’s third law (section 2.2; see also in2 Physics @ Preliminary section 13.4) but involves the combined mass of both stars. It also requires r, the true distance between the stars, with allowance made for any tilt of the orbit relative to the plane of the sky. This means that the distance to the system must be known. If r1, the distance from star 1 to the centre of mass can also be determined, then the masses m1 and m2 can be calculated individually by using a slight rearrangement of the earlier result for r1. The problem is that only rarely do we have enough information to do this.

Worked examples Question The α Centauri system is 1.338 pc away. The A and B components orbit each other with a period of 79.92 years and an average distance of 23.7 AU. What is the total mass of the system?

Solution Use

m1 + m2 = m1 + m2 =

4π 2r 3 GT 2

but ensure the correct SI units are used.

(

)

4π 2 3.55 × 1012 m

binary status In binary or multiple star systems we generally call the brightest star the primary and assign it the letter A. The secondary star is B and any other stars in the system are C, D etc. Unlike the system in Figure 23.1.1, a binary system is often drawn centred on the primary, which is considered stationary. For example, in Figure 23.1.3 the system is centred on β Centauri A, with the very close companion β Centauri B in orbit around it. Notice that the orbit is elliptical, although not the same ellipse you would draw if the motions were represented relative to the centre of mass (as in Figure 23.1.1). However, even this elliptical orbit is only the apparent orbit on the ‘plane of the sky’. The ‘real’ elliptical orbit is tilted relative to the plane of the sky.

3

(6.672 ×10 )(2.522 ×10 s) −11

Try this!

9

2

= 4.16 × 1030 kg ≈ 2.1 solar masses



This is usually written with a special symbol for ‘solar mass’ as 2.1 M. In fact the orbit is highly elliptical, but the result is still valid.

Question Measurements indicate that α Centauri A has an average distance of 11.2 AU from the system’s centre of mass. What is the mass of each star?

409

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Stellar companions and variables

Solution Use

m2 1.68 × 1012 m  = 3.55 × 1012 m  4.16 × 1030 kg 

r1  m2  =  r  (m1 + m2 ) 

(

)

  

Therefore m2 = 1.97 × 1030 kg ≈ 0.99 M, leaving m1 = 2.19 × 1030 kg ≈ 1.1 M. See the examples in Chapter 22 for further information on these two stars.

PHYSICS FEATURE

5. Current issues, research and developments in physics

β Centauri

Orbit Period: 357.00 ± 0.07 days Semi-major axis: 25.30 ± 0.19 milli-arc seconds Eccentricity: 0.821 ± 0.003 Inclination to the plane of the sky: 67.4 ± 0.3 degrees

Primary

Mass: m1 = 9.1 ± 0.3 M

Absolute V magnitude:

Secondary

M1 = –3.85 ± 0.05

Mass: m2 = 9.1 ± 0.3 M

Absolute V magnitude:

M2 = –3.70 ± 0:05

The high accuracy of these results offers an important starting point for understanding the structure of the stars themselves. –25

Position (milli-arc seconds)

β

Centauri has long been known as a visual binary  with the two stars separated by more than 1 arc second. Early observations with an interferometer suggested that the brighter star was itself a very close binary—much too close to be seen as separate through a conventional telescope. Figure 23.1.3 shows the orbit of β Centauri B. Its orbit has been determined from high resolution observations using the Sydney University Stellar Interferometer (see section 21.5). The stars are very close, so angles on the sky are measured in milli-arc seconds (one-thousandth of an arc second). Combining this with spectroscopic data yields all the parameters of the system.

E

–20 –15

orbit of β Centauri B

N

–10 –5 0 5 –10

β Centauri A

–5

0

5

10

15

20

25

Position (milli-arc seconds)

Distance  102.3 ± 1.7 pc Figure 23.1.3

Observations and orbit of β Centauri B relative to β Centauri A

Checkpoint 23.1 1 2

410

Draw a diagram of a binary system that shows the two stars, the centre of mass of the system and their elliptical orbits around the centre of mass. Determine where the centre of mass is located between Jupiter and its moon Callisto. MJupiter = 1.90 × 1027 kg MCallisto = 1.08 × 1023 kg Average distance of Callisto from Jupiter = 1 883 000 km

astrophysics

23.2 Doubly different Although most stars are in binary or multiple systems, they don’t all reveal themselves in the same way. We can classify the types of binaries according to how they are observed.

Describe binary stars in terms of the means of their detection: visual, eclipsing, spectroscopic and astrometric.

Visual binaries A visual binary (or multiple system) can be resolved into two (or more) stars by a telescope under sufficiently good seeing conditions. Like α Centauri (Figure 23.2.1), many of the brighter stars are ‘doubles’ well known to amateur astronomers. A few are merely chance alignments of unrelated stars, but most are genuine binaries. A simple calculation of angles will reveal that if a star is to be seen as a double in a telescope with angular resolution about 1 arc second, the stars must be at least tens of AU apart and the system must be relatively close. Such stars orbit each other slowly; but over many years the orbital motion may become apparent. More than 100 000 visual double stars are catalogued, but the orbits of only a few thousand are known.

B a

Figure 23.2.1 α Centauri A and B 1990

1980

Astrometric binaries Some stars are sufficiently close that their motion through space is apparent as A few of these stars reveal motion across the sky, that is, as proper motion. their binary nature by the wobbling of their paths across the sky (Figure 23.2.2). The centre of mass of such an astrometric binary follows a straight path, but the individual stars appear to wobble as they orbit. Few binaries are discovered in this way, since it usually requires long-term observations of nearby stars. The Hipparcos astrometry mission (section 22.1) revealed many new examples, because its very high precision position measurements revealed the motions of stars much more quickly.

Spectroscope binaries

1970 1960

Sirius B

1950 1940

Sirius A

1930 1920 1910

Most binary systems are too distant to appear as either visual or astrometric 1900 centre of mass However, the spectrum of an unresolved binary will have light binaries. contributed by both stars. As the stars orbit, one will typically have a component Figure 23.2.2 Sirius is a visual and an of its motion towards us, while the other is moving away. The light from the two astrometric binary. stars will show small blue and red Doppler A is moving towards the shifts that change as the stars orbit. Provided observer: lines blue-shifted A B the stars are close in brightness, the result is B is moving away from the a doubled-lined spectroscope binary in A BA B A B observer: lines red-shifted which each spectral line will appear doubled by these small shifts in wavelength (Figure B A and B are moving across 23.2.3). At other times, the motions of the the observer’s line of sight: stars may be entirely across the line of sight no Doppler shift A A&B A&B A&B and show no Doppler shift and therefore display only a single set of lines. A is moving away from the If the stars are significantly different in observer: lines red-shifted B A brightness, the spectrum of the brighter star B is moving towards the will flood the fainter one and only one set observer: lines blue-shifted B AB A B A of moving spectral lines may be visible, Figure 23.2.3 The changing pattern of spectral lines in a spectroscopic binary forming a single-lined spectroscopic binary. (very exaggerated)

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Stellar companions and variables To be visible as a spectroscopic binary, the component of the orbital velocity measured by the Doppler shift must be relatively large. This means that the stars must be close—much closer than most visual binaries, so they orbit quickly. Also, the orbit must be orientated so that the orbital motion has a component in the line of sight. This is not enough to really understand the system since, generally, the tilt of the orbit to the line of sight is unknown. The brighter star of β Centauri is a singled-lined spectroscopic binary that has now been resolved by the SUSI interferometer (see Physics Feature p 410). This combination of spectroscopic and astrometric data is rare and valuable, since it allows all the parameters of the system to be derived, in particular the mass.

Eclipsing binaries

PRACTICAL EXPERIENCES Activity 23.1

Activity Manual, Page 185

Visual magnitude

orbit of secondary 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6

primary secondary primary eclipse

Figure 23.2.4

Figure 23.2.5

412

secondary eclipse

2.867321 days

If the stars of a binary system are close together and orientated so that the orbital plane is close to edge-on, then it may be seen as an eclipsing binary. In these systems the stars regularly eclipse one another, periodically blocking out some of the light from the system. To be well enough aligned with our line of sight means that the stars are usually very close together, with orbital periods of just a few hours or days. They are therefore usually also spectroscopic binaries. They may be so close together that the stars are distorted by their mutual gravitational forces. The change in brightness is apparent in the light curve of the systems—a plot of the apparent magnitude of the system versus time (Figure 23.2.4) that repeats with every orbit. If the stars are different, the two eclipses will not be identical. A primary eclipse will primary result in a greater loss of light than the secondary eclipse eclipse, with the details determined by the tilt of the orbit, the relative size of the stars, their surface temperatures and even the structure of their atmospheres. The importance of eclipsing binaries lies in the wealth of information we can glean from observations, including masses and distances.

Algol (β Persei), a famous example of an eclipsing binary system, and its corresponding light curve

This artist’s impression of Circinus X-1 shows a binary system composed of an ordinary star losing material to the accretion disc around a neutron star.

Cataclysmic variables and X-ray binaries Some binaries are so close that they exchange mass between the stars. When one of the stars is a compact stellar remnant such as a white dwarf, a neutron star or a black hole (see section 24.5), it may accrete gas from its companion. This gas releases gravitational potential energy, becomes very hot and emits high-energy radiation. If the compact remnant is a white dwarf, its light output may vary dramatically as the system suffers one or more outbursts as a cataclysmic variable. If the remnant is a neutron star or black hole, it may be apparent as a source of X-rays from the infalling gas. It is then known as an X-ray binary (Figure 23.2.5).

astrophysics

The mass—luminosity relationship 1 000 000 100 000

Luminosity (solar luminosities)

O

ne important product of our ability to determine the masses of some stars in binary systems is the mass–luminosity relationship (Figure 23.2.6). This is a plot of the mass M of a main sequence star versus its luminosity L, which can be approximately fitted by the relationship L ∝ M 3.5.   It indicates that the luminosity of a star increases very rapidly with its mass. As luminosity is based on the consumption of hydrogen fuel in the core of the star (see in2 Physics @ Preliminary section 15.4), this indicates that fuel consumption increases much more rapidly than fuel availability. The result is that high-mass stars on the main sequence have shorter lifetimes than low-mass stars.

10 000 1000 100 10 1 0.1

L ∝ M 3.5

0.01 0.001 0.1

1

10

100

Mass (M/M )

Figure 23.2.6

The mass–luminosity relation for main sequence stars

Checkpoint 23.2 1 2

Recall the different types of binary systems and outline the characteristics of each type. Explain the significance of the mass–luminosity relationship.

Classify variable stars as either intrinsic or extrinsic and periodic or non-periodic.

23.3 Variable stars Eclipsing binaries and cataclysmic variables can also classified as variable stars. They vary in brightness because they are part of a close binary system. Other stars vary by themselves, whether part of a binary system or not, as a normal stage in their lives. More than 30 000 variable stars have been catalogued, and many thousands more are suspected to be variable. In fact, all stars vary in brightness to some degree. The Sun varies by about 0.1% within an 11-year cycle because of the solar activity cycle (see in2 Physics @ Preliminary section 16.3). Also, rapid variations are associated with tiny oscillations in the Sun measured by helioseismology (see in2 Physics @ Preliminary section 16.2). These are both just part of the normal behaviour of a Sun-like star, and the Sun is not regarded as a variable star. Other stars vary more in brightness and we track these changes with a light curve. There are many types of variable stars and the main groupings are shown in Figure 23.3.1.

GROUP

TYPE

CLASS Cepheids PULSATING STARS

INTRINSIC VARIABLE

Type II W Virginis

RV Tauri Long-period variables

ERUPTIVE (cataclysmic) STARS

VARIABLE STARS

RR Lyrae

Type I Classical

Mira type

Semiregular Supernovae Novae Dwarf novae Symbiotic stars Flare stars R Coronae Borealis T Tauri stars

ECLIPSING BINARIES EXTRINSIC VARIABLE ROTATING VARIABLES

Figure 23.3.1

Different types of variable star systems 413

23

Stellar companions and variables Effective temperature (K) 30 000 10 000

7 000

6 000

4 000

–10

5

10

–6

Type I

–4 Absolute magnitude (Mv)

RV Tauri Semiregular stars

Classical Cepheids

–2

Long-period variables

Type II

0

RR Lyrae

2

Ma

W Virginis

4

2

10 10

equ

enc

e

T Tauri stars

1

Instability

–1

strip

8

3

10

Miras

in s

6

4

10

10

Luminosity compared to Sun

–8

–2

10

10

12

Flare stars

14

–3

10

–4

10 O5 B0

–0.5

Figure 23.3.2

A0

F0 G0 Spectral class

K0

0.0

+0.3 +0.6 Colour index

+0.8

M0 +1.4 +2.0

The location of some types of variable stars on the HR diagram

It divides variables stars into extrinsic variables and intrinsic variables. Extrinsic variables are perhaps less interesting, at least as variable stars, because their variation is due to a process external to the body of the star itself. Eclipsing binaries are the best example, although the group also includes stars that vary because of their rotation (e.g. the effect of ‘spots’ on their surfaces). Intrinsic variables vary because of physical changes in the star or the stellar system. The HR diagram (see in2 Physics @ Preliminary section 15.3) summarises the properties of stars, and therefore different types of intrinsic variables occupy specific regions of the HR diagram (Figure 23.3.2). Intrinsic variables can be split into non-periodic variables and periodic variables. As the names suggests, the difference is whether their variation repeats at reasonably regular intervals.

Non-periodic variables The non-periodic variables cover a wide range of different types of stars that are physically very different, as illustrated by the systems described in Table 23.3.1.

Table 23.3.1  Major classes of non-periodic variables Variable type

Brightness change

Physical description

Stars in binary systems Type I Supernovae Novae

Dwarf novae

Symbiotic stars

~20 magnitudes within hours, then gradually fades over weeks Between 7 and 16 magnitudes, usually within a few days, then fades overs years to its initial brightness; some have been seen to brighten again Between 2 and 5 magnitudes, repeating semi-regularly; different types behave differently Vary semi-regularly over a range of about 3 magnitudes

Accretion of gas onto a white dwarf from its companion, leading to a runaway nuclear explosion A close binary composed of a Sun-like star leaking gas onto a white dwarf that eventually will accumulate enough material to generate a surface nuclear explosion A close binary composed of a Sun-like star leaking gas onto a white dwarf; instability in the accretion of gas appears to produce the outbursts A close binary composed of a red giant and a white dwarf; outbursts from the red giant fall onto the white dwarf

Single stars (may be in a binary system) Type II supernovae Flare stars

R Coronae Borealis

T Tauri

414

~20 magnitudes within hours, then gradually fade over weeks Typically 1 or 2 magnitudes and fade within hours Usually at maximum brightness but fade at irregular intervals by up to 9 magnitudes, returning to normal over months Vary irregularly in brightness

Core collapse of a massive star blows off its outer layers Solar-like flares on the surface of faint red dwarfs contribute dramatically to the overall visible light output of the star Carbon-rich dust clouds obscure the surface of the yellow supergiant star

Very young stars with an accretion disc, still approaching the main sequence

astrophysics

Periodic variables Periodic variables show regular or semi-regular changes in brightness, with periods ranging from hours to hundreds of days. The major types are shown in Table 23.3.2. Their brightness changes as the stars pulsate in size, surface temperature and colour. The pulsation arises from a slight instability in the balance between the inward pull of gravity and the outward pressure of the gas and radiation (see in2 Physics @ Preliminary section 15.4). This instability only occurs in the outer layers of the stars and does not affect the energy production in the core. Conditions in pulsating stars are just right to allow the pulsations to continue instead of dying away, as they would in most stars. These conditions are found in stars in the ‘instability strip’ on the HR diagram, with another similar zone of ‘long-period variables’ among the red giants (Figure 23.3.2). Table 23.3.2  Major classes of pulsating variables Variable type

Brightness change

Physical description

Cepheid

Between 0.5 and 2 magnitudes with periods from 1 to 70 days Up to 2 magnitudes with periods of less than 1 day Alternating deep and shallow minima with periods from 20 to 100 days Between 2 and 10 magnitudes with periods from 80 to 1000 days Up to 2 magnitudes with periods from 80 to 1000 days, but with irregularities

Luminous yellow supergiants Old giants stars with MV ≈ +0.6 Yellow supergiants

RR Lyrae RV Tauri Mira Semi-regular

Red giants and supergiants Red giants and supergiants

Naming of stars

S

tar names and designations follow many different systems for different types of stars. There are also many historical names and catalogues with designations still commonly in use. For example, the brightest star in the sky has many designations such as Sirius, α Canis Majoris, 9 Canis Majoris, HD 48915 and Hipparcos 32349 to name a few. Variable stars have one of the oddest systems. The first variable discovered in a constellation is called R, e.g. R Canis Majoris. Next comes, S, T, U … Z. Then what to do? We use RR, RS, RT … all the way to ZZ, and then AA, AB … to QZ. Finally, if we have more than 334 variables in a constellation, we start a sensible system with V335, V336 etc! Different classes of variable stars often take their name from the first one identified, for example R Coronae Borealis stars.

