Physics for Engineers and Scientists -- 3rd Ed. Vol. 3

W. W. Norton & Company, Inc. • www.NortonEbooks.com

PHYSICS FOR ENGINEERS AND SCIENTISTS THIRD EDITION Volume Three

Hans C. Ohanian, John T. Markert

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Physics for Engineers and Scientists Third Edition Volume 3 (Chapters 36–41) R E L AT I V I T Y, Q U A N TA , A N D PA RT I C L E S

HANS C. OHANIAN, UNIVERSITY OF VERMONT

J O H N T. M A R K E R T, U N I V E R S I T Y O F T E X A S AT A U S T I N

W • W • NORTON & COMPANY

B

NEW YORK • LONDON

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To Susan Ohanian, writer, who gently tried to teach me some of her craft.—H.C.O. To Frank D. Markert, a printer by trade; to Christiana Park, for her thirst for new knowledge; and to Erin, Ryan, Sean, and Gwen, for their wonder and clarity.—J.T.M.

Copyright © 2007 by W.W. Norton & Company, Inc. All rights reserved Printed in the United States of America Third Edition Composition: Techbooks Manufacturing: RR Donnelley & Sons Company Editor: Leo A. W. Wiegman Media Editor: April E. Lange Director of Manufacturing—College: Roy Tedoff Senior Project Editor: Christopher Granville Photo Researcher: Kelly Mitchell Editorial Assistant: Lisa Rand, Sarah L. Mann Copy Editor: Richard K. Mickey Book designer: Sandy Watanabe Layout artist: Paul Lacy Illustration Studio: J. B. Woolsey and Penumbra Design, Inc. Cover Illustration: John Belcher, inter alia. Cover Design: Joan Greenfield Library of Congress Cataloging-in-Publication Data has been applied for. ISBN 978-0-393-11103-3 (ebook)

W. W. Norton & Company, Inc., 500 Fifth Avenue, New York, N.Y. 10110 www.wwnorton.com W. W. Norton & Company Ltd., Castle House, 75/76 Wells Street, London W1T 3QT 1234567890 W. W. Norton & Company has been independent since its founding in 1923, when William Warder Norton and Mary D. Herter Norton first published lectures delivered at the People’s Institute, the adult education division of New York City’s Cooper Union. The Nortons soon expanded their program beyond the Institute, publishing books by celebrated academics from America and abroad. By mid-century, the two major pillars of Norton’s publishing program—trade books and college texts— were firmly established. In the 1950s, the Norton family transferred control of the company to its employees, and today—with a staff of four hundred and a comparable number of trade, college, and professional titles published each year—W. W. Norton & Company stands as the largest and oldest publishing house owned wholly by its employees.

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Brief Contents Chapters 1–21 appear in Volume 1; Chapters 22–36 appear in Volume 2; Chapters 36–41 appear in Volume 3.

PREFACE xiii OWNER’S MANUAL xxv PRELUDE: THE WORLD OF PHYSICS xxxv

PART I MOTION, FORCE, AND ENERGY 1

PART IV ELECTRICITY AND MAGNETISM 692

1. SPACE, TIME, AND MASS 2

22. ELECTRIC FORCE AND ELECTRIC CHARGE 694

2. MOTION ALONG A STRAIGHT LINE 28

23. THE ELECTRIC FIELD 721

3. VECTORS 69

24. GAUSS’ LAW 756

4. MOTION IN TWO AND THREE DIMENSIONS 94

25. ELECTROSTATIC POTENTIAL AND ENERGY 789

5. NEWTON’S LAWS OF MOTION 130

26. CAPACITORS AND DIELECTRICS 828

6. FURTHER APPLICATIONS OF NEWTON’S LAWS 173

27. CURRENTS AND OHM’S LAW 858

7. WORK AND ENERGY 204

28. DIRECT CURRENT CIRCUITS 887

8. CONSERVATION OF ENERGY 235

29. MAGNETIC FORCE AND FIELD 926

9. GRAVITATION 271 10. SYSTEMS OF PARTICLES 305

30. CHARGES AND CURRENTS IN MAGNETIC FIELDS 964

11. COLLISIONS 338

31. ELECTROMAGNETIC INDUCTION 993

12. ROTATION OF A RIGID BODY 365

32. ALTERNATING CURRENT CIRCUITS 1030

13. DYNAMICS OF A RIGID BODY 394 14. STATICS AND ELASTICITY 429

PART V WAVES AND OPTICS 1068 33. ELECTROMAGNETIC WAVES 1070

PART II OSCILLATIONS, WAVES, AND FLUIDS 466 15. OSCILLATIONS 468

34. REFLECTION, REFRACTION, AND OPTICS 1111 35. INTERFERENCE AND DIFFRACTION 1168

16. WAVES 507 17. SOUND 536 18. FLUID MECHANICS 565

PART VI RELATIVITY, QUANTA, AND PARTICLES 1214 36. THE THEORY OF SPECIAL RELATIVITY 1216

PART III TEMPERATURE, HEAT, AND THERMODYNAMICS 600 19. THE IDEAL GAS 602

37. QUANTA OF LIGHT 1254 38. SPECTRAL LINES, BOHR’S THEORY, AND QUANTUM MECHANICS 1286

20. HEAT 628

39. QUANTUM STRUCTURE OF ATOMS, MOLECULES, AND SOLIDS 1320

21. THERMODYNAMICS 661

40. NUCLEI 1354 41. ELEMENTARY PARTICLES AND COSMOLOGY 1396 APPENDICES A-1 iii

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Table of Contents Chapters 1–21 appear in Volume 1; Chapters 22–36 appear in Volume 2; and Chapters 36–41 appear in Volume 3.

PREFACE xiii

3.3

OWNER’S MANUAL xxv 3.4

PRELUDE: THE WORLD OF PHYSICS xxxv

PART I MOTION, FORCE, AND ENERGY 1 1. SPACE, TIME, AND MASS 2 1.1 1.2 1.3 1.4 1.5 1.6

Coordinates and Reference Frames 3 The Unit of Length 5 The Unit of Time 9 The Unit of Mass 11 Derived Units 13 Significant Figures; Consistency of Units and Conversion of Units 14

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 20

2. MOTION ALONG A STRAIGHT LINE 28 2.1 2.2 2.3 2.4 2.5 2.6 2.7*

Average Speed 29 Average Velocity for Motion along a Straight Line 32 Instantaneous Velocity 35 Acceleration 39 Motion with Constant Acceleration 42 The Acceleration of Free Fall 49 Integration of the Equations of Motion 54

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 57

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 87

4. MOTION IN TWO AND THREE DIMENSIONS 94 4.1 4.2 4.3 4.4 4.5 4.6

5. NEWTON’S LAWS OF MOTION 130 5.1 5.2 5.3 5.4 5.5 5.6

3.2

The Displacement Vector and Other Vectors 70 Vector Addition and Subtraction 72

Newton’s First Law 131 Newton’s Second Law 133 The Combination of Forces 138 Weight; Contact Force and Normal Force 141 Newton’s Third Law 144 Motion with a Constant Force 151

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 159

6. FURTHER APPLICATIONS OF NEWTON’S LAWS 173 6.1 6.3 6.4*

3.1

Components of Velocity and Acceleration 95 The Velocity and Acceleration Vectors 98 Motion with Constant Acceleration 102 The Motion of Projectiles 104 Uniform Circular Motion 112 The Relativity of Motion and the Addition of Velocities 115

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 118

6.2

3. VECTORS 69

The Position Vector; Components of a Vector 76 Vector Multiplication 81

Friction 174 Restoring Force of a Spring; Hooke’s Law 182 Force for Uniform Circular Motion 184 The Four Fundamental Forces 191

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 192

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7. WORK AND ENERGY 204

12.2 12.3

7.1 7.2 7.3 7.4

Work 205 Work for a Variable Force 211 Kinetic Energy 214 Gravitational Potential Energy 218

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 224

12.4* 12.5

Rotation about a Fixed Axis 367 Motion with Constant Angular Acceleration 374 Motion with Time-Dependent Angular Acceleration 376 Kinetic Energy of Rotation; Moment of Inertia 378

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 384

8. CONSERVATION OF ENERGY 235 13. DYNAMICS OF A RIGID BODY 394 8.1 8.2 8.3 8.4* 8.5

Potential Energy of a Conservative Force 236 The Curve of Potential Energy 244 Other Forms of Energy 248 Mass and Energy 251 Power 253

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 259

9. GRAVITATION 271 9.1 9.2 9.3 9.4 9.5

Newton’s Law of Universal Gravitation 272 The Measurement of G 277 Circular Orbits 278 Elliptical Orbits; Kepler’s Laws 282 Energy in Orbital Motion 288

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 293

13.1 13.2 13.3 13.4*

Work, Energy, and Power in Rotational Motion; Torque 395 The Equation of Rotational Motion 399 Angular Momentum and its Conservation 406 Torque and Angular Momentum as Vectors 410

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 417

14. STATICS AND ELASTICITY 429 14.1 14.2 14.3 14.4

Statics of Rigid Bodies 430 Examples of Static Equilibrium 433 Levers and Pulleys 441 Elasticity of Materials 445

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 450

10. SYSTEMS OF PARTICLES 305 10.1 10.2 10.3 10.4

Momentum 306 Center of Mass 313 The Motion of the Center of Mass 323 Energy of a System of Particles 327

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 328

PART II OSCILLATIONS, WAVES, AND FLUIDS 466 15. OSCILLATIONS 468 15.1

11. COLLISIONS 338

15.2 15.3

11.1 11.2 11.3 11.4*

Impulsive Forces 339 Elastic Collisions in One Dimension 344 Inelastic Collisions in One Dimension 348 Collisions in Two and Three Dimensions 351

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 354

15.4 15.5*

Simple Harmonic Motion 469 The Simple Harmonic Oscillator 476 Kinetic Energy and Potential Energy 480 The Simple Pendulum 484 Damped Oscillations and Forced Oscillations 488

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 494

16. WAVES 507 12. ROTATION OF A RIGID BODY 365 16.1 12.1

Motion of a Rigid Body 366

16.2

Transverse and Longitudinal Wave Motion 508 Periodic Waves 509

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16.3 16.4

The Superposition of Waves 516 Standing Waves 520

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 524

21. THERMODYNAMICS 661 21.1 21.2 21.3

17. SOUND 536 17.1 17.2 17.3 17.4 17.5*

Sound Waves in Air 538 Intensity of Sound 540 The Speed of Sound; Standing Waves 543 The Doppler Effect 574 Diffraction 553

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 555

18. FLUID MECHANICS 565 18.1 18.2 18.3 18.4 18.5 18.6

Density and Flow Velocity 567 Incompressible Steady Flow; Streamlines 569 Pressure 573 Pressure in a Static Fluid 575 Archimedes’ Principle 580 Fluid Dynamics; Bernoulli’s Equation 582

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21.4

The First Law of Thermodynamics 663 Heat Engines; The Carnot Engine 665 The Second Law of Thermodynamics 675 Entropy 677

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 681

PART IV ELECTRICITY AND MAGNETISM 692 22. ELECTRIC FORCE AND ELECTRIC CHARGE 694 22.1 22.2 22.3 22.4 22.5

The Electrostatic Force 695 Coulomb’s Law 698 The Superposition of Electrical Forces 703 Charge Quantization and Charge Conservation 706 Conductors and Insulators; Charging by Friction or by Induction 708

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 712

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 587

23. THE ELECTRIC FIELD 721

PART III TEMPERATURE, HEAT, AND THERMODYNAMICS 600

23.1 23.2 23.3

19. THE IDEAL GAS 602 19.1 19.2 19.3 19.4

The Ideal-Gas Law 603 The Temperature Scale 609 Kinetic Pressure 613 The Internal Energy of an Ideal Gas 616

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 619

23.4 23.5

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 745

24. GAUSS’ LAW 756 24.1 24.2

20. HEAT 628 20.1 20.2 20.3 20.4 20.5 20.6*

Heat as a Form of Energy Transfer 629 Thermal Expansion of Solids and Liquids 633 Thermal Conduction 638 Changes of State 642 The Specific Heat of a Gas 644 Adiabatic Expansion of a Gas 647

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 650

The Electric Field of Point Charges 722 The Electric Field of Continuous Charge Distributions 729 Lines of Electric Field 736 Motion in a Uniform Electric Field 740 Electric Dipole in an Electric Field 742

24.3 24.4 24.5

Electric Flux 757 Gauss’ Law 762 Applications of Gauss’ Law 763 Superposition of Electric Fields 772 Conductors and Electric Fields 774

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 778

25. ELECTROSTATIC POTENTIAL AND ENERGY 789 25.1 25.2

The Electrostatic Potential 790 Calculation of the Potential from the Field 798

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25.3 25.4 25.5

Potential in Conductors 803 Calculation of the Field from the Potential 806 Energy of Systems of Charges 811

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 816

26. CAPACITORS AND DIELECTRICS 828 26.1 26.2 26.3 26.4

Capacitance 829 Capacitors in Combination 834 Dielectrics 838 Energy in Capacitors 844

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 847

27. CURRENTS AND OHM’S LAW 858 27.1 27.2 27.3 27.4

Electric Current 859 Resistance and Ohm’s Law 863 Resistivity of Materials 868 Resistances in Combination 872

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 878

28. DIRECT CURRENT CIRCUITS 887 28.1 28.2 28.3 28.4 28.5 28.6* 28.7* 28.8*

Electromotive Force 888 Sources of Electromotive Force 890 Single-Loop Circuits 893 Multi-Loop Circuits 897 Energy in Circuits; Joule Heat 901 Electrical Measurements 903 The RC Circuit 907 The Hazards of Electric Currents 913

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 914

29. MAGNETIC FORCE AND FIELD 926 29.1 29.2 29.3 29.4 29.5

The Magnetic Force 928 The Magnetic Field 931 Ampére’s Law 938 Solenoids and Magnets 943 The Biot-Savart Law 948

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 951

30. CHARGES AND CURRENTS IN MAGNETIC FIELDS 964 30.1 30.2 30.3 30.4 30.5*

Circular Motion in a Uniform Magnetic Field 965 Force on a Wire 969 Torque on a Loop 972 Magnetism in Materials 976 The Hall Effect 980

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 984

31. ELECTROMAGNETIC INDUCTION 993 31.1 31.2 31.3 31.4 31.5 31.6*

Motional EMF 994 Faraday’s Law 997 Some Examples; Lenz’ Law 1001 Inductance 1008 Magnetic Energy 1013 The RL Circuit 1015

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 1018

32. ALTERNATING CURRENT CIRCUITS* 1030

Resistor Circuit 1013 32.2 Capacitor Circuit 1035 32.3 Inductor Circuit 1038 32.4 * Freely Oscillating LC and RLC Circuits 1041 32.5 * Series Circuits with Alternating EMF 1046 32.6 The Transformer 1053 32.1

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 1057

PART V WAVES AND OPTICS 1068 33. ELECTROMAGNETIC WAVES 1070 33.1 33.2* 33.3 33.4 33.5 33.6*

Induction of Magnetic Fields; Maxwell’s Equations 1071 The Electromagnetic Wave Pulse 1075 Plane Waves; Polarization 1079 The Generation of Electromagnetic Waves 1088 Energy of a Wave 1092 The Wave Equation 1096

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 1099

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34. REFLECTION, REFRACTION, AND OPTICS 1111 34.1 34.2 34.3 34.4 34.5 34.6*

Huygens’ Construction 1113 Reflection 1114 Refraction 1117 Spherical Mirrors 1128 Thin Lenses 1135 Optical Instruments 1144

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 1152

38. SPECTRAL LINES, BOHR’S THEORY, AND QUANTUM MECHANICS 1286 38.1 38.2 38.3 38.4 38.5

Spectral Lines 1287 Spectral Series of Hydrogen 1291 The Nuclear Atom 1293 Bohr’s Theory 1295 Quantum Mechanics; The Schrödinger Equation 1302

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 1312

35. INTERFERENCE AND DIFFRACTION 1168 35.1 35.2* 35.3 35.4 35.5 35.6

Thin Films 1169 The Michelson Interferometer 1174 Interference from Two Slits 1177 Interference from Multiple Slits 1183 Diffraction by a Single Slit 1190 Diffraction by a Circular Aperture; Rayleigh’s Criterion 1196

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 1199

39. QUANTUM STRUCTURE OF ATOMS, MOLECULES, AND SOLIDS 1320 39.1 39.2 39.3* 39.4 39.5

Principal, Orbital, and Magnetic Quantum Numbers; Spin 1321 The Exclusion Principle and the Structure of Atoms 1328 Energy Levels in Molecules 1333 Energy Bands in Solids 1336 Semiconductor Devices 1340

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 1345

PART VI RELATIVITY, QUANTA, AND PARTICLES 1214

40. NUCLEI 1354 40.1

36. THE THEORY OF SPECIAL RELATIVITY 1216 36.1 36.2 36.3 36.4 36.5 36.6 36.7*

The Speed of Light; the Ether 1218 Einstein’s Principle of Relativity 1220 Time Dilation 1224 Length Contraction 1230 The Lorentz Transformations and the Combination of Velocities 1232 Relativistic Momentum and Energy 1239 Mass and Energy 1242

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 1244

40.2 40.3 40.4 40.5 40.6* 40.7

37.1 37.2 37.3 37.4 37.5 37.6

Blackbody Radiation 1255 Energy Quanta 1258 Photons and the Photoelectric Effect 1264 The Compton Effect 1269 X Rays 1273 Wave vs. Particle 1276

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 1279

Isotopes 1355 The Strong Force and the Nuclear Binding Energy 1359 Radioactivity 1365 The Law of Radioactive Decay 1372 Fission 1377 Nuclear Bombs and Nuclear Reactors 1379 Fusion 1384

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 1386

41. ELEMENTARY PARTICLES AND COSMOLOGY 1396 41.1

37. QUANTA OF LIGHT 1254

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41.2 41.3 41.4 41.5 41.6

The Tools of High-Energy Physics 1397 The Multitude of Particles 1403 Interactions and Conservation Laws 1405 Fields and Quanta 1409 Quarks 1412 Cosmology 1416

SUMMARY / QUESTIONS / PROBLEMS / REVIEW PROBLEMS / ANSWERS TO CHECKUPS 1424

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APPENDICES APPENDIX 1: GREEK ALPHABET A-1

APPENDIX 2: MATHEMATICS REVIEW A-1

APPENDIX 6: THE INTERNATIONAL SYSTEM OF UNITS (SI) A-21 A6.1 A6.2

A2.1 A2.2 A2.3 A2.4 A2.5 A2.6 A2.7

Symbols A-1 Powers and Roots A-1 Arithmetic in Scientific Notation A-2 Algebra A-3 Equations with Two Unknowns A-5 The Quadratic Formula A-5 Logarithms and the Exponential Function A-5

APPENDIX 3: GEOMETRY AND TRIGONOMETRY REVIEW A-7 A3.1 A3.2 A3.3 A3.4 A3.5

Perimeters, Areas, and Volumes A-7 Angles A-7 The Trigonometric Functions A-8 The Trigonometric Identities A-9 The Laws of Cosines and Sines A-10

APPENDIX 4: CALCULUS REVIEW A-10 A4.1 A4.2 A4.3 A4.4 A4.5 A4.6

Derivatives A-10 Important Rules for Differentiation A-11 Integrals A-12 Important Rules for Integration A-15 The Taylor Series A-18 Some Approximations A-18

APPENDIX 5: PROPAGATING UNCERTAINTIES A-19

A6.3

Base Units A-21 Derived Units A-23 Prefixes A-23

APPENDIX 7: BEST VALUES OF FUNDAMENTAL CONSTANTS A-23

APPENDIX 8: CONVERSION FACTORS A-26

APPENDIX 9: THE PERIODIC TABLE AND CHEMICAL ELEMENTS A-31

APPENDIX 10: FORMULA SHEETS A-33

Chapters 1–21 A-33 Chapters 22–41 A-34 APPENDIX 11: ANSWERS TO ODD-NUMBERED PROBLEMS AND REVIEW PROBLEMS A-35

PHOTO CREDITS A-39

INDEX A-41

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Preface

Our aim in Physics for Engineers and Scientists, Third Edition, is to present a modern view of classical mechanics and electromagnetism, including some optics and quantum physics. We also want to offer students a glimpse of the practical applications of physics in science, engineering, and everyday life. The book and its learning package emerged from a collaborative effort that began more than six years ago. We adapted the core of Ohanian’s earlier Physics (Second Edition, 1989) and combined it with relevant findings from recent physics education research on how students learn most effectively. The result is a text that presents a clear, uncluttered explication of the core concepts in physics, well suited to the needs of undergraduate engineering and science students.

O r g a n i z a t i o n o f To p i c s The 41 chapters of the book cover the essential topics of introductory physics: mechanics of particles, rigid bodies, and fluids; oscillations, wave motion, heat and thermodynamics; electricity and magnetism; optics; special relativity; and atomic and subatomic physics. Our arrangement and treatment of topics are fairly traditional with a few deliberate distinctions. We introduce the principle of superposition of forces early in Chapter 5 on Newton’s laws of motion, and we give the students considerable exposure to the vector superposition of gravitational forces in Chapter 9. This leaves the students well prepared for the later application of vector superposition of electric and magnetic forces generated by charge or current distributions. We place gravitation in Chapter 9 immediately after the chapters on work and energy, because we regard gravitation as a direct application of these concepts (instructors who prefer to postpone gravitation can, of course, do so). We introduce forces on stationary electric charges in a detailed, complete exposition in Chapter 22, before proceeding to the less obvious concept of the electric field in Chapter 23. We start the study of magnetism in Chapter 29 with the force on a moving charged particle near a current, instead of the more common practice of starting with a postulate about the magnetic field in the abstract. With our approach, the observed magnetic forces on moving charges lead naturally to the magnetic field, and this progression from magnetic force to

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magnetic field will remind students of the closely parallel progression from electric force to electric field. For efficiency and brevity, we sometimes combine in one chapter closely related topics that other authors elect to spread over more than one chapter. Thus, we cover induction and inductance together in Chapter 31 and interference and diffraction together in Chapter 35.

C o n c i s e Wr i t i n g w i t h S h a r p F o c u s o n C o r e C o n c e p t s Our goal is concise exposition with a sharp focus on core concepts. Brevity is desirable because long chapters with a large number of topics and excessive verbiage are confusing and tedious for the student. In our writing, we obey the admonitions of Strunk and White’s Elements of Style: use the active voice; make statements in positive form; use definite, specific, concrete language; omit needless words. We strove for simplicity in organizing the content. Each chapter covers a small set of core topics—rarely more than five or six—and we usually place each core topic in a section of its own. This divides the content into manageable segments and gives the chapter a clear and clean outline. Transitional sentences at the beginning or end of sections spell out the logical connections between each section and the next. Within each section, we strove for a seamless narrative leading from the discussions of concepts to their applications in Example problems. We sought to avoid the patchy, cobbled structure of many texts in which the discussions appear to serve as filler between one equation and the next.

Emphasis on the Atomic Structure of Matter Throughout the book, we encourage students to keep in mind the atomic structure of matter and to think of the material world as a multitude of restless electrons, protons, and neutrons. For instance, in the mechanics chapters, we emphasize that all macroscopic bodies are systems of particles and that the equations of motion for macroscopic bodies emerge from the equations of motion of the individual particles. We emphasize that macroscopic forces are the result of a superposition of the forces among the particles of the system, and we consider atoms and their bonds in the qualitative discussions of elasticity, thermal expansion, and changes of state. By exposing students to the atomic structure of matter in the first semester, we help them to grasp the nature of the charged particles that play a central role in the treatment of electricity and magnetism in the second semester. Thus, in the electricity chapters, we introduce the concepts of positive and negative charge by referring to protons and electrons, not by referring to the antiquated procedure of rubbing glass rods with silk rags. We try to make sure that students are always aware of the limitations of the nineteenth-century fiction that matter and electric charge are continua. Blind reliance on this old fiction has often been justified by the claim that, although engineering students need physics as a problem-solving tool, the atomic structure of matter is of little concern to them. This supposition may be adequate for a superficial treatment of mechanical engineering. Yet much of modern engineering—from materials science to electronics—hinges on understanding the atomic structure of matter. For this purpose, engineers need a physicist’s view of physics.

R e a l - Wo r l d E x a m p l e s B e g i n E a c h C h a p t e r Each chapter opens with a “Concepts in Context” photograph illustrating a practical application of physics. The caption for this photo explores various core concepts in a concrete real-world context. The questions included in the caption are linked to several solved Examples or discussions later in the chapter. Such revisiting of the

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chapter-opening application provides layers of learning, as new concepts are carefully built upon a foundation firmly planted in the real world. The emphasis on real-world data is also evident throughout other Examples and in the end-of-chapter problems. By exposing students to realistic data, we give them confidence to apply physics in their later science or engineering courses.

Conceptual Discussions Precede and Motivate the Math Only after a careful exposition of the conceptual foundations in a qualitative physical context does each section proceed to the mathematical treatment. Thus, we ensure that the mathematical formulas and their consequences and variations are rooted in a firm conceptual foundation. We were very careful to provide clear, thorough, and accurate explanations and derivations of all mathematical statements, to ensure that students acquire a good intuition about why particular equations are applied. Immediately after such derivations, we provide solved Examples to establish a firm connection between theory and concrete practical applications.

E x a m p l e s E n l i v e n t h e Te x t We devote significant portions of each chapter to carefully selected Examples of solved problems—about 390 altogether or 9 on average per chapter. These Examples are concrete illustrations of the preceding conceptual discussions. They build cumulatively upon each other, from simple to more complicated as the chapter progresses. To enliven the text, we employ realistic data in the Examples, such as students would actually encounter outside the classroom. The solved Examples are designed to cover most variations of possible problems, with solutions that include both general approaches and specific details on how to extract the important information for the given problem. For instance, when such keywords as initially or at rest occur in a solved Example, we are careful to point out their importance in the problem-solving process. Comments appended to some Examples draw attention to limitations in the solution or to wider implications.

Checkup Questions Implement Active Learning We conclude each section of a chapter with a series of brief Checkup questions. These permit students to test their mastery of core concepts, and they can be of great help in clearing up common misconceptions. Checkup questions include variations and “flip sides” of simple concepts that often occur to students but are rarely addressed. We give detailed answers to each Checkup question at the back of the chapter. The entire book contains roughly 5 Checkup questions per section—comprising a total of about 800 Checkup questions. The final Checkup question of each section is always in multiple-choice forrmat— specifically designed for interactive teaching. At the University of Texas, instructors use such multiple-choice questions as classroom concept quizzes for welcome breaks in conventional lecturing. When more than one answer is popular, the instructor and class immediately know that more discussion or more examples are needed. Such occasions lend themselves well to peer instruction, in which the students explain to one another their reasoning before responding. This pedagogy implements an active, participatory alternative to the traditional lecture format. In addition, several supplements to the textbook, including the Student Activity Workbook, Online Concept Tutorials, Smartwork online homework, and PhysiQuizzes also implement active learning and a mastery-based approach.

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P r o b l e m - S o l v i n g Te c h n i q u e s Many chapters have inserts in the form of boxes devoted to Problem-Solving Techniques. These 39 skill boxes summarize the main steps or approaches for the solution of common classes of problems. Often deployed after several seemingly disparate Examples, the Problem-Solving Techniques boxes underscore the unity and generality of the techniques used in the Examples. The boxes list the steps or approaches to be taken, providing a handy reference and review.