Checkpoint 23.3 1 2

Distinguish between extrinsic and intrinsic variable stars. Explain how the pulsation of periodic variables can tell us about the structure of those stars.

23.4 Cepheid variables Of all the variable stars, the Cepheid variables are the most important, followed by their cousins, the RR Lyrae variables. Both serve as ‘standard candles’ in distance measurements, because their absolute magnitudes can be estimated from their pulsation properties. The Cepheids are named for the prototype of the class, the northern nakedeye star δ (delta) Cephei. They vary regularly and have a characteristic light curve. As they oscillate in size they change in surface temperature and therefore spectral class (Figure 23.4.1).

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Stellar companions and variables

Brightness

PERIOD

Time

Figure 23.4.1

Explain the importance of the period–luminosity relationship for determining the distance of cepheids

In the early 20th century, Henrietta Leavitt (1868–1921) was studying Cepheids in the Magellanic Clouds, two satellite galaxies to the Milky Way. The Magellanic Clouds are relatively small compared with their distance, so all Cepheids in each Cloud are at essentially the same distance from us. The different apparent magnitudes of these stars therefore reflect true differences in luminosity (or absolute With this simplification, Leavitt realised that Cepheids with magnitude). longer periods were more luminous than Cepheids with shorter periods. This is the period–luminosity relationship for Cepheids (Figure 23.4.2). It was discovered later that there are actually two classes of Cepheids. Type I (or classical) Cepheids are massive young stars crossing the instability strip as they evolve. Type II Cepheids (or W Virginis stars) are much older stars also crossing the instability strip. The power of the period–luminosity relationship lies in the fact that period can be determined by following the brightness variation of any Cepheid we can see. Then its absolute magnitude can be estimated from the period–luminosity curve (Figure 23.4.2) and the distance determined by comparing the apparent and absolute magnitudes (see section 22.4). However, reality is always a bit more difficult; for example, interstellar dust can make the stars appear dimmer than they should. Because they are supergiant stars, Cepheids can be identified in some nearby galaxies. As a result, Cepheid distances are a fundamental stepping stone in measuring much larger distances in the universe. RR Lyrae variables are not as bright as Cepheids, but they are simpler since they all appear to have about the same luminosity (Figure 23.4.2). Once a star is recognised as an RR Lyrae variable, its distance can be determined by comparing its apparent and absolute magnitudes.

Type I (Classical) Cepheids

104

Luminosity (L )

Variations in the brightness, size and colour of a Cepheid variable during its pulsation

Type II (W Virginis) Cepheids

103

102

RR Lyrae 1 0.5

Figure 23.4.2 1

3

5

10

Period (days)

416

30 50

100

The period–luminosity relations for Cepheid and RR Lyrae variable stars. The green arrows indicate the luminosity relative to the Sun of a type I Cepheid that has a period of 10 days.

astrophysics Worked examples Question ζ (zeta) Geminorum is a Type I Cepheid variable star in the constellation of Gemini. It varies in brightness between 3.7 and 4.2 magnitudes every 10.2 days. Using the period–luminosity relation, estimate its absolute magnitude.

Solution Using the arrows on Figure 23.4.2, a period of 10.2 days indicates a luminosity of approximately 3500L. This can be converted to a magnitude difference using: I  mB − m A = 2.50 log10  A   IB  but working with absolute magnitudes. MV for the Sun is +4.85, so: 4.85 – MA = 2.50 log10 (3500) and thus MA is –4.0.

Question Using ζ Geminorum’s average apparent magnitude, estimate its distance.

Solution If we assume the average apparent magnitude is +3.95 then, using the relation between absolute and apparent magnitude: d 3.95 − ( −4.0) = 5 log10    10  d 7.95 5 Then   = 10 and d ≈ 390 pc  10  This is close to the accepted value.

Checkpoint 23.4 1 2 3

Recall why Cepheids and RR Lyrae variables are so important in astrophysics. Outline how Leavitt discovered the period–luminosity relationship. Describe the characteristics of the two types of Cepheids.

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Stellar companions and variables

PRACTICAL EXPERIENCES

CHAPTER 23

This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 23.1: Eclipsing binaries Go to the website given to observe the light curves of eclipsing binary stars. Record your observations and use them to interpret any light curves. Discussion questions 1 Identify the basic features of the light curve of an eclipsing binary. 2 Sketch a light curve and explain the positions of the stars at specific points of the curve.

Perform an investigation to model the light curves of eclipsing binaries using computer simulation.

Chapter summary • • • •

• •

•

•

418

A binary star system is composed of two stars orbiting their common centre of mass. The major importance of binary stars is their use in determining the masses of the stars. 4π 2r 3 Kepler’s third law in the form m1 + m2 = GT 2 involves the combined mass of both stars. With more information, the individual masses can be determined. A visual binary (or multiple system) can be resolved into two (or more) stars by a telescope under sufficiently good seeing conditions. An astrometric binary reveals its binary nature by the wobbling of its path across the sky. In a doubled-lined spectroscope binary each spectral line will periodically appear doubled by the Doppler shift of the individual stars. In a single-lined spectroscope binary the spectrum of the brighter star will flood the spectrum of the fainter one and only one set of moving spectral lines will be visible. If the stars of a binary system are close together and orientated so that the orbital plane is close to edge-on, it may be seen as an eclipsing binary in which the stars regularly eclipse one another.

•

•

•

• • • • • • • •

The light curve of an eclipsing binary or other variable star is a plot of the apparent magnitude of the system versus time. Some binaries are close enough to exchange mass between the stars, and become apparent as cataclysmic variables or X-ray binaries. The mass–luminosity relationship is a plot of the mass M of a main sequence star versus its luminosity L, which can be approximately fitted by the relationship L ∝ M 3.5. Variable star systems vary in brightness. Variation in extrinsic variables is due to a process external to the body of the star itself. Variation in intrinsic variables is due to physical changes in the star or the stellar system. Non-periodic variables cover a wide range of different types of stars that are physically very different. Periodic variables show regular or semi-regular changes in brightness due to pulsation. Cepheid variables and RR Lyrae variables serve as ‘standard candles’ in distance measurements. Cepheid variables obey a period–luminosity relationship that can be used to determine their distance. RR Lyrae variables all appear to be about the same luminosity, which can be used to determine their distance.

Review questions

astrophysics

Physically Speaking Complete the passage below by filling in the missing words from this list: brightness, companion, eclipse, eclipsing binaries, eruption, extrinsic, intrinsic, magnitudes, periods, pulsating variables, pulsation, rotation, variability

Variable stars are stars that change ______________. This change can range from 0.001 to as much as 20 ______________ over ______________ of a fraction of a second to years. There are a number of reasons why the brightness of variable stars will change. ______________ swell and shrink due to internal forces, while a star in an ______________ will dim when it is eclipsed by a faint ______________ and then brighten when the occulting star moves out of the way. Variable stars are classified as either ______________, when variability is caused by physical changes such as ______________ or ______________ in the star or stellar system, or ______________ when ______________ is caused by the ______________ of one star by another or by the effects of stellar ______________.

Reviewing 1 Define what is meant by the term binary star. 2 Discuss why binary star systems are important in astronomy. 3 Draw a typical light curve that would result from an eclipsing binary system that consists of two identical stars.

4 Outline the characteristics that can be determined if two stars form a spectroscopic binary.

5 Explain why relatively few binary systems are discovered as astrometric binaries.

6 On the basis of the other types of binary systems described, propose an idea of why you might describe a system as an ‘interferometric binary’ system.

7 Define what is meant by the term variable star. 8 List examples of variable star types that would be called extrinsic or intrinsic.

9 List the properties of a star that vary in a pulsating variable star. 10 Construct a block diagram indicating the relationship between all the types of binary stars and variable stars described in this chapter.

11 Explain the properties that distinguish a Cepheid variable star. 12 Outline the importance of the periodic change in luminosity of Cepheid variables.

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Stellar companions and variables

Solving problems 13 If a binary system consists of a 0.5 M red dwarf star and a 3 M red giant star separated by 20 AU, calculate the location of the centre of mass.



Solve problems and analyse information by applying:



4π r GT 2

m1 + m2 =

2 3

14 Sketch the orbits around the centre of mass of the 0.5 M and 3 M stars in the previous question.

15

a Sirius, the brightest star in the sky is a binary composed of a main sequence star and a white dwarf. It is 2.631 pc away. The A and B components orbit each other with a period of 50.1 years and an average distance of 19.8 AU. Calculate the total mass of the system. b Measurements indicate that Sirius A has an average distance of 6.5 AU from the system’s centre of mass. Calculate the mass of each star.

16 V Puppis is a very closely spaced eclipsing binary in the southern constellation of Puppis. Its light curve is shown in Figure 23.5.1. Estimate the period of the orbit and the amplitude of the eclipses, in magnitudes.

Apparent magnitude

4.2 4.4 4.6 4.8 5.0 –0.3

0.0

0.3

0.6

0.9

1.2

1.5

1.8

2.1

2.4

Time (days)

Figure 23.5.1

Light curve of V Puppis

17 β Doradus is a Type I Cepheid variable in the southern constellation of Dorado. It varies in magnitude from 3.46 to 4.08 in 9.94 days. Calculate its distance.

18 The star RR Lyrae is spectral class F5 with an average apparent magnitude

Re

iew

420

Q uesti o

n

s

v

of 7.1. Calculate its distance.

astrophysics

PHYSICS FOCUS Variables and supergiants by Jonathon Nally For most of the 20th century, professionals and amateurs were separated by a wide gap in technology. Only the professionals could afford the large telescopes and complex support gear required for modern astronomy and astrophysics. That gap has now closed to some degree. Amateurs now have affordable access to sophisticated telescopes, electronic detectors, software, and all the other paraphernalia needed to contribute to many areas of astronomy. But it’s great to see that there’s still a place for the good old human eye. And this is nowhere more evident than in the field of variable star research. Some variable stars change brightness over a period of days, weeks, months or even years. Many change their brightness with perfect reliability, while others are totally unpredictable. This means that constant, longterm monitoring is needed, which is the sort of thing that professional astronomers really are unable to do— but it’s a job perfectly suited to amateurs. Every night, all over the world, scores of dedicated variable star observers are outside, making observations and brightness readings of hundreds of variable stars. They tabulate their data and send them in to one of several umbrella organisations scattered around the world. The data is then made available to the professionals who, it cannot be stressed enough, rely so heavily on this work by the amateurs. Usually, it takes many years of patient monitoring for enough data to be built up to make a difference in research programs. But sometimes things can happen much, much faster. On the night of 6 January 2002, Aussie amateur astronomer Nick Brown spotted something unusual. A certain star was much brighter than it should have been—V838 Monocerotis had gone into outburst. It was a nova, and Nick was the first person to spot it. He quickly reported it, and it came to the attention of a bunch of professional astronomers who were interested in just this particular kind of star. For the past year, this team has been studying V838, and

20 May 2002

2 September 2002

28 October 2002

17 December 2002

Figure 23.5.2

Hubble images of the ‘light’ echo surrounding V838 Monocerotis

have found it to be very unusual indeed. It had expanded into a supergiant star over a period of just a few months, brightening three times before fading dramatically. At first they thought the fade was caused by obscuring dust, but now it turns out that the star has actually cooled down. In fact, it has become the coolest supergiant star ever found and all because of a single observation made by a dedicated amateur astronomer in Australia. Source: Transcript from ABC Science Show, Radio National, 10 May 2003

1 Outline the advantages of an electronic detector over the human eye for variable star observation. 2 Describe why it takes years of observations of a variable star to build a useful data set. Is this true of all types of variables?

Extension 3 V838 Monocerotis is now well known because of images taken by the Hubble Space Telescope (Figure 23.5.2) showing a ‘light echo’ from the outburst. Define the term light echo. 4 Recount what has happened to V838 Monocerotis since the time of the radio broadcast in 2003. 5 Propose some other fields of astronomy in which you think amateur astronomers can make valuable contributions.

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Birth, life and death Stellar evolution: How do we know?

interstellar medium, nebula, emission nebulae, dark nebulae, reflection nebulae, giant molecular clouds, protostar, zero-age main sequence, accretion disc, jets, proton– proton chain, carbon–nitrogen–oxygen cycle, red giant, triple alpha process, helium flash, horizontal branch (HB), asymptotic giant branch, planetary nebula, supernova, supernova remnant, neutron star, black hole, pulsar

Astronomers talk about the ‘evolution’ of a star, but not quite in the sense a biologist uses the word. Astronomers mean that an individual star transforms itself during its lifetime as it consumes its nuclear fuel. However, stars have also evolved over generations of birth and death. Stars born today are somewhat different from stars born closer to the time of the Big Bang. How do we know about the evolution of the stars? It’s a little like biology in that we almost never see evolution happening in our lifetimes, but the evidence is there to be found. Some of the ancient stars are still around to see today.