Math Help We have placed a Math Help box wherever students encounter a mathematical concept or technique that may be difficult or unfamiliar. These 6 skill boxes briefly review and summarize such topics as trigonometry, derivatives, integrals, and ellipses. Students can find more detailed help in Appendix 2 on basic algebra, 3 on trigonometry and geometry, 4 on calculus, and 5 on propagation of uncertainties.

Physics in Practice Many chapters have a short essay on Physics in Practice that illustrates an application of physics in engineering and everyday life. These 27 essay boxes discuss practical topics, such as ultracentrifuges, communication and weather satellites, magnetic levitation, etc. Each of these essays provides a wealth of interesting detail and offers a practical supplement to some of the chapter topics. They have been designed to be engaging, yet sufficiently qualitative to provide some respite from the more analytical discussions, Examples, and Questions.

Figures and Balloon Captions Over 1,500 figures illustrate the text. We made every effort to assemble a visual narrative as clear as the verbal narrative. Each figure in a sequence carefully builds upon the visual information in the figure that precedes it. Many figures in the text contain a caption in “balloon” that points to important features within the figure. The balloon caption is a concise and informative supplement to the conventional figure caption. The balloons make immediately obvious some details that would require a long, wordy explanation in the conventional caption. Often the balloon captions are arranged so that some cause-effect or other sequential thought process becomes immediately evident. All drawn figures are available to instructors in digital form for use in the course.

End–of–Chapter Summar y Each chapter narrative closes with several support elements, starting with a brief Summary. The Summary contains the essential physical laws, quantities, definitions, and key equations introduced in the chapter. A page reference, key equation number, and often a thumbnail figure accompany these laws, definitions, and equations. The Summary does not include repetition of the detailed explanations of the chapter. The Summary is followed by Questions for Discussion, Problems, Review Problems, and Answers to Checkups.

Questions for Discussion After the chapter’s Summary, we include a large selection of qualitative Questions for Discussion — about 700 in the entire book or roughly 17 per chapter. We intend these qualitative end-of-chapter Questions to stimulate student thinking. Some of these questions are deliberately formulated so as to have no unique answer, which is intended to promote class discussion.

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Problems After the chapter’s qualitative Questions, we include computational Problems grouped by chapter section — about 3000 in the entire book, or roughly 73 per chapter. Each problem’s level of difficulty is indicated by no asterisk, one asterisk (*), or two asterisks (**). Most no-asterisk Problems are easy and straightforward, only requiring students to “plug in” the correct values to compute answers or to retrace the steps of an Example. One-asterisk Problems are of medium difficulty. They contain a few complications requiring the combination of several concepts or the manipulation of several formulas. Two-asterisk Problems are difficult and challenging. They demand considerable thought and perhaps some insight, and occasionally require appreciable mathematical skill. When an Online Concept Tutorial (see below) is available for help in mastering the concepts in a given section, a dagger footnote (†) tells students where to find the tutorial. We tried to make the Problems interesting for students by drawing on realistic examples from technology, science, sports, and everyday life. Many of the Problems are based on data extracted from engineering handbooks, car repair manuals, Jane’s Book of Aircraft, The Guinness Book of World Records, newspaper reports, research and industrial instrumentation manuals, etc. Many other Problems deal with atoms and subatomic particles. These Problems are intended to reinforce the atomistic view of the material world. In some cases, experts will perhaps consider the use of classical physics somewhat objectionable in a problem that really ought to be handled by quantum mechanics. But we believe that the advantages of familiarization with atomic quantities and magnitudes outweigh the disadvantages of an occasional naive use of classical mechanics. Among the Problems are a smaller number of somewhat contrived, artificial Problems that make no pretense of realism (for example, “A block slides on an inclined plane tied by string...”). Such unrealistic Problems are sometimes the best way to bring an important concept into sharp focus. Some Problems are formulated as guided problems, with a series of questions that take the student through an important problemsolving procedure, step by step.

Review Problems After the Problems section of each chapter, we offer an extra selection of Review Problems — about 600 in the entire book or roughly 15 per chapter. We wrote these Review Problems specifically to help students prepare for examinations. Hence, Review Problems often test comprehension by requiring students to apply concepts from more than one section of the chapter and occasionally from prior, related chapters. Answers to all odd-numbered Problems and Review Problems are given in Appendix 11.

Units and Significant Figures We use the SI system of units exclusively throughout the text. In the abbreviations for the units we follow the recommendations of the International Committee for Weights and Measures (CIPM), although we retain some traditional units, such as revolution and calorie that have been discontinued by the CIPM. In addition, for the sake of clarity we spell out the name of the unit in full whenever the abbreviation is likely to lead to ambiguity and confusion (for instance, in the case of V for volt, which is easily confused with V for potential; or in the case of C for coulomb, which might be confused with C for capacitance). We try to use realistic numbers of significant figures, with most Examples and Problems using two or three. In cases where it is natural to employ some data with two significant figures and some with three, we have been careful to propagate the appropriate number of significant figures to the result.

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For reference purposes, we give the definitions of the British units. Currently only the United States, Bangladesh, and Liberia still adhere to these units. In the United States, automobile manufacturers have already switched to metric units for design and construction. The U. S. Army has also switched to metric units, so soldiers give distances in meters and kilometers (in army slang, the kilometer is called a “klick,” a usage that is commendable itself for its brevity). British units are not used in examples or in problems, with the exception of a handful of problems in the early chapters. In the definitions of the British units, the pound (lb) is taken to be the unit of mass, and the pound force (lbf ) is taken to be the unit of force. This is in accord with the practice approved by the American National Standards Institute (ANSI), the Institute of Electrical and Electronics Engineers (IEEE), and the U. S. Department of Defense.

Optional Sections and Chapters We recognize course content varies from institution to institution. Some sections and some chapters can be regarded as optional and can be omitted without loss of continuity. These optional sections are marked by asterisks in the Table of Contents.

Mathematical Prerequisites In order to accommodate students who are taking an introductory calculus course concurrently, derivatives are used slowly at first (Chapter 2), and routinely later on. Likewise, the use of integrals is postponed as long as possible (Chapter 7), and they come into heavy use only in the second volume (after Chapter 21). For students who need a review of calculus, Appendix 4 contains a concise primer on derivatives and integrals.

Acknowledgments We have had the benefit of a talented author team for our support resources. In addition to their primary role in the assembly of the learning package, they all have also made substantial contributions to the accuracy and clarity of the text. Stiliana Antonova, Barnard College Charles Chiu, University of Texas-Austin William J. Ellis, University of California-Davis Mirela Fetea, University of Richmond Rebecca Grossman, Columbia University David Harrison, University of Toronto Prabha Ramakrishnan, North Carolina State University Hang Deng-Luzader, Frostburg State University Stephen Luzader, Frostburg State University Kevin Martus, William Paterson University David Marx, Illinois State University Jason Stevens, Deerfield Academy Brian Woodahl, Indiana University–Purdue University-Indianapolis Raymond Zich, Illinois State University And at Sapling Systems and Science Technologies in Austin, Texas, for content, James Caras, Ph.D.; Jon Harmon, B.S.; Kevin Nelson, Ph.D.; John A. Underwood, Ph.D.; and Jason Vestuto, M.S. and for animation and programming, Jeff Sims and Nathan Wheeler. Our manuscript was subjected to many rounds of peer review. The reviewers were instrumental in identifying myriad improvements, for which we are grateful:

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Yildirim Aktas Patricia E. Allen Steven M. Anlage B. Antanaitis Laszlo Baksay Marco Battaglia Lowell Boone Marc Borowczak Amit Chakrabarti D. Cornelison Corbin Covault Kaushik De William E. Dieterle James Dunne R. Eagleton Gregory Earle William Ellis Mark Eriksson Morten Eskildsen Bernard Feldman Mirela Fetea J. D. Garcia U. Garg Michael Gurvitch David Harrison John Hernandez L. Hodges Jean-Pierre Jouas Kevin Kimberlin Sebastian Kuhn Tiffany Landry Dean Lee Frank Lee Stephen Luzader Kevin Martus M. Matkovich David McIntyre Rahul Mehta Kenneth Mendelson Laszlo Mihaly Richard Mistrick Rabindra Mohapatra Philip P. J. Morrison Greg Mowry David Murdock Anthony J. Nicastro Scott Nutter Robert Oerter Ray H. O’Neal, Jr. Frederick Oho, Paul Parris Ashok Puri Michael Richmond John Rollino David Schaefer Joseph Serene H. Shenton Jason Stevens

University of North Carolina–Charlotte Appalachian State University University of Maryland Lafayette College Florida Institute of Technology University of California-Berkeley University of Evansville Walsh University Kansas State University Northern Arizona University Case Western Reserve University University of Texas at Arlington California University of Pennsylvania Mississippi State University California Polytechnic University-Pomona University of Texas-Dallas University of California-Davis University of Wisconsin-Madison University of Notre Dame University of Missouri–St. Louis University of Richmond University of Arizona University of Notre Dame State University of New York at Stony Brook University of Toronto University of North Carolina–Chapel Hill Iowa State University United Nations International School Bradley University Old Dominion University Folsom Lake College North Carolina State University George Washington University Frostburg State University William Paterson University Oakton Community College Oregon State University University of Central Arkansas Marquette University State University of New York at Stony Brook Pennsylvania State University University of Maryland University of Texas at Austin University of Saint Thomas Tennessee Technological University West Chester University Northern Kentucky University George Mason University Florida A & M University Winona State University University of Missouri–Rolla University of New Orleans Rochester Institute of Technology Rutgers University–Newark Towson State University Georgetown University University of Delaware Deerfield Academy

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Jay Strieb John Swez Devki N. Talwar Chin-Che Tin Tim Usher Andrew Wallace Barrett Wells Edward A.P. Whittaker David Wick Don Wieber J. William Gary Suzanne Willis Thomas Wilson William. J. F. Wilson Brian Woodahl Hai-Sheng Wu

Villanova University Indiana State University Indiana University of Pennsylvania Auburn University California State University-San Bernardino Angelo State University University of Connecticut Stevens Institute of Technology Clarkson University Contra Costa College University of California-Riverside Northern Illinois University Marshall University University of Calgary Indiana University–Purdue University-Indianapolis Mankato State University

We thank John Belcher, Michael Danziger, and Mark Bessette of the Massachusetts Institute of Technology for creating the cover image. It illustrates the magnetic field generated by two currents in two copper rings. This is one frame of a continuous animation; at the instant shown, the current in the upper ring is opposite to that in the lower ring and is of smaller magnitude. The magnetic field structure shown in this picture was calculated using a modified intregration technique. This image was created as part of the Technology Enabled Active Learning (TEAL) program in introductory physics at MIT, which teaches physics interactively, combining desktop experiments with visualizations of those experiments to “make the unseen seen.” We thank the several editors that supervised this project: first Stephen Mosberg, then Richard Mixter, John Byram, and finally Leo Wiegman, who had the largest share in the development of the text, and also gave us the benefit of his incisive lineby-line editing of the proofs, catching many slips and suggesting many improvements. We also thank the editorial staff at W. W. Norton & Co., including Chris Granville, April Lange, Roy Tedoff, Rubina Yeh, Rob Bellinger, Kelly Mitchell, Neil Hoos, Lisa Rand, and Sarah Mann, as well as the publishing professionals whom Norton engaged, such as Paul Lacy, Richard K. Mickey, Susan McLaughlin, and John B. Woolsey for their enthusiasm and their patience in dealing with the interminable revisions and corrections of the text and its support package. In addition, JTM is grateful to Robert W. Christy of Dartmouth University for various pointers on textbook writing.

HANS C. OHANIAN

JOHN T. MARKERT

Burlington, Vermont [email protected]

Austin, Texas [email protected]

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Publication Formats Physics for Engineers and Scientists comprises six parts. The text is published in two hardcover versions and several paperback versions.

H a r d c o v e r Ve r s i o n s Third Extended Edition, Parts I–VI, 1450 pages, ISBN 0-393-92631-1 (Chapters 1–41 including Relativity, Quanta and Particles) Third Edition, Parts I–V, 1282 pages, ISBN 0-393-97422-7 (Chapters 1–36, including Special Relativity)

P a p e r b a c k Ve r s i o n s Volume 1, (Chapters 1–21) 778 pages, ISBN 0-393-93003-3 Part I Motion, Force, and Energy (Chapters 1–14) Part II Oscillations, Waves, and Fluids (Chapters 15–18) Part III Temperature, Heat, and Thermodynamics (Chapter 19–21) Volume 2, (Chapters 22–36) 568 pages, ISBN 0-393-93004-1 Part IV Electricity and Magnetism (Chapters 22–32) Part V Waves and Optics (Chapters 33–35 and Chapter 36 on Special Relativity) Volume 3, (Chapters 36–41) 250 pages, ISBN 0-393-92969-8 Part VI Relativity, Quanta, and Particles In addition, to explore customized versions, please contact your Norton representative.

Tw o N o r t o n e b o o k O p t i o n s Physics for Engineers and Scientists is available in a Norton ebook format that retains the content of the print book. The ebook offers a variety of tools for study and review, including sticky notes, highlighters, zoomable images, links to Online Concept Tutorials, and a search function. Purchased together, the SmartWork with integrated ebook bundle makes it easy for students to check text references when completing online homework assignments. The ebook may also be purchased as a standalone item. The downloadable PDF version is available for purchase from Powells.com.

Package Options Each version of the text purchased from Norton—with or without SmartWork—will come with free access to our website at Norton’s StudySpace that includes the valuable Online Concept Tutorials. Each version of the text may be purchased as a stand-alone book or as a package that includes—each for a fee—Norton’s new SmartWork online homework system or the Student Activity Workbook by David Harrison and William Ellis. Hence, several optinal packages are available to instructors: • Textbook–StudySpace–Online Concept Tutorials + Student Activity Workbook • Textbook–StudySpace–Online Concept Tutorials + SmartWork/ebook • Textbook–StudySpace–Online Concept Tutorials + SmartWork/ebook + Student Activity Workbook

nortonebooks.com

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The Suppor t Program To enhance individual learning and also peer instruction, a carefully integrated support program accompanies the text. Each element of the support program has two goals. First, each support resource mirrors the text’s emphasis on sharply focused core concepts. Second, treatment of a core concept in a support resource offers a perspective that is different from but compatible with that of the text. If a student needs help beyond the text, he or she would more likely benefit from a fresh presentation on the same concept rather than from one that simply repeats the text presentation. Hence, the text and its support package offers three or more different approaches to the core concepts. For example, Newton’s First and Second Laws are rendered with interactive animations in the Online Concept Tutorial “Forces,” with pencil-and- paper exercises in Chapter 5 of the Student Activity Workbook crafted by David Harrison and William Ellis, and with concept test inquiries in PhysiQuiz questions written by Charles Chiu and edited by Jason Stevens. Both printed and digital resources are offered within the support program. Outstanding web-based resources for both instructors and students include tutorials and a homework system.

S m a r t Wo r k O n l i n e H o m e w o r k S y s t e m

www.wwnorton.com/physics

SmartWork—Norton’s online homework management system—provides ready-made automatically graded assignments, including guided problems, simple feedback questions, and animated tutorials—all specifically designed to extend the text’s emphasis on core concepts and problem-solving skills. Developed in collaboration with Sapling Systems, SmartWork features an intuitive, easy-to-use interface that offers instructors flexible tools to manage assignments, while making it easy for students to compose mathematical expressions, draw vectors and graphs, and receive helpful and immediate feedback. Two different types of questions expand upon the exposition of concepts in the text: Simple Feedback Problems present students with problems that anticipate common misconceptions and offer prompts at just the right moment to help them discover the correct solution. Guided Tutorial Problems addresses more challenging topics. If a student answers a problem incorrectly, SmartWork guides the student through a series of discrete tutorial steps that lead to a general solution. Each step is a simple feedback question that the student answers, with hints if necessary. After completing all of the tutorial steps, the student returns to the original problem ready to apply this newly-obtained knowledge. SmartWork problems use algorithmic variables so two students are unlikely to see exactly the same problem. Instructors can use the problem sets provided, or can customize these ready-made questions and assignments, or use SmartWork to create their own. SmartWork is available bundled with the Norton ebook of Physics for Engineers and Scientists. Where appropriate, SmartWork prompts students to review relevant sections in the textbook. Links to the ebook make it easy for students to consult the text while working through problems online.

O n l i n e C o n c e p t Tu t o r i a l s

www.wwnorton.com/physics

Developed in collaboration with Science Technologies specifically for this course, these 45 tutorials feature interactive animations that reinforce conceptual understanding and develop students’ quantitative skills. In-text icons alert students to

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the availability of a tutorial. All Online Concept Tutorials are available on the free StudySpace web site and are integrated into SmartWork. Tutorials can also be accessed from a CD-ROM that requires no installation, browser tune-ups, or plug-ins.

S t u d y S p a c e We b s i t e www.wwnorton.com/physics. This free and open website is the portal for both public and premium content. Free content at StudySpace includes the Online Concept Tutorials and a Study Plan for each chapter in Physics for Engineers and Scientists. Premium content at StudySpace includes links to the online ebook and to SmartWork.

STUDYSPACE

We b A s s i g n Selected end-of-chapter problems from Physics for Engineers and Scientists are available in WebAssign, with additional problems available to adopting instructors by request to WebAssign.

Additional Instructor Resources by Mirela Fetea, University of Richmond; Kevin Martus, William Paterson University; and Brian Woodahl, Indiana University-Purdue University-Indianapolis. The Test Bank offers approximately 2000 multiple-choice questions, available in ExamView, WebCT, BlackBoard, rich-text, and printed format.

TEST BANK

by Stephen Luzader and Hang-Deng Luzader, both of Frostburg State University, and David Marx of Illinois State University. The Instructor Solution Manual offers solutions to all end-of-chapter Problems and Review Problems, checked for accuracy and clarity.

INSTRUCTOR SOLUTIONS MANUAL

by Charles Chiu, University of Texas at Austin, with Jason Stevens, Deerfield Academy. The PhysiQuiz multiple-choice questions are designed for use with classroom response, or “clicker”, systems. The 300 PhysiQuiz questions are available as PowerPoint slides, in printed format, and as transparency masters.

PHYSIQUIZ “CLICKER” QUESTIONS

NORTON MEDIA LIBRARY INSTRUCTOR CD-ROM The Media Library for instrutors includes

selected figures, tables, and equations from the text in JPEG and PowerPoint formats, PhysiQuiz “clicker” questions, and PowerPoint-ready offline versions of the Online Concept Tutorials. offers a guide to the support package with descriptions of the Online Concept Tutorials, information about the SmartWork homework problems available for each chapter, printed PhysiQuiz “clicker” questions, and instructor notes for the workshop activities in the Student Activity Workbook.

INSTRUCTOR RESOURCE MANUAL

TRANSPARENCY ACETATES

Approximately 200 printed color acetates of key figures from

the text. Course Cartridges for BlackBoard and WebCT include access to the Online Concept Tutorials, a Study Plan for each chapter, multiple-choice tests, plus links to the premium, password-protected contents of the Norton ebook and SmartWork.

BLACKBOARD AND WEBCT COURSE CARTRIDGES

www.wwnorton.com/physics

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Additional Student Resources by David Harrison, University of Toronto, and William Ellis, University of California Davis. The Student Activity Workbook is an important part of the learning package. For each chapter of Physics for Engineers and Scientists, the Workbook’s Activities break down a physical condition into constituent parts. The Activities are pencil and paper exercises well suited to either individual or small group collaboration.The Activities include both conceptual and quantitative exercises. Some Activities are guided problems that pose a question and present a solution scheme via follow up questions. The Workbook is available in two paperback volumes: Volume 1 comprises Chapters 1–21 and Volume 2 comprises Chapters 22–41.

STUDENT ACTIVITY WORKBOOK

by Stephen Luzader and Hang-Deng Luzader, both of Frostburg State University, and David Marx of Illinois State University. The Student Solutions Manual contains detailed solutions to approximately 25% of the problems in the book, chosen from the odd-numbered problems whose answers appear in the back of the book. The Manual is available in two paperback volumes: Volume 1 comprises Chapters 1–21 and Volume 2 comprises Chapters 22–41.

STUDENT SOLUTIONS MANUAL

ONLINE CONCEPT TUTORIALS CD-ROM The

45 Online Concept Tutorials (see above) can also be accessed from an optional CD-ROM that requires no installation, browser tune-up, or plug-in.

About the Authors Hans C. Ohanian received his B.S. from the University of California, Berkeley, and his Ph.D from Princeton University, where he worked with John A. Wheeler. He has taught at Rensselaer Polytechnic Institute, Union College, and the University of Vermont. He is the author of several textbooks spanning all undergraduate levels: Physics, Principles of Physics, Relativity: A Modern Introduction, Modern Physics, Principles of Quantum Mechanics, Classical Electrodynamics, and, with Remo Ruffini, Gravitation and Spacetime. He is also the author of dozens of articles dealing with gravitation, relativity, and quantum theory, including many articles on fundamental physics published in the American Journal of Physics, where he served as associate editor for some years. He lives in Vermont. [email protected]

John T. Markert received his B.A. in physics and mathematics from Bowdoin College (1979), and his M.S. (1984) and Ph.D. (1987) in physics from Cornell University, where he was recipient of the Clark Award for Excellence in Teaching. After postdoctoral research at the University of California, San Diego, he joined the faculty at the University of Texas at Austin in 1990, where he has received the College of Natural Sciences Teaching Excellence Award and is currently Professor of Physics and Department Chair. His introductory physics teaching methods emphasize context-based approaches, interactive techniques, and peer instruction. He is author or coauthor of over 120 journal articles, including experimental condensed-matter physics research in superconductivity, magnetism, and nanoscience. He lives in Austin, Texas, with his spouse and four children. [email protected]

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Owner’s Manual for Physics for Engineers and Scientists These pages give a brief tour of the features of Physics for Engineers and Scientists and its study resources. Some resources are found within the book. Others are located in accompaning paperback publications or at the StudySpace web portal. Features on the text pages shown here come chiefly from the discussion of friction in Chapter 6, but are common in other chapters. The learning resources listed below help students study by offering alternative explanations of the core concepts found in the text. These student resources are briefly described at the end of this owner’s manual: • Online Concept Tutorials

• Student Activity Workbook

• SmartWork Online Homework

• Student Solutions Manual

• StudySpace

Each chapter of the textbook starts with a real-world example of a core concept. Chapter 6 opens with the concept of friction and uses automobile tires as an example of friction that is revisited in several different conditions. The opening photograph, it’s caption and the caption’s closing questions all discuss this example.

Most chapters have six or fewer sections. Most sections are four or five pages in length and cover one major topic.

In this chapter, the rubber tires of an automobile are revisited to explore concepts in friction on pages 176, 179, 180, and 186, as indicated.

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In mathematical expressions, such as ma=F, the bold type indicates a vector and italic indicates variables that are not vectors. The icon indicates an Online Concept Tutorial is available for a key concept. Each such icon includes the identification number of the tutorial—8, in this case. These tutorials offer a visual guide and selfquiz for the concept at hand. Find all the Tutorials at www.wwnorton.com/physics.

Text in italic type indicates major definitions of laws or statements of general principles. Text in bold type highlights the first use of a key term and is generally accompanied by an explanation.

Key concepts or important variants of these concepts have a key-term label in the margin.

Highlighted equations are key equations that express central physics concepts mathematically.

Short biographical sketches appear in the margins of this text. Each offer a brief glimpse into the life of some major contributor to our knowledge about the physical world—in this case, Italian artist and engineer Leonardo da Vinci.

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Examples are a critical part of each chapter. • Examples provide concrete illustrations of the concepts being discussed. • As the chapter unfolds, Examples progress from simple to more complex. Throughout the text, figures often build on each other with a new layer of information. • Balloon comments often point out components of special note in the figure.

(b) y

(a) A push at an angle has both horizontal and vertical components. 30

P fk

P

The Concept in Context icon here indicates the chapter-opening example —automobile tires—is being revisited. In this Example, we explore the slowing down of a skidding automobile with a specific coefficient of kinetic friction for a rubber tire.

N

Px Py

O

w

x

Solutions in Examples may cover both general approaches and specific details on how to extract the information from the problem statement.

Comments occasionally close an Example to point out the particular limitations and broader implications of a Solution.

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A Checkup appears at the end of each section within a chapter. • Each Checkup is a self-quiz to test the reader’s mastery of the concepts in the preceding section. • Each Checkup has an answer (see below).

Problem-Solving Techniques boxes appear in relevant places throughout the book and offers tips on how to approach problems of a particular kind—in this case, problems involving the use of friction or centripetal force.

Answers to Checkups appear at the very back of each chapter, after the Review Problems.

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Math Help boxes offer specific mathematical guidance at the initial location in the text where that technique is most relevant. • In this case in Chapter 9, ellipses are important in studying orbits. • Additional math help is available in Appendices 2, 3, 4, and 5 at the back of the textbook.

Throughout the text, Physics in Practice boxes offer specific details on a real-world application of the concept under discussion—in this case, forces at work in automobile collisions in Chapter 11.

The text frequently offers tables of typical values of physical quantities. • Such tables usually are labeled “Some ...,” as in this case, from Chapter 5. • These tables give some impression of the magnitudes encountered in the real world.

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Each chapter closes with a Summary followed by Questions for Discussion, Problems, Review Problems, and Answers to Checkups.

A Summary lists the subjects and page references for any special content in this chapter—such as Math Help, Problem-Solving Techniques, or Physics in Practice boxes. • Next the Summary lists the chapter’s core concepts in the order they are treated. The concept appears on the left in bold. • The mathematical expression for the concept appears in the middle column with an equation number on the far right.

About 15 or more Questions for Discussion follow the Summary in each chapter. • These questions require thought, but not calculation; e.g. “Why are wet streets slippery?” • Some of these questions are intended as brain teasers that have no unique answer, but will lead to provocative discussions.

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About the Book

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About 70 Problems and 15 Review Problems follow each chapter’s Questions for Discussion. • The Problem’s statement contains data and conditions upon which a solution will hinge. • Problems are grouped by chapter section and proceed from simple to more complex within each section. • Many Problems employ real-world data and occasionally may introduce applications beyond those treated in the chapter.

Review Problems are specifically designed to help students prepare for examinations. • Review Problems often test comprehension of concepts from more than one section within the chapter. • Review Problems often take a guided approach by posing series of questions that build on each other.

The dagger footnote (†) that accompanies a Problem heading—in this case, “6.1 Friction”—indicates the availability of an Online Concept Tutorial on this specific topic and states its web address.

Problems and Review Problems are marked by level of difficulty: • Those without an asterisk are the most common and require very little manipulation of existing equations; or they may merely require retracing the steps of a worked Example. • Problems marked with one asterisk (*) are of medium difficulty and may require use of several concepts and manipulation more than one equation to isolate and solve for the unknown variable. • Problems marked with two asterisks (**) are challenging, demand considerable thought, may require significant mathematical skill, and are the least common.

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OWNER’S MANUAL

O n l i n e C o n c e p t Tu t o r i a l s www.wwnorton.com/physics

A n Online Concept Tutorial accompanies many central topics in this textbook. When a Tutorial is available, its numbered icon appears at section heading within the chapter and a dagger footnote appears in the end-of-chapter Problems section as reminder. These Tutorials are digitally delivered, either via the Internet or via a CD-ROM for those without Internet access.

Many Tutorials contain online experiments— in this case, determining how the kinetic friction force varies with the normal force and with the choice of materials.

The Online Concept Tutorials listed here indicate each textbook section supported by the tutorial (in paratheses).

1 Unit Conversion 1.5, 1.6 2 Significant Digits 1.6 3 Acceleration 2.4, 2.5, 2.6 4 Vector Addition and Vector Components 3.1, 3.2, 3.3 5 Projectile Motion 4.4 6 Forces 5.4 7 “Free-Body” Diagrams 5.3, 5.5, 5.6 8 Friction 6.1 9 Work of a Variable Force 7.1, 7.2, 7.4 10 Conservation of Energy 8.1, 8.2, 8.3 11 Circular Orbits 9.1, 9.3 12 Kepler’s Laws 9.4 13 Momentum in Collisions 11.1, 11.3 14 Elastic and Inelastic Collisions 11.2, 11.3 15 Rotation about a Fixed Axis 12.2 16 Oscillations and Simple Harmonic Motion 15.1 17 Simple Pendulum 15.4 18 Wave Superposition 16.3, 16.4 19 Doppler Effect 17.4 20 Fluid Flow 18.1, 18.2, 18.6 21 Ideal-Gas Law 19.1 22 Specific Heat and Changes of State 20.1, 20.4

The online experiments allow students to change independent variables—in this case, mass and material. • Students may collect and display data in a built-in lab notebook. • Each Tutorial includes an interactive self-quiz.

23 Heat Engines 21.2 24 Coulomb’s Law 22.2 25 Electric Charge 22.1, 22.5 26 Electric Force Superposition 22.3 27 Electric Field 23.1, 23.3 28 Electric Flux 24.1 29 Gauss’ Law 24.2, 24.3 30 Electrostatic Potential 25.1, 25.2, 25.4 31 Superconductors 27.3 32 DC Circuits 28.1, 28.2, 28.3, 28.4, 28.7 33 Motion in a Uniform Magnetic Field 30.1 34 Electromagnetic Induction 31.2, 31.3 35 AC Circuits 32.1, 32.2, 32.3, 32.5 36 Polarization 33.3 37 Huygens’ Construction 34.1, 34.2, 34.3 38 Geometric Optics and Lenses 34.4, 34.5 39 Interference and Diffraction 35.3, 35.5 40 X-ray Diffraction 35.4 41 Special Relativity 36.1, 36.2 42 Implications of Special Relativity 36.2, 36.3 43 Bohr Model of the Atom 38.1, 38.2, 38.4 44 Quantum Numbers 39.1, 39.2 45 Radioactive Decay 40.4

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About the Support Resources

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S martWork is a subscription-based online homework-management system that makes www.wwnorton.com/physics

Simple Feedback Problems anticipate common misconceptions and offer prompts at just the right moment to help students reach the correct solution.

Guided Tutorial Problems address challenging topics. • If a student solves one of these problems incorrectly, she is presented with a series of discrete tutorial steps that lead to a general solution. • Each step includes hinting and feedback. After working through these remedial steps, the student returns to a restatement of the original problem, ready to apply this newly obtained knowledge.

SmartWork is available as a standalone purchase, or with an integrated ebook version of Physics for Engineers and Scientists. • Where appropriate, SmartWork prompts students to review relevant sections of the text. • Links to the ebook make it easy for students consult the text while working through problems.

it easy for instructors to assign, collect and grade end-of-chapter problems from Physics for Engineers and Scientists. Built-in hinting and feedback address common misperceptions and help students get the maximum benefit from these assignments.

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OWNER’S MANUAL

S t u d e n t A c t i v i t y Wo r k b o o k T he Student Activity Workbook is available in two paperback volumes: Volume 1 comprises Chapters 1–21 and Volume 2 comprises Chapters 22–41.

For each chapter of the textbook, the Student Activity Workbook offers Activities designed to break down a physical condition into constituent parts. • The Activities are unique to the Workbook and not found in the textbook. • The Activities are pencil and paper exercises well suited to either individual or small group collaboration. • The Activities include both conceptual and quantitative questions.

Student Solution Manual T he Student Solution Manual is available in two paperback volumes: Volume 1 comprises Chapters 1–21 and Volume 2 comprises Chapters 22–41.

StudySpace

The Student Solutions Manual contains worked solutions for about 50% of the odd-numbered Problems and Review Problems in the text. • Appendix 11 in the back of the textbook contains only the final answer for odd-numbered problems in the chapters, not the intermediate steps of the solutions.

w w w. w w n o r t o n . c o m / p h y s i c s

T he StudySpace website is the free and open portal through which students access the resources that accompany this text. • 45 Online Concept Tutorials—at no additional cost.

STUDY PLANS

• 41 Study Plans, one for each chapter—at no additional cost. • Smartwork online homework system—a subscription service. • ebook links to textbook chapters—as part of subscription service.

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Prelude The World of Physics

P

hysics is the study of matter. In a quite literal sense, physics is the greatest of all natural sciences: it encompasses the smallest particles, such as electrons and quarks; and it also encompasses the largest bodies, such as galaxies and the entire Universe. The smallest particles and the largest bodies differ in size by a factor of more than ten thousand billion billion billion billion! In the pictures on the following pages we will survey the world of physics and attempt to develop some rough feeling for the sizes of things in this world. This preliminary survey sets the stage for our explanations of the mechanisms that make things behave in the way they do. Such explanations are at the heart of physics, and they are the concern of the later chapters of this book. Since the numbers we will be dealing with in this prelude and in the later chapters are often very large or very small, we will find it convenient to employ the scientific notation for these numbers. In this notation, numbers are written with powers of 10; thus, hundred is written as 102, thousand is written as 103, ten thousand is written as 104, and so on. A tenth is written as 101, a hundredth is written as 102, a thousandth is written as 103, and so on. The following table lists some powers of ten: 10  101 100  102 1000  103 10000  104 100000  105 1000000  106

0.1  1/10  101 0.01  1/100 102 0.001  1/1000  103 0.0001  1/10000  104 0.00001  1/100000  105 0.000001  1/1000000  106 etc.

Note that the power of 10, or the exponent on the 10, simply tells us how many zeros follow the 1 in the number (if the power of 10 is positive) or how many zeros follow the 1 in the denominator of the fraction (if the power of 10 is negative). In scientific notation, a number that does not coincide with one of the powers of 10 is written as a product of a decimal number and a power of 10. For example, in this notation, 1500000000 is written as 1.5  109. Alternatively, this number could be written as 15  108 or as 0.15  1010; but in scientific notation it is customary to place the decimal point immediately after the first nonzero digit. The same rule applies to numbers smaller than 1; thus, 0.000 015 is written as 1.5  105. The pictures on the following pages fall into two sequences. In the first sequence we zoom out: we begin with a picture of a woman’s face and proceed step by step to pictures of the entire Earth, the Solar System, the Galaxy, and the Universe. This ascending sequence contains 27 pictures, with the scale decreasing in steps of factors of 10. Most of our pictures are photographs. Many of these have become available only in recent years; they were taken by high-flying aircraft, Landsat satellites, astronauts, or sophisticated electron microscopes. For some of our pictures no photographs are available and we have to rely, instead, on drawings.

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PA RT I : T H E L A R G E - S C A L E W O R L D

Fig. P1 SCALE 1:1.5 This is Erin, an intelligent biped of the planet Earth, Solar System, Orion Spiral Arm, Milky Way Galaxy, Local Group, Local Supercluster. Erin belongs to the phylum Chordata, class Mammalia, order Primates, family Hominidae, genus Homo, species sapiens. She is made of 5.4  1027 atoms, with 1.9  1028 electrons, the same number of protons, and 1.5  1028 neutrons.

0

0

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10−1 m

100 m

Fig. P2 SCALE 1:1.5  10 Erin has a height of 1.7 meters and a mass of 57 kilograms. Her chemical composition (by mass) is 65% oxygen, 18.5% carbon, 9.5% hydrogen, 3.3% nitrogen, 1.5% calcium, 1% phosphorus, and 1.2% other elements. The matter in Erin’s body and the matter in her immediate environment occur in three states of aggregation: solid, liquid, and gas. All these forms of matter are made of atoms and molecules, but solid, liquid, and gas are qualitatively different because the arrangements of the atomic and molecular building blocks are different. In a solid, each building block occupies a definite place. When a solid is assembled out of molecular or atomic building blocks, these blocks are locked in place once and for all, and they cannot move or drift about except with great difficulty. This rigidity of the arrangement is what makes the aggregate hard—it makes the solid “solid.” In a liquid, the molecular or atomic building blocks are not rigidly connected. They are thrown together at random and they move about fairly freely, but there is enough adhesion between neighboring blocks to prevent the liquid from dispersing. Finally, in a gas, the molecules or atoms are almost completely independent of one another. They are distributed at random over the volume of the gas and are separated by appreciable distances, coming in touch only occasionally during collisions. A gas will disperse spontaneously if it is not held in confinement by a container or by some restraining force.

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Fig. P3 SCALE 1:1.5  102 The building behind Erin is the New York Public Library, one of the largest libraries on Earth. This library holds 1.4  1010 volumes, containing roughly 10% of the total accumulated knowledge of our terrestrial civilization.

0

0.5 × 101

101 m

0

0.5 × 102

102 m

Fig. P4 SCALE 1:1.5  103 The New York Public Library is located at the corner of Fifth Avenue and 42nd Street, in the middle of New York City, with Bryant Park immediately behind it.

Fig. P5 SCALE 1:1.5  104 This aerial photograph shows an area of 1 kilometer  1 kilometer in the vicinity of the New York Public Library. The streets in this part of the city are laid out in a regular rectangular pattern. The library is the building in the park in the middle of the picture. The photograph was taken early in the morning, and the high buildings typical of New York cast long shadows. The photograph was taken from an airplane flying at an altitude of a few thousand meters. North is at the top of the photograph.

0

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Fig. P6 SCALE 1:1.5  105 This photograph shows a large portion of New York City. We can barely recognize the library and its park as a small rectangular patch slightly above the center of the picture. The central mass of land is the island of Manhattan, with the Hudson River on the left and the East River on the right. This photograph and the next two were taken by satellites orbiting the Earth at an altitude of about 700 kilometers.

0

0.5 × 104

104 m

Fig. P7 SCALE 1:1.5  106 In this photograph, Manhattan is in the upper middle. On this scale, we can no longer distinguish the pattern of streets in the city. The vast expanse of water in the lower right of the picture is part of the Atlantic Ocean. The mass of land in the upper right is Long Island. Parallel to the south shore of Long Island we can see a string of very narrow islands; they almost look man-made. These are barrier islands; they are heaps of sand piled up by ocean waves in the course of thousands of years.

0

0.5 × 105

105 m

Fig. P8 SCALE 1:1.5  107 Here we see the eastern coast of the United States, from Cape Cod to Cape Fear. Cape Cod is the hook near the northern end of the coastline, and Cape Fear is the promontory near the southern end of the coastine. Note that on this scale no signs of human habitation are visible. However, at night the lights of large cities would stand out clearly. This photograph was taken in the fall, when leaves had brilliant colors. Streaks of orange trace out the spine of the Appalachian mountains.

0

0.5 × 106

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Fig. P9 SCALE 1:1.5  108 In this photograph, taken by the Apollo 16 astronauts during their trip to the Moon, we see a large part of the Earth. Through the gap in the clouds in the lower middle of the picture, we can see the coast of California and Mexico. We can recognize the peninsula of Baja California and the Gulf of California. Erin’s location, the East Coast of the United States, is covered by a big system of swirling clouds on the right of the photograph. Note that a large part of the area visible in this photograph is ocean. About 71% of the surface of the Earth is ocean; only 29% is land. The atmosphere covering this surface is about 100 kilometers thick; on the scale of this photograph, its thickness is about 0.7 millimeter. Seen from a large distance, the predominant colors of the planet Earth are blue (oceans) and white (clouds).

0

0.5 × 107

107 m

0

0.5 × 108

108 m

Fig. P10 SCALE 1:1.5  109 This photograph of the Earth was taken by the Apollo 16 astronauts standing on the surface of the Moon. Sunlight is striking the Earth from the top of the picture. As is obvious from this and from the preceding photograph, the Earth is a sphere. Its radius is 6.37  106 meters and its mass is 5.98  1024 kilograms.

Fig. P11 SCALE 1:1.5  1010 In this drawing, the dot at the center represents the Earth, and the solid line indicates the orbit of the Moon around the Earth (many of the pictures on the following pages are also drawings). As in the preceding picture, the Sun is far below the bottom of the picture. The position of the Moon is that of January 1, 2000. The orbit of the Moon around the Earth is an ellipse, but an ellipse that is very close to a circle. The solid red curve in the drawing is the orbit of the Moon, and the dashed green curve is a circle; by comparing these two curves we can see how little the ellipse deviates from a circle centered on the Earth. The point on the ellipse closest to the Earth is called the perigee, and the point farthest from the Earth is called the apogee. The distance between the Moon and the Earth is roughly 30 times the diameter of the Earth. The Moon takes 27.3 days to travel once around the Earth.

Jan.

1

2 n.

17

Ja

Jan.

apogee

Jan

.9

Jan. 25

13

perigee

, .1 Jan 000 2

Jan. 5

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Jan. 2

Jan. 1, 2000

Dec. 31

0.5 × 1010

0

Fig. P12 SCALE 1:1.5  1011 This picture shows the Earth, the Moon, and portions of their orbits around the Sun. On this scale, both the Earth and the Moon look like small dots. Again, the Sun is far below the bottom of the picture. In the middle, we see the Earth and the Moon in their positions for January 1, 2000. On the right and on the left we see, respectively, their positions for 1 day before and 1 day after this date. Note that the net motion of the Moon consists of the combination of two simultaneous motions: the Moon orbits around the Earth, which in turn orbits around the Sun.

1010 m

Fig. P13 SCALE 1:1.5  1012 Here we see the orbits of the Earth and of Venus. However, Venus itself is beyond the edge of the picture. The small circle is the orbit of the Moon. The dot representing the Earth is much larger than what it should be, although the artist has drawn it as minuscule as possible. On this scale, even the Sun is quite small; if it were included in this picture, it would be only 1 millimeter across.

0.5 × 1011

0

1011 m

Halley

elion

perih

Earth Venus Mars

Mercury

on

apheli

0

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Fig. P14 SCALE 1:1.5  1013 This drawing shows the positions of the Sun and the inner planets: Mercury, Venus, Earth, and Mars. The positions of the planets are those of January 1, 2000. The orbits of all these planets are ellipses, but they are close to circles. The point of the orbit nearest to the Sun is called the perihelion and the point farthest from the Sun is called the aphelion. The Earth reaches perihelion about January 3 and aphelion about July 6 of each year. All the planets travel around their orbits in the same direction: counterclockwise in our picture. The marks along the orbit of the Earth indicate the successive positions at intervals of 10 days. Beyond the orbit of Mars, a large number of asteroids orbit around the Sun; these have been omitted to prevent excessive clutter. Furthermore, a large number of comets orbit around the Sun. Most of these have pronounced elliptical orbits. The comet Halley has been included in our drawing.

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The Sun is a sphere of radius 6.96  108 meters. On the scale of the picture, the Sun looks like a very small dot, even smaller than the dot drawn here. The mass of the Sun is 1.99  1030 kilograms. The matter in the Sun is in the plasma state, sometimes called the fourth state of matter. Plasma is a very hot gas in which violent collisions between the atoms in their random thermal motion have fragmented the atoms, ripping electrons off them. An atom that has lost one or more electrons is called an ion. Thus, plasma consists of a mixture of electrons and ions engaging in frequent collisions. These collisions are accompanied by the emission of light, making the plasma luminous.

Ha

Fig. P15 SCALE 1:1.5  1014 This picture shows the positions of the outer planets of the Solar System: Jupiter, Saturn, Uranus, Neptune, and Pluto. On this scale, the orbits of the inner planets are barely visible. As in our other pictures, the positions of the planets are those of January 1, 2000. The outer planets move slowly and their orbits are very large; thus they take a long time to go once around their orbit. The extreme case is that of Pluto, which takes 248 years to complete one orbit. Uranus, Neptune, and Pluto are so far away and so faint that their discovery became possible only through the use of telescopes. Uranus was discovered in 1781, Neptune in 1846, and the tiny Pluto in 1930. Pluto is now known as one of several dwarf planets.

lley Sat

n ur

us

er pit

Ju

Ur P lu

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an

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to

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ptu

ne

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Fig. P16 SCALE 1:1.5  1015 We now see that the Solar System is surrounded by a vast expanse of space. Although this space is shown empty in the picture, the Solar System is encircled by a large cloud of millions of comets whose orbits crisscross the sky in all directions. Furthermore, the interstellar space in this picture and in the succeeding pictures contains traces of gas and of dust. The interstellar gas is mainly hydrogen; its density is typically 1 atom per cubic centimeter.

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Fig. P17 SCALE 1:1.5  1016 More interstellar space. The small circle is the orbit of Pluto.

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Fig. P18 SCALE 1:1.5  1017 And more interstellar space. On this scale, the Solar System looks like a minuscule dot, 0.1 millimeter across.

α Centauri A and B 0

Proxima Centauri

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10 16 m

Fig. P19 SCALE 1:1.5  1018 Here, at last, we see the stars nearest to the Sun. The picture shows all the stars within a cubical box 1017 meters  1017 meters  1017 meters centered on the Sun: Alpha Centauri A, Alpha Centauri B, and Proxima Centauri. All three are in the constellation Centaurus, in the southern sky. The star closest to the Sun is Proxima Centauri. This is a very faint, reddish star (a “red dwarf ”), at a distance of 4.0  1016 meters from the Sun. Astronomers like to express stellar distances in light-years: Proxima Centauri is 4.2 light-years from the Sun, which means light takes 4.2 years to travel from this star to the Sun.

Sun

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Fig. P20 SCALE 1:1.5  1019 This picture displays the brightest stars within a cubical box 1018 meters  1018 meters  1018 meters centered on the Sun. There are many more stars in this box besides those shown—the total number of stars in this box is close to 2000. Sirius is the brightest of all the stars in the night sky. If it were at the same distance from the Earth as the Sun, it would be 28 times brighter than the Sun.

Capella Altair

Vega Sun

Castor Procyon Pollux

Arcturus

Caph

Fomalhaut

Sirius α Cent. Alnair

Menkent

Denebola

Alderamin

0.5 × 1018

0

Fig. P21 SCALE 1:1.5  1020 Here we expand our box to 1019 meters  1019 meters  1019 meters, again showing only the brightest stars and omitting many others. The total number of stars within this box is about 2 million. We recognize several clusters of stars in this picture: the Pleiades Cluster, the Hyades Cluster, the Coma Berenices Cluster, and the Perseus Cluster. Each of these has hundreds of stars crowded into a fairly small patch of sky. In this diagram, Starbursts signify single stars, circles with starbursts indicate star clusters, and a circle with a single star indicate a star cluster with its brightest star.

1018 m

Alkaid Dubhe Eltanin El Nath

PERSEUS Mirfak

COMA

PLEIADES Bellatrix

Hadar

Spica

Antares Acrux

Alioth Regulus

Mimosa

0

Sun

HYADES Achernar

Canopus Miaplacidus Schaula

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10 19 m

Fig. P22 SCALE 1:1.5  1021 This photograph shows a view of the Milky Way in the direction of the constellation Sagittarius. Now there are so many stars in our field of view that they appear to form clouds of stars. There are about a million stars in this photograph, and there are many more stars too faint to show up distinctly. Although this photograph is not centered on the Sun, it is similar to what we would see if we could look toward the Solar System from very far away.

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Fig. P23 SCALE 1:1.5  1022 This is the spiral galaxy NGC 2997. Its clouds of stars are arranged in spiral arms wound around a central bulge. The bright central bulge is the nucleus of the galaxy; it has a more or less spherical shape. The surrounding region, with the spiral arms, is the disk of the galaxy. This disk is quite thin; it has a thickness of only about 3% of its diameter. The stars making up the disk circle around the galactic center in a clockwise direction. Our Sun is in a spiral galaxy of roughly similar shape and size: the Milky Way Galaxy. The total number of stars in this galaxy is about 1011. The Sun is in one of the spiral arms, roughly one-third inward from the edge of the disk toward the center.

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Fig. P24 SCALE 1:1.5  1023 Galaxies are often found in clusters of several galaxies. Some of these clusters consist of just a few galaxies, others of hundreds or even thousands. The photograph shows a cluster, or group, of galaxies beyond the constellation Fornax. The group contains an elliptical galaxy like a luminous yellow egg (center), three large spiral galaxies (left), and a spiral with a bar (bottom left). Our Galaxy is part of a modest cluster, the Local Group, consisting of our own Galaxy, the great Andromeda Galaxy, the Triangulum Galaxy, the Large Magellanic Cloud, plus 16 other small galaxies. According to recent investigations, the dark, apparently empty, space near galaxies contains some form of distributed matter, with a total mass 20 or 30 times as large as the mass in the luminous, visible galaxies. But the composition of this invisible, extragalactic dark matter is not known. Fig. P25 SCALE 1:1.5  1024 The Local Group lies on the fringes of a very large cluster of galaxies, called the Local Supercluster. This is a cluster of clusters of galaxies. At the center of the Local Supercluster is the Virgo Cluster with several thousand galaxies. Seen from a large distance, our supercluster would present a view comparable with this photograph, which shows a multitude of galaxies beyond the constellation Fornax, all at a very large distance from us. The photograph was taken with the Hubble Space Telescope coupled to two very sensitive cameras using an exposure time of almost 300 hours. All these distant galaxies are moving away from us and away from each other. The very distant galaxies in the photo are moving away from us at speeds almost equal to the speed of light. This motion of recession of the galaxies is analogous to the outward motion of, say, the fragments of a grenade after its explosion. The motion of the galaxies suggests that the Universe began with a big explosion, the Big Bang, that launched the galaxies away from each other.

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Fig. P26 SCALE 1:1.5  1025 On this scale a galaxy equal in size to our own Galaxy would look like a fuzzy dot, 0.1 millimeter across. Thus, the galaxies are too small to show up clearly on a photograph. Instead we must rely on a plot of the positions of the galaxies. The plot shows the positions of about 200 galaxies. The dense cluster of galaxies in the lower half of the plot is the Virgo Cluster. Since we are looking into a volume of space, some of the galaxies are in the foreground, some are in the background; but our plot takes no account of perspective. The luminous stars in the galaxies constitute only a small fraction of the total mass of the Universe. The space around the galaxies and the clusters of galaxies contains dark matter, and the space between the clusters contains dark energy, a strange form of matter that causes an acceleration of the expansion of the Universe. 0

0.5 × 10 24

10 24 m

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Fig. P27 SCALE 1:1.5  1026 This plot shows the positions of about 100,000 galaxies in a patch of the sky at distances of up to 1  109 light years from the Earth. The false color in this image indicates the distance–red for shorter distances, blue for larger distances. The visible galaxies plotted here contribute only about 5% of the total mass in the universe. The dark matter near the galaxies contribute another 25%. The remaining 70% of the total mass in the universe is in the form of dark energy, which is uniformly distributed over the vast reaches of intergalactic space. This is the last of our pictures in the ascending series. We have reached the limits of our zoom out. If we wanted to draw another picture, 10 times larger than this, we would need to know the shape and size of the entire Universe. We do not yet know that.

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Fig. P28 SCALE 1:1.5 We now return to Erin and zoom in on her eye. The surface of her skin appears smooth and firm. But this is an illusion. Matter appears continuous because the number of atoms in each cubic centimeter is extremely large. In a cubic centimeter of human tissue there are about 1023 atoms. This large number creates the illusion that matter is continuously distributed—we see only the forest and not the individual trees. The solidity of matter is also an illusion. The atoms in our bodies are mostly vacuum. As we will discover in the following pictures, within each atom the volume actually occupied by subatomic particles is only a very small fraction of the total volume.

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Fig. P29 SCALE 1:1.5  101 Our eyes are very sophisticated sense organs; they collect more information than all our other sense organs taken together. The photograph shows the pupil and the iris of Erin’s eye. Annular muscles in the iris change the size of the pupil and thereby control the amount of light that enters the eye. In strong light the pupil automatically shrinks to about 2 millimeters; in very weak light it expands to as much as 7 millimeters.

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Fig. P30 SCALE 1:1.5  102 This false-color photograph shows the delicate network of blood vessels on the front surface of the retina, the light-sensitive membrane lining the interior of the eyeball. The rear surface of the retina is densely packed with two kinds of cells that sense light: cone cells and rod cells. In a human retina there are about 6 million cone cells and 120 million rod cells. The cone cells distinguish colors; the rod cells distinguish only brightness and darkness, but they are more sensitive than the cone cells and therefore give us vision in faint light (“night vision”). This and the following photographs were made with various kinds of electron microscopes. An ordinary microscope uses a beam of light to illuminate the object; an electron microscope uses a beam of electrons. Electron microscopes can achieve much sharper contrast and much higher magnification than ordinary microscopes. 0

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Fig. P31 SCALE 1:1.5  103 Here we have a false-color photograph of rod cells prepared with a scanning electron microscope (SEM). For this photograph, the retina was cut apart and the microscope was aimed at the edge of the cut. In the top half of the picture we see tightly packed rods. Each rod is connected to the main body of a cell containing the nucleus. In the bottom part of the picture we can distinguish tightly packed cell bodies of the cell.

0

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Fig. P32 SCALE 1:1.5  104 This is a close-up view of a few rods cells. The upper portions of the rods contain a special pigment—visual purple—which is very sensitive to light. The absorption of light by this pigment initiates a chain of chemical reactions that finally trigger nerve pulses from the eye to the brain.

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Fig. P33 SCALE 1:1.5  105 These are strands of DNA, or deoxyribonucleic acid, as seen with a transmission electron microscope (TEM) at very high magnification. DNA is found in the nuclei of cells. It is a long molecule made by stringing together a large number of nitrogenous base molecules on a backbone of sugar and phosphate molecules. The base molecules are of four kinds, the same in all living organisms. But the sequence in which they are strung together varies from one organism to another. This sequence spells out a message—the base molecules are the “letters” in this message. The message contains all the genetic instructions governing the metabolism, growth, and reproduction of the cell. The strands of DNA in the photograph are encrusted with a variety of small protein molecules. At intervals, the strands of DNA are wrapped around larger protein molecules that form lumps looking like the beads of a necklace.

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Fig. P34 SCALE 1:1.5  106 The highest magnifications are attained by a newer kind of electron microscope, the scanning tunneling microscope (STM). This picture was prepared with such a microscope. The picture shows strands of DNA deposited on a substrate of graphite. In contrast to the strands of the preceding picture, these strands are uncoated; that is, they are without protein encrustations.

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Fig. P35 SCALE 1:1.5  107 This close-up picture of strands of DNA reveals the helical structure of this molecule. The strand consists of a pair of helical coils wrapped around each other. This picture was generated by a computer from data obtained by illuminating DNA samples with X rays (Xray scattering).