24.1 The ISM Within the Milky Way galaxy, the space between the stars is not quite empty. It is filled by a patchy medium of gas and dust called the interstellar medium (ISM). The medium is visible as a nebula (meaning cloud), when it interacts with starlight. In visible light images such as Figure 24.1.1, we see three types of nebulae: • bright emission nebulae in which the gas is energised by hot young stars to emit an emission spectrum (see section 22.2) • dark nebulae in which the dust scatters starlight, reddening or completely blocking our view of background stars • reflection nebulae in which we see the light scattered by the dust, especially at blue wavelengths. The importance of gas and dust in our galaxy is apparent by looking at its distribution in other similar galaxies (Figure 24.1.2), although it makes up only about 1% of the mass of all the stars. Figure 24.1.1 422

The Trifid nebula, composed of an emission nebula (pink) and a reflection nebula (blue), crossed by dark dust lanes

astrophysics

The gas is about 70% hydrogen and 28% helium (if measuring mass), with traces of other elements and molecules. This is spread through several different components of the ISM with very different properties (Table 24.1). Of particular interest are the cold giant molecular clouds in which the gas is most dense, and simple molecules can form. Individual molecular clouds have masses up to millions of solar masses. They make up only about 1% of the volume of the ISM but contain 90% of the mass and are the sites of star formation. Where massive stars have formed in a giant molecular cloud, their intense ultraviolet (UV) radiation eats into the cloud and creates an emission nebula. Elsewhere the embedded dust may be revealed by dark nebulae and reflection nebulae. The dust is just a fraction of a millimetre in size and represents about 1% of the mass of the gas. Its characteristics vary with location, but it seems to consist of silicate or carbon grains, sometimes with a coating of various ices. At least some of it is formed in the cool outer atmospheres of red supergiant stars and blown outwards by the star’s stellar wind. The ISM is part of a giant recycling system that includes the stars (Figure 24.1.3). Gas from the ISM forms stars, is processed by nuclear reactions inside the stars, and some is returned to the ISM when the largest stars age and die. Table 24.1.1 Major components of the ISM, with the air around you included for comparison Component

Temperature (K)

Atmosphere at Earth’s surface Molecular clouds Neutral atomic clouds Partly ionised intercloud medium Highly ionised coronal gas

300 20–50 50–150 103–104 105–106

Figure 24.1.2

neutral atom clouds ionised clouds

molecular clouds

ISM

Density (atoms per m3) 25

~10 109–1011 106–109 ~104 2 10 –103

stellar outflows and explosions

Checkpoint 24.1 1 2 3

IR image of the galaxy M81, with the spiral arms traced by emission from dust

Recall the three types of nebulae and their characteristics. State the main components of ISM. Outline the significance of giant molecular clouds.

star formation

nuclear fusion in stars

Figure 24.1.3

The galactic recycling system

24.2 Star birth The first step in the recycling scheme is star formation. If a part of a giant molecular cloud is sufficiently cool and dense, it can be pushed into gravitational collapse. Likely triggers include the shockwave from the explosion of a nearby star, or the ‘density waves’ that sweep through the disc of the galaxy. Once the trigger has boosted the density sufficiently, gravity takes over and that piece of the cloud starts to collapse because of the gravitational forces between its own particles. Initially the gas falls freely inward and the density

Describe the processes involved in stellar formation.

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increases quickly at one or more centres to form cores with more slowly collapsing envelopes. –10 As each core collapses, the infalling gas releases 5 10 –8 gravitational potential energy (see in2 Physics @ 4 –6 10 15M . Preliminary section 4.1). Some of this energy is ° –4 3 converted to kinetic energy and heats the gas while the 9M 104 years 10 ° –2 rest is radiated as infra-red light. As each core gets 5M 105 years 2 ° 10 0 hotter, the gas pressure increases, slowing the collapse 3M ° Ma 2 in s 106 years 10 of the core while the surrounding gas continues to fall equ enc inwards. The collapsing fragment of the cloud is now a e 4 1 protostar, probably one of many, perhaps destined to 1M 6 ° –1 10 form part of a binary system or even an entire cluster 8 ZAMS 7 –2 of stars. It is a luminous source of infra-red light 10 10 10 0.5M years ° tracing a path on the HR diagram (Figure 24.2.1), but –3 12 10 buried deep in the larger cloud. 14 –4 10 As the collapse continues, the core O5 B0 A0 F0 G0 K0 M0 temperature rises, eventually passing 10 million K and Spectral class initiating the nuclear fusion of hydrogen into helium –0.5 0.0 +0.3 +0.6 +0.8 +1.4 +2.0 Colour index (see in2 Physics @ Preliminary section 15.4)—the star ‘turns on’! The increased energy production slows and Figure 24.2.1 Evolutionary tracks for protostars of various masses eventually stops the collapse of the core. Lower mass stars like the Sun go through a highly active phase as a T Tauri variable star, but eventually settle into a stable equilibrium between the inward pull of gravity and the outward force of gas and radiation pressure. The star is ‘born’ on the zero-age main sequence (ZAMS) of the HR diagram, a line along the lower edge of the broader main sequence band on Figure 24.2.1. The time taken to reach this stage is a function of the mass of the collapsing fragment of cloud. More massive protostars heat up more quickly and begin nuclear reactions sooner. Very low mass stars will reach the ZAMS as red dwarfs long after their higher mass counterparts have lived their entire lives and died! Objects with mass below about 0.08M form brown dwarfs and will never fuse hydrogen into helium. Just how many stars form in each mass range is uncertain and undoubtedly varies. However, for every one 10 M star, there are typically about 300 stars like the Sun formed and many more M-type red dwarfs. Effective temperature (K) 6 000

4 000

Luminosity compared to Sun

7 000

Absolute magnitude (Mv)

30 000 10 000

Discs and jets

S

urrounding the protostar is an accretion disc that forms naturally as the protostar collapses and spins faster. Infalling material accumulates in the disc, but it is accompanied by jets of outflowing material—a situation often seen in astronomy. When the stars ‘turns on’, the system is swept clean by radiation from the star and may reveal a planetary system that formed within the disc.

jet protostar outflow

inflow

accretion disc

jet

Figure 24.2.2 424

The accretion disc and jet surrounding a protostar

astrophysics

Checkpoint 24.2 1 2 3 4

Outline some of the events that trigger the birth of stars. Describe a protostar. Describe the ZAMS. Recall the forces that are in balance when a star is in equilibrium.

24.3 Stars in the prime of life A place on the ZAMS is a starting point for the main part of any star’s life. That place, how long the star took to get there and everything that happens afterwards are determined almost entirely by the mass of the star. This arises because mass determines the gravitational forces within the star and therefore the pressure that must resist those forces if the star is to be stable. The pressure is generated by the high temperatures inside a star as a result of the energy produced by the nuclear reactions in its core. So the stability of the star depends on it having enough nuclear fuel available and processing it fast enough. Stars change very slowly as they age, so they spend most of their lives on the main sequence, not far from the ZAMS, and that is where most stars are found in the HR diagram. How long they can maintain this peaceful existence depends on how much hydrogen fuel they start with and how quickly they consume it. The mass–luminosity relationship for main sequence stars (section 23.2) tells Despite their larger us that high-mass stars spend their fuel reserves recklessly. supply, more massive stars have shorter lives on the main sequence before their fuel runs low and they need to evolve (Table 24.3.1). In comparison, the Sun is a G2 star with an expected main sequence lifetime of about 11 billion years. Table 24.3.1  Properties of main sequence stars relative to the Sun Spectral class

Mass (m)

O5 B0 A0 F0 G0 K0 M0

40 15 3.5 1.7 1.1 0.8 0.5

Luminosity (l) 400 000 13 000 80 6.4 1.4 0.46 0.08

Surface temperature (K)

Radius (r)

Time on main sequence (million years)

40 000 28 000 10 000 7 500 6 000 5 000 3 500

13 4.9 3.0 1.5 1.1 0.9 0.8

1.0 11 440 3 000 8 000 17 000 56 000

To understand why the high-mass stars are so rash with their fuel, compared to their more frugal lower mass cousins, we need to consider the two processes at work in main sequence stars converting hydrogen into helium: the proton– proton chain and the carbon–nitrogen–oxygen cycle (see in2 Physics In both cases the net reaction is a combination @ Preliminary section 15.4). of four hydrogen nuclei (protons) into one helium nucleus with the release of energy and some other light particles. Energy is available because the mass of the four hydrogen nuclei is more than that of one helium nucleus. The lost mass is converted to energy according to Einstein’s famous equation E = mc2.

Describe the types of nuclear reactions involved in Main Sequence and post-Main Sequence stars. Discuss the synthesis of elements in stars by fusion.

425

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Birth, life and death

1H 1 2H 1

1 1H

1 1H

3 2He

1H 1 1 1H

4He 2

3 2He

1H 1

1 1H

2H 1

proton neutron positron neutrino gamma ray

1 1H

Figure 24.3.1

In cooler stars like the Sun, the dominant reaction is the proton–proton chain (Figure 24.3.1). Several pathways are possible but, for core temperatures from 10 to 14 million K, the main reaction sequence is:

A series of reactions must occur to form helium in the proton–proton chain.

12 6 C

1H 1

4 2 He

12

C re

cyc

led

12 6 C

13 7 N

1H 1

15 7 N

13 6 C

positron neutrino gamma ray

15 8O 1H 1

Figure 24.3.2

14 7 N 1H 1

3 2

H + 11 H → 21 H + e + + ν

2 1

H + 11 H → 23 He + γ

He + 23 He → 42 He + 2 × 11H

The reaction emits e+ (positrons, i.e. positive electrons), ν (neutrinos—elusive particles with very small mass) and γ-rays. At temperatures below 10 million K, this reaction barely works. At higher temperatures the energy production rises quickly (proportional to ~T 4). In more massive main sequence stars, the core temperature is even higher and the carbon–nitrogen–oxygen (CNO) cycle takes over as the dominant reaction. This process also produces helium from hydrogen but uses carbon nuclei as a catalyst, as illustrated in Figure 24.3.2. 12 6

C + 11 H → 137 N + γ 13 7

13 6 14 7

A series of reactions starting with carbon make up the carbon–nitrogen–oxygen cycle.

1 1

C + 11 H → 147 N + γ

N + 11 He → 158 O + γ 15 8

15 7

N → 136 C + e + + ν

O → 157 N + e + + ν

N + 11 He → 126 C + 42 He

CNO energy production is negligible below about 13 million K, but dominates proton–proton energy production by 20 million K. The energy production rises dramatically with temperature (proportional to ~T 20!), with some other reaction sequences also possible. The enormous increase in reaction rate produced by small changes in temperature, especially in the CNO cycle, explains why high-mass stars expend the reserves of hydrogen in their cores so quickly. It also explains the differences in later evolution of low- and high-mass stars. These different evolutionary paths are illustrated in Figure 24.3.3 and described in the following sections. As it ages, a main sequence star builds up helium in its core. This dampens the reaction rate, leading to a loss of pressure in the core, which responds by contracting a little. This releases gravitational potential energy that raises the temperature and therefore boosts the reaction rate. As the reaction is so sensitive to temperature, the overall effect is to increase the energy output and inflate the whole star. This moves the star on the HR diagram and gives the main sequence some width (Figure 24.3.4). 426

astrophysics

ionised clouds

neutral atom clouds molecular clouds

ISM

STAR FORMATION High mass 8M ≤ M

°

Main sequence star H fusion in core via CNO cycle

Major mass loss via stellar wind

Red supergiant He fusion in core via triple-α process, H fusion in shell

Major mass loss via stellar wind

AGB red supergiant C fusion in core, He, H fusion in shells, finally multiple shells of heavy element fusion

Dramatic mass loss, including heavy elements via supernova remnant

Type II supernova produces heaviest elements

Black hole M ≥ 3M No fusion reactions°

°

Small mass loss via stellar wind

Larger mass loss via stellar wind

Red giant He fusion in core via triple-α process, H fusion in shell

Significant mass loss via planetary nebula shell

Planetary nebula White dwarf M ≤ 1.4M No fusion °reactions, slowly cooling

Figure 24.3.3

A summary of the evolution of single stars of various masses. Gas is returned to the ISM as raw material for future stars.

Effective temperature (K) 30 000 10 000

7 000

6 000

4 000

–10 2 million years initial 30M .

–8

5 10

Supergiants (I)

°

–6

4 10

30 million years initial 10M .

–4 Absolute magnitude (Mv)

T

he Sun is 4.6 billion years into its main sequence lifetime. It is now about 5% bigger and brighter and 200 K warmer at the surface than it was when it landed on the ZAMS. Over the remaining 6 billion years of its main sequence life, the Sun will gradually double in luminosity and get 25% larger. This is a small change on the HR diagram (Figure 24.3.4), but one that will dramatically alter the Earth, eventually boiling away its oceans.

Brown dwarf No ongoing fusion

AGB red giant Possible C fusion in core, He fusion in shell, H fusion in shell

Increasing mass loss via stellar wind

Neutron star 1.4M ≤ M ≤ 3M ° reactions ° No fusion

Climate prediction: getting hotter!

°

°

Main sequence star H fusion in core via p–p chain and CNO cycle

3 10

°

180 million years

–2 initial 5M

initial 3M

°

2

2 10

640 million years

°

0

Giants (II, III)

Ma

in s

4

Subgiants (IV) enc e (V 11 billion years ) present sun

6

initial 1M

equ

1 4.6 billion years 56 billion –1 10 years ZAMS

°

8 10

White dwarfs

12

initial 0.5M

°

(VII)

14

Figure 24.3.4

The motion of a star on the HR diagram during its life on the main sequence

10

Luminosity compared to Sun

Significant mass loss via stellar wind

Very low mass M ≤ 0.08M

Low to medium mass 0.08M ≤ M ≤ 8M

–2 10 –3 10 –4 10

O5 B0 –0.5

A0

F0 G0 Spectral class

K0

0.0

+0.3 +0.6 Colour index

+0.8

M0 +1.4 +2.0

Checkpoint 24.3 1 2 3

Explain why the lifetime on the main sequence is different for stars of different mass. Write the net nuclear reaction equations that summarise the proton–proton chain and CNO cycle. If the CNO cycle produces negligible energy at 13 million K, explain why it can dominate energy production at 20 million K. 427

24

Birth, life and death

24.4 Where to for the Sun? Outline the key stages in a star’s life in terms of the physical processes involved.

Eventually a main sequence star runs very low on hydrogen fuel in its core. For a star like the Sun, there is no mixing of new gas into the core and no increase in the reaction rate can counteract the lack of fuel. The core collapse accelerates, raising the temperature as gravitational potential energy is released, and igniting hydrogen fusion in a shell around the helium core. Computer modelling shows that the outer layers of the star respond by expanding outwards and cooling. The plotted position on the HR diagram rapidly moves to the right across the sub-giant range (Figure 24.4.1). The luminosity of the star then begins to grow and the outer layers inflate further as the star climbs the red giant branch. About 1.3 billion years after leaving the main sequence, the G class V main sequence Sun becomes a K or M class III giant, about 200 times its current size. The star now has a dense core that is perhaps one-third its original size, but the grossly extended outer layers reach out to nearly envelope the orbit of the Earth. These outer layers are of very low density and only weakly held. As a main sequence star, the Sun looses mass at a rate of ~10–14 M per year as solar wind (see in2 Physics @ Preliminary section 16.4). As a red giant, the mass loss will be perhaps 107 times larger, and it will simply blow away a significant fraction of its mass. In the core of the red giant the temperature eventually rises to about 100 million K and the fusion of helium into carbon begins via the triple alpha process: 4 He + 42 He → 48 Be + γ 2 8 4

Be + 42 He → 126 C + γ

Adding another 42 He can produce 168 O. For stars of less than about 2.6 M, the onset of helium fusion occurs as the helium flash, since conditions in the cores of these stars cause the entire core to begin helium fusion almost at once. Slightly more massive stars begin helium fusion a little more sedately, because the physical state of the core is a little different. We believe that less massive stars that begin life as red dwarfs probably never go through a red giant phase. They never attain the temperatures needed to fuse helium. It’s hard to be sure, as their time on the main sequence is much greater than the current age of the universe and so there are no evolved red dwarfs to look at. In stars that become red giants, the new energy supply from helium fusion causes a reduction in the size of the star and the surface temperature begins to rise again. Stars fusing helium in their cores, with hydrogen fusion in a surrounding shell, appear as smaller yellow G- and K-type giants. The Sun may spend 100 million years in this horizontal branch (HB) phase, but its evolution is accelerating. Helium fusion occurs much faster than the hydrogen reactions and, all too quickly, the helium in the core is consumed and The outer layers expand the core must again begin to collapse and heat. again and the star moves up the asymptotic giant branch (AGB) in the HR diagram. The star now has a core of carbon and oxygen surrounded by shells of helium and hydrogen fusion. In higher mass stars, other elements are formed by