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Fig. P36 SCALE 1:1.5  108 This picture shows a layer of palladium atoms on surface of graphite as seen with an STM. Here we have visual evidence of the atomic structure of matter. The palladium atoms are arranged in a symmetric, repetitive hexagonal pattern. Materials with such regular arrangements of atoms are called crystals. Each of the palladium atoms is approximately a sphere, about 3  1010 meter across. However, the atom does not have a sharply defined boundary; its surface is somewhat fuzzy. Atoms of other elements are also approximately spheres, with sizes that range from 2  1010 to 4  1010 meter across. At present we know of more than 100 kinds of atoms or chemical elements. The lightest atom is hydrogen, with a mass of 1.67  1027 kilogram; the heaviest is element 114, ununquadium, with a mass about 289 times as large. 0

Fig. P37 SCALE 1:1.5  109 The drawing shows the interior of an atom of neon. This atom consists of 10 electrons orbiting around a nucleus. In the drawing, the electrons have been indicated by small dots, and the nucleus by a slightly larger dot at the center of the picture. These dots have been drawn as small as possible, but even so the size of these dots does not give a correct impression of the actual sizes of the electrons and of the nucleus. The electron is smaller than any other particle we know; maybe the electron is truly pointlike and has no size at all. The nucleus has a finite size, but this size is much too small to show up on the drawing. Note that the electrons tend to cluster near the center of the atom. However, the overall size of the atom depends on the distance to the outermost electron; this electron defines the outer edge of the atom. The electrons move around the nucleus in a very complicated motion, and so the resulting electron distribution resembles a fuzzy cloud, similar to the STM image of the previous picture. This drawing, however, shows the electrons as they would be seen at one instant of time with a hypothetical microscope that employs gamma rays instead of light rays to illuminate an object; no such microscope has yet been built. The mass of each electron is 9.11  1031 kilogram, but most of the mass of the atom is in the nucleus; the 10 electrons of the neon atom have only 0.03% of the total mass of the atom.

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Fig. P38 SCALE 1:1.5  1010 Here we are closing in on the nucleus. We are seeing the central part of the atom. Only two electrons are in our field of view; the others are beyond the margin of the drawing. The size of the nucleus is still much smaller than the size of the dot at the center of the drawing.

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Fig. P39 SCALE 1:1.5  1011 In this drawing we finally see the nucleus in its true size. At this magnification, the nucleus of the neon atom looks like a small dot, 0.5 millimeter in diameter. Since the nucleus is extremely small and yet contains most of the mass of the atom, the density of the nuclear material is enormous. If we could assemble a drop of pure nuclear material of a volume of 1 cubic centimeter, it would have a mass of 2.3  1011 kilograms, or 230 million metric tons! Our drawings show clearly that most of the volume within the atom is empty space. The nucleus occupies only a very small fraction of this volume.

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Fig. P40 SCALE 1:1.5  1012 We can now begin to distinguish the nuclear structure. The nucleus has a nearly spherical shape, but its surface is slightly fuzzy.

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Fig. P41 SCALE 1:1.5  1013 At this extreme magnification we can see the details of the nuclear structure. The nucleus of the neon atom is made up of 10 protons (white balls) and 10 neutrons (red balls). Each proton and each neutron is a sphere with a diameter of about 2  1015 meter, and a mass of 1.67  1027 kilogram. In the nucleus, these protons and neutrons are tightly packed together, so tightly that they almost touch. The protons and neutrons move around the volume of the nucleus at high speed in a complicated motion.

Magnification 6.7  1012 

0

Fig. P42 SCALE 1:1.5  1014 This final picture shows three pointlike bodies within a proton. These pointlike bodies are quarks—each proton and each neutron is made of three quarks. Recent experiments have told us that the quarks are much smaller than protons, but we do not yet know their precise size. Hence the dots in the drawing probably do not give a fair description of the size of the quarks. The quarks within protons and neutrons are of two kinds, called up and down. The proton consists of two up quarks and one down quark joined together; the neutron consists of one up quark and two down quarks joined together. This final picture takes us to the limits of our knowledge of the subatomic world. As a next step we would like to zoom in on the quarks and show what they are made of. According to a speculative theory, they are made of small snippets or loops of strings, 1035m long. But we do not yet have any evidence for this theory.

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Relativity, Quanta, and Particles CONTENTS 36 The Theory of Special Relativity 37 Quanta of Light 38 Spectral Lines, Bohr’s Theory, and Quantum Mechanics 39 Quantum Structure of Atoms, Molecules, and Solids 40 Nuclei 41 Elementary Particles and Cosmology

The solar cells in the 73-meter long photovoltaic arrays on the International Space Station convert solar energy into electrical power. The arrays are fitted with gimbals that angle the arrays toward the Sun at all times so as to maximize the power supplied to the Space Station.

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CHAPTER

36

The Theor y of Special Relativity

CONCEPTS IN CONTEXT 36.1 The Speed of Light; the Ether 36.2 Einstein’s Principle of Relativity 36.3 Time Dilation 36.4 Length Contraction 36.5 The Lorentz Transformations and the Combination of Velocities 36.6 Relativistic Momentum and Energy 36.7 Mass and Energy

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Determination of latitude and longitude by means of radio signals from Global Positioning System (GPS) satellites, such as the one shown here, requires measurement of the travel time from several satellites to the relevant point on the ground. GPS satellites are in relative motion with respect to the Earth’s surface, but the speed of their radio signals is not affected by this motion. In our study of the theory of Special Relativity we will study the propagation of light in different reference frames in relative motion, and we can then ask:

? How are distances calculated from GPS signals, and how do we know that the speed of light is unaffected by the motion of the satellite or by the translational motion of the Earth? (Section 36.1, page 1219)

Concepts in Context

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? Relative to clocks on the surface of the Earth, clocks on GPS satellites are in motion at a somewhat high speed. How does this affect the rate of the clocks? (Example 2, page 1227)

? How does the motion of a GPS satellite relative to the Earth affect the frequency of the radio signal? (Example 3, page 1228)

A

s we saw in Chapter 5, Newton’s laws of motion are equally valid in every inertial reference frame. All inertial reference frames are unaccelerated, but they can differ in their uniform translational motion. Since Newton’s laws make no distinction between different inertial reference frames, no mechanical experiment can detect a uniform translational motion of one inertial reference frame by itself. Such motion can be detected only as a relative motion of one reference frame with respect to another reference frame. This is the Newtonian principle of relativity. For a concrete illustration of this principle, consider the reference frame of a cruise ship moving steadily away from the shore, without acceleration, and consider the reference frame of the shore. Both of these reference frames are inertial, and the behavior of a ball used in a game of tennis on the ship is not different from the behavior of a similar ball on shore—balls on the ship and on the shore accelerate in the same way when subjected to forces, and they obey the same laws of conservation of energy, conservation of momentum, etc.1 Thus, experiments with such balls aboard the ship will not reveal the uniform motion of the ship relative to the shore. To detect this motion, the crew of the ship must take sightings of points on the shore or use some other navigational technique that fixes the position and velocity of the ship relative to the shore. Hence, in regard to mechanical experiments, uniform translational motion of our inertial reference frame is always relative motion—it can be detected only as motion of our reference frame with respect to another reference frame. There is no such thing as absolute motion. The question naturally arises whether the relativity of motion indicated by mechanical experiments also applies to electric, magnetic, optical, and other experiments. Do any of these experiments permit us to detect an absolute motion or our reference frame? In 1905, Albert Einstein proposed that no experiment of any kind should ever permit us to detect such motion. He postulated a principle of relativity for all the laws of physics. This postulate serves as the foundation for Einstein’s theory of Special Relativity, widely regarded as one of the greatest achievements of twentieth-century physics. The theory of relativity requires a drastic revision of our concepts of space and time, and it also requires a drastic revision of Newton’s laws. At high speeds—near the speed of light— particles obey new, relativistic laws which are quite different from Newton’s laws. However, at low speeds—small compared with the speed of light—the differences between Einstein’s and Newton’s theories are usually undetectable. Hence Newton’s laws are adequate for describing the behavior of particles and of other bodies at the relatively low speeds we encounter in our everyday experience. Before we deal with the details of Einstein’s theory of relativity, we will briefly describe why nonmechanical experiments—and, especially, experiments with light— might be expected to detect absolute motion, which mechanical experiments cannot detect.

1

This assumes the ship moves steadily. If the ship lurches forward, or pitches, or rolls, it ceases to be an inertial reference frame, and ball will behave in a manner “inconsistent” with Newton’s laws.

ALBERT EINSTEIN (1879–1955) German (and Swiss, and American) theoretical physicist, professor at Zurich, at Berlin, and at the Institute for Advanced Study at Princeton. Einstein was the most celebrated physicist of the twentieth century. He formulated the theory of Special Relativity in 1905 and the theory of General Relativity in 1916. Einstein also made incisive contributions to modern quantum theory, for which he received the Nobel Prize in 1921. Einstein spent the last years of his life in an unsuccessful quest for a unified theory of forces that was supposed to incorporate gravity and electricity in a single set of equations.

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3 6 . 1 T H E S P E E D O F L I G H T; T H E E T H E R

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(a)

Flash signal travels with speed of light c relative to spaceship.

3108 m/s

(b)

1108 m/s When spacecraft speeds toward Earth…

4108 m/s?

…Galilean addition rule for velocity implies a faster speed of light relative to Earth.

FIGURE 36.1 Velocities according to the Galilean addition rule in (a) reference frame of a spaceship and (b) reference frame of the Earth. A speed of light different from 3  108 m/s is at odds with Maxwell’s equations.

ether

Since the laws of mechanics are the same in all inertial reference frames, it might seem quite natural to assume that the laws of electricity, magnetism, and optics are also the same in all inertial reference frames. But this assumption immediately leads to a paradox concerning the speed of light. As we know from Chapter 33, light is an oscillating electric and magnetic disturbance propagating through space, and Maxwell’s equations permit us to deduce that the speed of propagation of this disturbance must always be2 3.00  108 m/s. The trouble with this deduction is that, according to the Galilean addition rule for velocity [Eq. (4.53)], the speed of light ought not to be the same in all reference frames. For instance, imagine that an alien spaceship approaching the Earth with a speed of, say, 1.00  108 m/s flashes a light signal toward the Earth. If this signal has a speed of 3.00  108 m/s in the reference frame of the spaceship, then the Galilean addition rule tells us that it ought to have a speed of 4.00  108 m/s in the reference frame of the Earth (see Fig. 36.1). To resolve this paradox, either we must give up the notion that the laws of electricity and magnetism (and the values of the speed of light) are the same in all inertial reference frames, or else we must give up the Galilean addition rule for velocities. Both alternatives are unpleasant: the former means that we must abandon all hope for a principle of relativity embracing electricity and magnetism, and the latter means that together with the Galilean addition rule we must abandon the transformation rule for position vectors measured in different reference frames [Eq. (4.50)] as well as the intuitively “obvious” notions of absolute time and length from which these rules are derived. Since the failure of a relativity principle embracing electricity and magnetism might seem to be the lesser of two evils, let us first explore this alternative. Let us assume that there exists a preferred inertial reference frame in which the laws of electricity and magnetism take their simplest form, that is, the form expressed in Maxwell’s equations, Eqs. (33.4)–(33.7). In this reference frame, the speed of light has its standard value of c  3.00  108 m/s, whereas in any other reference frame it is larger or smaller according to the Galilean addition rule. The propagation of light is then analogous to the propagation of sound. There exists a preferred reference frame in which the equations for the propagation of sound waves in, say, air take their simplest form: the reference frame in which the air is at rest. In this reference frame, sound has its standard speed of 331 m/s. In any other reference frame, the equations for the propagation of sound waves are more complicated, but the speed of propagation can always be obtained directly from the Galilean addition rule. For instance, if a wind of 40 m/s (a hurricane) is blowing over the surface of the Earth, then sound waves have a speed of 331 m/s relative to the air, but their speed relative to the ground depends on direction— downwind the speed is 371 m/s, whereas upwind it is 291 m/s. This analogy between the propagation of light and of sound suggests that there might exist some pervasive medium whose oscillations bring about the propagation of light, just as the oscillations of air bring about the propagation of sound. Presumably this ghostly medium fills all of space, even the interplanetary and interstellar space, which is normally regarded as a vacuum. The physicists of the nineteenth century called this hypothetical medium the ether, and they attempted to describe light waves as oscillations of the ether, analogous to sound waves as oscillations of the air. The preferred reference frame in which light has its standard speed is then the reference frame in which the ether is at rest. The existence of such a preferred reference frame would

The exact value of the speed of light is 2.997 924 58  108 m/s, but throughout this chapter we will round this off to 3.00  108 m/s.

2

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The Speed of Light; the Ether

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imply that velocity is absolute—the ether frame …corresponds to flow of the ether past Earth, would set an absolute standard of rest, and the velocMotion of the Earth forming the ether wind. (a) (b) ity of any body could always be referred to this frame. relative to a hypothetical medium, the ether,… For instance, instead of describing the velocity of the V Earth relative to some other material body, such as ether the Sun, we could always describe its velocity relative wind to the ether. V Presumably the Earth has some nonzero velocity relative to the ether. Even if the Earth were at rest in the ether at one instant, this condition could not last, since the Earth continually changes its ether motion as it orbits around the Sun. The motion of the ether past the Earth was called the ether wind by the FIGURE 36.2 (a) Motion of Earth in refnineteenth-century physicists (Fig. 36.2). If the Sun is at rest in the ether, then the erence frame of ether. (b) Motion of ether in ether wind would have a velocity opposite to the velocity of the Earth around the reference frame of Earth. Sun—about 30 km/s; if the Sun is in (steady) motion, then the ether wind would vary with the seasons—it would reach a maximum when the orbital motion of the Earth is parallel to the motion of the Sun, and a minimum when antiparallel. Experimenters attempted to detect this ether wind by its effects on the propagation of light. A light wave in a laboratory on the Earth would have a greater speed when moving downwind and a smaller speed when moving upwind or across the wind. If the speed of the ether wind “blowing” through the laboratory is V, then the speed of light in this laboratory would be c  V for a light signal with downwind motion, c  V for upwind motion, and 2c 2  V 2 for motion perpendicular to the wind (see Fig. 36.3). With a value of 300 000 km/s for c and a value of approximately 30 km/s for V, the increase or decrease of the speed of light amounts to only about 1 part in 10 000, and a very sensitive apparatus is required for the detection of this small change. In a famous experiment first performed in 1881 and often repeated thereafter, A. Concepts A. Michelson and E. W. Morley attempted to detect small changes in the speed of in Context light by means of an interferometer. The results of their experiment were negative. As discussed in Section 35.2, the sensitivity of the original experiment of Michelson and Morley was such that an ether wind of 5 km/s would have produced a detectable effect. Michelson-Morley experiment Since the expected wind is about 30 km/s, the experimental result contradicts the ether theory of the propagation of light. Later, more refined versions of the experiment established that if there were an ether wind, its speed would certainly have to be less than 3 m/s. The experimental evidence therefore establishes conclusively that the motion of the Earth has no effect on the propagation of light. As the Earth moves around the Sun, its velocity is first in one direction, then in another, and another; and the Earth is first in one inertial reference frame, then in another, and another. But all these inertial reference frames appear to be completely equivalent in regard to the propagation of light. There is no preferred reference frame. There is no ether.

Light moving downwind through (a) the ether…

c

V

(b)

(c) V c

cV cV cV cV …would have a greater speed on Earth…

…and light moving upwind…

V c

c V  c2 V 2 …or across the wind would have a lower speed.

FIGURE 36.3 The velocity of light is c relative to the ether, and the velocity of the ether is V relative to the laboratory. The velocity of light relative to the laboratory is then the vector sum c  V. (a) If c and V are parallel, the magnitude of the vector sum is c  V. (b) If c and V are antiparallel, the magnitude of the vector sum is c  V. (c) If c and V form this right triangle, the magnitude of the vector sum is 2c 2  V 2.

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Concepts in Context

The Theory of Special Relativity

Today, we take it for granted that the speed of light is unaffected by the motion of the Earth. Many experiments and instruments rely on the fact that the travel time for an electromagnetic signal between an emitter and a receiver depends only on the distance between the two; the speed of light, and thus the travel time, does not depend on the motion of the emitter or of the receiver. For example, accurate determination of a position on the ground using the Global Positioning System (GPS) is achieved by precisely measuring the travel time of signals from several satellites. When a GPS receiver calculates the distances to several satellites, it assumes that the speed of light is unaffected by the motion of the satellites and by the translational motion of the Earth.

✔

Checkup 36.1

You are in an open automobile (a convertible) traveling at 80 km/h. Does the speed of sound waves (relative to you) depend on direction? QUESTION 2: At the equator of the Earth, the rotational speed is 0.46 km/s. Was the original Michelson–Morley experiment capable of detecting the corresponding ether wind? QUESTION 3: If Michelson and Morley had detected an ether wind of 30 km/s opposite to the motion of the Earth around the Sun, what could they have concluded about the absolute motion of the Earth? The Sun? QUESTION 4: In 1887, after observations lasting a few days, a null result was obtained in the Michelson–Morley experiment with 5 km/s sensitivity. To be sure that there was no ether, Michelson and Morley had to repeat the experiment (A) With a different color of light (B) At 1 km/s resolution (C) At a different time of year QUESTION 1:

36.2 EINSTEIN’S PRINCIPLE O F R E L AT I V I T Y

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Neither the laws of mechanics nor the laws for the propagation of light reveal any intrinsic distinction between different inertial reference frames. This motivated Einstein to take a bold step and to propose a general hypothesis concerning all the laws of physics. This hypothesis is the Principle of Relativity: Principle of Relativity

All the laws of physics are the same in all inertial reference frames. In addition, Einstein proposed the Principle of the Universality of the Speed of Light:

Principle of Universality of Speed of Light

The speed of light (in vacuum) is the same in all inertial reference frames; it always has the value c  3.00  108 m/s. These deceptively simple principles form the foundation of the theory of Special Relativity. As we pointed out in the preceding section, the universality of the speed of light conflicts with the Galilean addition rule for velocity. We will therefore have to dis-

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Einstein’s Principle of Relativity

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card this rule, and we will also have to discard the transformation rule for position vectors or coordinates on which it is based (see Section 4.6). The universality of the speed of light also requires that we give up some of our intuitive, everyday notions of space and time. Obviously, the fact that a light signal always has a speed of 3.00  108 m/s, regardless of how hard we try to move toward it or away from it in a fast aircraft or spaceship, does violence to our intuition. This strange behavior of light is only possible because of a strange behavior of length and time in relativistic physics. As we will see later in this chapter, neither length nor time is absolute—they both depend on the reference frame in which they are measured, and they suffer contraction or dilation when the reference frame changes. Before we can inquire into the consequences of Einstein’s two principles, we must carefully describe the construction of reference frames and the synchronization of their clocks. A reference frame is a coordinate grid erected around some given origin and a set of clocks (see Fig. 36.4), which can be used to determine the space and time coordinates of any event. In relativity, as in everyday life, an event is an occurrence that happens at one point of space at one point of time (for example, a climber stepping onto the summit of Mt. McKinley). An event is therefore represented by one point of space and time. The space coordinates of the event are directly given by the coordinate grid intersections at the event. The time coordinate of the event is the time registered by the clock at the event. Different choices of reference frame result in different space coordinates and different time coordinates for the event, and in Section 36.5 we will see how these different coordinates in different reference frames are related. The clocks of any chosen reference frame must be synchronized with each other and with the master clock sitting at its origin of coordinates. Einstein proposed that this synchronization can be accomplished by sending out a flash of light from a point exactly midway between the clock at the origin and the other clock (see Fig. 36.5). The two clocks are synchronized if both show exactly the same time when the light from the midpoint reaches them. Note that this synchronization procedure hinges on the universality of the speed of light. If the speed of light were not a universal constant, but were dependent on the reference frame and on the direction of propagation (say, faster toward the right in Fig. 36.5 and slower toward the left), then we could not achieve synchronization by the simple procedure with a flash of light from the midpoint.

A reference frame consists of a coordinate grid and a set of clocks.

A flash of light is sent out from a point midway between two clocks. y

y

event

wave front

x

x

O

In any chosen reference frame, all clocks must be synchronized.

FIGURE 36.4 A reference frame.

Synchronized clocks show exactly the same time when light from midpoint reaches them.

FIGURE 36.5 Synchronization procedure for a pair of clocks. The leading wave front reaches the clocks at the lower left corner and the upper right corner simultaneously.

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One immediate consequence of our synchronization procedure is that simultaneity is relative, that is, the simultaneity of two events depends on the reference frame. Einstein illustrated this with the following concrete example. Suppose that a train is traveling at high speed along a straight track and two bolts of lightning strike the front end and the rear end of the train, leaving scorch marks on the train and on the ground (the existence of these marks helps us to remember exactly where the lightning struck in each reference frame). Suppose that these two strokes of lightning are exactly simultaneous in the reference frame of the Earth. Then in the reference frame of the train, the two strokes of lightning will not be simultaneous—as judged by the clocks on the train, the stroke at the front end of the train occurs slightly earlier than the stroke at the rear end. To see how this difference between the two reference frames comes Lightning strikes train and ground about, let us apply our procedure for testing simultaneity. Suppose that in at front and rear. When strikes on the reference frame of the Earth, an observer stands near the track at the ground are simultaneous, … midpoint between the two scorch marks the lightning made on the ground (see Fig. 36.6); she will then receive flashes of light from the lightning at her left and her right at the same instant. Thus, this observer will confirm that in the reference frame of the Earth, the lightning was simultaneous. In the reference frame of the train, an observer can likewise test for simultaneity by placing himself exactly at the midpoint between the scorch …an observer on ground at midpoint of strikes will receive two marks the lightning made at the front and rear ends of the train (see Fig. light flashes at the same instant. 36.7) and waiting for the arrival of the flashes of light. Will he receive the flashes of light from the front and the rear of the train at the same instant? FIGURE 36.6 Observer on the ground at the midpoint We can answer this by examining the motion of this observer and the between the scorch marks on the track watches for the propagation of the flashes of light in the reference frame of the ground arrival of flashes of light. (see Fig. 36.8). This observer is traveling toward the flash of light approaching him from the front end of the train, and he is traveling away from the For an observer in train at midpoint, flash of light trying to catch up with him from the back end of the train. lightning flashes originate at equal Thus, this observer will encounter the flash of light from the front end distances and travel at equal speeds. before the flash of light from the rear end can catch up. In the reference frame of the ground, this delay between the flashes of light seen by the observer on the train is attributed to his motion toward one flash and away from the other. But in the reference frame of the train, the observer cannot attribute the delay to such a difference in motion—the light flashes from the front and the rear of the train originated at exactly equal distances from him, and, according to the Principle of Universality of the Speed of FIGURE 36.7 Observer in the train at the midpoint Light, they traveled at equal speeds. Hence this observer will conclude between the front and rear ends, where lightning has made that the stroke of lightning at the front end of the train occurred earlier and scorch marks. the stroke of lightning at the rear end occurred later. Although this qualitative argument shows that simultaneity depends In reference frame of ground, on the reference frame, it does not tell us by how much. A quantitative calmoving observer on train encounters culation shows that for a train traveling at ordinary speed the delay is insignifflash approaching from front first… icant, 1013 s or less. But the delay increases with the speed and it also increases with the distance between the flashes of lightning. For instance, consider a fast spaceship traveling by the Earth at 90% of the speed of light, and consider two flashes of lightning at two points on the Earth separated by a fairly large distance, say, Boston and New York, separated by a distance of 300 km. If these flashes are simultaneous in the reference frame of the Earth, they will differ by 0.001 s in the reference frame of the spaceship. If simultaneity is relative, then the synchronization of clocks is also rela…and flash catching up from rear tive. In the reference frame of the Earth, all the clocks of this reference later. On train, observer concludes that flash from front occurred earlier. frame are synchronized, that is, the hands of these clocks reach the, say, noon FIGURE 36.8 Observer on a train moving toward position simultaneously. But when observed from the reference frame of the right.

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36.2

Einstein’s Principle of Relativity

Moving Earth clocks as viewed from spaceship are not synchronized.

Clocks on moving spaceship as viewed from Earth are not synchronized.

y'

y

x

O

FIGURE 36.9 Clocks of the reference frame of the Earth as observed at one instant of spaceship time.

1223

O

Boston

New York Clocks further ahead in direction of clock motion are “later.”

x' Clocks further ahead in direction of clock motion are “later.”

FIGURE 36.10 Clocks of the reference frame of the spaceship as observed at one instant of Earth time.

the train or the spaceship, the clocks in the Earth’s reference frame that are further ahead in the direction of clock motion show earlier times. Consequently, they reach the noon position later than those on the right side—they are “late” in the same way as the lightning of the left side is late. This can be seen in Fig. 36.9, which shows the clocks belonging to the reference frame of the Earth as observed at one instant of time from a fast spaceship traveling in the direction from Boston to New York. The effect is symmetric. In the reference frame of the spaceship, all the clocks onboard are synchronized. But, as observed from the reference frame of the Earth, the clocks on the front part of the spaceship are late. Figure 36.10 shows the clocks belonging to the reference frame of the spaceship as observed at one instant from the Earth. Note that in Fig. 36.9 we are viewing the reference frame of the Earth moving past the spaceship, and in Fig. 36.10 we are viewing the reference frame of the spaceship moving past the Earth. In either case, the clocks on the leading edge of the reference frame are late. The relativity of synchronization is a direct consequence of the universality of the speed of light, since our procedure for testing simultaneity depends crucially on the speed of light. The failure of an absolute synchronization valid for all reference frames implies that there exists no absolute time. Each reference frame has its own way of reckoning time—time is relative. Instead of the single absolute time coordinate t we used in Newtonian physics, we now have to use a separate time coordinate for each individual reference frame.

✔

Checkup 36.2

A spaceship approaches the Earth at 1.00  108 m/s and sends a light signal toward the Earth, as in Fig. 36.1. According to Einstein, what is the speed of this light signal relative to the Earth? QUESTION 2: Figure 36.10 shows the clocks of the reference frame of a spaceship in motion relative to the Earth. The clocks at the front of the spaceship are late. Does this mean that the crew of the spaceship find that their clocks are out of synchronization? QUESTION 3: An earthquake occurs in San Francisco, and simultaneously (in Earth time) another earthquake occurs in New York. These earthquakes are not simultaneous as seen in the reference frame of a fast spaceship traveling westward in the direction from New York to San Francisco. Which is late? QUESTION 4: A satellite, moving away from you at speed v, emits a pulse of radio waves in your direction. The speed of the waves relative to you is QUESTION 1:

(A) c  v

(B) c  v

(C) 2c 2  v2

(D) c

(E) v

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3 6 . 3 T I M E D I L AT I O N

Online Concept Tutorial

42

L

In reference frame of spaceship, light makes a round trip, traveling a distance 2L in time t'  2L/c.

FIGURE 36.11 Spaceship with a “racetrack” for a light pulse.

The relativity of time shows up not only in the synchronization of clocks, but also in the rate of clocks. When experimenters on Earth observe a clock onboard the moving spaceship, they find that it suffers time dilation relative to the clocks on Earth: the rate of the moving clock is slow compared with the rate of identically manufactured clocks at rest on Earth. To see how this comes about, imagine that experimenters in the spaceship set up a racetrack of length L perpendicular to the direction of motion of the spaceship (see Fig. 36.11). If the experimenters in the spaceship use one of their clocks to measure the time of flight ¢t of a light signal that goes from one end of the track to the other and returns to its starting point, they will then find that the light signal takes a time of ¢t  2L c to complete the round trip. But experimenters on the Earth see that the light signal has concurrent vertical and horizontal motions (see Fig. 36.12). For the experimenters on Earth, the light signal travels a total distance larger than 2L. Since the speed must still be the standard speed of light c, they will find that according to their clocks the light signal now takes a time ¢t longer than 2L/c to complete the trip. Thus, a given time interval ¢t registered by a clock on the spaceship is registered as a longer time interval ¢t by the clocks on the Earth. This means that the clock on the spaceship runs slow when judged by the clocks on the Earth. Note that the experiment involves one clock on the spaceship (the clock at the starting point of the track), but several clocks on the Earth, because the light signal does not return to the point at which it started on Earth, and the observers on Earth will have to use one (stationary) clock at the starting point and another (stationary) clock at the end point to measure the time of flight. For a quantitative evaluation of the time dilation, we note that in Fig. 36.12 the upward portion of the path of the light signal is the hypotenuse of a right triangle of sides L and V ¢t 2, where V is the speed of the spaceship relative to the Earth. The total length of the path that the light signal has to cover in the reference frame of the Earth is therefore 2  2L2  (V¢t2)2

In reference frame of the Earth, light making a round trip travels a distance greater than 2L.