428

astrophysics

Effective temperature (K) 7 000

6 000

4 000

–10

5 10

Supergiants (I) planetary nebula

–8

AGB

–6 –4

H → He

–2 initial 5M 0

He → C H → He

helium ignition RGB

HB

helium flash

AGB Giants (II, III) He → C HB H → He Ma in s equ Subgiants (IV) enc e (V ) H → He

°

2 4

RGB

H → He

8 White dwarfs (VII)

12

no nuclear reactions

2 10

1

°

10

3 10

10

initial 1M

6

4 10

initial 0.5M

°

14

–1 10

Luminosity compared to Sun

30 000 10 000

Absolute magnitude (Mv)

nuclei capturing neutrons. The Mira-type long-period variable stars are AGB giants. After a further 20 million years these shells become unstable and begin to rapidly switch on and off, because the helium reaction is spectacularly dependent on the temperature (proportional to ~T 40!). The star has been losing mass throughout the giant phases, but now the pulsations gently puff off the outer layers, exposing deeper layers of the star. The intense UV light from the remains of the star lights up the ejected gas as a planetary nebula (Figure 24.4.2). The nebula returns gas to the ISM that has been enriched in carbon, nitrogen and oxygen by its passage through the nuclear fires of the star. At the centre of the planetary nebula, the remains of the star contract and heat up to about 100 000 K on the surface, as the final fusion of hydrogen occurs in a shell around the The fusion quickly ends and the carbon–oxygen core. star, now a white dwarf, begins a long, slow cooling that will last for tens of billions of years. With surface temperatures of about 10 000 K, but a size comparable to that of the Earth, the white dwarfs are small and faint and appear near the bottom of the HR diagram (Figure 24.4.1). The star has reached its stellar graveyard with only about half the mass it had on the main sequence and having only consumed about 12% of its original supply of hydrogen fuel. Half its mass has been returned to the ISM, somewhat enriched in heavier elements, as raw material for the next generation of stars.

–2 10 –3 10 –4 10

O5 B0 –0.5

A0

F0 G0 Spectral class

K0

0.0

+0.3 +0.6 Colour index

+0.8

Figure 24.4.1

M0 +1.4 +2.0

Evolution of stars of low to medium mass after they leave the main sequence

Explain the concept of star death in relation to: • planetary nebula • white dwarfs.

PHYSICS FEATURE When is a star not a star?

W

hat is a star? A sensible definition would say that a star produces energy from nuclear reactions in its core and is stable because of the balance between the inward pull of gravity and the outward forces of gas and radiation pressure. By that standard, a white dwarf is not a star! A white dwarf is a glowing ember. There are no nuclear reactions to generate energy, despite a core temperature as high as 10 million K. The energy trapped within the white dwarf from its energetic past slowly leaks out through its relatively small surface area. With as much as 1.4M of material in an object the size of the Earth, the density of a white dwarf is enormous—about 109 kg m–3— a million times the density of water. Gas pressure cannot support the white dwarf against the correspondingly enormous gravitational force. The balancing force comes from a new physical effect: degenerate electron pressure. Essentially, the electrons reach a state in which they refused to be packed more tightly.

Figure 24.4.2

An example of planetary nebulae 429

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Birth, life and death

Checkpoint 24.4 1 2 3

Outline the stages in the evolution of a solar mass star from main sequence star to white dwarf. Describe the processes by which a star loses sufficient mass to end its life as a white dwarf. Recall the characteristics of a white dwarf.

24.5 The fate of massive stars

430

Luminosity compared to Sun

Absolute magnitude (Mv)

For any star the basic problem is the same throughout its life. To be stable, it must generate enough energy to ensure that internal pressure balances the gravitational force. In a massive star, this means higher core temperatures to generate the energy required and a correspondingly shorter lifetime. For stars that start life with more mass than the Sun, but still less than about 8 M (the middle of spectral class B), evolution is faster than for lower mass stars (Figure 24.4.1), but they have the same fate in store. They lose more mass via stellar winds and all end up below the 1.4 M limit for the mass of a stable white dwarf. Stars starting life more than about 8 M evolve much more quickly Describe the types of nuclear and have a different ending (Figure 24.3.3). Their main sequence lifetimes are reactions involved in measured only in millions of years (Table 24.3.1). Even at this stage, their stellar Main Sequence and post-Main winds lead to significant mass loss (perhaps 10–6 M per year for a 60 M star). Sequence stars. Their evolutionary path is flatter on the HR diagram (Figure 24.5.1), indicating Discuss the synthesis of elements in stars by fusion. that their size changes dramatically but their luminosity does not. These stars become supergiants, perhaps 1000 times the size of the Sun. They cross back and forth across the instability strip, becoming Cepheid variables for part of their lives. In their cores, the onset of helium burning is gradual, Effective temperature (K) but as the temperature increases to more than 300 million K 30 000 10 000 7 000 6 000 4 000 –10 supernova the fusion of carbon with helium to produce oxygen AGB 5 helium 10 He → C –8 ignition becomes the dominant reaction. When the temperature H → He HB RGB 4 –6 climbs above 500 million K, carbon nuclei can fuse together 10 Supergiants (I) H → He –4 initial 10M to produce sodium, neon and magnesium. All the carbon is 3 ° 10 –2 consumed quickly and the core collapses further, pushing 2 10 the temperature ever higher. Ultimately, fusion of silicon 0 Giants (II, III) Ma in s produces iron at core temperatures of 7 billion K. 2 10 equ Subgiants (IV) enc The core of the supergiant resembles the layers of e 4 (V) 1 an onion in which an iron core is surrounded by shells in 6 –1 which silicon and sulfur, oxygen and carbon, helium and 10 8 hydrogen are all undergoing fusion reactions (Figure –2 10 10 White 24.5.2). However this is a fleeting phase. A 25 M star will dwarfs –3 12 10 (VII) fuse hydrogen on the main sequence for about 7 million 14 –4 10 years. It then spends 500 000 years also consuming helium, O5 B0 A0 F0 G0 K0 M0 600 years consuming carbon, half a year consuming oxygen Spectral class and just a day consuming silicon. –0.5 0.0 +0.3 +0.6 +0.8 +1.4 +2.0 Colour index Synthesis of iron is the limit of normal fusion processes, since any further reaction costs the star energy rather than Figure 24.5.1 Evolution of high-mass stars after they leave generating it, accelerating its collapse. Within seconds a the main sequence

astrophysics portion of the core the size of the Earth collapses to just a few kilometres across, achieving densities of 1017 kg m–3. The speed Fe core core of collapse causes the core to bypass the electron degeneracy region At 1012 K, degenerate pressure that supports a white dwarf. neutron pressure finally halts the collapse at the centre; however, the surrounding layers hit the core and bounce back, blowing the star apart in a supernova explosion. In a few seconds 1046 J of energy is released, 99% of it as elusive neutrinos. Just 1% of the H shell supergiant star Si, S shell He shell energy appears as visible light, but this is sufficient for it to C, O shell outshine all other stars in the galaxy for a few days! Figure 24.5.2 The shell structure in the core of a high-mass star During the explosion, iron nuclei are ripped apart and just before it explodes a large number of neutrons are released. These neutrons can be captured by existing heavy nuclei to produce heavier nuclei than Explain the concept of star could be produced by any process earlier in the star’s history. death in relation to: The blast wave and several solar masses of gas ejected by the supernova • supernovae explosion speed outward at several thousand km s–1, sweeping up gas blown away • neutron stars/pulsars earlier by the stellar wind (Figure 24.5.3). The result is a supernova remnant • black holes. (Figure 24.5.4), glowing across the full range of wavelengths and carrying gas enriched in a full range of heavy elements back into the ISM. After perhaps 100 000 years, the expanding remnant fades and merges into the ISM, but it may have already triggered a new burst of star formation nearby. Back at the site of the explosion, some of material very quickly falls back onto Computer simulations remain very the core of degenerate neutrons. uncertain, but they indicate that stars with initial masses of between 8 M and roughly 25 M will produce a core of less than 3 M. The core will then be stable as a neutron star. Much rarer stars with initial masses of between roughly 25 M and 40 M will have more material falling back. The core mass will exceed the 3 M limit of stability for a neutron star. No known force will stop the collapse and the result is the formation of a black hole. Computer simulations suggest that even more massive stars may form a black hole before an explosion can occur!

Figure 24.5.3

1994.4

2000.9

2005.2

2008.3

The expanding remnant from supernova 1987A brightens in its radio emission as the outrushing gas crashes into the surrounding medium.

Figure 24.5.4

The 14 light-year diameter remnant from Kepler’s 1609 supernova is shown in this image assembled from images taken by the Hubble, Spitzer and Chandra space telescopes. 431

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PHYSICS FEATURE

rotation axis

magnetic axis beamed radiation

Neutron stars: Pulsars

A

neutron star packs a mass about that of the Sun into an object magnetic field typically only 10 km across. As the core shrinks to this size, the rotation of the neutron star increases to a rate that is neutron star typically dozens of times a second! The acceleration due to gravity at the surface is possible about 1012 m s–2 (compared with 9.8 m s–2 line of at the surface of the Earth) and its magnetic sight to Earth field near the surface is about 108 T! beamed Neutron stars were first proposed in the radiation 1930s, but no-one expected to be able to Figure 24.5.5 The ‘lighthouse’ model sweeps beams of radiation see such tiny objects. In 1967 radio emission around the sky as the pulsar rotates. was detected from a rapidly pulsing source that was quickly recognised to be a rotating neutron very precise rate that reduces only very slowly as the star—a pulsar. pulsar loses energy. Pulsars produce two beams of radiation from A few pulsars have been observed at visible, X-ray near their magnetic poles. As these poles are usually and γ-ray wavelengths, but most of the 1700 known not aligned with the rotation axis, these beams sweep pulsars have been discovered by radio telescopes. around the sky as the pulsar rotates rapidly (Figure Many of these, including the first binary pulsar 24.5.5). As it spins, the beams may sweep across the system, have been found by the 64 m Parkes radio Earth, to be seen as a pulse. The pulses arrive at a telescope (Figure 21.4.1).

Black holes

B

lack holes are even more bizarre than neutron stars. With no known force able to resist the collapse, all the mass is concentrated at a point and surrounded by an intense gravitational field. They are ‘black’ because not even light can escape once it has crossed a theoretical boundary called the ‘event horizon’. At the event horizon, the escape velocity (see section 1.3) is the speed of light. Black holes have many intriguing physical properties, both within the event horizon and outside it. Many of these do not directly affect their behaviour as astronomical objects, as their results cannot be observed. Others enable us to detect the hole, because of the effect on surrounding matter, despite the tiny size and ‘blackness’ of the hole. A black hole in a close binary system may draw gas off its companion. The gas will be heated as it falls onto an accretion disc (something like that around a protostar) around the black hole. The gas emits highenergy radiation at UV, X-ray and γ-ray wavelengths. 432

Figure 24.5.6 An artist’s impression of an accretion disc around a black hole with jets of emission

Stellar mass black holes are thought to range in mass from 3M to about 15M. A supermassive black hole of 3 to 4 million M is be believed to mark the centre of the Milky Way galaxy, and even larger black holes are thought to be present in some distant galaxies.

astrophysics

Checkpoint 24.5 1 2 3

Outline what happens in a supernova explosion. If iron is the heaviest element made through normal fusion reactions, explain how heavier elements are made. Outline the properties of a neutron star.

24.6 How do we know? How do we know that the picture of stellar evolution presented in this chapter is correct, when almost every step takes far longer than a human lifetime? Our understanding is based on computer modelling of the structure of stars. This takes the physical laws we know and carefully applies them to the situation inside stars. Comparison of our models with reality gives us confidence that our understanding of stellar structure and evolution is largely correct. A key set of evidence lies in the HR diagrams of star clusters. Open (or galactic) star clusters typically contain a few hundred relatively young stars in a loose grouping about 10 light-years across (Figure 24.6.1). In contrast, a globular star cluster usually contains hundreds of thousands of old stars in a sphere about 100 light-years across (Figure 24.6.2). Stars in an open cluster are expected to have formed from a single cloud with the same initial composition. We therefore expect a well-defined ZAMS line to mark the birth of the cluster stars on an HR diagram. Plotting that diagram is made easier since all the stars are at essentially the same distance from us, so the apparent magnitudes of the stars reflect their true relative luminosity.

Figure 24.6.1

The open star cluster M44

Figure 24.6.2

Explain how the age of a globular cluster can be determined from its zero-age main sequence plot for a HR diagram.

The globular star cluster M92

433

24

Birth, life and death

Figure 24.6.3a is an HR diagram for the open cluster M44. The line on the diagram estimates the ZAMS. Most of the stars are clearly still on the main sequence. However, stars at the high-mass end are starting to evolve away from the main sequence, and a few more-evolved supergiants are apparently also present. This is consistent with our expectations that stars move off the main sequence to the right on the HR diagram and massive stars evolve most quickly. Figure 24.6.3b is a rather different HR diagram for the globular cluster M92. The ‘turn-off point’ is much lower down the main sequence, and a well-developed The turn-off point is an indicator of the age of the giant branch is apparent. cluster; the further down the main sequence the turn-off point is, the older the cluster. The M44 cluster is estimated to be about 600 million years old, while the M92 cluster was formed early in the universe—about 13 billion years ago. a

4.0

b

12

6.0

14

8.0 V

V 10.0

Activity Manual, Page 188

0

16

2

18

12.0

PRACTICAL EXPERIENCES Activity 24.1

–2

ZAMS

4

ZAMS 20

6

14.0 22 0.0 0.4 0.8 1.2 1.6 2.0 B–V

0

0.4 0.8 1.2 B–V

1.6

2.0

Figure 24.6.3  Colour index versus apparent magnitude for stars in the (a) M44 cluster and (b) M92 cluster

PHYSICS FEATURE A final thought

S

imon Newcomb (1835–1909) was a CanadianAmerican astronomer and mathematician who, in the late 19th century, said: ‘We are probably nearing the limit of all we can know about astronomy.’ One hundred and twenty years later his comments seem a bit premature, since much of our modern

understanding of the stars was unknown to Newcomb. We should take this as a warning not to think that our impressive knowledge today is complete. There is much we don’t understand and probably even more we don’t know anything about. That’s what makes astrophysics so exciting!

Checkpoint 24.6 1 2

434

Explain how the HR diagrams of star clusters support our understanding of stellar evolution. If black holes emit no light, explain how we can observe them.

PRACTICAL EXPERIENCES

astrophysics

CHAPTER 24 This is a starting point to get you thinking about the mandatory practical experiences outlined in the syllabus. For detailed instructions and advice, use in2 Physics @ HSC Activity Manual.

Activity 24.1: Star clusters on the HR diagram Revise what you learnt in the Module 4 ‘The Cosmic Engine’ of in2 Physics @ Preliminary by briefly reviewing the properties of different types of stars. Discussion questions 1 Explain the differences in properties of open clusters and globular clusters. 2 Describe how these differences are revealed on an HR diagram. 106 105

10R

°

100R Blue giants °

°

Luminosity (L )

1000R

°

Supergiants (I)

104 103 1R

Red giants

Dwarfs (V)

°

Giants (III)

102 101

Present information by plotting Hertzsprung–Russell diagrams for: • nearby or brightest stars • stars in a young open cluster • stars in a globular cluster.

0.1R

°

100 10–1

0.01R

°

10–2 10–3

Red dwarfs

White dwarfs 0.001R

10–4 40 000

°

20 000

10 000

5000

2300

Temperature (K)

Figure 24.7.1

Simulation of HR diagram

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24 • • • • •

• • •

• • •

•

•

•

Chapter summary

Birth, life and death

The space between the stars is filled by the gas and dust of the interstellar medium (ISM). The gas of emission nebulae is energised by hot young stars to emit an emission spectrum. The dust of dark nebulae scatters starlight, reddening or completely blocking our view. The light of reflection nebulae is scattered by dust, especially at blue wavelengths. Giant molecular clouds are regions where the gas is most dense and simple molecules can form. They are the sites of star formation. Gas from the ISM forms stars, is processed inside the stars, and some is then returned to the ISM. Giant molecular clouds can be pushed into gravitational collapse by ‘triggers’ such as supernovae. Protostars warm up as infalling gas releases gravitational potential energy and raises the central temperature until nuclear fusion of hydrogen into helium begins. Surrounding the protostar is an accretion disc and jets of outflowing material. Stars are ‘born’ onto the zero-age main sequence (ZAMS). The location of a star on the ZAMS, the time to get there and its subsequent evolution are almost entirely determined by the mass of the star. The proton–proton chain and the carbon–nitrogen– oxygen cycle both combine four hydrogen nuclei into one helium nucleus, with the release of energy and some other light particles. When a main sequence star runs very low on hydrogen fuel in its core, its core shrinks and is surrounded by a shell of hydrogen fusion, while the outer layers inflate to create a red giant. When the temperature of the core of a red giant is greater than 100 million K, the fusion of helium into

•

• • •

•

•

• •

•

• •

•

carbon begins via the triple alpha process, explosively in some stars as the helium flash. A star moves up the asymptotic giant branch (AGB) when it has a core of carbon and oxygen surrounded by shells of helium and hydrogen fusion. A planetary nebula is puffed off by a low- to mediummass star and returns gas to the ISM. A white dwarf cools slowly for tens of billions of years, supported by electron degeneracy pressure. Stars starting life with more than about 8 M evolve much more quickly and have strong stellar winds, leading to significant mass loss. The core of a red supergiant briefly resembles the layers of an onion in which an iron core is surrounded by shells of silicon and sulfur, oxygen and carbon, and helium and hydrogen fusion. Core collapse and bounce back by the surrounding layers blows a massive star apart in a supernova explosion. The blast wave and ejected gas from a supernova sweep up other gas to form a supernova remnant. Stars with initial masses of between 8M and roughly 25 M will produce a neutron star that packs a mass of up to 3 M into a rapidly spinning, highly magnetised object typically about 10 km across. Pulsars are neutron stars that produce beams of radiation from near their magnetic poles that sweep across the Earth to be seen as very precisely timed pulses. Stars with initial masses greater than roughly 25 M will produce a black hole with mass greater than 3 M. Black holes are ‘black’ because not even light can escape. They are detected because of their effect on the surrounding matter. The ‘turn-off point’ on the main sequence is an indicator of the age of a star cluster.

Review questions Physically Speaking

Match each term to the best definition.

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Term

Definition

Planetary nebula

Pre-main sequence stage of a star

Supernova

Final state of a low mass star

Giant molecular cloud

Relatively gentle ejection of the outer layers of a star

Protostar

Location in space of the ‘birth’ of a star

ZAMS

Possible final state of a high mass star

White dwarf

Location on the HR diagram of the ‘birth’ of a star

Neutron star

Explosive ejection of the outer layers of a star

astrophysics 14 Compare the HR diagrams of an open cluster and a globular cluster.

1 Stars A and B form in the same giant molecular

Analyse information from an HR diagram and use available evidence to determine the characteristics of a star and its evolutionary stage.

15 For each of the stars marked on the HR diagram in Figure 24.7.2, identify the spectral class and luminosity class and describe its future.

2 Why do stars need core temperatures in excess of 10 million K before fusion commences? sits on the main sequence.

Effective temperature (K) 30 000 10 000

–6

on the main sequence.

6 Construct a table of the major nuclear reactions in the life of the Sun. In the table, list: a the temperature needed for the process to occur b the stage in the Sun’s life at which it occurs.

7 Recall the nuclear reactions that occur during different stages of the life of a 5 M star.

8 Explain how a white dwarf can produce light with no nuclear fusion in the core.

9 Outline the properties of a neutron star. 10 Outline the properties of a black hole. Present information by plotting on an HR diagram the pathways of stars of 1, 5 and 10 solar masses during their life cycle.

11 Construct a flow chart that maps the evolution of stars of masses 1, 5 and 10M. Add as much detail as possible.

12 On HR diagrams, construct the complete evolutionary path from birth to death of stars of 1, 5 and 10M.

13 Estimate the time associated with the different stages in the lifetime of stars of 1, 5 and 10M. Construct a timeline to scale of the life of each star.

Absolute magnitude (Mv)

branch, giant molecular cloud, red giant branch, protostar, helium flash, main sequence, T Tauri star

4 000

C

A

5 10

Supergiants (I)

4 10

–4

5 A 1M star goes through several stages in its life,

planetary nebula, white dwarf, asymptotic giant

6 000

–8

4 Explain how the mass of a star determines its lifetime

many of which are listed here but are out of order. Construct a list of these terms in the correct time sequence:

7 000

–10

3 10

D

–2 0

Ma

B

2 4

2 10

Giants (II, III)

in s

equ

10

enc Subgiants (IV) e (V )

1

6

–1 10

8

–2 10

White dwarfs (VII)

10 12

–3 10

E

14 O5 B0 –0.5

Luminosity compared to Sun

3 Explain how the mass of a star determines where it

–4 10

A0

F0 G0 Spectral class

K0

0.0

+0.3 +0.6 Colour index

+0.8

M0 +1.4 +2.0

Figure 24.7.2

Solving problems 16 The Sun is mainly powered by nuclear reactions of the proton–proton cycle. Each reaction contributes 4.2 × 10–12 J to the solar luminosity of 3.84 × 1026 W. a Calculate how many hydrogen atoms fuse to form helium every second to produce this luminosity. b If the mass of the proton is 1.67 × 10–27 kg, calculate how much mass this represents. c Assuming 70% of the Sun’s initial mass of 1.99 × 1030 kg was hydrogen, calculate how many years it will take to convert all the hydrogen to helium. d Contrast this result to the expected lifetime of the Sun. Explain any discrepancy. iew

Q uesti o

n

s

v

cloud. Star A is 1M, star B is 5M. a Predict which star will reach the main sequence first. b Predict which wavelength bands you would best use to observe the early stages of the formation of the stars.

Re

Reviewing

437

6

The review contains questions that address the key concepts developed in this module and will assist you to prepare for the HSC Physics examination. Please note that the questions on the HSC examination that address the option modules are different in structure and format from those for the core modules. Past exam papers can be found on the Board of Studies NSW website.

Multiple choice (1 mark each) 1 Which of the following telescopes must be used above Earth’s atmosphere to observe astronomical objects? A Optical B Radio C X-ray D Infra-red

2

3

4



438

The star Canopus (α Carinae) has an apparent visual magnitude of –0.7 and the star Deneb (α Cygni) has an apparent visual magnitude of +1.3. Which of the following statements about Canopus is true? A 2.0 times fainter than Deneb B 2.0 times brighter than Deneb C 6.3 times fainter than Deneb D 6.3 times brighter than Deneb The spectra of two stars indicate that they are of the same spectral type, but star A has relatively broad absorption lines while star B has very narrow lines. What does this tell us about the two stars? A Star A is more massive than star B. B Star A has a more dense atmosphere than star B. C Star A is older than star B. D Star A is moving away from us faster than star B. Consider two single stars with different masses, both on the zero-age main sequence. What properties would you expect for the more massive star, relative to the less massive star? A Lower luminosity and shorter lifetime B Lower luminosity and longer lifetime C Higher luminosity and longer lifetime D Higher luminosity and shorter lifetime

5

What properties would you expect to be approximately common between stars within an open star cluster? A Age B Luminosity C Age and initial chemical composition D Age, initial chemical composition and luminosity

Short response 6 Describe the problems associated with viewing astronomical objects through ground-based optical telescopes.  (2 marks)

7 Explain why the limit of accurate parallax measurement from Earth is different to that from space.  (2 marks)

8 The apparent magnitude of star A is 2.4 while that of star B is 0.3. a Calculate the brightness ratio of star A to star B. b Explain which star is brighter.  (2 marks)

9 Outline how to calculate the distance to a Cepheid variable star.  (2 marks)

10 Describe the evolution of a 1 M star. Make sure to include the various stages reached and nuclear reactions taking place.  (2 marks)

11 Explain why the HR diagram of a globular cluster is different from the HR diagram of an open cluster. (2 marks)

astrophysics

Extended response 12 Outline an experiment you undertook in class to observe emission, absorption and continuous spectra. Describe the differences you observed between the spectra, and identify an object that produces each type of spectra.  (3 marks)

13 For each of the stars marked on the HR diagram in Figure 24.8.1, list the properties and probable stage of evolution.  (3 marks) Effective temperature (K) 30 000 10 000

7 000

6 000

4 000

–10

5

10

–8 A

4

10

Absolute magnitude (Mv)

–4

3

10

–2

C

0

Ma

in s

2

equ

4

2

10

E

10

enc

e

D

6

1 –1

10

8

Luminosity compared to Sun

–6

–2

10

10

–3

12

B

14

F

10

–4

10 O5 B0

–0.5

A0

F0 G0 Spectral class

K0

M0

0.0

+0.3 +0.6 Colour index

+0.8

+1.4

+2.0

Figure 24.8.1

439

7 Context

Figure 25.0.1

440

It is important to fine tune your skills in order to get the most out of your experiment.

skills

In the preliminary text you were introduced to many skills that are needed to get through Physics. In the HSC year it is important to further develop these skills, and to make sure you know how to use the skills without thinking twice. You will now be able to fine tune and add to what you have learnt. One of the most important skills for a physicist is being unambiguous, because the mathematical nature of physics demands that everything be defined strictly and clearly. It’s also important as a scientist to have the skill of communicating clearly the results of research to other scientists. The HSC year is about clearly getting across what you have learnt. It is important not to be ambiguous, so you must learn how to interpret questions and construct your answers. Throughout this module we will focus on the skills you need to make sure what you have learnt is conveyed clearly.

Inquiry Activity The Bungee jumper Let’s say you are responsible for getting the bungee cord length (lo) correct for the next jumper. The elasticity (k) of the cord is always known. For simplicity, assume that the energy conversion is close to 100% efficient. You are given the chance to take any measurements of the jumper. How could you calculate the length of the cord you need to attach to the jumper if the bridge stands 100 m above the river below and the jumper is to stop between 10 and 100 cm from the water for maximum thrill. Not too close for safety, but not too far for thrill.

Figure 25.0.2 

A bungee jumper

Hints A sketch of the situation (Figure 25.0.3) has been given to you to help you along the way. 1 List the types of energy that is available at the top of the jump 1 22 EPE== kx kx  (this will include gravitational U = mgh, elastic potential energy EPE 2 1 22 = kx and kinetic energy EEPE = mv  ) and the values of things that are known. K 2 2 Using the law of conservation of energy, write an equation that includes all the types of energies you have listed. 3 hi is the maximum height that the cord can stretch to. It is made of the original length of the cord plus the stretched length, i.e. hi = lo + x. Figure 25.0.3  A sketch of the situation 4 Rearrange the equation to solve for the bungee cord length (lo). Think about how to do this and then have a go at doing it in a scaled-down experiment. What will you need to know, or take measurements of? How accurate will you need to be? Set up your experiment to test it.

Questions 1 List any assumptions that you have made in your calculations. 2 Determine how it is possible for you to test your prediction accurately. 3 List problems involved with your experiment and how it is possible to reduce them. 441

25

Skills stage 2 Skills introduction

prefix, systematic error, random error, accuracy, precision

Skills are a very important component of the Year 12 syllabus. In Year 11 you started learning them and in Year 12 you will develop them further. The skills that are important to develop include planning and carrying out investigations, researching and communicating what you have found, and problem solving. Most of these skills are developed while you learn the material in the modules, but there are a few specific concepts that will be discussed separately in this chapter, to aid with your development.

25.1 Metric prefixes In the preliminary course, you were taught to express numbers using scientific notation. This is a way of representing a large or small number in an unambiguous, compact way. To add to this, we will introduce the idea of prefixes. Although you may not know the term prefix, you are probably very familiar with some of these in everyday life, for example kilogram and millimetre and centimetre. They are usually printed on the back of exercise books. Prefixes help reduce the written size of the unit of measurement. A list of common prefixes (and some less common ones) is given in Table 25.1.1. For example, the basic SI unit of distance is the metre (m). A centimetre (cm) is then 1/100th of a metre. The prefixes that are commonly used in your course have an asterisk placed next to them. You are expected to know the values of these, as they will be used in questions and their value will not be given in any exam you undertake, so it is in your best interests to become familiar with them.

442

SKILLS Sometimes you will need to convert from one prefix to another. The easiest way to carry out the conversions is to follow the steps set out in the worked example below.

Worked example Question Convert 5.49 mm to µm. Step 1: Express the number in scientific notation. 5.49 × 10–3 m

Step 2: Look at the prefix that you need to convert the number to. Express the prefix in scientific notation.

micro = 10–6

Step 3: Look at the difference between the two. –3

Number

Prefix

10–1

deci-

d

10 10–3 10–6 10–9

centimillimicronano-

c m µ n

10–12

pico-

p

10–15

femto-

f

10

–18

atto-

a

10

–21

zepto-

z

10–24

yocto-

y

1

–2

Solution

Table 25.1.1  Common metric prefixes

–6

Symbol



10  m to 10  m

10

deka-

da



To change the exponent to 10–6, the decimal place must be moved 3 places to the right.

102

hecto-

h

3

10 106

kilomega-

k M

109

–3



5.490 × 10



5490 × 10–6 = 5490 µm

Checkpoint 25.1 1

2 3

Express the following masses using prefixes in terms of the specified unit. a 0.45 g in mg b 345 g in kg c 5 600 000 g in Mg Convert 4.5 km to Mm. (Note the difference between Mm and mm.) Convert 0.54 nm to km.

giga-

G

12

10

tera-

T

1015

peta-

P

10

exa-

E

1021

zeta-

Z

yotta-

Y

18

24

10

* * * *

* *

25.2 Numerical calculations In terms of exam performance, exam technique can be just as important as understanding the physics that you have been studying. If you cannot decipher the question and communicate your answer clearly, you cannot show that you understand physics. Chapter 26 ‘Revisiting the BOS key terms’ will go through the technique of answering questions that require a written response. Here we will look at numerical answers. When attempting any question involving numerical calculations, follow this series of steps: 1 Highlight numerical values within the question. 2 Identify the unknown that must be determined. 3 Write the relevant equation. In each of steps 1–3, write the item down at the top of your answer, then: 4 Show all working. 5 Write your answer with units. 443

25

Skills stage 2

Worked example Question A driver of a car travelling at 16.0 m s–1 slams on the brakes when he sees a ball on the road in front of him. The ball is 25 m away at the time the brakes are applied. Will the car stop in time, given that the deceleration is –2.5 m s–2?