(36.1)

The time taken to cover this distance is

V t/2

¢t  L

2  2L2  (V¢t2)2 c

(36.2)

If we square both sides of this equation, we obtain (¢t)2 

4L2  V 2(¢t)2 c2

(36.3)

which we can solve for (¢t)2 and then for ¢t: Boston

New York

(¢t)2 

Total distance traveled is 2  L2  (V t/2)2 .

FIGURE 36.12 The trajectory of the light pulse as observed from the Earth. The light pulse has both a vertical motion (up or down) and also a horizontal motion (toward the right).

4L2c2

1  V 2c 2

and ¢t 

2Lc

21  V 2c 2

(36.4)

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36.3

Time Dilation

1225

Since, in the reference frame of the spaceship, 2L c  ¢t, this gives us

¢t 

e

¢t 21  V c 2

2

clock at rest in spaceship registers ¢t clocks in Earth reference frame measure ¢t

(36.5)

This is the time-dilation formula. It shows that the time registered by the clocks on the Earth is longer than the time registered by the clock on the spaceship by a factor of 1 21  V 2 c 2, that is, the clock on the spaceship runs slow when measured with the clocks on the Earth. Figure 36.13 is a plot of the time-dilation factor 1  21  V 2 c 2 for speeds in the range from 0 to c. At low speeds the time-dilation effect is insignificant, but at speeds near c, it becomes quite large. The slowing down of the rate of lapse of time applies to all physical processes— atomic, nuclear, biological, etc. Thus, the astronauts on the spaceship will perform all their tasks in “slow motion,” and they will age slower than normal when measured with the clocks on the Earth. But they themselves will be unaware of this. From their point of view, they are in an inertial reference frame in which the usual laws of physics are valid, and the physical processes in their reference frame proceed at the normal rate, without any indication of anything unusual. The time-dilation effect is symmetric: as measured by the clocks on the spaceship, a clock on the Earth runs slow by the same factor:

¢t 

¢t 21  V c 2

2

e

clock at rest on Earth registers ¢t (36.6) clocks in spaceship reference frame measure ¢t

The derivation of Eq. (36.6) can be based on an argument similar to that given above, with a racetrack for light at rest on the Earth. Incidentally: in these arguments we have implicitly assumed that the length of the racetrack is unaffected by the motion of the spaceship or the Earth, that is, we have assumed that the length is absolute. As we will see in the next section, this is true for lengths perpendicular to the direction of motion, although it is not true for lengths along the direction of motion.

Very drastic time-dilation effects have been observed in the decay of short-lived elementary particles. For instance, a muon particle (see Chapter 41) usually decays in about 2.2  106 s; but if it is moving at high speed through the laboratory, then the internal processes that produce the decay will slow down as judged by the clocks in our laboratory, and the muon lives a longer time. In accurate experiments performed at the European Organization for Nuclear Research (CERN) accelerator near Geneva, muons with a speed of 99.94% of the speed of light were found to have a lifetime 29 times as long as the lifetime of muons at rest. Is this dilation of the lifetime in agreement with Eq. (36.5)?

EXAMPLE 1

SOLUTION: We can regard the muon as a clock at rest in the reference frame of an (imaginary) spaceship with a speed of V  0.9994c. Over the lifetime of the muon, this clock registers a time interval of ¢t  2.2  106 s. However, the experimenters in the laboratory see the spaceship moving at V  0.9994c, and over the lifetime of the muon they see their own clocks register a larger time ¢t than

time dilation

1/ 1–V 2/c 2

5.0

4.0

3.0

Time-dilation effect is insignificant at low speeds…

2.0 …but becomes overwhelming at speeds close to c.

1.0

0

0.2c

0.4c 0.6c 0.8c

1.0c

V

FIGURE 36.13 Time-dilation factor as a function of speed V.

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the time ¢t registered by the moving clock. According to Eq. (36.5), the times ¢t and ¢t differ by the time-dilation factor 1>21  V 2 c 2, that is, 1 21  V

2

c

2



1 21  (0.9994c) c 2

2



1 2

21  (0.9994)



1  29 0.035

This time-dilation factor is in agreement with the experimental result.

At everyday speeds, the time-dilation effect is extremely small. For example, consider a clock aboard an airplane traveling at 300 m/s over the ground. The time-dilation factor is then 1 21  V

twin paradox

2

c

2



1 21  (300) (3.00  10 ) 2

8 2



1 21  1012

(36.7)

Evaluation of this gives 1.000 000 000 000 5, which means that a clock in the airplane will slow down by only 5 parts in 1013! However, detection of such a small change is not beyond the reach of modern atomic clocks. In a notable experiment, scientists from the National Institute of Standards and Technology placed portable atomic clocks aboard a commercial airliner and kept them flying for several days, making a complete round trip around the world. Before and after the trip, the clocks were compared with an identical clock that was kept on the ground. The flying clocks were found to have lost time—in one instance, the total time lost because of the motion of the clock was about 107 s. The time-dilation effect leads to the famous twin paradox, which we can state as follows: a pair of identical twins, Terra and Stella, celebrate their, say, twentieth birthday on Earth. Then Stella boards a spaceship that carries her at a speed of V  0.99c to Proxima Centauri, 4 light-years away; the spaceship immediately turns around and brings Stella back to Earth. According to the clocks on Earth, this trip takes about 4 years each way, so Terra’s age will be 28 years when the twins meet again. But Stella has benefited from time dilation—relative to the reference frame of the Earth, the spaceship clocks run slow by a factor 1 21  V

2

c

2



1 2

21  (0.99)



1 0.14

(36.8)

Hence 8 years of travel registered by the Earth clocks amount to only 8  0.14  1.1 years according to the spaceship clocks, and Stella’s biological age on return will be only 21 years. Stella will be younger than Terra.

In reference frame of the Earth, the spaceship moves, and its clocks experience time dilation. (a)

FIGURE 36.14 Time-dilation effect symmetry.

(b)

In reference frame of the spaceship, the Earth moves, and its clocks experience time dilation.

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36.3

Time Dilation

1227

The paradox arises when we examine the elapsed times from the point of view of the reference frame of the spaceship. In this reference frame, the Earth is moving away from the spacecraft (see Fig. 36.14). Hence in this reference frame, the Earth clocks run slow—and Terra should be younger than Stella. The resolution of this paradox hinges on the fact that our time-dilation formula is valid only if the time of a moving clock is measured from the point of view of an inertial reference frame. The reference frame of the Earth is (approximately) inertial, and therefore our calculation of the time dilation of the spaceship clocks is valid. But the reference frame of the spaceship is not inertial—the spaceship must decelerate when it reaches Proxima Centauri, stop, and then accelerate toward the Earth. If the reference frame is not inertial, the Principle of Relativity does not apply. Therefore, we cannot use the simple time-dilation formula to find the time dilation of the Earth clocks from the point of view of the spaceship reference frame. The “paradox” results from the misuse of this formula. A detailed analysis of the behavior of the Earth clocks from the point of view of the spaceship reference frame establishes that the Earth clocks indeed do also run slow as long as the spaceship is moving with uniform velocity, but that the Earth clocks run fast when the spaceship is undergoing its acceleration to turn around at Proxima Centauri. The time that the Earth clocks gain during the accelerated portions of the trip more than compensates for the time they lose during the other portions of the trip. This confirms that Stella will be younger than Terra, even from the point of view of the spaceship reference frame.

The speed of a GPS satellite relative to a point on the Earth’s surface is typically V  3.9  103 m/s. Assume that the clock on a GPS satellite is synchronized with the clocks of the Earth’s reference frame at one instant. By how much do they differ 1.0 hour later? What is the corresponding distance error for a radio signal? In this calculation, ignore the rotation of the Earth and treat the satellite motion the same as motion with uniform velocity. EXAMPLE 2

SOLUTION: A time interval ¢t measured on Earth will differ from the interval

¢t measured on the GPS satellite by the time-dilation factor: ¢t 

¢t 21  V 2c 2

We wish to find the difference ¢t  ¢t when ¢t  1.0 hour. Since the speed of the satellite is much less than the speed of light, we can simplify the timedilation factor using the expansion (1  x)n  1  nx for small x. Here, with n  12 and x  V 2c 2, we have 1 21  V 2c 2

 a1 

V2 c2

b

12

1

1 V2 2 c2

(36.9)

Inserting this into the time-dilation expression, we have ¢t  ¢t  a 1 

1 V2 b 2 c2

or ¢t  ¢t  ¢t 

1 V2 2 c2

(36.10)

Concepts in Context

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With ¢t  1.0 h  3600 s and V  3.9  103 m/s, we find that the clocks will differ by ¢t  ¢t  3600 s 

1 3.9  103 m/s 2  a b  3.0  107 s 8 2 3.00  10 m/s

If such a timing error were uncorrected and the satellite clock were used to determine the distance of a point on the Earth by light travel time, then the error in this distance would be ¢x  c  (¢t  ¢t)  3.00  108 m/s  3.0  107 s  90 m (36.11) Since the accuracy of GPS positioning is required to be much better than that, corrections for the time dilation must be applied, in addition to many more mundane corrections, for example, those due to refraction and the decrease in the speed of radio waves in the atmosphere.

Finally, we point out that the time dilation of relativistic physics affects the Doppler shift of the frequency of light waves whenever the emitter is in motion relative to the receiver. Since we want to find the frequency detected by the receiver, we consider the reference frame of the receiver, and we pretend that this reference frame serves as the “medium” in which the light propagates. In Newtonian physics the Doppler shift for an emitter moving at speed V through a medium in which the emitted wave has a speed c is simply f   f (1 ; Vc), where f is the frequency radiated by the emitter, f  is the frequency detected by the receiver, and the positive sign applies if the emitter is receding, the negative sign if approaching [see the derivation of Eq. (17.17) for the case of sound waves]. In relativistic physics, the Doppler shift must also include the time dilation of the emitter. The time dilation of the period of the waves corresponds to a reduction in the detected frequency by the additional factor 21  V 2c 2. Including this factor, we have f

21  V 2 c 2 f 1 ; Vc

(36.12)

where now V is to be interpreted as the speed of the emitter relative to the receiver. Since 1  V 2 c 2  (1  Vc)(1  Vc), Eq. (36.12) can be simplified by appropriate cancellations:

f relativistic Doppler shift

1  Vc f B 1  Vc

1  Vc f f B 1  Vc

Concepts in Context

for receding emitter (36.13) for approaching emitter

A GPS satellite transmits radio (microwave) signals at a frequency of 1.575 GHz. Assume for simplicity that a GPS satellite is directly approaching your location on the Earth’s surface with a relative speed of 3.9  103 m/s. By what factor is the frequency that you detect on Earth increased EXAMPLE 3

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36.3

Time Dilation

due to the ordinary Doppler shift of Newtonian physics? Due to the time dilation of relativistic physics? SOLUTION: For an approaching emitter, the Newtonian result for the upward-

shifted frequency that you receive is given solely by the denominator in Eq. (36.12) [see also Eq. (17.17)]: f

1 f 1  Vc

Taking the ratio of received to emitted frequencies and inserting the values, we obtain f f



1 1  (3.9  103 m/s)(3.00  108 m/s)

 1.000 013

This is an upward shift of 13 parts per million, or of about (1.3  105)  (1.575 GHz)  20 kHz. The time dilation factor is

B

1

V2 c2

1

1 V2 2 c2

which shifts the received frequency downward by a factor that differs from unity by (3.9  103 m/s)2 1 V2 1    8.4  1011 8 2 2 c2 2 (3.00  10 m/s) or less than a tenth of a part per billion. Thus for such a “low” speed, the time dilation contribution to the frequency shift is completely negligible compared with the ordinary Doppler effect. We already used this fact when we examined police radar guns in Example 17.7.

✔

Checkup 36.3

Distinguish between the relativity of the synchronization of clocks and the relativity of the rates of clocks. QUESTION 2: Consider the plot of the time-dilation factor given in Fig. 36.13. If we increase the speed of a clock by a factor of 2, do we increase the time-dilation factor by a factor of 2, or by less, or by more? QUESTION 1:

QUESTION 3: An astronaut aims a beam of light from a green laser pointer at you. If you approach the astronaut at V  0.10c, do you detect blue light or yellow light? QUESTION 4: A spaceship moves away from Earth at high speed. How do experimenters on the Earth measure a clock in the spaceship to be running? How do those in the spaceship measure a clock on the Earth to be running? (A) Slow; fast (B) Fast; slow (C) Slow; slow (D) Fast; fast

1229

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36.4 LENGTH CONTRACTION

Boston

New York

Measuring rod is at rest in reference frame of spaceship.

FIGURE 36.15 Spaceship with a measuring rod oriented along the direction of its motion.

y'

x'

O

Lengths on spaceship along direction of motion measured by observers on Earth are short.

FIGURE 36.16 Reference frame of the spaceship as observed at one instant of Earth time (including length contraction; this length contraction was not included in Fig. 36.10).

In the preceding sections we have seen that time is relative—both the synchronization of clocks and the rate of clocks depend on the reference frame. Now we will see that length is also relative. A measuring rod, or any other body, onboard the spaceship suffers length contraction along the direction of motion. The length of the moving measuring rod will be short when compared with the length of an identically manufactured measuring rod at rest on the Earth. The reason for this is that the length measurement of a moving body depends on simultaneity, and since simultaneity is relative, so is length. Suppose that the spaceship, traveling from Boston toward New York, carries a measuring rod that has a length of, say, 300 km in the reference frame of the spaceship (see Fig. 36.15). To measure the length of this rod in the reference frame of the Earth, we station observers in the vicinity of New York and Boston with instructions to ascertain the positions of the front and the rear ends of the measuring rod at one instant of time, say, at noon. But when the observers on the Earth do this, the observers on the spaceship will claim that the position measurements were not done simultaneously, and that the observers in the vicinity of Boston measured the position of the rear end at a later time. In the extra time, the rear end moves an extra distance to the right, and hence the distance between the positions measured for the rear and the front ends will be reduced. From the point of view of the observers on the spaceship, it is therefore immediately obvious that the length measured by the observers on the Earth will be short. Figures 36.16 and 36.17 show the reference frame of the spaceship moving past the Earth and the reference frame of the Earth moving past the spaceship, respectively. In these figures the length contraction has been included (it was left out in Figs. 36.9 and 36.10). As illustrated in these figures, the length-contraction effect is symmetric: a body at rest in the spaceship will suffer contraction when measured in the reference frame of the Earth, and a body at rest on the Earth will suffer contraction when measured in the reference frame of the spaceship. We can obtain a formula for the length contraction by exploiting the formula for the time dilation. Consider a rod of length L at rest in the spaceship. According to the spaceship clocks, an observer on the Earth takes a time ¢t  LV to travel from one end of the rod to the other. Taking time dilation into account, we then conclude that the observer on the Earth will judge that this takes a shorter time of only

y

¢t  x

O Boston

New York Lengths on the Earth along direction of motion measured by observers on spaceship are short.

FIGURE 36.17 Reference frame of the Earth as observed at one instant of spaceship time (including length contraction; this length contraction was not included in Fig. 36.9).

B

1

V2 c

2

¢t 

B

1

V 2 L c2 V

(36.14)

However, the observer on the Earth cannot attribute this reduction of the travel time to a slowing of his clock—in his own reference frame his clock runs at a normal rate. Instead, he must attribute the reduction of travel time to a contraction of length of the rod. Measured in the reference frame of the Earth, the rod in the spaceship has some contracted length L, and it moves at speed V. Hence, the rod passes by a point on the Earth in a time LV. This time must agree with the time ¢t calculated in Eq. (36.14), so L V 2 L  1 2 V B c V or

(36.15)

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36.4

L

B

1

V2 c2

L

Length Contraction

for a rod L at rest in spaceship

(36.16)

This is the formula for length contraction. According to this formula, the length of a rod or any body in motion relative to a reference frame is shortened by a factor of 21  V 2 c 2. Figure 36.18 is a plot of the length-contraction factor 21  V 2 c 2 as a function of speed. As already mentioned, this contraction effect is symetric: if the rod is at rest on the Earth and is measured in the reference frame of the spaceship, the formula for the length contraction is

L 

B

1

V2 c

2

L for a rod L at rest on Earth

(36.17)

1231

length contraction

1–V 2/c 2

1.0

0.8

0.6

0.4

The length contraction has not been tested directly by experiment. There is no practical method for a high-precision measurement of the length of a fast-moving body. Our best bet might be high-speed photography, but this is nowhere near accurate enough, since the contraction is extremely small even at the highest speeds that we can impart to a macroscopic body. Note, however, that the experimental evidence for time dilation can be regarded as indirect evidence for length contraction, since, as we saw above, the former implies the latter. The contraction effect applies only to lengths along the direction of motion of the body. Lengths perpendicular to the direction of motion are unaffected. The proof of this is by contradiction: imagine that we have two identically manufactured pieces of pipe, one at rest on the Earth, one at rest on the spaceship (see Fig. 36.19). If the motion of the spaceship relative to Earth were to bring about a transverse contraction of the spaceship pipe, then, by symmetry, the motion of the Earth relative to the spaceship would bring about a contraction of the Earth pipe. These contraction effects are contradictory, since in one case the spaceship pipe would fit inside the Earth pipe, and in the other case it would fit outside.

A proton is passing by the Earth at a speed of 0.50c. In the reference frame of the proton, what is the length of the diameter of the Earth in a direction parallel to that of the motion of the proton? By how much is this shorter than the diameter in the reference frame of the Earth?

EXAMPLE 4

SOLUTION: We can regard the reference frame of the proton as the reference

frame of a spaceship, and we can regard the diameter of the Earth as a rod at rest in the reference frame of the Earth. Relative to the proton, or the spaceship, the Earth has a speed V  0.50c. In the reference frame of the Earth, the diameter has the familiar value L  1.3  107 m. But in the reference frame of the proton, or the spaceship, the Earth has a speed of V  0.50c, and the rod is observed to have a shorter length L. According to Eq. (36.16),the lengths L and L differ by the length contraction factor 21  V 2 c 2, that is, 21  V 2 c 2  21  (0.50c)2 c 2  21  0.25  0.87

Length-contraction effect is insignificant at low speeds…

0.2

0

0.2c

0.4c 0.6c 0.8c

1.0c

V

…but becomes severe at speeds close to c.

FIGURE 36.18 Length-contraction factor as a function of speed V.

This piece of pipe is at rest in reference frame of the Earth…

…and this one is at rest in reference frame of spaceship.

Only length along direction of motion is contracted.

FIGURE 36.19 Two identical pieces of pipe.

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Since the diameter of the Earth in its own reference frame is L  1.3  107 m, the length L in the reference frame of the proton is L  0.87  L  0.87  1.3  107 m  1.1  107 m This is shorter than 1.3  107 m by 2  106 m, or 2000 km! C O M M E N T S : The dimensions of the Earth perpendicular to the direction of

motion do not contract. This implies that, in the reference frame of the proton, the Earth is not a sphere, but an ellipsoid.

From the length contraction of a three-dimensional body we can deduce the volume contraction. The volume of the Earth, which is calculated by taking a product of the dimension parallel to the motion and the two dimensions perpendicular to the motion, will be contracted by just one factor of 21  V 2 c 2, that is, a factor of 0.87 in the case of Example 4.

✔

Checkup 36.4

QUESTION 1: A cannonball is perfectly round in its own reference frame. Describe the shape of the cannonball in a reference frame relative to which it is moving at a high speed, say, 0.5c. QUESTION 2: Could we use the argument based on the two identical pieces of pipe (see Fig. 36.19) to prove that lengths along the direction of motion are not affected? Why, or Why not? QUESTION 3: In view of the length contraction, does the density of a material depend on its speed? QUESTION 4: A track is 100 m long. A particle moves parallel to the track with speed V such that 1  V 2 c 2  0.25. In the reference frame of the particle, the length of the track is

(A) 25 m

(B) 50 m

(C) 100 m

(D) 200 m

(E) 400 m

3 6 . 5 T H E L O R E N T Z T R A N S F O R M AT I O N S A N D T H E C O M B I N AT I O N O F V E L O C I T I E S Suppose that we measure the space and time coordinates of an event—such as the impact of lightning on a point on the ground—in the reference frame of the Earth and also in the reference frame of a moving spaceship. We will then obtain different values of these coordinates in the Earth and in the spaceship reference frames, but these different values of the coordinates are related by transformation formulas. In Einstein’s physics, the transformation formulas for the coordinates are fairly complicated, because they are designed so as to keep the speed of light the same in all reference frames, and they incorporate the length contraction and the time dilation. Before dealing with these complicated formulas, let us examine the much simpler transformation formulas for coordinates in Newton’s physics, where there is no length contraction and no time dilation.

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36.5

y'

y

The Lorentz Transformations and the Combination of Velocities

1233

The x' –y' grid moves with speed V along the x axis of the x –y grid.

P V y'

y O

O' Vt

x'

x x'

x

FIGURE 36.20 The coordinate grid x–y belonging to the reference frame of the Earth and the coordinate grid xy belonging to the reference frame of the spaceship. The coordinates of the point P are x, y in the first grid and x, y in the second grid.

Figure 36.20 shows the coordinate grids x–y and xy of the first (Earth) and the second (spaceship) reference frames. The second reference frame is moving with velocity V along the x axis of the first reference frame. We assume that at time t  0, the origins of the two coordinate grids coincide; at time t, the distance between the origins is then Vt. The coordinates of the point P are x, y in the first reference frame and x, y in the second reference frame. By inspection of Fig. 36.20, we see that the distance x equals the sum of the distances x and Vt : x  x  Vt

(36.18)

x  x  Vt

(36.19)

Hence

Furthermore, the distance y equals the distance y y  y

(36.20)

Equations (36.19) and (36.20) are the transformation equations for the x and y coordinates in Newton’s physics. These two equations are merely the x and y components of the general vector equation r  r  Vt for the transformation of the position vector we found in Chapter 4 [see Eq. (4.50)]. We could have obtained our equations for the transformation of the x and y coordinates from the general vector equation; but it is just as easy to re-derive these results by inspection of Fig. 36.20. Note that although Fig. 36.20 makes the equations for the transformation of the x and y coordinates seem selfevident, these equations hinge on the absolute character of length in Newton’s physics. Absolute length means that the observers in the two reference frames agree on the measurement of any length or any distance between two points. If the observers disagreed on the values of the distances x or x—for example, if one observer claimed that the distance x was 3.0 m and the other observer claimed that this distance x was contracted to 2.5 m—then Eq. (36.18) would not be valid. The left side of Eq. (36.18) is a distance defined at one instant in the first reference frame, whereas the right side is a sum of a distance x defined at one instant in the second reference frame and a distance (Vt) defined at one instant in the first reference frame, and such a sum makes no sense unless the observers agree on the values of these distances.

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Furthermore, in Newton’s physics time is absolute. This means that the times registered by the clocks in the two reference frames are always equal, t  t Galilean transformation equations

(36.21)

Taken together, Eqs. (36.19)–(36.21) are called the Galilean transformation equations; they relate the space and time coordinates in one reference frame to those in the other. From these equations we can deduce the Galilean addition rule for the components of the velocity. For instance, if the x coordinate changes by dx in a time dt, then Eqs. (36.19) and (36.21) give us dx  dx  V dt

(36.22)

dt  dt

(36.23)

Dividing these two equations side by side, we obtain dx dx  V dt dt

(36.24)

Here, dxdt is the x velocity of the particle, light signal, or whatever, measured in the second reference frame; and dxdt is the x velocity measured in the first reference frame. Hence Eq. (36.24) says vx  vx  V

(36.25)

This, of course, is simply the Galilean addition rule for the x component of the velocity [see Eq. (4.53)]. In Einstein’s physics, the Galilean formulas for the transformation of the coordinates and for the addition of velocities must be replaced by more complicated formulas, designed in such a way as to keep the speed of light the same in all reference frames. The transformation equations that accomplish this trick are called the Lorentz transformations. If the new reference frame moves, again, with velocity V along the x axis of the first reference frame, and if the origins coincide at time t  0, the Lorentz transformations take the form Lorentz transformations

x 

x  Vt 21  V 2 c 2 y  y

t 

t  Vxc2

21  V 2 c 2

(36.26)

(36.27)

(36.28)

These equations cannot be obtained by simple inspection of Fig. 36.20, because the distances displayed in this figure are not absolute in Einstein’s physics, and they cannot simply be added as in Newton’s physics. We will not bother with a formal derivation of the Lorentz transformation equations, because we can achieve a clear understanding of the various terms and factors in these equations by comparing them with the Galilean transformation equations. Equation (36.26) differs from the Galilean equation (36.19) only by the factor 1 21  V 2 c 2; this factor represents the length contraction. Equation (36.27) is identical to the Galilean equation, because lengths perpendicular to the direction of motion

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36.5

The Lorentz Transformations and the Combination of Velocities

remain unchanged. And Eq. (36.28) differs from the Galilean equation in two ways: it contains an extra factor 1  21  V 2c 2 representing the time dilation, and it contains an extra term (Vxc 2) 21  V 2c 2 representing the relativity of synchronization discussed in Section 36.2. We already gave quantitative treatments of the time dilation and the length contraction, and we therefore do not need to reexamine these here. But we did not yet give a quantitative treatment of the relativity of synchronization, and we will now examine this, to justify the presence of the extra term in Eq. (36.28). Suppose that observers in the Earth frame send a light signal in the positive x direction from the origin O (with coordinate x  0) to some point P (with coordinate x 0). The signal leaves the origin O at time t  0 and arrives at the point P at time t. According to the observers in the Earth frame, the arrival time is t  xc, since the light signal travels at speed c. We want to know the arrival time t, as seen by the observers in the spaceship frame. For these observers, the light signal moves in the positive x direction at speed c, while simultaneously the Earth frame and the point P move in the negative x direction at speed V. Hence the “closing speed” of the light signal and the point P is c  V (this closing speed is larger than c, but that is not objectionable; it merely reflects the fact that if the target and light signal are both moving toward each other, they will meet sooner than if the target is at rest). To calculate the time of arrival of the light signal at the point P, the spaceship observers have to divide the length OP by the closing speed c  V. But since the length OP is a moving length, these observers must take the length contraction into account: the length between O and P is not x, but is x 21  V 2 c 2. Accordingly, the observers in the spaceship frame find that the arrival time of the light signal is t 

x21  V 2c 2 cV

Before we compare this with the value of t given by the Lorentz transformation (36.28), let us multiply the numerator and denominator by 21  V 2 c 2 and rearrange the result: t   

x (1  V 2 c 2) x21  V 2c 2 21  V 2 c 2   cV 21  V 2 c 2 c(1  Vc)21  V 2 c 2 x (1  Vc)(1  V c)

c (1  V c)21  V 2 c 2 xc  Vxc 2

21  V 2 c 2

(36.29)

Here, the term xc in the numerator is simply the time t that the light signal takes to arrive according to observers in the Earth frame. The term V xc2 in the numerator represents the relativity of synchronization. Comparing Eqs. (36.28) and (36.29), (with xc  t), we see they agree exactly—and this agreement provides the justification of the extra term in Eq. (36.28).

A measuring rod of length ¢x is at rest along the x axis of the spaceship frame, which is moving in the positive x direction with speed V relative to the Earth frame. What is the length of the measuring rod in the Earth frame according to the Lorentz transformation equations?

EXAMPLE 5

1235

HENDRIK ANTOON LORENTZ (1853–1928) Dutch theoretical physicist, professor at Leiden. He investigated the relationship between electricity, magnetism, and mechanics. In order to explain the observed effect of magnetic fields on emitters of light (Zeeman effect), he postulated the existence of electric charges in the atom, for which he was awarded the Nobel Prize in 1902. He derived the Lorentz transformation equations by examining Maxwell’s equations, but he was not aware that this leads to a new concept of space and time.

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SOLUTION: We begin with Eq. (36.26) written as an equation for differences:

¢x 

¢x  V¢t 21  V 2 c 2

(36.30)

In the Earth frame, the length of the rod is measured at one instant t of time, so ¢t  0. Hence Eq. (36.30) reduces to ¢x 

¢x 21  V 2 c 2

which gives ¢x  ¢x 

B

1

V2 c2

This is the expected length-contraction formula (36.16).

A clock is at rest in the Earth frame. If a time ¢t elapses as shown by this clock, how much time elapses according to the clocks of a spaceship moving at speed V relative to the Earth frame?

EXAMPLE 6

SOLUTION: Again, we begin by writing the Lorentz transformation equation

(36.28) in terms of differences: ¢t 

¢t  V ¢xc2 21  V 2c 2

(36.31)

For the clock at rest in the Earth frame, ¢x  0, and therefore ¢t 

¢t 21  V 2c 2

which is the expected time-dilation formula, Eq. (36.6).

inverse Lorentz transformations

COMMENT: Note that both the Lorentz transformation equations (36.26) and (36.28) have factors of 1 21  V 2 c 2, even though the length contraction has a factor of 21  V 2 c 2, not a factor of 1  21  V 2c 2. The reason becomes clear from inspection of these two examples: the Lorentz transformation equations (36.26) and (36.28) incorporate the time dilation for a clock at rest on Earth, but Eq. (36.26) provides the length of a rod at rest in the spaceship frame. If we want the length contraction for a rod at rest in the Earth frame, we would need to use the inverse Lorentz transformation equations, that is, the equations for x, t in terms of x, t. These can be obtained by solving Eqs. (36.26) and (36.28) for x and t, that is, by pretending x and t are unknowns, to be evaluated by combining these equations to obtain each of them in terms of x and t  (see Problem 33). The resulting equations have exactly the same form as Eqs. 36.26 and 36.28 with primed and unprimed space and time coordinates exchanged and with V replaced by V.

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The Lorentz Transformations and the Combination of Velocities

1237

Note that if the relative velocity between the two reference frames is small compared with the speed of light, then Vc in Eqs. (36.26) and (36.28) is small, and any term involving this quantity can be omitted in the equations. The Lorentz transformations then reduce to x  x  Vt y  y t  t Thus, for low speeds, the Lorentz transformations reduce to the Galilean transformations (36.19)–(36.21). The crucial feature of the Lorentz transformation equations is that they leave the speed of light unchanged. To verify this, we need to find the relativistic combination rule for velocity. If the x coordinate changes by dx in a time dt, then the Lorentz transformation equations tell us that dx  dt 

dx  Vdt 21  V 2 c 2 dt  Vdxc 2

21  V 2c 2

(36.32) (36.33)

and dividing these two equations side by side, we find dx dx  Vdt  dt dt  Vdxc 2

(36.34)

On the right side we can divide both the numerator and the denominator by dt, with the result dx dt  V dx  dt 1  V(dxdt)c 2

(36.35)

In this expression, dxdt is the x component of the velocity of a light signal or particle measured in the first reference frame and dxdt is the x component of the velocity measured in the second reference frame. Hence Eq. (36.35) may be written vx 

vx  V 1  vxVc 2

(36.36)

This is the relativistic combination law for the x components of the velocity (there are somewhat different formulas for the combination of the y and z components of the velocity; see Problem 41). It is instructive to compare the relativistic combination rule for velocities with the Galilean addition rule vx  vx  V

(36.37)

It is the denominator in Eq. (36.36) that makes all the difference. For instance, suppose that vx is the velocity of a light signal propagating along the x axis of the first reference frame. Then vx  c, and Eq. (36.36) yields vx 

cV 1  cVc

2



c (1  Vc) c 1  Vc

(36.38)

relativistic velocity combination

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Light signal has speed 3  108 m/s in reference frame of spaceship speeding toward the Earth.

(a) 3  108 m/s

(b) 3  108 m/s

Light signal has the same speed of 3  108 m/s in reference frame of the Earth.

FIGURE 36.21 Addition of velocities according to the relativistic combination rule. A spaceship speeding toward the Earth emits a light signal toward the Earth. (a) Reference frame of the spaceship. (b) Reference frame of the Earth.

Spaceship approaches the Earth at speed of 0.40c.

The Theory of Special Relativity

Thus, as required by the Principle of Universality of the Speed of Light, the velocity of the light signal in the second reference frame has exactly the same magnitude as in the first reference frame (see Fig. 36.21). The relativistic combination rule for light velocities has been explicitly tested in an experiment at CERN, on the French-Swiss border; involving a beam of very fast pions. These particles decay spontaneously by a reaction that emits a flash of very intense, very energetic light (gamma rays). Hence, such a beam of pions can be regarded as a high-speed light source. In the experiment, the velocity of the pions relative to the laboratory was V  0.999 75c. The Galilean addition law for velocity would have predicted laboratory velocities of 1.999 75c for light emitted in the forward direction and 0.000 25c for light emitted in the backward direction. But the experiments confirmed the relativistic combination rule—the laboratory velocity of the light had the same magnitude c in all directions.

An alien spaceship approaching the Earth at a speed of 0.40c fires a rocket at the Earth (see Fig. 36.22). If the velocity of the rocket is 0.80c in the reference frame of the spaceship, what is its velocity in the reference frame of the Earth?

EXAMPLE 7

SOLUTION: The equation for the combination of velocities is easiest to use if vx

is treated as known and vx as unknown. Accordingly, we take vx to be the velocity in the reference frame of the spaceship, and we take vx to be the velocity in the reference frame of the Earth. The x axis is directed from the spaceship toward the Earth, and the velocity of the rocket in the reference frame of the spaceship is vx  0.80c. The velocity V must be taken to be that of the Earth relative to the spaceship; this velocity is negative, V  0.40c. Then Eq. (36.36) gives vx 

0.40c

0.80c

Velocity of rocket is 0.80c in reference frame of spaceship.

What is velocity of rocket in reference frame of the Earth?

✔

1  vxVc

2



0.80c  (0.40c) 1  (0.80c)(0.40c)c 2

 0.91c

Checkup 36.5

QUESTION 1: Suppose that the new reference frame moves in the direction of the negative x axis of the first reference frame. What are the Lorentz transformation equations in this case? QUESTION 2: How do we know that the Lorentz transformation equations are consistent with the requirement that the speed of light is left unchanged?

If the spaceship in Example 7 is moving away from the Earth instead of approaching the Earth, how does this change the answer for vx ? QUESTION 4: A radioactive nucleus approaches Earth at v  c  2 and emits an electron toward the Earth at v  c2 relative to the nucleus. What is the electron’s speed relative to the Earth? (A) c4 (B) 4c5 (C) 23 4c (D) c (E) 4c3 QUESTION 3:

FIGURE 36.22 A spaceship launches a rocket toward Earth.

vx  V

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Relativistic Momentum and Energy

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3 6 . 6 R E L AT I V I S T I C M O M E N T U M AND ENERGY The drastic revision that the theory of Special Relativity imposes on Newton’s concepts of space and time implies a corresponding revision of the concepts of momentum and energy. The formulas for momentum and energy and the equations expressing their conservation are intimately tied to the transformation equations of the space and time coordinates. To see that this is so, we briefly examine the Newtonian (nonrelativistic) case. In Newton’s physics, the momentum of a particle of mass m and velocity v is p  mv

(36.39)

To find the momentum of this particle in a new reference frame, we note that the Galilean transformation equation for the velocity vector is v  v  V

(36.40)

Multiplying this by the mass, we find the transformation equation for the momentum: p  mv  m v  mV

(36.41)

From this we see that the momentum p in the new reference frame differs from the momentum p in the old reference frame by only the constant quantity mV (a quantity independent of the velocity v of the particle). Hence, if the total momentum of a system of colliding particles is conserved in one reference frame, it will also be conserved in the other reference frame—and the Law of Conservation of Momentum obeys the Principle of Relativity. This shows that the nonrelativistic formula for momentum and the nonrelativistic Galilean formula for the addition of velocities match in just the right way. According to the relativistic physics of Einstein, we must replace the Galilean addition formula for velocities by the relativistic combination rule. If the Law of Conservation of Momentum is to obey the Principle of Relativity, we must then design a new relativistic formula for momentum that matches the new relativistic combination rule for velocities. The required relativistic formula for momentum is p

mv 21  v 2 c 2

(36.42)

We will not give a proof of this formula. If the speed of the particle is small compared with the speed of light, then 21  v 2 c 2  1 and Eq. (36.42) becomes approximately p  mv

(36.43)

This shows that for low speeds, the relativistic and the Newtonian formulas for the momentum agree. We can therefore regard the Newtonian formula for the momentum as a simple and useful approximation for low speeds. This approximation is quite adequate for the description of all the phenomena we encounter in everyday life and (almost) all the phenomena we encounter in the realm of engineering, such as the phenomena we dealt with in the earlier chapters of this book. But at high speeds, the formulas differ drastically. We must then abandon the Newtonian formula, and rely entirely on the relativistic formula. Note that the relativistic momentum becomes infinite as the speed of the particle approaches the speed of light. Figure 36.23 is a plot of the magnitude of the momentum as a function of the speed.

relativistic momentum

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An electron in the beam of a TV tube has a speed of 1.0  108 m/s. What is the magnitude of the momentum of this electron?

EXAMPLE 8

SOLUTION: For this electron, v  c  (1.0  108 m/s)  (3.0  108 m/s)  0.33.

According to Eq. (36.42), the magnitude of the momentum is then p

mv 21  v c 2

2



9.11  1031 kg  1.0  108 m/s 2

21  (0.33)

 9.7  1023 kg m/s

COMMENTS: Note that if we had calculated the momentum according to the

nonrelativistic formula p  mv, we would have obtained 9.1  1023 kg m/s, and we would have been in error by about 6%.

We also need a new relativistic formula for kinetic energy. This formula is

K

relativistic kinetic energy

mc2 21  v2 c2

 mc2

(36.44)

For low speeds, this relativistic formula for kinetic energy can be shown to agree approximately with the nonrelativistic formula K  12 mv2. The relativistic kinetic energy becomes infinite as the speed of the particle approaches the speed of light. This indicates that, for any particle with mass (and for any body), the speed of light is unattainable, since it is impossible to supply a particle with an infinite amount of energy. Figure 36.24 is a plot of the kinetic energy vs. the speed.

p

K

4mc

4mc2

3mc

3mc2

2mc

Momentum p ≈ mv is proportional to velocity at low speeds…

2mc2

mc2

mc

0

0.2c 0.4c 0.6c 0.8c 1.0c

v

…but increases much more quickly with v at speeds close to c.

FIGURE 36.23 Momentum of a particle as a function of speed v.

0

Kinetic energy K ≈ 12 mv 2 is proportional to square of velocity at low speeds…

0.2c 0.4c 0.6c 0.8c 1.0c

v

…but increases much more quickly with v at speeds close to c.

FIGURE 36.24 Kinetic energy of a particle as function of speed v.

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36.6

Relativistic Momentum and Energy

The maximum speed that electrons achieve in the Stanford Linear Accelerator (SLAC) is 0.999 999 999 67c. What is the kinetic energy of an electron moving at this speed?

EXAMPLE 9

2 2 SOLUTION: The relativistic formula (36.44) contains a factor 21  v  c . Since

vc for these electrons is extremely close to 1, most calculators are unable to evaluate 21  v2 c 2. To get around this difficulty, we write

B

1

V2 c

2



B

1

V v v 1   22 1  c c cB B

(36.45)

and we evaluate 1  vc “by hand,” 1vc  1  0.999 999 999 67  3.3  1010 The rest of the calculation is within the reach of our calculator: K  mc2 a

1 21  v c 2

2

 1 b  mc 2 a

1 2221  vc

 9.11  1031 kg  (3.00  108 m/s)2  a

 1b 1

2223.3  1010

 1b

 3.2  109 J

Although the theory of Special Relativity requires a revision of the basic equations of mechanics, it does not require any revision of the basic equations of electricity and magnetism. Maxwell’s equations are already relativistic, that is, they match the relativistic behavior of length and of time in just the right way. This concordance between Maxwell’s equations and the requirements of relativity is no accident. Maxwell’s equations incorporate a universal speed of light—they imply c  1 1m00 in every reference frame. Einstein’s search for a theory of relativity was motivated by his faith in Maxwell’s equations and his recognition that if Maxwell’s equations were right then the Galilean coordinate transformations had to be wrong.

✔

Checkup 36.6

For a given speed, is the relativistic value of the momentum always larger than the Newtonian value? By what factor? QUESTION 2: Is the relativistic momentum always in the direction of the velocity of the particle? QUESTION 3: In science fiction stories, spaceships routinely reach speeds equal to or in excess of the speed of light. What is wrong with this? QUESTION 4: According to the relativistic formula (36.44) for the kinetic energy, what is the kinetic energy of a particle of zero speed? QUESTION 1:

(A) 0 (D) 2mc 2

(B) mc 2 (E) Infinite

(C) mc 2

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36.7 MASS AND ENERGY One of the great discoveries that emerged from relativity is that energy can be transformed into mass, and mass can be transformed into energy. Thus, mass is a form of energy. The amount of energy contained in an amount m of mass at rest is given by Einstein’s famous formula E  mc 2

rest mass energy

(36.46)

The quantity mc2 is called the rest-mass energy.3 The formula (36.46) can be derived from the theory of relativity, but, as with some other equations in this chapter, we will not give the derivation. The most spectacular demonstration of Einstein’s mass–energy formula is found in the annihilation of matter and antimatter (see Chapter 41). If a proton collides with an antiproton, or an electron with an antielectron, the two colliding particles react violently and they annihilate each other in an explosion that generates an intense flash of very energetic light. According to Eq. (36.46), the annihilation of just 1000 kg of matter and antimatter (500 kg of each) would release an amount of energy E  mc 2  1000 kg  (3.00  108 m/s)2  9.00  1019 J

(36.47)

This is enough energy to satisfy the needs of the United States for a full year. Unfortunately, antimatter is not readily available in large amounts. On Earth, antiparticles can be obtained only from reactions induced by the impact of beams of high-energy particles on a target. These collisions occasionally result in the creation of a particle–antiparticle pair. Such pair creation is the reverse of pair annihilation. The creation process transforms some of the kinetic energy of the collision into mass, and a subsequent annihilation merely gives back the original energy. But the relationship between energy and mass in Eq. (36.46) also has another aspect. Energy has mass. Whenever the internal energy stored in a body is changed, its rest mass (and weight) is changed. The change in rest mass that accompanies a given change of energy is ¢m  ¢Ec 2

mass and energy changes

(36.48)

For instance, in the fission of uranium, the nuclear material loses energy, and correspondingly its mass (and weight) decreases. The complete fission of 1.0 kg of uranium releases an energy of 8.2  1013 J, and correspondingly the mass of the nuclear material decreases by ¢m  (8.2  1013 J)c 2  0.000 91 kg, or about 0.1%. The fact that energy has mass indicates that energy is a form of mass. Conversely, as we have seen above, mass is a form of energy. Hence mass and energy must be regarded as different aspects of essentially the same thing. The laws of conservation of mass and conservation of energy are therefore not two independent laws—each implies the other. For example, consider the fission reaction of uranium inside the reactor vessel of a nuclear power plant (for details, see Chapter 40). The reaction conserves

3

Throughout this section, mass means the mass that a body has when at rest or nearly at rest; to emphasize this, we use the term rest mass. The definition and measurement of mass for a body in motion at high (relativistic) speeds is rather tricky, because Newton’s equation ma  F fails and the direction of the acceleration is not necessarily the direction of the force. The only kind of mass that is unambiguously defined in Einstein’s physics is the mass that the body has when at rest, and this is the only kind of mass we will consider.

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36.7

Mass and Energy

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energy—it merely transforms nuclear energy into heat, light, and kinetic energy but does not change the total amount of energy. The reaction also conserves mass—if the reactor vessel is hermetically sealed and thermally insulated from its environment, then the reaction does not change the mass of the contents of the vessel. However, if the vessel has an opening that lets some of the heat and light escape, then the mass of the residues will not match the mass of the original amount of uranium. The mass of the residues will be about 0.1% smaller than the original mass of the uranium. This mass defect represents the mass carried away by the energy that escapes. Thus, the nuclear fission reaction merely transforms energy into new forms of energy and mass into new forms of mass. In this regard, a nuclear reaction is not fundamentally different from a chemical reaction. The mass of the residues in an exothermic chemical reaction is slightly less than the original mass. The heat released in such a chemical reaction carries away some mass, but, in contrast to a nuclear reaction, this amount of mass is so small as to be quite immeasurable. The total energy of a free particle in motion is the sum of its rest-mass energy (36.46) and its kinetic energy (36.44): E  mc 2  K  mc 2 

mc 2 21  v c 2

2

 mc 2

(36.49)

This leads to a simple formula for the relativistic total energy of the particle: E

mc 2 21  v2 c 2

(36.50)

It is easy to verify (see Problem 66) that the relativistic energy can be expressed as follows in terms of the relativistic momentum: E  2c 2 p2  m2c4

(36.51)

For an ultra-relativistic particle, moving at a speed close to that of light, the first term (c2p2) within the square root is much larger than the second term (m2c4). Hence, for such a particle we can ignore the second term, and we then obtain the simple result E  2c 2p2 or E  cp

(ultra-relativistic particle)

(36.52)

Thus, the momentum and the energy of an ultra-relativistic particle are directly proportional.

EXAMPLE 10

Consider an electron of speed 0.999 999 999 67c, as in Example 9. What is the momentum of such an electron?

SOLUTION: Such as electron is ultra-relativistic. Its kinetic energy is much larger

than its rest-mass energy, and the total energy is therefore approximately equal to the kinetic energy, which we have already calculated in Example 9: E  mc2  K  K  3.2  109 J

relativistic total energy

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Hence Eq. (36.47) yields p

✔

3.2  109 J E   1.1  1017 kg m/s c 3.0  108 m/s

Checkup 36.7

QUESTION 1: What is the rest-mass energy of a 1.0-kg piece of stone? Why can’t we exploit this energy? QUESTION 2: Does kinetic energy have mass? For instance, does the kinetic energy of the particles of a gas contribute to the overall mass of the gas? QUESTION 3: For an ultra-relativistic particle (of speed near c), the momentum and the energy are proportional. Is this also true for a particle of lower speed, say, 0.9c or lower? QUESTION 4: We must add energy to a hydrogen atom to ionize it and thus obtain a proton and an electron. A neutron, on the other hand, will spontaneously decay to provide a moving proton and a moving electron. From this information, which has a greater mass, the hydrogen atom or the neutron? Or do they have the same mass? (A) Hydrogen atom (B) Neutron (C) Both have same mass

S U M M A RY PRINCIPLE OF RELATIVITY

All the laws of physics are the same in

all inertial reference frames. PRINCIPLE OF UNIVERSALITY OF SPEED OF LIGHT

The speed of

light is the same in all inertial reference frames. (¢t registered by clock in its own reference frame.)

TIME DILATION

¢t 

¢t 21  V 2c2

1/ 1–V 2/c 2

(36.5)

5.0 4.0 3.0 2.0 1.0 0 0.2c 0.4c 0.6c 0.8c 1.0c

RELATIVISTIC DOPPLER SHIFT

f

1  Vc f B 1  Vc

1  Vc f f B 1  Vc

V

for receding emitter (36.13) for approaching emitter

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L  21  V 2 c 2L

LENGTH CONTRACTION (L is length of body in its own reference frame.)

(36.16)

y'

x'

O

Lengths on spaceship along direction of motion measured by observers on Earth are short.

LORENTZ TRANSFORMATIONS

x 

x  Vt

(36.26)

21  V 2 c 2

y  y t 

RELATIVISTIC COMBINATION OF VELOCITIES

RELATIVISTIC MOMENTUM

vx 

p

(36.27) t  Vxc 2

(36.28)

21  V 2 c 2 vx  V

(36.36)

1  vxVc 2

mv 21  v2c 2

p

(36.42)

Momentum p ≈ mv is proportional to velocity at low speeds…

4mc 3mc

…but increases much more quickly with v at speeds close to c.

2mc mc

0 0.2c 0.4c 0.6c 0.8c 1.0c

RELATIVISTIC KINETIC ENERGY

K

mc 2 21  v c 2

2

v

K

 mc 2

(36.44)

4mc2 3mc2 2mc2 mc2

Kinetic energy K ≈ 12 mv 2 is proportional to square of velocity at low speeds…

…but increases much more quickly with v at speeds close to c.

0 0.2c 0.4c 0.6c0.8c 1.0c

RELATIVISTIC TOTAL ENERGY

E

mc 2 21  v 2c 2

v

(36.50)

E  2c 2 p2  m2c4 REST-MASS ENERGY

E  mc 2

(36.46)

MASS AND ENERGY CHANGES

m  Ec 2

(36.48)

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QUESTIONS FOR DISCUSSION 10. According to the arguments of Section 36.3, a light signal traveling along a track placed perpendicular to the direction of motion of the spaceship (see Fig. 36.11) takes a longer time to complete a round trip when measured by the clocks on the Earth than when measured by the clocks on the spaceship. Would the same be true for a light signal traveling along a track placed parallel to the direction of motion? Explain qualitatively.

1. An astronaut is inside a closed space capsule coasting through interstellar space. Is there any way the astronaut can measure the speed of the capsule without looking outside? 2. Why did Michelson and Morley use two light beams, rather than a single light beam, in their experiment? 3. When Einstein was a boy he wondered about the following question: A runner holds a mirror at arm’s length in front of his face. Can he see himself in the mirror if he runs at (almost) the speed of light? Answer this question both according to the ether theory and according to the theory of Special Relativity.

11. A cannonball is perfectly round in its own reference frame. Describe the shape of this cannonball in a reference frame relative to which it has a speed of 0.95c. Is the volume of the cannonball the same in both reference frames?

4. Consider the piece of paper on which one page of this book is printed. Which of the following properties of the piece of paper are absolute, that is, which are independent of whether the paper is at rest or in motion relative to you? (a) The thickness of the paper, (b) the mass of the paper, (c) the volume of the paper, (d) the number of atoms in the paper, (e) the chemical composition of the paper, (f ) the speed of light reflected by the paper, and (g) the color of the colored print on the paper.

12. A rod at rest in the ground makes an angle of 30 with the x axis in the reference frame of the Earth. Will the angle be larger or smaller in the reference frame of a spaceship moving along the x axis? 13. In the charming tale “City Speed Limit” by George Gamow,4 the protagonist, Mr. Tompkins, finds himself riding a bicycle in a city where the speed of light is very low, roughly 30 km/h. What weird effects must Mr. Tompkins have noticed under these circumstances?

5. Two streetlamps, one in Boston and the other in New York City, are turned on at exactly 6:00 P.M. Eastern Standard Time. Find a reference frame in which the streetlamp in New York was turned on late.

14. A long spaceship is accelerating away from the Earth. In the reference frame of the Earth, are the instantaneous speeds of the nose and of the tail of the spaceship the same?

6. According to the theory of Special Relativity, the time order of events can be reversed under certain conditions. Does this mean that a sparrow might fall from the sky before it leaves the nest?

15. Suppose that a very fast runner holding a long horizontal pole runs through a barn open at both ends. The length of the pole (in its rest frame) is 6 m, and the length of the barn (in its rest frame) is 5 m. In the reference frame of the barn, the pole will suffer length contraction and, at one instant of time, all of the pole will be inside the barn. However, in the reference frame of the runner, the barn will suffer length contraction and all of the pole will never be inside the barn at one instant of time. Is this a contradiction?

7. Because of the rotational motion of the Earth about its axis, a point on the equator moves with a speed of 460 m/s relative to a point on the North Pole. Does this mean that a clock placed on the equator runs more slowly than a similar clock placed on the pole? 8. According to Jacob Bronowski, author of The Ascent of Man, the explanation of time dilation is as follows: If you are moving away from a clock tower at a speed nearly equal to the speed of light, you keep pace with the light that the face of the clock sent out at, say, 11 o’clock. Hence, if you look toward the clock tower, you always see its hands at 11 o’clock. Is this explanation correct? If not, what is wrong with it?

16. Why can a spaceship not travel as fast as or faster than the speed of light? 17. If the beam from a revolving searchlight is intercepted by a distant cloud, the bright spot will move across the surface of the cloud very quickly, with a speed that can easily exceed the speed of light. Does this conflict with our conclusion of Section 36.6, that the speed of light is unattainable?

9. Suppose you wanted to travel into the future and see what the twenty-fifth century is like. In principle, how could you do this? Could you ever return to the twenty-first century?

4

George Gamow, Mr. Tompkins in Wonderland.

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Problems

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PROBLEMS †

36.1 The Speed of Light; the Ether

††

5. If a moving clock is to have a time-dilation factor of 10, what must be its speed?

1. Consider the case where the Sun moves at a high speed v through the hypothetical ether. What are the minimum and maximum ether-wind speeds on Earth when the Sun moves through the ether at (a) 30 km/s and (b) 60 km/s? Assume the orbital speed of the Earth is 30 km/s.

6. Neutrons have an average lifetime of 15 minutes when at rest in the laboratory. What is the average lifetime of neutrons of a speed of 25% of the speed of light? 50%? 90%?

*2. A Michelson–Morley interferometer determines the shift of two waves traveling in perpendicular directions (see also Section 35.2 and Fig. 35.9).

7. Consider an unstable particle, such as a pion, which has a lifetime of only 2.6  108 s when at rest in the laboratory. What speed must you give such a particle to make its lifetime twice as long as when at rest in the laboratory?

(a) Assume that one wave travels a distance L along the ether wind with speed c  V to a mirror, and back with speed c  V, as in Figs. 36.3a–b. Show that the round-trip time can be written t‘ 

8. The speed of the Sun around the center of our Galaxy is 200 km/s. Clocks in the Solar System will therefore run slow as compared with clocks at rest in the Galaxy. By what factor are the Solar System clocks slow?

V 2 1 2L a1  2 b c c

9. The orbital speed of the Earth around the Sun is 30 km/s. In one year, how many seconds do the clocks on the Earth lose with respect to the clocks of an inertial reference frame at rest relative to the Sun? [Hint: If Vc is small, the approximation 21  (Vc)2  1  12 (Vc)2 is valid.]