SOlution Step 1: The data have been highlighted in the question. Step 2: List them below.

Assume the direction of initial velocity is positive.



u = +16.0 m s–1



v=0



a = –2.5 m s–2



s = +25 m



The unknown is actual displacement s taken to stop.



Even though our list shows that there is a displacement, this is not the actual displacement of the car, but the initial displacement of the ball from the car.



s=?

Step 3: v 2 = u 2 + 2as Step 4:

0 = (16.0)2 + 2(–2.5)s



s = 256/5



s = 51.2 m

Step 5: This is more than 25 m, therefore the car will not stop in time.

Recall from the preliminary course that the answer to a calculation needs to be expressed correctly in order to reflect the accuracy of the measurements. First you must look at the calculation itself. The calculation should be completed with all the available digits in each value. When multiplication or division is used, the answer should be expressed to the least number of significant figures used in the values in the question. With addition and subtraction, the answer is expressed to the lowest number of decimal places within the question. Remember that a calculator can ‘invent’ accuracy. Just because the calculator gives you an answer with 10 digits doesn’t mean that the answer has become that accurate. Refer back to the question and reduce your answer accordingly.

Worked example Question Correctly express the answer from the worked example above.

Solution The equation has multiplication and division in it, therefore we will need to look at the answer in terms of significant figures. The least number of significant figures in the question is 2. Therefore the answer to the calculation is 51 m. 444

SKILLS

Checkpoint 25.2 1 2

An aeroplane has touched down on the runway at 290 km h–1. The braking ability allows it to decelerate at a rate of –11 m s–2. The runway is 3.9 km long. Will the aeroplane stop in time? A satellite needs to be launched into a circular near-Earth orbit. At what height above the Earth’s surface would the satellite need to be placed in order to maintain a period of 90 minutes?

25.3 Sourcing experimental errors There are errors involved in all measurements. The magnitude of the error and how many errors are present can be defined by the equipment you choose to use in an experiment as well as the method you use. So how can you minimise the effect of errors and make sure your experiment is both valid and reliable? A valid experiment is one that actually tests what the aim has stated. This sounds obvious but it can be easy to stray from the task at hand. A reliable experiment is one that can be repeated each time with the same results, no matter who is carrying it out. How do you achieve this? • Make clear statements in your method. • Express clearly the need for repetition and averages. • Use the most accurate equipment available. It is also important to understand the difference between systematic and random uncertainities (often simply called ‘errors’). Systematic error is something within the experiment that causes the readings to be consistently high or low; they are always the same magnitude and sign no matter how often the experiment is repeated. This includes things such as errors in calibration, or a repetitive mistake in experimental technique. Systematic error can be minimised by careful experimental techniques and proper understanding of the experimental apparatus. Random error is the unpredictable fluctuation of results, due to rapidly changing external influences, limitations of the readability or sloppiness of an instrument or method. If a measurement is taken repeatedly, it would be noted that the values would be readings scattered around a number. Although random error cannot be eliminated, it can be reduced by good experimental practice and by repeating the measurements many times and taking an average. Two words closely related to these ideas are accuracy and precision. An accurate measurement is one with small systematic error, while a precise measurement is one with small random error (see Figure 25.3.1). It is not always clear if an uncertainty (error) is systematic or random. Some causes could be considered partly systematic and random. Consider the effect on the result and how best it fits in the definitions before applying the rule blindly. Remember to record your values to the accuracy of your instruments. If you have equipment that is accurate to 1/1000th of a unit, record it in its entirety. Remember to place any measured zeros at the end if appropriate (e.g. 7.600 m). If you don’t, you have lost that accuracy and use of the equipment is pointless. Then perform any calculations to the full number of figures available in the data and only reduce the number of figures expressed in the answer at the end.

precise but inaccurate

accurate but imprecise

inaccurate and imprecise

accurate and precise

Figure 25.3.1

Accuracy and precision

445

25

Skills stage 2

You may be asked in an exam to express how you made sure that your experiment was valid and reliable, so make sure you do this when you are doing the experiment (refer to Chapter 17 ‘Physics skills’ in in2 Physics @ Preliminary).

Checkpoint 25.3 1

You are required to carry out an experiment that determines the reaction time of your lab partner. Hold a ruler just above your partner’s fingers (see Figure 25.3.2). Your partner must catch the ruler after you drop it without warning. Record how far the ruler has dropped by the time it is caught. Use your knowledge of physics to determine the reaction time of your lab partner.



Repeat the process five times at an interval of 30 s. Does your partner become better?



After a day or so, do a similar experiment, this time using a data-logger sensor. Does your partner become better? 2 3 4

Determine if the experiment is reliable. Is the experiment valid? How can the experiment be made more reliable and valid?

Figure 25.3.2

Your partner must catch the ruler after you drop it without warning.

25.4 Presenting research for an exam During the course of your study you were required to gather a lot of data from primary and secondary sources. It is important to know how much of this you need in order to accurately answer questions based on this material. In the activity manual you have found templates to help guide your search so that you had information that covered the content to the right depth. The discussion questions in the student book also help guide you in reaching the depth that is required. If you can answer these from your research you are on the right track. Once you have extracted the information, practise questions on the topic from past exam papers so you can get a feel for what can be asked. The biggest mistake you can make is storytelling. Remember to answer the question, don’t just regurgitate everything that you have researched. How to answer questions as they are asked will be looked at in more detail in Chapter 26 ‘Revisiting the BOS key terms’.

Checkpoint 25.4 During your study of Module 1 ‘Space’, you were required to research a scientist in relation to their contribution to space exploration. Use this information to answer the following question: How has your scientist influenced those who have come after him/her?

446

SKILLS

25.5 Australian scientist One skill that you need to undertake in Physics is the study of an Australian scientist. In this student book there is information about Australian scientists. Pick one that interests you to study in detail. Things that you are looking to know about will include their name, the area of physics in which their current research is being carried out, what the research is, where they are currently working and what area of your study relates to this scientist’s work.

Checkpoint 25.5 For your chosen scientist, state the research carried out and explain how the research relates to the Physics course.

25.6 Linearising a formula When looking for a relationship between two variables it is usually simpler to deal with a linear relationship. What happens if the relationship between two variables in a formula is not a straight line? It is not necessarily a problem, as sometimes we can rearrange the formula in order to graph the two variables so that they relate to each other in a linear way. Confused? Let’s look at the steps to follow with an example.

Worked example Question A student is looking at how the velocity v of a ball relates to its displacement s as it falls through the air when dropped. To do this, the student drops a ball and records the start velocity, the displacement it goes through and the end velocity. What does the student need to graph in order to get a linear relationship?

Solution The formula that we can use to relate these two variables is: v 2 = u 2 + 2as From this equation, we can see that there is not a linear relationship between v and s, but there is a linear relationship between v 2 and s. Graphing v 2 against s would produce a straight line.

Although it is a good skill to know, linearising is less important today when Excel (and similar applications) can fit other functions such as power laws directly (a power law with power 2 in this example). The fit may not be identical because the least squares fitting will be minimising slightly different values.

Checkpoint 25.6 1 2

L . g What variables would need to be placed on the axes of a graph if a straight line was to result?

Determine the linear relationship between length and period from the formula T = 2π

447

26

Revisiting the BOS key terms Verbs in action In the Preliminary course, you were introduced to the idea of grouping the BOS key terms in order to help you answer questions. Below is a table with the BOS key terms in the allocated groups. From working through the examples in the Preliminary course, you should be very familiar with how to formalise your answers. In the HSC year you will not have as much of an opportunity to fill in a table and analyse your answer as completely as you have up to now. Because you are now familiar with the structure of the answers, we will introduce the short examination technique to help you. If you are still not confident with structuring your answers, continue to work as outlined in in2 Physics @ Preliminary.

Table 26.1.1  BOS key terms in their allocated groups Group 1: Knowledge

Group 2: Comprehension

Group 3: Application

Define

Account

Apply

Extract

Clarify

Calculate

Identify

Compare

Classify

Outline

Contrast

Demonstrate

Recall

Describe

Examine

Recount

Discuss Distinguish Extrapolate Interpret Predict

Group 4: Analysis

Group 5: Synthesis

Group 6: Evaluation

Analyse

Construct

Appreciate

Explain

Propose

Assess

Investigate

Summarise

Deduce

Synthesise

Evaluate Justify Recommend

448

SKILLS

26.1 Steps to answering questions 1 Highlight the verb. 2 Highlight the main topics that make the question. 3 Recall the group the verb falls into and determine the depth of information needed. – Group 1: List information. – Group 2: Interpret details of topic. – Group 3: Apply knowledge. – Group 4: Look at relationships and organisation. – Group 5: Put together information to decipher. – Group 6: Make judgements and draw conclusions based on fact. 4 Record in point form in the margin what needs to be stated—include diagrams. This should be a brief outline to organise your ideas so the order is correct and you don’t go off topic as you progress in your writing. 5 Elaborate on each point in the space provided—refer to your diagrams. The best way to make sure you can write a good answer is to practise writing good answers. What follows is a set of worked examples for each of the verb groups.

Group 1: Knowledge

Group 2: Comprehension

Worked example 1

Worked example 2

Question

Question

Define orbital velocity with reference to satellites.

During an experiment carried out in class, students had two currentcarrying wires, one placed near an electronic balance, the other on it. They noticed that the readings fluctuated as they changed the current moving through one of the wires. Discuss how these readings changed in relation to the current.

Notes Define falls into group 1, therefore this requires you to list information. Orbital velocity: tangential velocity of satellite when moving in an orbit around the Earth The speed of the satellite is given by the following formula: Gm r This shows that the velocity is affected by the mass of the central object—the Earth—and the radius from the centre of the object. v=

Notes Discuss falls into group 2, therefore this requires you to list information and interpret details of topic. – A current-carrying wire has a magnetic field around it. – Two current-carrying wires near each other will attract or repel, depending on the direction of the current.

Solution

– One wire is on the balance; balance will register the weight of it.

Orbital velocity of an object is the tangential velocity of an object that is in orbital motion around another object.

– If the wires are repelling, the reading also registers the extra force.

If the orbit is circular, the satellite must have a velocity that is determined by the formula: Gm v= r The speed of the satellite will be determined by the radius from the centre of the Earth as well as the mass of the Earth. The velocity will vary if the orbit is eliptical.

– If the size of the current or the direction of the current is changed, the reading will change by the amount the force is changing. – Force is determined by: II F =k 12 l d 449

26

Revisiting the BOS key terms

Solution Two current-carrying wires will either attract or repel. If the current is travelling in the same direction in both wires, the wires will attract. If the current is in different directions, then they will repel. The size of the force is determined by: II F =k 12 l d As one wire is resting on the balance, the balance is registering both the weight of the wire and the force created by the interaction of the wires.

r3 2

86 400

=

r=

6.67 × 10−11 × 6.0 × 1024 4 π2 3

6.67 × 10−11 × 6.0 × 1024 × 86 4002



= 4.2 × 107 m



≈ 42 000 km

4 π2

Group 4: Analysis Worked example 4

Force is proportional to the amount of current. If there is a repulsive force originally, then the registered reading will decrease as the current is decreased and increase as the current is increased. If there is an attractive force originally, then the registered reading will decrease as the current is increased and increase as the current is decreased.

Notes

If the direction of the current is reversed, then the force will change from attractive to repulsive or vice versa, with a change in the readings as mentioned above.

Explain falls into group 4, therefore this requires you to list information and interpret details of topic and apply knowledge and consider relationships and organisation.

Worked example 3

– English versus German research.

Question Calculate the radius of orbit for a geostationary satellite.

Notes Calculate falls into group 3, therefore this requires you to list information and interpret details of topic and apply knowledge. – Geostationary = 24 h, orbit = T – It is in orbit around the Earth, therefore M = 6.0 × 1024 kg. – Equation that applies is T

2

=

Solution r3 T2

=

GM 4 π2

r=? M = 6.0 × 1024 kg T = 24 × 60 × 60 = 86 400 s G = 6.67 × 10–11

450

Explain how the debate over the apparent inconsistency in behaviour of cathode rays was finalised.

– Inconsistent behaviour refers to cathode rays displaying both wave and particle properties.

Group 3: Application

r3

Question

GM 4 π2

– German findings supported wave theory as electric field did not cause deflection. – Hertz showed that cathode rays could pass through thin layers of metal such as gold and silver. – Resolved when JJ Thomson could deflect the ray with an electric field.

Solution Both Hertz and Crookes observed properties of the cathode rays that showed properties of both waves and particles. The main reason for Hertz believing that the ray was a wave was from two experiments: one showed that when an electric field was applied there was no evidence of the ray moving, and the second showed that the rays could pass through thin metal. This resembled the behaviour of light, which was known to be a wave. It was not until JJ Thomson performed the same experiment again that the debate was put to rest. Thomson produced a nearperfect vacuum, eliminating any trace atoms within the tube, and coated the end of the tube with a fluorescent screen. It was noted that when the electric field was applied the ray was bent in the direction in which a negative particle would move.

SKILLS

Group 5: Synthesis

Group 6: Evaluation

Worked example 5

Worked example 6

Question

Question

Propose an experimental method to be able to determine Planck’s constant.

Justify the extensive safety precautions that are evident on the Space Shuttle in order to protect the astronauts.

Notes:

Notes

Propose falls into group 5, therefore this requires you to list information and interpret details of topic and apply knowledge and consider relationships and organisation and put together information to decipher.

Justify falls into group 6, therefore this requires you to list information and interpret details of topic and apply knowledge and consider relationships and organisation and put together information to decipher and make judgement and draw conclusion based on fact.

– E = hf – W = qV so measuring V will allow you to get a measure of E. – f can be varied by using coloured filters. – Set up apparatus as shown. – Measure the stopping voltage for each frequency. – Produce a graph of V against f. – Convert V to E by multiplying by 1.6 × 10–19. – Determine the gradient of the graph. This is equal to h.

Solution • Set up the equipment as shown. (Draw a diagram like Figure 9.3.1.) • Measure the stopping voltage for each frequency of light. (Stopping voltage is the voltage to stop current flowing in the circuit.) • Produce a graph of E against f. To determine E multiply V by 1.6 × 10–19. (Draw a graph like that shown below.) • The gradient of the graph is h. EK

EK = hf – φ slope = h

–φ

threshold frequency

Frequency

– Heat shields: heat energy not allowed to travel into the Shuttle – Lying down on take off: improves tolerance of g-forces – Parachutes: remaining kinetic energy is removed on landing – Entry angle – Nose cone: rounded to change kinetic to heat energy efficiently

Solution The Space Shuttle was the first reusable space transport. It was designed to fly many missions and be reused quickly while protecting those inside. On take off, the g-forces that are applied to the astronauts are massive. As they need to be able to withstand the g-force, the astronauts lie down during take off, so the blood does not rush away from their vital organs and they remain conscious throughout the launch. Most of the precautions are taken for the re-entry. On the approach the astronauts must aim the shuttle to come in at an angle of between 1° and 2°, in order to avoid burning up in the atmosphere or skipping off the atmosphere. The burning is due to friction. A small amount of this has to occur, and so heat shields are used to minimise the effects on the astronauts. The underside of the surface of the Shuttle is covered with tiles that can be superheated and dissipate the energy quickly. This avoids extensive heat inside and converts the massive amounts of KE to heat quickly, slowing the Shuttle. The nose cone of the Shuttle is fairly blunt. This also aids in the conversion of KE to heat and therefore helps to slow the shuttle. A pointed nose would be more likely to melt. There is a limit to how much energy can be converted to heat safely, so the final means of slowing down is the use of parachutes. These parachutes are deployed on landing and allow the Shuttle to slow and stop safely. The main purpose of the Shuttle is to transport humans safely. These precautions must be taken in order for the occupants to survive.