(b) Assume the other wave travels the same distance perpendicular to the ether wind with speed 2c 2  V 2 (see Fig. 36.3c). Show that its round-trip time is t 

V 2 12 2L a1  2 b c c

10. In 1961, the cosmonaut G. S. Titov circled the Earth for 25 h at a speed of 7.8 km/s. According to Eq. (36.5), what was the time-dilation factor of his body clock relative to the clocks on Earth? By how many seconds did his body clock fall behind during the entire trip? (Hint: Use the approximation given in Problem 9.)

(c) Use the expansion (1  x)n  1  nx for small x to show that the difference in arrival times is ¢t  t ‘  t 

LV 2 c3

11. At a speed V, the time-dilation factor has some value. Suppose that at speed 2V, the time-dilation factor has twice the previous value. What is the speed V ?

(d) What fraction of a full period is this shift for light with l  500 nm? Use the values L  11 m and V  30 km/s. *3. Ordinarily, the two arms of a Michelson–Morley interferometer cannot be set exactly equal, and instead have two values, L1 and L2. Insert these respective values into the results of Problem 2a and b and obtain a new expression for ¢t (see Problem 2c). Note that this result alone cannot be used to determine the ether-wind speed, since the difference between L1 and L2 is not accurately known. In an actual experiment, the entire apparatus is rotated 90 (thus interchanging L1 and L2). Obtain an expression for the net shift by subtracting the differences in arrival times for the two orientations. †,††

12. An astronaut traveling at V  0.80 c taps her foot 3.0 times per second. What is the frequency of taps determined by an observer on the Earth? 13. An atomic clock aboard a spaceship runs slow compared with an Earth-based atomic clock at a rate of 1.0 second per day. What is the speed of the spaceship? 14. A spaceship equipped with a chronometer is sent on a roundtrip to Alpha Centauri, 4.4 light-years away. The spaceship travels at 0.10 c, and returns immediately. (a) According to clocks on the Earth, how long does this trip take?

3 6 . 2 E i n s t e i n ’s P r i n c i p l e o f R e l a t i v i t y

4. A spaceship traveling at speed 12 c relative to the Earth ejects a spacepod traveling in the forward direction at speed 14 c relative to the spaceship. The spacepod emits a light signal toward the Earth at speed c relative to the spacepod. What is the speed of the light signal relative to the spaceship? What is the speed of the light signal relative to the Earth? Which observer (on the spaceship or on the spacepod) determines that the light strikes the Earth earlier?

36.3 Time Dilation

(b) According to the chronometer on the spaceship, how long does this trip take? 15. Consider the Doppler-shift formula for a receding source. By what factor does the frequency decrease for V  0.50c? For V  0.70c? For V  0.90c? †

For help, see Online Concept Tutorial 41 at www.wwnorton.com/physics For help, see Online Concept Tutorial 42 at www.wwnorton.com/physics

††

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16. The frequencies of light received from distant galaxies and quasars are shifted due to the Doppler effect. Frequencies 5.0 times smaller than expected for a stationary source have been detected from receding quasars. What is Vc for such a quasar? *17. In a test of the relativistic time-dilation effect, physicists compared the rates of vibration of nuclei of iron moving at different speeds. One sample of iron nuclei was placed on the rim of a high-speed rotor; another sample of similar nuclei was placed at the center. The radius of the rotor was 10 cm, and it rotated at 35 000 rev/min. Under these conditions, what was the speed of the rim of the rotor relative to the center? According to Eq. (36.5), what was the time-dilation factor of the sample at the rim compared with the sample at the center? (Hint: Use the approximation given in Problem 9.) *18. If cosmonauts from the Earth wanted to travel to the Andromeda galaxy in a time of no more than 10 years as reckoned by clocks aboard their spaceship, at what (constant) speed would they have to travel? How much time would have elapsed on Earth after 10 years of time on the spaceship? The distance to the Andromeda galaxy is 2.2  106 light-years. *19. Because of the rotation of the Earth, a point on the equator has a speed of 460 m/s relative to a point at the North Pole. According to the time-dilation effect of Special Relativity, by what factor do the rates of two clocks differ if one is located on the equator and the other at the North Pole? After 1.00 year has elapsed, by how many seconds will the clocks differ? Which clock will be ahead? (Although the special-relativistic time dilation slows one clock at the equator, there is an additional gravitational time dilation that slows the other clock. These two time-dilation effects balance, and the two clocks actually run at the same rate.) **20. The star Alpha Centauri is 4.4 light-years away from us. Suppose that we send a spaceship on an expedition to this star. Relative to the Earth, the spaceship accelerates at a constant rate of 0.10g until it reaches the midpoint, 2.2 light-years from Earth. The spaceship then decelerates at a constant rate of 0.10g until it reaches Alpha Centauri. The spaceship performs the return trip in the same manner. (a) What is the time required for the complete trip according to the clocks on the Earth? Ignore the time that the spaceship spends at its destination. (b) What is the time required for the complete trip according to the clocks on the spaceship? Assume that the instantaneous time-dilation factor is still 21  V 2c 2 even though the speed V is a function of time.

36.4 Length Contraction 21. A meterstick is moving by an observer in a direction parallel to its length. The speed of the meterstick is 0.50c. What is its measured length in the reference frame of the observer? 22. According to the manufacturer’s specifications, a spaceship has a length of 200 m. At what speed (relative to the Earth) will

this spaceship have a length of 100 m in the reference frame of the Earth? 23. A cannonball flies through our laboratory at a speed of 0.30c. Measurement of the transverse diameter of the cannonball gives a result of 0.20 m. What can you predict for the measurement of the length, or the longitudinal diameter, of the cannonball? 24. What is the percent length contraction of an automobile traveling at 96 km/h? (Hint: Use the approximation given in Problem 9.) 25. A hangar for housing spaceships is 100 m long. How fast must a 200-m-long spaceship be traveling to (briefly) fit in the hangar? 26. A right triangle of sheet metal with two 45 angles lies in the x–y plane, with one of its sides along the x axis (see Fig. 36.25). The length of each side is 0.20 m, and the length of the hypotenuse is 12  0.20 m. Suppose that this triangle is observed from an x, y reference frame moving at 0.80c along the x axis. What are the lengths of the sides and of the hypotenuse in this reference frame? What are the angles? y

45° 0.20 m

45°

x 0.20 m

FIGURE 36.25 A triangle.

*27. Two identical spaceships are traveling in the same direction. An observer on Earth measures the first to have speed 0.80c and observes the second to be 1.50 times as long as the first one. What is the speed of the second spaceship? *28. Suppose that a meterstick at rest in the reference frame of the Earth lies in the x–y plane and makes an angle of 30 with the x axis. Suppose that one end of the meterstick is at the origin. At a fixed time t, what are the x and y components of the displacement from this end of the meterstick to the other? At a fixed time t, what are the x and y components of the displacement from one end of the meterstick to the other in a new reference frame moving with velocity V  0.70c in the positive x direction? What is the angle the meterstick makes with the x axis of this new reference frame? *29. Electric charge is uniformly distributed throughout a sphere; the charge density is 2.0  106 C/m3. If this sphere is put in motion relative to the laboratory at a speed of 0.80c, what will be the charge density? Keep in mind that the total amount of electric charge is unchanged by the motion of the sphere.

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Problems

*30. It can be shown that when a point charge moves at uniform velocity of relativistic magnitude, its pattern of electric field lines is contracted by the usual length-contraction factor 21  V 2c 2 in the longitudinal direction and is unchanged in the transverse direction. Figure 36.26 shows the resulting pattern of field lines for a speed V  0.60c. Draw a similar picture for a speed of 0.80c. 0.60c

1249

34. A spaceship has a length of 300 m, measured in its own reference frame. It is traveling in the positive x direction at a speed of 0.80c relative to the Earth. A strobe light at the nose of the spaceship sends a pulse of light toward the tail of the spaceship. (a) As measured in the reference frame of the spaceship, how long does this light pulse take to reach the tail? (b) As measured in the reference frame of the Earth, how long does this light pulse take to reach the tail? 35. A spaceship is moving at a speed of 0.60c toward the Earth. A second spaceship, following the first one, is moving at a speed of 0.90c. What is the speed of the second spaceship as observed in the reference frame of the first? 36. Find the inverse of Eq. (36.36); that is, express vx in terms of vx. 37. The captain of a spaceship traveling away from Earth in the x direction at V  0.80c observes that a nova explosion occurs at a point with spacetime coordinates t  6.0  108 s, x  1.9  1017 m, y  1.2  1017 m as measured in the reference frame of the spaceship. He reports this event to the Earth via radio without delay.

FIGURE. 36.26 Electric field lines of a charge moving at V  0.60c. *31. A flexible drive belt runs over two flywheels whose axles are mounted on a rigid base (see Fig. 36.27). In the reference frame of the base, the horizontal portions of the belt have a speed v and therefore are subject to length contraction, which tightens the belt around the flywheels. However, in a reference frame moving to the right with the upper portion of the belt, the base is subject to length contraction, which ought to loosen the belt around the flywheels. Resolve this paradox by a qualitative argument. (Hint: Consider the lower portion of the belt as seen in the reference frame of the upper portion.) v

(a) What are the spacetime coordinates of the explosion in the reference frame of the Earth? Assume that the master clock of the spaceship coincides with the master clock of the Earth at the instant t  t  0 when the midpoint of the spaceship passes by the Earth, and that the origin of the spaceship x, y coordinates is at the midpoint of the spaceship. (b) Will the Earth receive the captain’s report before or after astronomers on the Earth see the nova explosion in their telescopes? No calculation is required for this question. *38. Consider the situation described in Problem 37. Since light takes some time to travel from the nova to the spaceship, the space and time coordinates that the captain reports are not directly measured but, rather, deduced from the time of arrival and the direction of the nova light reaching the spaceship. (a) At what time (t time) did the nova light reach the spaceship?

FIGURE 36.27 A drive belt and two flywheels.

3 6 . 5 T h e L o r e n t z Tr a n s f o r m a t i o n s a n d t h e C o m b i n a t i o n o f Ve l o c i t i e s 32. In the reference frame of the Earth, a firecracker is observed to explode at x  6.0  108 m, at t  4.0 s. According to the Lorentz transformation equations, what are the x and t coordinates of this event as observed in the reference frame of a spaceship traveling in the x direction at a speed of 0.50c? According to the Galilean transformation equations? 33. Obtain the inverse Lorentz transformation equations by solving Eqs. (36.26) and (36.28) for x and t, each in terms of x and t.

(b) If the captain sends a report to Earth via radio as soon as he sees the nova, at what time (t time) does the Earth receive the report? (c) At what time do Earth astronomers see the nova? *39. At 11h0m0.0000s A.M. a boiler explodes in the basement of the Museum of Modern Art in New York City. At 11h0m0.0003s A.M. a similar boiler explodes in the basement of a soup factory in Camden, New Jersey, at a distance of 150 km from the first explosion. Show that, in the reference frame of a spaceship moving at a speed greater than V  0.60c from New York toward Camden, the first explosion occurs after the second. *40. A radioactive atom in a beam produced by an accelerator has a speed 0.80c relative to the laboratory. The atom decays and ejects an electron of speed 0.50c relative to itself. What is the speed of the electron relative to the laboratory, if ejected in the forward direction? If ejected in the backward direction?

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*41. In a manner similar to the procedure of Eqs. (36.32)–(36.36), show that relativistic combination formula for the y component of the velocity is vy 

vy 21  V 2c 2 1  vxV

2

c 2

*42. Consider two speeds vx and V, each of which is less than the speed of light. Show that if these speeds are combined by the relativistic combination formula, the result is always less than the speed of light. *43. The speed of light with respect to a medium is vx  cn, where n is the index of refraction. Suppose that the medium, say, flowing water, is moving past a stationary observer in the same direction as the light with speed V. Show that the observer measures the speed of light to be approximately vx 

1 c  a1  2bV n n

This effect was first observed by Fizeau in 1851. **44. The acceleration of a particle in one reference frame is ax  dvx dt, where the particle has instantaneous velocity vx in that frame. Consider a reference frame moving with speed V parallel to the positive x axis of the first frame. Show that the acceleration in the second frame is given by ax 

dvx dt

 ax

(1  V 2c 2 )32

*52. What is the percent difference between the Newtonian and the relativistic values for the momentum of a meteoroid reaching the Earth at a speed of 72 km/s? *53. Consider three accelerators that produce high-energy particles: the Large Hadron Collider, which will soon produce protons with kinetic energy 7 TeV; the Stanford Linear Accelerator, which produces electrons with kinetic energy 50 GeV; and the Relativistic Heavy Ion Collider, which produces gold nuclei (mass 197 u) with kinetic energy 20 TeV. In each case, calculate the difference c  v between the speed of light and the speed of the particle. *54. The most energetic cosmic-ray particles have energies of about 50 J. Assume that such a cosmic ray consists of a proton. By how much does the speed of such a proton differ from the speed of light? Express your answer in meters per second. [Hint: Use the approximation given in Eq. (36.45)]. *55. Consider the electrons of a speed of 0.999 999 999 67c produced by the Stanford Linear Accelerator. What is the magnitude of the momentum of such an electron? *56. At the Fermilab accelerator, protons are given kinetic energies of 1.6  107 J. By how many meters per second does the speed of such a proton differ from the speed of light? What is the magnitude of the momentum of such a proton? *57. A mass M at rest decays into two particles of masses m1 and m2. Use Eq. (36.51) to show that the magnitude of the momentum of each of the two particles is

(1  vxVc 2 )3

36.6 Relativistic Momentum and Energy 45. Consider a particle of mass m moving at a speed of 0.10c. What is its kinetic energy according to the relativistic formula? What is its kinetic energy according to the Newtonian formula? What is the percent deviation between these two results? 46. Suppose you want to give a rifle bullet of mass 0.010 kg a speed of 1.0% of the speed of light. What kinetic energy must you supply? 47. The yearly energy expenditure of the United Stated is about 8  1019 J. Suppose that all of this energy could be converted into kinetic energy of an automobile of mass 1000 kg. What would be the speed of this automobile? 6

48. The speed of an electron in a hydrogen atom is 2.6  10 m/s. For this speed, does your calculator show any difference between the kinetic energies calculated according to the relativistic formula and the Newtonian formula?

p

2M 2  (m1  m 2)2 2M 2  (m1  m 2)2 c 2M

*58. A particle of mass m1 is at rest. A second particle of mass m2 and kinetic energy K strikes the first particle and sticks to it, a perfectly inelastic collision. Use Eq. (36.51) to show that the mass M of the composite particle is M

B

(m1  m2) 2 

2m1K c2

*59. Show that the velocity of a relativistic particle can be expressed as follows: v

cp 2 2

2m c  p2

**60. At the Brookhaven AGS accelerator, protons of kinetic energy 5.3  109 J are made to collide with protons at rest.

49. What is the speed of an electron if its kinetic energy is 1.6  1013 J?

(a) What is the speed of a moving proton in the laboratory reference frame?

50. What is the momentum and what is the kinetic energy of an electron moving at a speed of one-half the speed of light?

(b) What is the speed of a reference frame in which the two colliding protons have the same speed (and are moving in opposite directions)?

51. Show that the momentum of a particle can be expressed in the concise form p

Ev c2

(c) What is the total energy of each proton in the latter reference frame?

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Review Problems

36.7 Mass and Energy 61. How much energy will be released by the annihilation of one electron and one antielectron (both initially at rest)? Express your answer in electron-volts. 62. The atomic bomb dropped on Hiroshima had an explosive energy equivalent to that of 20 000 tons of TNT, or 8.4  1013 J. How many kilograms of rest mass must have been converted into energy in this explosion? 63. The mass of the Sun is 2.0  1030 kg. The thermal energy in the Sun is about 2  1041 J. How much does the thermal energy contribute to the mass of the Sun? Express your answer in percent. 64. Combustion of one gallon of gasoline releases 1.3  108 J of energy. How much mass is converted to energy? Compare this with 2.8 kg, the mass of one gallon of gasoline.

1251

65. The masses of the proton, electron, and neutron are 1. 672 623  1027 kg, 9.11  1031 kg, and 1.674 929  1027 kg, respectively. When a neutron decays into a proton and an electron, how much energy is released (other than the energy of the rest mass of the proton and electron)? Compare this extra energy with the energy of the rest mass of the electron. 66. From Eqs. (36.42) and (36.50) show that the relativistic energy and the relativistic momentum are related by E 2  c 2 p2  m2c4 **67. A K0 particle at rest decays spontaneously into a p particle and a p particle. What will be the speed of each of the latter? The mass of the K0 is 8.87  1028 kg, and the masses of the p and p particles are 2.49  1028 kg each.

REVIEW PROBLEMS *68. Muons are unstable particles which—if at rest in a laboratory—decay after a time of only 2.2  106 s. Suppose that a muon is created in a collision between a cosmic ray and an oxygen nucleus at the top of the Earth’s atmosphere, at an altitude of 20 km above sea level. (a) If the muon has a downward speed of v  0.990c relative to the Earth, at what altitude will it decay? Ignore gravity in this calculation. (b) Without time dilation, at what altitude would the muon have decayed? *69. Suppose that a special breed of cat (Felis einsteinii) lives for exactly 7.0 years according to its own body clock. When such a cat is born, we put it aboard a spaceship and send it off at V  0.80c toward the star Alpha Centauri. How far from the Earth (reckoned in the reference frame of the Earth) will the cat be when it dies? How long after the departure of the spaceship will a radio signal announcing the death of the cat reach us? The radio signal is sent out from the spaceship at the instant the cat dies. *70. Suppose that a proton speeds by the Earth at v  0.80c along a line parallel to the axis of rotation of the Earth. (a) In the reference frame of the proton, what is the polar diameter of the Earth? The equatorial diameter? (b) In the reference frame of the proton, how long does the proton take to travel from the point of closest approach to the North Pole to the point of closest approach to the South Pole? In the reference frame of the Earth, how long does this take?

71. A spaceship travels in the positive x direction with speed 0.80c. A man on Earth makes these observations: At t  0, x  0, a photon with l  400 nm is emitted at the rear of the ship moving toward the front. At t  1.00 ms, x  960 m, a photon with l  600 nm is emitted at the front of the spaceship moving toward the rear. A woman on the spaceship observes the same events. (a) What time interval does she measure between the two events? Which happens earlier? (b) What wavelengths does she measure for the two photons? (c) What is the length of the ship (as determined by the woman on it)? *72. An observer on Earth sees one spaceship traveling away to the west at speed 0.40c and a second spaceship, also traveling away but to the east, at 0.70c. Each spaceship emits a signal in its own reference frame at 2.00 GHz. What frequency does the Earth observer measure for each signal? What frequency does each spaceship measure for the signal from the other? *73. Consider a cube measuring 1.0 m  1.0 m  1.0 m in its own reference frame. If this cube moves relative to the Earth at a speed of 0.60c, what are its dimensions in the reference frame of the Earth? What are the areas of its faces? What is its volume? Assume that the cube moves in a direction perpendicular to one of its faces. *74. A spaceship has a length of 200 m in its own reference frame. It is traveling at 0.95c relative to the Earth. Suppose that the tail of the spaceship emits a flash of light.

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The Theory of Special Relativity

(a) In the reference frame of the spaceship, how long does the light take to reach the nose?

(a) What is the kinetic energy of the meteoroid in the reference frame of the spaceship?

(b) In the reference frame of the Earth, how long does this take? Calculate the time directly from the motions of the spaceship and the flash of light; then compare it with the results calculated by applying the Lorentz transformations to the result obtained in (a).

(b) In the collision all of this kinetic energy suddenly becomes available for inelastic processes that damage the spaceship. The effect on the spaceship is similar to an explosion. How many tons of TNT will release the same explosive energy? One ton of TNT releases 4.2  109 J.

75. Suppose that a spaceship is moving at a speed of V  0.20c relative to the Earth and a meteoroid is moving at a speed of vx  0.10c relative to the Earth; in the same direction as the spaceship. What is the speed vx of the meteoroid relative to the spaceship according to Eq. (36.36)? What is the percent difference between this relativistic result and the Galilean result? 76. A collision between two gamma rays creates an electron and an antielectron that travel away from the point of creation in opposite directions, each with a speed of 0.95c in the laboratory. What is the speed of the antielectron in the reference frame of the electron, and vice versa? *77. A spaceship traveling at 0.70c away from the Earth launches a projectile of muzzle speed 0.90c (relative to the spaceship). What is the speed of the projectile relative to the Earth if it is launched in the forward direction? In the backward direction? *78. Three particles are moving along the positive x axis, in the positive direction. The first particle has a speed of 0.60c relative to the second, the second has a speed of 0.80c relative to the third, and the third has a speed of 0.50c relative to the laboratory. What is the speed of the first particle relative to the laboratory? 79. What is the kinetic energy of a spaceship of rest mass 50 metric tons moving at a speed of 0.50c? How many metric tons of matter–antimatter mixture would have to be consumed to make this much energy available? 80. A particle has a kinetic energy equal to its rest-mass energy. What is the speed of this particle? 81. Suppose that a spaceship traveling at 0.80c through our Solar System suffers a totally inelastic collision with a small meteoroid of mass 2.0 kg.

*82. At the Brookhaven AGS accelerator, protons of kinetic energy 5.3  109 J are made to collide with protons at rest. (a) What is the speed of one of these moving protons in the laboratory reference frame? (b) What is the magnitude of the momentum? *83. At the SSC accelerator that was to be built in the United States, protons would have been given kinetic energies of 3.2  106 J. What is the value of c  v for such a proton, that is, by how many meters per second does the speed differ from the speed of light? *84. Free neutrons decay spontaneously into a proton, an electron, and an antineutrino: nSp  e  n The neutron has a rest mass of 1.6749  1027 kg; the proton, 1.6726  1027 kg; the electron, 9.11  1031 kg; and the antineutrino nearly zero. Assume that the neutron is at rest. Other than the rest-mass energy of the proton and electron, what is the energy released in this decay? **85. A K0 particle moving at a speed of 0.60c through the laboratory decays into a muon and an antimuon. (a) In the reference frame of the K0, what is the speed of each muon? The mass of the K0 is 8.87  1028 kg, and the masses of the muon and the antimuon are 1.88  1028 kg each. (b) Assume that the muon moves in a direction parallel to the original direction of motion of the K0 and that the antimuon moves in the opposite direction. What are the speeds of the muon and the antimuon with respect to the laboratory?

Answers to Checkups Checkup 36.1 1. Yes. Sound waves propagate in a medium (air), and thus the

speed of sound waves relative to you depends on your speed relative to the air. The speed is largest for sound waves traveling opposite to your motion, from the front to the back of the convertible automobile. 2. An ether wind due only to rotation means that the Earth

remains translationally at rest in the ether (the ether moves

with the Earth). The original experiment had a sensitivity of about 5 km/s, and thus could not detect an ether wind due only to the rotational speed of 0.46 km/s. 3. Such an observation would have led to the conclusion that the

Earth has an absolute motion of 30 km/s relative to the ether. Since this is the same as the velocity of the Earth relative to the Sun, such a result, if observed year round, would have also implied that the Sun is at rest with respect to the ether.

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Answers to Checkups

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4. (C) At a different time of year. A single null result could have

3. Yes. Since the volume decreases with increasing speed, the

implied that the Earth was (coincidentally) nearly at rest with respect to the ether. Repeating the experiment when the Earth’s velocity was in a different direction ensures that this was not the case.

density, or mass per unit volume, increases with increasing speed. 4. (B) 50 m. The 100-m track is contracted by the factor

21  V 2c 2  10.25  0.50, that is, to a length of 0.50  100 m  50 m.

Checkup 36.2 1. The speed of light in vacuum always has the same value, 8

3.00  10 m/s. 2. No; the clocks shown in Fig. 36.10 are as observed from the

Earth. For the crew or anyone in the reference frame of the spaceship, the clocks are all synchronized. 3. Relative to a spaceship traveling westward, the Earth reference

frame is traveling eastward. Thus, New York is at the leading edge of this reference frame, so the New York clocks, and the New York earthquake, are late. 4. (D) c. The speed of an electromagnetic wave is the speed of

light; it does not depend on the speed of the emitter or receiver.

Checkup 36.3 1. Relativity of synchronization means that the times indicated

by different clocks in a moving reference frame are different but all these clocks run at the same rate; relativity of rates means that the rate of the clocks in the moving reference frame differs from that of the clocks on the ground. 2. At low speeds, a factor of 2 increase in velocity has a negligible

effect on the time-dilation factor, which remains nearly equal to unity. At higher speeds, the time-dilation factor increases more quickly; for example, an increase from 0.45c to 0.90c increases the time-dilation factor from 1.1 to 2.3, more than a factor of 2, and higher speeds result in larger-factor increases. 3. An approaching receiver is the same as an approaching emit-

ter, since only relative motion matters; and the lower relation in Eq. (36.13) applies. Thus the frequency increases. So the wavelength decreases, and you detect blue light. 4. (C) Slow; slow. The time-dilation effect is symmetric, so

observers in each reference frame measure a clock in the other reference frame to be running slow.

Checkup 36.4 1. Like the shape of the Earth moving at high speed relative to

the reference frame of the proton in Example 4, the cannonball is an ellipsoid, flattened along the direction of motion, in a reference frame relative to which it is moving at 0.5c. 2. No; contraction along the direction of motion does not affect

which pipe fits inside which. Moreover, any apparent contradiction can be explained in terms of differences in simultaneity at any two different positions along the direction of motion.

Checkup 36.5 1. The Lorentz transformation equations will be the same as

Eqs. (36.26)–(36.28), but now with a negative value of V. 2. We know the Lorentz transformation equations are consistent

with a speed of light that is unchanged because they lead directly to the relativistic velocity combination law, Eq. (36.36), which gives vx  c when vx  c. 3. This is the same as if the Earth is moving away from the

spaceship, that is, V is now positive, and so the relativistic velocity combination rule gives vx  (0.80c  0.40c) (1  0.80  0.40)  0.59c. 4. (B) 4c5. As in Example 7, we can obtain the relative speed

from the velocity combination rule, Eq. (36.36): vx  [(c2)  (c2)] [1  (12)  (12)] c(54) (45)c.

Checkup 36.6 1. Yes; other than for speed v  0, the relativistic value of the

momentum, given by Eq. (36.42), is always larger than the Newtonian value, by the factor 1 21  v2c2. 2. Yes, the momentum vector p is proportional to the velocity v,

and so p is always in the same direction as v. 3. For any mass, attaining the speed of light would require infi-

nite kinetic energy, and this is impossible. 4. (A) 0. With v  0 in the first term of Eq. (36.44), the two

terms in the realistic kinetic energy cancel.

Checkup 36.7 2

8

2

1. From E  mc we have 1.0 kg  (3.00  10 m/s)  9.0  16

10 J. This energy can’t be converted to useful forms of energy, except by annihilation, which would require an equal amount of (unavailable) antimatter. 2. Yes. For example, the mass of a warm container of gas is

greater than the mass of a cold container of gas, due to the additional kinetic energy 3. No. This is only true when v  c. For example, for v 0.99c,

the energy becomes very nearly proportional to the momentum. 4. (B) Neutron. Since the neutron produces a proton, an electron, and some kinetic energy, its total energy (its mass) must be greater than the hydrogen atom, which requires added energy just to separate the proton and electron.