451

Numerical answers

Numerical answers These are selected numerical answers only. A complete set of answers can be found in the Teacher Resource.

Module 1 Space Checkpoints:  1.1 1 ah = 0, av = 9.8 m s–2 down 2 θ = 45°, θ = 90° 3 drag 4 both at the same time. 6 see Table 1.1.1 1.2  1 F = Gm1m2)/d 2 3 1/d 2 1.3  2 EP = (–Gm1m2)/r 3 0 Review questions:  3 2.5 s 5 a 2Fi b 4Fi c Fi /4 8 11.2 km s–1 10 a 1.38 s b 4.37 m above ground c 18.0 m (right) d 14.3 m s–1, 25º below horizon 11 a 25.3 m s–1, 0.883 s b 17.9 m s–1, 1.25 s 13 5.35 m s–1 14 260.5 N, 89.99º from x-axis 15 7.0 m 17 5.30 × 108 J 18 10 300 m s–1 19 a 42 100 m s–1

Chapter 2 Checkpoints:  2.1  1 Sergey Korolyov, Wernher von Braun 6 a g-force = 1 b g-force = 0 c g-force = 0 2.2  3 ac = v 2/R 8 True 9 a = r 10 Mercury 2.3  1 circle, ellipse, hyperbola and parabola 2 hyperbola 2.4  2 hyperbolic 3 they are equal 4 ∆vmax = 2VP 2.5  2 false Review questions:  15 –6.25 × 107 J, 2.12 × 1012 m 16 a 10 690 kg s–1 b 5.2 m s–2, 12.1 m s–2 17 a w = mg b N = 0 c ac = g d v > 7.00 m s–1 e 0 19 32.7 h–1 20 16.8 days 22 a 4.22 × 107 m b 2.66 × 107 m 23

KE

Geostationary 9.47 × 109 J

Review questions:  16 1380 Nm 17 5.4 × 10–4 Nm 18 b 2.7 c nBA cos θ (= nBA when 0 = 0º) d 30 T

Chapter 7

Chapter 1

Type of orbit

Chapter 6

− GmM 2a

Two-body PE

ME = KE + PE

–1.887 × 1010 J

–9.40 × 109 J –9.44 × 109 J

Checkpoints:  7.2  4 1200 V Review questions:  22 3600 V 23 8% 24

Coils in primary

Primary voltage (V)

Coils in secondary

Secondary voltage (V)

100 320 50 000 25

6 240 393

200 66.7 30 500

12 50 240

Current in primary coil (A)

Voltage in primary coil (V)

Current in secondary coil (A)

Voltage in secondary coil (V)

5 0.1 0.5

6 240 200

0.125 2 0.1

240 12 1000

Extended questions:  10 a up b F = BIl sin θ = 5 × 10–4 N 11 b 2 × 10–4 N c 2.5 × 10–3 Nm

Module 3 From Ideas to Implementation Chapter 8 Checkpoints:  8.2 2 0.86 N C–1 8.3

2 9.6 × 10–19 N

–1

24 hyperbolic, no 26 a 2.22 × 1011 J b 2.45 × 1011 J

Chapter 3

Chapter 9

Review questions:  1 D 13 1.73 m s–2 16 a 80 kg b 121 kg 17 0.781c, 5.60 years 18 1.64 × 10–13 J 19 6.91 MeV 20 6.3 × 1013 J 21 v = 0.994c

Module 1 review Multiple choice:  1 B 2 D 3 A 4 A 5 C

Review questions:  19 3.0 × 10–19 J 20 b 0.25 × 1015 Hz c 1.66 × 10–19 J f 6.4 × 10–34 J s

Chapter 10 Review questions:  14 a 3.06 × 10–19 J b 3.27 × 1015 15 1.09 × 10–6 m 16 5.94 × 10–20 J = 0.37 eV

Short response:  7 –2.36 × 1010 J 8 0.9679c 10 greater by 1.08 × 10–8 kg ≈ 11 µg at equator

Chapter 11

Module 2 Motors and Generators

Module 3 review

Chapter 4 Checkpoints:  4.3 1 at 90° to each other 2 F ∝ B, F ∝ I, F ∝ l, F ∝ sin θ Review questions:  13 4.3 Ω 17 3 N 18 b gradient = B/F = 2.47 c 1/gradient = 0.4, Il  = 0.4, 100%

Chapter 5 There are no numerical answers for this chapter.

452

Turns ratio

Multiple choice:  1 A 2 D 3 A 4 C 5 B

Review questions:  12 2 × 10  N C 13 0.3 J 14 a 100q J b 1.0 × 103 N c q × 103 N d 1.6 × 10–16 N 15 EK = q × E × s 16 1.76 × 1014 m s–2 17 1.88 × 106 m s–1 18 5.1 × 10–14 N 19 a 0.22 m b 3.4 × 10–12 N 20 9.09 × 10–31 kg.

–1.49 × 1010 J –1.50 × 1010 J

Step-up Step-down Step-down

Module 2 review

3

Circular semi1.504 × 1010 J –2.99 × 1010 J synchronous

Step-up or step-down

Review questions:  19 2.53 × 10–6 m 20 a 75 K, 90 K Multiple choice:  1 C 2 C 3 A 4 B 5 B Extended response:  12 a 4.5 × 1014 Hz b 3.0 × 10–19 J

0.025 20 0.2

Numerical answers

Module 4 Quanta to Quarks Chapter 12 Checkpoints:  12.4  5 6.6 × 10–7 m 6 Balmer series Review questions:  13 a 9.61 × 10–19 J b 1.45 × 1015 Hz c 2.07 × 10–7 m 14 a –0.85 eV b 12.75 eV c 3.08 × 1015 Hz d 9.73 × 10–8 m e Lyman series 15 a 1.22 × 10–7 m c n = 2 and n = 1 d 1.22 × 10–7 m or 122 nm e Lyman series 17 a n = 1 to n = 5 b n = 5 to n = 1 c n = 3, n = 4, n = 5 to n = 2 d n = 5 to n = 4 18 b 9.38 × 10–8 m or 94 nm c ni = 4 and nf = 2 d 4.86 × 10–7 m or 486 nm 19 a ni = 6 c Balmer series 20 c 2.09 × 10–18 J d 9.50 × 10–8 m

Chapter 13 Checkpoints:  13.2  2 2.4 × 10–9 m 3 6.56 × 10–25 kg m s–1 13.5  2 1.93 × 10–10 m Review questions:  12 b 2.4 × 10–11 m 13 a 7.3 × 10–4 m s–1 14 a i 1.2 × 10–9 m ii 6.6 × 10–13 m 15 2.87 × 105 m s–1 16 a i 4 × 10–36 m ii 3 × 10–34 m iii 3 × 10–38 m 17 a + infinity b no 18 a 1.1 m

Chapter 14 Checkpoints:  14.2  1 a factor of ~ 1035 greater 14.6  2 5.2 × 10–27 kg 3 280 MeV Review questions:  14 a Fe = 231 N b Fg = 5.6 × 10–34 N 15 0.034348 amu 16 a 28.4 MeV b 0.03049 amu c 4.0026 amu 17 c 8.9 MeV per nucleon 18 a 102 (yttrium) b Uranium 7.6 MeV/nucleon, 39 Y iodine 8.5 MeV/nucleon, yttrium 8.6 MeV/nucleon 1 0n

65 66 23 + 29 19Cu a a→= 29 b Na + 11Cu

22 10 Ne

0 20 +1 17.406 → e + v MeV

Chapter 15

Chapter 19 Review questions: 17 a U-238 → Th-234: α Th-234 → Pa-234: β Pa-234 → U-234: β U-234 → Th-230: α Th-230 → Ra-226: α Ra-226 → Rn-222: α Rn-222 → Po-218: α Po-218 → Pb-214: α, Po-218 → At-218: β At-218 → Bi-214: α Pb-214 → Bi-214: β Bi-214 → Tl-210: α, Bi-214 → Po-214: β TI-210 → Pb-210: β Po-214 → Pb-210: α Pb-210 → Bi-210: β Bi-210 → Po-210: β Po-210 → Pb-206: α 18 a 3 years

Chapter 20 Review questions:  18 6 19 1.41 × 10–26 J 20 2.8 × 1010 Hz

Module 5 review Multiple choice:  1 C 2 B 3 B 4 B 5 B

Module 6 Astrophysics Chapter 21 Checkpoints:  21.3  2 40× 21.5 2 ≈ 0.1 arc second Review questions:  13 ¼ 14 ~0.2 ly 17 ~90 ly

Chapter 22

Multiple choice:  1 C 2 C 3 B 4 D 5 A

Review questions:  13 ≈ 3.0857 × 1019 m 14 59 pc, 190 ly + 6.6 pc 15 59 pc, 34.5 − 5.1 16 a 656.3 nm, 486.2 nm, 434.1 nm, 410.2 nm, 397.0 nm 17 ~5 × 1010 20 d ≈ 200 pc

Short response:  1 b 1.03 × 10–7 m or 103 nm

Chapter 23

There are no numerical answers for this chapter.

Module 4 review

Module 5 Medical Physics Chapter 16 Checkpoints:  16.4  3 40% Review questions:  22 8.16 × 10 6 rayl 23 1039.7 kg m–3 24 0.034%

Chapter 17 There are no numerical answers for this chapter.

Chapter 18 Review questions:  9 26.7° 10 a 43.98° b 67.5° 11 n2 = 1.33, water

Review questions:  13 3 AU from the centre of the red giant 15 a 6.16 × 1030 kg (~3MSun) b m1 = 4.14 × 1030 kg ≈ 2.1 MSun, m2 = 2.02 × 1030 kg ≈ 1.0 MSun 16 T = 1.5 days; amplitude (main eclipse) = 0.6 magnitudes, (secondary eclipse) = 0.5 magnitudes 17 360 pc or ~1170 ly 18 d ≈ 220 pc or ~720 ly

Chapter 24 Review questions:  16 a 3.66 × 1038 b 6.11 × 1011 kg c 72 billion years d ~6 times the expected lifetime of the Sun

Module 6 review Multiple choice:  1 C 2 D 3 B 4 D 5 C Short response:  8 a 1.4 b star B

Module 7 Skills Chapter 25 Checkpoints:  25.1 1 a 450 mg (milligrams) b 0.345 kg (kilograms) c 5.6 Mg (megagrams) 2 0.0045 Mm 3 0.00000000000054 km 25.2  2 300 km above surface

 T 2 25.6  1  g = L 2 s and t 2  2π 

Chapter 26

There are no numerical answers for this chapter.

453

Numerical answers

Glossary

2D real-time scan a scan in which a convex array transducer launches a rapid series of beams along different directions (sector scan) to build up a fan-like two-dimensional image of B-mode scans in real time 3D ultrasound a rapid series of sector scans are performed (oriented over a range of angles) to yield a three-dimensional image

alternating current (AC) electric current that changes direction periodically A-mode scan an ultrasound scan in which the wavefront is launched in a single direction, yielding a one-dimensional oscilloscope plot of echo intensity versus depth data angular resolution resolution that describes the ability to resolve details separated by very small angles in the sky

4D ultrasound 3D ultrasound images are acquired so rapidly that a series of three-dimensional images can be displayed as a movie

anode in physics, the positive electrode. (Caution: The definition in chemistry is more complicated.)

ablation excess heat is carried away from a heat shield by material in the heat shield melting or vaporising

anode glow small luminous region in a discharge tube adjacent to the anode

absolute magnitude the apparent brightness of an astronomical object (usually a star) as it would appear in our sky if it were moved to a standard distance of 10 pc (32.6 ly), measured in magnitudes and often represented as M

anti-nodes the points on a standing wave where the amplitude is maximum

absorption line spectrum a spectrum with dark emission lines at wavelengths characteristic of the elements present in the light source, usually produced in astrophysical situations by a source of a continuous spectrum source viewed through a cool, low-density gas; most stars produce an absorption line spectrum acceptor energy level an extra energy level is created in the energy gap near the valence band of a semiconductor as a result of the introduction of acceptor impurities acceptor impurities produce a deficiency of electrons in the crystal lattice of a semiconductor

antiparallel when one vector points in the opposite direction to another antiparticle a particle with ‘opposite’ properties to another particle; for example, a positron is the antiparticle of an electron aphelion in a satellite’s orbit around the Sun, the farthest position from the Sun apoapsis in a two-body system, the farthest position of a satellite from the central mass apogee in a satellite’s orbit around the Sun, the farthest position from the Earth

accretion disc a disc of material around a compact object (e.g. a protostar or a black hole) formed from infalling gas

apparent magnitude the apparent brightness of an astronomical object as it appears in our sky, measured in magnitudes and often represented as m

accuracy an accurate measurement is one with small systematic error

armature the part of a motor or generator that contains the main current-carrying coils or windings

acoustic impedance is a measure of how easily a medium oscillates in response to a sound. It is given by the formula Z = ρv where ρ is density of the medium and v is the velocity of sound in the medium

Aston dark space the non-luminous region nearest the cathode in a discharge tube

active optics a system to detect and overcome the slowly changing effects of gravity and temperature that distort the mirror or structure of a telescope and degrade the image quality

astrometry the measurement of the positions and motions of astronomical objects

adaptive optics a system to detect and overcome the rapidly changing effects of turbulence in the Earth’s atmosphere (seeing) that are apparent as motion and blurring of the image produced by a telescope aether drag the hypothesised tendency for aether to be trapped and dragged along by mountains and valleys on the surface of the rotating, orbiting Earth aether wind the movement of the hypothetical aether relative to the surface of the rotating, orbiting Earth; also called ‘aether drift’. Airy disc the diffraction pattern caused by light passing through a circular aperture such as the objective lens or mirror of a telescope alpha decay a form of radioactive decay in which an atomic nucleus emits an alpha particle alpha particle a highly energetic helium nucleus (two protons plus two neutrons) produced by radioactive decay 454

astrometric binary stellar system that reveals its binary nature by the wobbling of its path across the sky

astronomical unit approximately the average distance between the Earth and the Sun: 1 AU ≈ 1.4960 × 1011 m asymptotic giant branch the portion of the evolutionary path of a red giant in which the star has a core of carbon and oxygen surrounded by shells of helium and hydrogen fusion atomic mass is the mass of an atom made up from the total mass of the protons, neutrons and electrons atomic mass unit (amu) it is equal to the mass of one twelfth of an atom of Carbon-12, which is approximately the mass of a proton or neutron atomic number the number of protons in a nucleus atomic pile the name given to a type of nuclear fission reactor that was constructed using a pile of graphite bricks back emf an emf that is produced in accordance with Faraday’s law within the coils of a motor and opposes supply emf

Glossary

ballistic trajectory the trajectory of a projectile subject only to gravity and air resistance ballistics the science of projectile motion

Bragg law mathematical relationship between wavelength, slit spacing and wavelength for constructive interference of electromagnetic radiation from a diffraction grating

Balmer series a set of hydrogen spectral lines containing all the visible spectral lines that are produced when an electron transitions to the 2nd energy level or orbital (n = 2)

Bremsstrahlung ‘braking radiation’; X-rays with a broad wavelength range that result from the conversion of kinetic energy of rapidly braking high-speed electrons directly into X-ray photons

band gap same as energy gap

brightness is a measure of the energy received in a certain time per unit of collecting area of a source (e.g. a star)—or power per unit area—W m–2 carbon–nitrogen–oxygen cycle the chain of reactions that dominates the conversion of hydrogen into helium in the hot cores of stars of higher mass than the Sun

baryon a family of particles that are made up of three quarks (protons and neutrons are both baryons) base part of a bipolar transistor BCS theory microscopic theory of superconductivity developed Bardeen, Cooper and Schrieffer beam splitter a partially silvered mirror that allows a fraction of a light beam to pass through and another fraction to be reflected beta decay a form of radioactive decay in which an atomic nucleus emits a beta particle beta particle an electron ejected from the nucleus of a radioactive nuclide binary star a system of two stars orbiting their common centre of mass biopsy a procedure in which a tissue sample is obtained for medical tests bipolar transistor a semiconductor device used to finely control the flow of electric current black body an idealised example of a hot object that produces a continuous spectrum accurately described by a black body curve; an object that absorbs or transmits all the wavelengths of the electromagnetic spectrum black body curve (or Planck curve) the distribution of light versus wavelength produced by a black body, an idealised example of a hot object; the shape, peak wavelength and intensity depend simply on the temperature of the object black hole a remnant of a supernova explosion with a mass greater than about three solar masses, and a gravitational field so strong that even light cannot escape blanking the blocking of the electron beam, in a cathode ray oscilloscope, on its way back to the left of the screen so that it does not retrace over itself B-mode scan or ‘brightness mode scan’; similar method to an A-mode scan except that the echo intensity versus depth data are represented as a one-dimensional line of pixels, with pixel intensity proportional to echo intensity, yielding a one-dimensional slice of an image bone density or bone mineral density, is related to the mass per unit volume of bone and depends on its mineral (calcium) content. The lower the calcium content, the lower the bone density and so the weaker the bone bone scan a diagnostic imaging technique in which a patient is given technetium-99m methylene diphosphonate, which accumulates preferentially in excessively active bone and can indicate the presence of cancer or other diseases boson a family of particles within the Standard Model and includes photons, gluons and the proposed Higgs particle

CAT (computed axial tomography) a form of imaging in which multiple X-ray radiographs (‘projections’) are collected at different angles through a patient, from which a computer reconstructs a 3D stack of slices of the patient or ‘tomogram’; also known as CT cataclysmic variable a binary system in which one star is so close to its white dwarf companion that it pours mass onto the white dwarf, causing the system’s light output to vary dramatically as the system suffers one or more outbursts cathode in physics, the negative electrode (Caution: The definition in chemistry is more complicated.) cathode glow first luminous region near the cathode in a discharge tube cathode ray tube glass tube that contains two electrodes (an anode and cathode) that produce electron beams cathode rays collimated beams of electrons in an evacuated vessel central body a large mass around which other much smaller masses orbit, via gravitational attraction centre of mass the ‘balance point’ that always lies between the two stars about which each of them can be considered to orbit Cepheid variables periodic variables that vary between 0.5 and 2 magnitudes with periods from 1 to 70 days. They are yellow supergiants that are important distance indicators to nearby galaxies characteristic X-rays X-rays of sharply defined wavelengths that are characteristic of each individual chemical element and are given off when a target is bombarded by accelerated electrons, resulting from rearrangements of electrons in the inner shells of target atoms cladding in a clad optical fibre, the lower index coating that protects the core from damage and ensures total internal reflection classical theory electromagnetic wave theory used to calculate the mathematical relationship for the black body radiation curve closed or stable orbit an orbit that repeats indefinitely. Its twobody mechanical energy is negative coherence length the nominal separation between Cooper pairs coherent fibre bundle a bundle of optical fibres in which the order of fibres is the same at each end, ensuring that an image projected on one end is transmitted accurately to the other coils loops of current-carrying wire in motors, generators and transformers collector part of a bipolar transistor collimate to make a beam of radiation more focused or parallel by using opaque barriers to remove components of the radiation travelling at the wrong angle 455

Glossary

collimator any kind of cylindrical, conical or planar barrier used to collimate radiation that can’t be focused using mirrors or lenses

critical field magnetic field that can destroy the superconducting state

colour Doppler imaging a sonogram employing the Doppler effect in which the colour of the pixel represents the velocity of the tissue being observed

critical mass the minimum mass of nuclear fuel which is required to produce sufficient neutrons to cause a sustained fission reaction

colour index the difference in the brightness of a star in magnitudes when measured through two different filters, indicating the colour of the star. The B – V or mB – mV colour index is most commonly used commutator brushes conducting contacts that connect the commutator to the external circuit in an electric motor commutator slip-ring a cylindrical metal contact in a simple AC motor or generator that provides a continuous connection between the coils and the external circuit commutator split-ring a segmented cylindrical metal contact in a simple DC motor or generator that reverses the connection between the coils and the external circuit conduction band range of energy levels of free electrons in a solid conduction level energy a valence electron must acquire to become free and contribute to electrical conduction constructive interference two identical waves combine to produce a wave of greater amplitude when their crests overlap continuous spectrum a spectrum with light of all wavelengths, usually produced in astrophysical situations by a hot, dense gas contrast agent substance that when injected into a patient increases the contrast for target organs or tissues in medical imaging techniques such as X-ray, CAT or MRI control rods in a nuclear fission reactor the control rods absorb neutrons and are adjusted so that the chain reaction proceeds at a constant rate controlled nuclear reaction the conditions for a controlled nuclear chain reaction are that the available neutrons which cause the fission are regulated, resulting in a constant rate of nuclear reactions conventional current the conventional direction assigned to a current is the direction of motion of positive particles, opposite to the actual flow of electrons convex array transducer a piezoelectric transducer with a convex front face; the most commonly used shape of transducer for medical ultrasound imaging coolant a substance, commonly water, used to transfer heat energy away from the core of a nuclear reactor Cooper pair the temporary binding (or coupling) of two electrons in a superconductor mediated by a lattice deformation (phonon)

critical temperature characteristic temperature below which superconductivity occurs CRO cathode ray oscilloscope Crookes dark space the second non-luminous region near the cathode of a discharge tube separated from the Aston dark space by the cathode glow crystal a three-dimensional regular arrangement of atoms in a solid current (I) the net flow of charges through a region per unit time in amperes current-carrying conductor a conducting material (typically copper wire) through which an electric current is flowing. current-carrying loop a circular coil of wire through which an electric current is flowing cut-off frequency the frequency of electromagnetic radiation below which no electrons are emitted from a photocathode cyclotron is a type of particle accelerator that accelerates particles in a spiral trajectory. For medical purposes, the cyclotron produces accelerated protons that can convert non-radioactive nuclei into short-lived radioisotopes, usually β– emitters cyclotron motion circular motion of a charged particle at right angles to the direction of the magnetic field dark nebulae a portion of the interstellar medium where the dust scatters starlight, reddening or completely blocking our view of background stars decay series a map or sequence of nuclear transformations from one element or isotope to another density mass of a material per unit volume depletion region the region in the centre of a p–n junction that is depleted of charge due to the recombination of electrons with holes destructive interference two identical waves will cancel to produce a resulting wave of zero amplitude when the crest of one wave coincides with the trough of the other diffraction the tendency of wavefronts to bend around the edges of small obstacles, instead of casting a sharp shadow diffraction grating an optical surface, often reflective, that uses fine lines ruled onto its surface to disperse visible light into its component colours using diffraction and interference diffraction pattern interference pattern from a diffraction grating

core in a clad optical fibre, the higher index central fibre that carries the light

diffusion movement of free particles from places of high to low concentration

core in a nuclear fission reactor the core houses the fuel rods, control rods, a coolant system and a moderator material; it is where fission occurs

diode a device that allows current to travel in only one direction

cosmic rays extremely energetic particles that originate from outer space

discharge tubes glass tubes in which there is conduction as a result of ionisation of the contained gas initiated by electron motion between imbedded electrodes

critical angle the smallest angle of incidence (approaching a boundary with a lower refractive index material) that will allow some of the incident light to pass through the boundary instead of being reflected by it 456

direct current (DC) an electric current that travels in only one direction along a conductor

distance modulus the difference (m – M) between the apparent magnitude m of an object and its absolute magnitude M; related to its distance

Glossary

donor energy level an extra energy level created in the energy gap just below the conduction band of a semiconductor as a result of the introduction of donor impurities

electronics method of finely controlling the flow of electrical current

donor impurities produce unbonded electrons in a semiconductor

emf a measure of the strength of a source of electrical potential energy that produces a separation of charge; typically used to describe the energy per unit charge that produces a potential difference in an open circuit

doping the introduction of impurities into the crystal lattice of a solid Doppler effect in which the frequency (perceived by an observer) of any kind of wave (including sound and light) is changed by the motion of the wave source (or a reflecting surface). Increased frequency means the observer and source are approaching; decreased frequency means they are moving apart

ellipse a ‘compressed’ or ‘stretched’ circle

emission line spectrum a spectrum that shows bright emission lines at wavelengths characteristic of the elements present in the light source; usually produced in astrophysical situations by a hot, lowdensity gas

Doppler ultrasound a form of sonogram in which detection of a change in reflected ultrasound frequency (due to the Doppler effect) is used to deduce the velocity of a tissue, usually blood

emission nebulae a portion of the interstellar medium where the gas is energised by hot young stars to emit an emission spectrum

drag the resistive force exerted on an object moving through any fluid (for example air or water). In air, it is also known as air resistance

endoscope a device employing optical fibres and/or small cameras to collect images of internal organs with minimal surgical invasion

drain one of the three components of a field effect transistor dual X-ray absorptiometry imaging technique that compares the X-ray absorption by a specimen using X-rays of two different energies, making it more sensitive to differences in bone density than a single energy X-ray dwarf novae a small outburst, repeating semi-regularly, produced by instabilities in the flow of gas from a Sun-like star onto a close white dwarf companion; classified as a non-periodic intrinsic variable star dynodes electrodes in a photomultiplier tube that are used to increase the number of electrons emitted from a photocathode eccentricity a measure of how far the position of the centre of a satellite’s orbit lies from the central mass. A measure of how elliptical an orbit is; a circle has zero eccentricity echocardiography a form of colour Doppler ultrasound used to diagnose conditions of the heart eclipsing binary binary systems in which the stars are close together and orientated so that the orbital plane is close to edgeon, causing the stars to regularly eclipse one another, periodically blocking of some of the light from the system eddy current circulating current in a conducting material caused by a changing magnetic field eddy current braking a braking system that utilises eddy currents to slow a moving object effectively weightless the effect of being in free-fall or orbit. The only significant force acting on a body is gravity. Because there are no other opposing forces acting, the body experiences no compression or tension and so behaves like a truly weightless object in an inertial reference frame electromagnet a magnet in which the magnetic field is produced by the flow of electric current electromagnetic induction the generation of an electric current by a changing magnetic field electromagnetic wave consists of oscillating electric and magnetic fields at right angles to each other and travelling at the speed of light electron volt an atomic-sized energy unit; 1 eV is the energy gained by an electron when accelerated through a potential of one volt (1 eV = 1.6 × 10–19 J)

emitter part of a bipolar transistor

endoscopy procedures involving an endoscope energy gap range of forbidden energy levels between the valence and conduction bands; energy is needed to destroy the superconducting state escape velocity the minimum velocity required for a projectile to permanently escape the gravitational field of an astronomical body exclusion principle the Pauli exclusion principle was formulated by Pauli in 1925 and states that no two fermions can possess the same quantum numbers in the same system; explains the energy states and orbitals in the atom exhaust velocity the velocity of the exhaust gas as it leaves the rocket nozzle extrinsic semiconductor one in which conduction is dominated by donor or acceptor impurities extrinsic variables variable stars whose variation in brightness is due to a process external to the body of the star itself, for example eclipsing binaries or stars that vary because of their rotation Faraday dark space a non-luminous region in a discharge tube between the negative glow and the positive column Faraday’s law the induced emf in a coil is proportional to the product of the number of turns and the rate at which the magnetic field changes within the turns fictitious forces apparent forces used for convenience to ‘explain’ apparent changes in velocity when a system is observed from within a non-inertial (accelerating) frame of reference field of view the area of object (e.g. the sky) you can see at any moment with, for example, a telescope field-effect transistor uses an electric field to control the flow of current through a conducting channel in a semiconductor filter a slab or sheet of material designed to selectively remove certain wavelengths, frequencies or energies from a beam of radiation fission the splitting of an atomic nucleus into two or more fragments flare star a faint red dwarf that flares in visible brightness because of solar-like flares on the surface; classified as a non-periodic intrinsic variable star

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Glossary

fluorescence the phenomenon by which a material absorbs photons of one energy (frequency) and immediately re-emits photons of lower energy (frequency)

gravity and acceleration. It is the ratio of the ‘effective weight’ to true weight on Earth. g-force is measured in units of Earth gravity g

fluoroscope a traditional apparatus that produces real-time moving X-ray images by exposing a fluorescent screen to a broad beam of X-rays passing through a patient or specimen. Modern digital versions use electronic detectors in place of the screen

giant molecular clouds a portion of the interstellar medium where the gas is most dense; simple molecules can form and star formation occurs

flux leakage loss of magnetic flux as it passes from the primary coil through the secondary coil, resulting in loss of induction in the secondary coil and, hence, loss of energy

gimbal a pivoted frame that allows rotation around more than one axis; common example is the mount that supports a gyroscope

flux pinning the stopping of the motion of vortices through a type II superconductor by crystal defects and boundaries

gradient coils coils in an MRI machine that apply gradients in the magnetic field across a patient so that variations in the detected Larmor frequency of relaxing nuclei can be used to obtain the position information needed to create a tomogram

focal length the distance between a lens or mirror and the image it forms of a distant object

gravimeter a device for measuring the local value of the gravitational field (or equivalently, the gravitational acceleration

forbidden energy gap see energy gap

gravitational field g a vector representing the direction and strength of gravity at a point in space. The magnitude g of the vector equals the acceleration due to gravity at that point

force (F) a push, pull or twist measured in newtons (N) forward bias connecting the positive and negative terminals of a power supply to the p and n sides respectively of a p–n junction fuel rod tubes filled with nuclear fuel and located in the centre of the reactor known as the core functional MRI a form of MRI in which the contrast agent is designed to concentrate in parts of the brain that are active during certain activities, making it possible to identify the functions of different regions of the brain fusion the joining of two nuclei to form a new nucleus Galilean transformation the formula for transforming velocities relative to a different frame of reference; vB (rel. to A) = vB – vA galvanometer a device used to measure the relative strength and direction of electric current gamma camera an array of collimated gamma detectors for imaging patients who have been given diagnostic gamma-emitting radiopharmaceutical gamma decay decay involving emission of a gamma ray gamma photons high-energy photons mostly emitted in nuclear reactions and positron emission tomography gamma rays a form of electromagnetic radiation (photons) with wavelength