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CHAPTER

37

Quanta of Light

CONCEPTS IN CONTEXT 37.1 Blackbody Radiation 37.2 Energy Quanta 37.3 Photons and the Photoelectric Effect 37.4 The Compton Effect 37.5 X Rays 37.6 Wave vs. Particle

The Sun emits thermal radiation consisting of electromagnetic waves of many wavelengths. For the Sun and many hot bodies, there are universal features in the distribution of the emitted wavelengths. The explanation of the wavelength distribution of such thermal radiation requires the revolutionary idea that electromagnetic waves are made up of small, indivisible packets of energy, called quanta of light, or photons. Our study of thermal radiation will enable us to ask:

? How do we know the temperature of the surface of the Sun? (Section 37.1, page 1257)

? At what wavelength is the Sun’s thermal radiation maximum? (Example 1, page 1261)

? How many photons are there in the sunlight that reaches the surface of the Earth? (Example 4, page 1264) 1254

Concepts in Context

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37.1

Blackbody Radiation

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I

n Chapter 35 we examined the wave properties of light. We saw that light exhibits interference and diffraction, in agreement with Maxwell’s theory, according to which light is a wave consisting of oscillating electric and magnetic fields with a smooth, continuous density of energy. In this chapter we will see that light has particle properties. We will discuss experimental evidence that establishes that a light beam consists of a stream of discrete, particle-like energy packets. These energy packets are called quanta of light or photons. The discovery of the quantization of light by Max Planck in 1900 initiated the modern era in physics. Physicists quickly came to recognize that quantization of energy is a pervasive feature of the atomic and subatomic realm. The energies of the atoms and the energies of the subatomic particles—electrons, protons, and neutrons—are quantized. As we will discuss in the next chapters, such a quantization of energy is in conflict with Newton’s laws, and physicists had to find new laws that govern the behavior of atoms and of subatomic particles. The new theory that governs the realm of the atom is called quantum physics. In contrast, the old theory of Newton is called classical physics. The fact that light has the dual attributes of wave and of particle indicates that neither the classical wave nor the classical particle concept we have used in earlier chapters gives an adequate description of light. We have to think of light as a wave– particle object, which sometimes behaves pretty much as a wave, sometimes pretty much as a particle, and sometimes as a bit of both. Furthermore, we will see in the next chapter that electrons, protons, neutrons, and all the other known “particles” also exhibit such dual attributes of wave and of particle.

quantum physics vs. classical physics

3 7 . 1 B L A C K B O D Y R A D I AT I O N The first hint of a failure of classical physics emerged around 1900 from the study of thermal radiation. When we heat a body to high temperature, it gives off a glow. For instance, when we heat a bar of iron to 1200 or 1300 K, it glows in a bright orange or yellow color. This glow is thermal radiation (“radiant heat”). The color of the thermal radiation depends on the temperature; an extremely hot iron bar glows in a nearly white color (“white-hot”); at an intermediate temperature it glows yellow; and at a lower temperatures it glows orange and then bright red to dull red (Fig. 37.1). You can observe this change of color with temperature by turning on the heating coil of a kitchen range; the coil first glows dull red, and then orange, but it never reaches white-hot. Bodies at room temperature also emit thermal radiation, but the glow is infrared, and not visible to the eye (see the thermogram of Fig. 37.2). The spectrum of thermal radiation is continuous—if we analyze the light emitted by a glowing body with a prism, we find that the energy of the light is smoothly distributed over all wavelengths. For a quantitative description of the distribution of energy over different wavelengths, we plot the energy flux (or the power per unit area) radiated by the surface of the glowing body vs. the wavelength of the radiation. We can think of such a plot as the intensity distribution seen in the spectrum that an (ideal) prism produces when we use it to analyze thermal radiation. Measurements of the thermal radiation emitted by a glowing body show that the energy flux at very long and at very short wavelengths is quite small, and that the energy flux has a maximum at some intermediate wavelength. The location of this maximum depends on the temperature. For

FIGURE 37.1 A hot, glowing iron bar. The tip is at the highest temperature (“white-hot”), and the temperature gradually decreases along the bar.

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Quanta of Light

28°C

FIGURE 37.2 A thermogram. Each “false” color represents a different wavelength of infrared electromagnetic radiation; unlike the visible spectrum, the brightest colors here are hottest (white, then red, yellow, green, blue, p).

20°C

Exposed skin radiates more than clothed parts of bodies.

example, Fig. 37.3 gives plots of the distribution of energy flux radiated by glowing bodies at 1000 K, 1250 K, and 1450 K. By comparing these plots, we see that an increase of temperature produces more radiation at all wavelengths (the 1450-K curve is everywhere higher than the other curves); and we also see that an increase of temperature shifts the location of the maximum to shorter wavelengths (the peak of the 1450-K curve is located at a shorter wavelength than the peaks of the other curves). The thermal radiation emerging from the surface of a glowing body is generated within the volume of the body by the random thermal motions of atoms and electrons. Before the radiation reaches the surface and escapes, it is absorbed and re-emitted many times and it attains thermal equilibrium with the atoms and electrons. This equilibration process distributes the radiation continuously over all wavelengths and shapes its spectrum, completely washing out all of the original spectral features that the radiation had when first emitted by the atoms in the body.

energy flux (arbitrary units)

Higher temperature shifts location of peak to shorter wavelength.

classical 1450 K 1450 K

1250 K

FIGURE 37.3 Distribution of the energy flux in the spectra of thermal radiation emitted by glowing bodies at 1000 K, 1250 K, and 1450 K. The maxima (peaks) of these curves lie in the infrared region. The maximum is at 2900 nm for 1000 K, at 2300 nm for 1250 K, and at 2000 nm for 1450 K. The dashed curve gives the prediction of classical physics according to Rayleigh’s calculation.

1000 K

0

2000

4000 wavelength

6000 nm

Higher temperature produces more intensity at all wavelengths.

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37.1

Blackbody Radiation

The flux of thermal radiation emerging from the surface of a glowing body depends to some extent on the characteristics of the surface. The surface usually permits the escape of only a fraction of the flux reaching it from the inside of the body. Correspondingly, if the body is irradiated with an equal flux of thermal radiation from the outside, the surface permits the ingress of only an equal fraction of this flux, reflecting the rest. This equality of the emissive and absorptive characteristics of the surface can be deduced by an argument based on thermodynamics. Thus we are led to a general rule: a good absorber is a good emitter; and a poor absorber is a poor emitter. With this rule we can understand how the silvered, mirrorlike glass walls of thermos bottles or dewars provide such excellent thermal insulation. These bottles are constructed with a double glass wall, and the space between these walls is evacuated (Fig. 37.4). Heat cannot flow across the evacuated space by conduction or convection; it can only flow by radiation. To inhibit radiation, the glass walls are silvered and thereby made into mirrorlike reflecting surfaces; these highly reflective surfaces are very poor absorbers and emitters of radiation. This keeps the heat transfer between the walls very small. A body with a perfectly absorbing (and emitting) surface is called a blackbody; when such a body is cold and emits no radiation of its own, it looks black because it does not reflect any of the illumination reaching it from the outside. But when a blackbody is hot, its surface emits more thermal radiation than any other hot body at the same temperature. A body covered with black soot is an approximate blackbody. Experimental physicists prefer to achieve the characteristics of an ideal blackbody by a trick: take a body with a cavity, such as a hollow cube, and drill a small hole in one side of the cube (Fig. 37.5). The hole then acts as a blackbody—any radiation incident on the hole from outside will be completely absorbed. Because of this equivalence between a blackbody and a hole in a cavity, the terms blackbody radiation and cavity radiation are used interchangeably. The curves plotted in Fig. 37.3 are based on measurements of radiation emerging from a small hole in a body with a cavity; thus, these curves represent the spectra of blackbody radiation. The Sun is an almost perfect blackbody radiator. A very small fraction of the Sun’s radiation is due to specific chemical features of the Sun; the overwhelming majority of sunlight makes up a spectrum of thermal radiation that precisely follows the shape given in Fig. 37.3. As we have seen, a blackbody spectrum depends on the temperature of the blackbody, so we can tell the temperature of the surface of the Sun from the spectrum of thermal radiation that the Sun sends to the Earth.

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blackbody

blackbody radiation or cavity radiation

Concepts in Context

Vacuum layer prevents heat flow by conduction or convection.

Since any radiation entering a small hole suffers multiple reflections and is trapped,… silvered glass

FIGURE 37.4 A thermos bottle.

Highly reflecting surfaces are poor absorbers of radiation.

…the hole is therefore a perfect absorber.

FIGURE 37.5 A cavity with a small hole.

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The blackbody plays a special role in the study of thermal radiation because the spectrum of its thermal radiation does not depend on the material of which it is made or on any other characteristics of the body. By an argument based on thermodynamics, it can be established that the spectrum depends only on the temperature of the blackbody. The mathematical formula for the distribution of energy in this spectrum is therefore a universal law, and in the last years of the nineteenth century, physicists engaged in an intensive experimental and theoretical effort to find this universal law of blackbody radiation.

✔

Checkup 37.1

A wood-burning stove is made of cast iron. How would the performance of the stove change if it were made of polished stainless steel? QUESTION 2: One house has a roof of dark-colored shingles; another has a roof of light-colored shingles. Which roof absorbs more heat during the day, in sunlight? QUESTION 3: For space “walks,” astronauts wear suits with a shiny, silvery surface layer. What is the purpose of this surface layer? QUESTION 4: Two ideal cavities emit thermal radiation. The spectrum emitted by the first cavity is most intense at a wavelength of 600 nm, and that of the second at 500 nm. Compared with the second cavity, the radiation emitted by the first cavity is (A) More intense at all wavelengths (B) More intense at short wavelengths and less intense at long wavelengths (C) Less intense at short wavelengths and more intense at long wavelengths (D) Less intense at all wavelengths QUESTION 1:

3 7 . 2 E N E R G Y Q U A N TA Before 1900, physicists made several attempts at a theoretical explanation of the distribution of energy in the spectrum of blackbody radiation, but they met with disaster. One of the best of these attempts was that of Lord Rayleigh. Since the energy flux of the radiation emerging from the hole in a cavity is directly proportional to the energy density of the radiation inside the cavity [compare Eqs. (33.17) and (33.22)], Rayleigh decided to calculate the latter quantity. He began by noting that the radiation in a cavity is made up of a large number of standing electromagnetic waves. Figure 37.6 shows some of these standing waves; they are analogous to the standing waves in an organ pipe closed at both ends. Each of these standing waves can be regarded as a mode of vibration of the cavity. Rayleigh then appealed to the equipartition theorem, according to which, at thermal equilibrium, each mode of vibration has an average thermal energy of kT, where k is Boltzmann’s constant (in Section 19.4 we stated a special case of the equipartition theorem for free translational or rotational motion of a gas molecule). Thus, each of the standing waves of Fig. 37.6 ought to have an energy kT, and from this we can calculate the energy distribution in the spectrum of the radiation (the dashed curve in Fig. 37.3 shows the energy distribution obtained from Rayleigh’s calculation). Although this calculation gave reasonable results at the long-wavelength end of the blackbody spectrum, it gave disastrous results at the shortwavelength end: the number of possible standing-wave modes of very short wavelength is infinitely large, and if each of these modes had an energy kT, the total energy

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37.2

Energy Quanta

1259

y

Radiation in cavity is made up of a large number of standing electromagnetic waves.

x

FIGURE 37.6 Some of the possible standing electromagnetic waves in a closed cavity. For the sake of simplicity, only waves with a horizontal direction of propagation are shown. The plots give the electric fields as a function of x at one instant of time.

There are few modes of long wavelength…

L Ey

…and an infinite number of possible modes of shorter wavelength.

Ey

0

L

(a)

x

Ey

L 0

x

(b)

in the cavity would be infinite! This disastrous failure of classical physics has been called the ultraviolet catastrophe. The correct formula for the distribution of energy in the spectrum of blackbody radiation was finally obtained by Max Planck in 1900. By some inspired guesswork, based on thermodynamics, Planck found a mathematical formula that gave a precise fit to the experimental curves of blackbody radiation, such as those plotted in Fig. 37.3. He then searched for a theoretical explanation for his formula. The search led Planck to a revolutionary discovery: the quantization of energy. This discovery was to bring about the overthrow of classical physics and the birth of quantum physics. For Planck, the postulate of the quantization of energy was “an act of desperation,” which he committed because “a theoretical explanation had to be found at any cost, whatever the price.” Planck’s derivation of the blackbody radiation formula involves some sophisticated statistical mechanics which we will not reproduce. We will merely give a sketchy outline of this derivation. Planck began by making a theoretical model of the walls of the cavity: he regarded the atoms in the walls as small harmonic oscillators of many different natural frequencies, that is, small masses (with electric charges) attached to springs of many different spring constants. Although this is a rather crude model of the atoms that make up the walls of the cavity, it was adequate for his purposes since, as described in the preceding section, the radiation in a cavity is known to be completely independent of the characteristics of the wall. The random thermal motions of the oscillators result in the emission of electromagnetic radiation. This radiation fills the cavity and acts back on the oscillators. When thermal equilibrium is attained, the average rate of emission of radiation energy by the oscillators matches their rate of absorption of radiation energy. Thus, the oscillators share their energy with the radiation in the cavity, and Planck was able to show that, under equilibrium conditions, the average radiation energy at some frequency f (or at a wavelength l  cf ) is directly proportional to the average energy of an oscillator of frequency f. These steps of Planck’s calculation involved nothing but classical mechanics. But in the next step of the calculation, Planck departed radically from classical physics.

0

L

(c)

ultraviolet catastrophe

x

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He postulated that the energy of the oscillators is quantized according to the following rule: In an oscillator of frequency f, the only permitted values of the energy are E  0, hf, 2hf, 3hf, …

(37.1)

All other values of the energy are forbidden. The constant h in Eq. (37.1) is a new fundamental constant, called Planck’s constant. The value of this constant is Planck’s constant energy quantum hf

energy of oscillator quantum number n

h  6.63  1034 Js

(37.2)

The energy hf is called an energy quantum; according to the quantization rule, the energy of an oscillator is always some multiple of the basic energy quantum hf : E  nhf

n  0, 1, 2, 3, …

(37.3)

The integer n is called the quantum number of the oscillator. With this quantization condition, Planck calculated the average energy of the oscillators; and from that he derived his formula for the energy flux radiated by a blackbody per unit wavelength, Sl 

2phc2 l5  (e hclkT  1)

(37.4)

The energy flux per unit wavelength means that if dl is a small interval of wavelength centered on a given wavelength l, then the energy flux dS of electromagnetic waves that have wavelengths in the interval dl is dS  Sl dl

MAX PLANCK (1858–1947) German theoretical physicist, professor at Berlin and President of the Kaiser Wilhelm Institute (later renamed the Max Planck Institute). Planck made important contributions to thermodynamics before he became involved with the problem of blackbody radiation. He was deeply troubled by the quantization of energy, because he recognized that this held disastrous consequences for classical mechanics and electromagnetism. He was awarded the Nobel Prize in 1918 for his discovery of energy quanta.

The distribution function (37.4) agrees with the experimentally measured distributions shown in Fig. 37.3. Although we cannot go into the details of this derivation, we can achieve a rough understanding of how Planck’s calculation avoids the ultraviolet catastrophe. The thermal energy of the walls of the cavity is shared at random among all the oscillators in these walls. Some of these oscillators have high frequencies, some have low frequencies. For an oscillator of very high frequency, the energy quantum hf is very large. If this oscillator is initially quiescent (n  0), it cannot begin to move unless it acquires one energy quantum; but since this energy quantum hf is very large, the random thermal disturbances will be insufficient to provide it—the oscillator will remain quiescent. Thus, the quantization of energy tends to inhibit the thermal excitation of the high-frequency oscillators. If the high-frequency oscillators remain quiescent, they will not supply energy to the corresponding high-frequency standing waves in the cavity. These waves will then not have the energy kT predicted by Rayleigh; instead they will have no energy at all. This avoids the ultraviolet catastrophe. Note that for an oscillator with a frequency of f  1015 Hz, which is typical for atomic vibrations, the energy quantum is hf  6.6  1034 Js  1015 Hz  6.6  1019 J. Since this is a very small amount of energy, quantization does not make itself felt at a macroscopic level. But quantization plays a pervasive role at the atomic level. Unfortunately, Planck could not offer any basic justification for his postulate of quantization of energy. His postulate gave him a blackbody radiation law which was in complete agreement with the experimentally measured distribution of energy over wavelength (as displayed in Fig. 37.3), but his postulate brought him into conflict with classical physics. Superficially, the quantization of energy is analogous to the quantization of electric charge—we know from Chapter 22 that the electric charge of any

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Energy Quanta

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particle or body is always some multiple of the fundamental charge e. However, whereas the quantization of charge is consistent with the laws of classical physics, the quantization of energy is inconsistent with these laws. The energy of any oscillator that obeys Newton’s laws—such as a mass on a spring—can be changed by a small amount by pushing on the oscillator with a very weak force, and it should therefore be possible to change the energy of the oscillator continuously by any amount we please, not just in discrete steps of one energy quantum. Thus, quantization of energy makes no sense in classical physics. Planck could only justify his postulate by its consequences; but, in theoretical physics, the end does not justify the means. A deeper explanation of the quantization of energy emerged only much later, with the development of quantum mechanics (see Chapter 38). From Planck’s formula (37.4) for the distribution of energy in the spectrum of blackbody radiation one can establish that the energy flux has a maximum at a wavelength lmax 

1 hc 1  4.965 k T

(37.5)

where c is, as always, the speed of light, k is Boltzmann’s constant (see Section 19.1), h is Planck’s constant, and T is the absolute temperature of the blackbody. Equation (37.5) is called Wien’s displacement Law. If we insert the numerical values of h, c, and k, Wien’s Law takes the simple form lmax 

2.90  103 mK T

(37.6)

Wien’s displacement Law

Wien’s Law asserts that lmax is inversely proportional to the temperature T. This means that an increase of temperature shifts the location of the maximum to shorter wavelengths, in agreement with the experimental results presented in Fig. 37.3. If the temperature is sufficiently high—6000 K or so—the maximum of the spectrum lies in the visible region. For instance, the Sun, with a surface temperature of 5800 K, emits its largest flux of thermal radiation in the visible region. According to Wien’s Law, at what wavelength does the thermal radiation emitted by the Sun have its maximum? At what wavelength does the thermal radiation emitted by the tungsten filament in a lightbulb have its maximum? Assume that the Sun and the tungsten filament are approximately blackbodies at temperatures of 5800 K and 3200 K, respectively.1 EXAMPLE 1

SOLUTION: With T  5800 K, Wien’s Law gives

lmax 

2.90  103 mK  5.0  107 m  500 nm 5800 K

and with T  3200 K, it gives lmax 

2.90  103 mK  9.1  107 m  910 nm 3200 K

1 The blackbody approximation is fairly good for the Sun (except at sunspots; see the chapter photo). But it is not good for an ordinary tungsten filament, because the tungsten surface is not a good absorber. The blackbody approximation becomes better if the filament is tightly coiled, like a solenoid, and if we examine the radiation in the interior of this coil (a cavity).

Concepts in Context

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Thus, the maximum of the thermal radiation from the Sun lies in the visible region (in the green), but the maximum of the thermal radiation from the lightbulb lies in the infrared.

Furthermore, from Planck’s formula (37.4) one can calculate the combined energy flux for all wavelengths radiated from the surface of a blackbody by integration of the intensity per unit wavelength, Sl, over all wavelengths. This gives a total intensity proportional to the fourth power of the temperature, S  sT 4

Stefan–Boltzmann Law

(37.7)

This is called the Stefan–Boltzmann Law. The constant of proportionality s in this law can, again, be expressed in terms of h, c, and k. With the appropriate numerical values, the Stefan–Boltzmann constant then has the value s  5.67  108 W(m2K4)

(37.8)

Both the Wien and Stefan–Boltzmann laws had been discovered empirically many years before Planck supplied their theoretical foundation. Note that the energy flux S in Eq. (37.7) depends only on the temperature of the blackbody; it does not depend on the material from which the body is made. Since S is the energy flux, or the power per unit area, we can obtain the net power radiated from the body by multiplying S by the surface area; thus, the net power depends on the size of the body.

On a clear night, the surface of the Earth loses heat by radiation. Suppose that the temperature of the ground is 10C and that the ground radiates like a blackbody. What is the rate of loss of heat per square meter? EXAMPLE 2

SOLUTION: The temperature of the ground is 10C, so the absolute temperature of the ground is 283 K. Hence, the Stefan–Boltzmann Law tells us that the radiated flux, or power per unit area, is

S  sT 4  5.67  108 W/(m2K4)  (283 K)4  364 W/m2 COMMENTS: This large radiative heat loss of the ground explains the sharp drop of temperature experienced during clear nights. The drop of temperature is much less severe if there is an overcast sky. The clouds then reflect most of the radiation back to the ground—they act like a blanket to keep the ground warm.

In a house, a room is heated by means of a radiator filled with hot water at 82C. The radiator consists of a large vertical panel. If you place your hand near the panel (see Fig. 37.7a), what is the rate at which thermal radiation is incident on your hand? The area of one side of your hand is 0.016 m2. EXAMPLE 3

SOLUTION: When your hand is very near the panel, the energy flux reaching your hand is the same as the energy flux, or intensity, emitted by the radiator panel. The power incident on your hand is the incident flux times the area A of your hand,

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37.2

Energy Quanta

P  S A  sT 4  A

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(a)

 5.67  108 W/(m2K4)  (82  273)4 K4  0.016 m2  14 W COMMENTS: The distance at which you place your hand from the panel is not crucial. If your hand is somewhat farther away (see Fig. 37.7b), then some of the radiation emitted by the portion of the panel directly facing your hand will miss your hand, but an equal amount of radiation from a more distant portion of the panel will now be able to reach your hand, and these amounts compensate. You can do a simple experiment to verify this: place your hand very near a radiator panel (or near a stove), and gradually move your hand away—you will hardly notice any difference for the first few centimeters of motion. For a radiator panel of finite size, the energy flux incident on your hand will decrease once you move far enough away. But if you are in a room all of whose walls, ceiling, and floor are lined with radiator panels, a compensation similar to that depicted in Fig. 37.7b is valid throughout the room—in such an environment, your hand receives the same flux no matter where it is located or in what direction it faces.

✔

Intensity reaching hand equals that emitted by radiator panel.

Hand is very near panel.

(b) Some intensity from nearest part of panel misses hand,…

Checkup 37.2

QUESTION 1: The caption for Fig. 37.3 gives the wavelengths of the maxima for different temperatures. What is the product of wavelength and temperature in each case? Is this in agreement with Wien’s Law? QUESTION 2: If we increase the temperature of a blackbody from 1000 K to 2000 K, by what factor do we change the wavelength of the maximum of the spectrum? By what factor do we change the total energy radiated from the surface? QUESTION 3: An oscillator has a quantized energy hf. Is its kinetic energy quantized? Is its potential energy quantized? QUESTION 4: The pendulum of a clock can be regarded as an oscillator. Is the energy of the pendulum quantized? Why don’t we notice the quantization? 2 Q U E S T I O N 5 : Suppose that P lanck’s constant was 6.6  10 Js instead of 34 Js. Would we notice the quantization of a pendulum? 6.6  10 QUESTION 6: Figure 37.8 shows photos of three stars. The light emitted by these stars is thermal radiation. Which of these stars is the hottest? The coolest? (A) Red; yellow (B) Red; blue (C) Yellow; blue (D) Blue; red (E) Blue; yellow

Different colors of thermal radiation from stars indicate different surface temperatures.

…but is mostly compensated by radiation from other parts.

Hand is somewhat away from panel.

FIGURE 37.7 (a) Hand placed near a flat radiator panel. (b) Hand placed farther away.

FIGURE 37.8 Betelgeuse, Bellatrix, and Rigel (red, yellow, and blue, respectively) in the constellation Orion.

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37.3 PHOTONS AND THE PHOTOELECTRIC EFFECT

photon

In 1905, Einstein showed that Planck’s formula could be understood much more simply in terms of a direct quantization of the energy of the radiation. Planck had postulated that the oscillators in the wall of the cavity have discrete quantized energies, but he had treated the electromagnetic radiation as a smooth, continuous density of energy, exactly as it is supposed to be according to classical electromagnetic theory. In contrast, Einstein proposed that electromagnetic radiation consists of discrete particle-like packets of energy. He regarded a wave of some given frequency f as a stream of more or less localized energy packets, each with one quantum of energy hf (see Fig. 37.9). The particle-like energy packets of magnitude hf are called photons. The wave then has an energy hf if it contains only one photon, 2hf if it contains two photons, and so on. The thermal radiation in a cavity, with waves traveling randomly in all directions, can then be regarded as a gas of photons. Einstein applied statistical mechanics to calculate the energy spectrum of this gas and he thereby obtained the energy spectrum of the cavity radiation. The essential difference between Planck’s and Einstein’s view of the cavity radiation is that Planck quantized only the exchange of radiation with the walls of the cavity, whereas Einstein quantized the radiation itself. Thus, in Einstein’s view, electromagnetic radiation is always quantized, regardless of where or how it is produced. Not only is the radiation quantized when it is produced by the oscillators in the walls of a cavity (as in the case of thermal radiation), but also when it is produced outside of a cavity, say, by the acceleration of electric charges on the antenna of a radio transmitter. (a)

FIGURE 37.9 (a) According to classical theory, the energy is smoothly distributed over an electromagnetic wave, although the energy density has maxima wherever the wave reaches maximum amplitude. (b) According to Einstein, the energy is localized in small energy packets, which move with the wave.

Concepts in Context

(b)

Classically, an electromagnetic wave has a continuously varying energy density.

In quantum theory, wave is made up of a large number of photons, each with energy E  hf.

c

c

The energy flux of sunlight reaching the surface of the Earth is 1.0  103 W/m2 at normal incidence. How many photons reach the surface of the Earth per square meter per second? For the purposes of this calculation assume that all the photons in sunlight have an average wavelength of 500 nm. EXAMPLE 4

SOLUTION: The energy of a photon of wavelength 500 nm is

E  hf  h

c l

 6.63  1034 Js 

3.00  108 m/s 7

5.0  10

m

 4.0  1019 J

(37.9)

The energy incident per square meter per second is 1.0  103 J. To obtain the number of photons, we must divide this by the energy per photon, 4.0  1019 J. That is,

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37.3

photons second

Photons and the Photoelectric Effect

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energy second



energy photon

For a square meter, this gives 1.0  103 J(sm2)

4.0  1019 Jphoton

 2.5  1021 photons per square meter per second

COMMENTS: Because the number of photons in sunlight and other common light sources is so large, at the macroscopic level we do not perceive the grainy character of the energy distribution in light.

With this concept of light as a stream of photons, Einstein was also able to offer an explanation of the photoelectric effect. In some early experiments on the production of radio waves by sparks, Hertz had noticed that light shining on an electrode tended to promote the formation of sparks. Subsequent careful experimental investigations demonstrated that the impact of light on an electrode can eject electrons, which trigger the sparks. The electrons emerge with a kinetic energy which increases directly with the frequency of the light. Figure 37.10 is a schematic diagram of the apparatus used in the investigation of the photoelectric effect. Light from a lamp illuminates an electrode of metal (C ) enclosed in an evacuated tube. Electrons ejected from this electrode travel to the collecting electrode (A ), and then flow around the external circuit. A galvanometer (G) detects this flow of electrons. The kinetic energy of the ejected photoelectrons can be determined by applying a potential difference V between the emitting and the collecting electrodes by means of an adjustable source of emf. With the polarity shown in the figure, the collector has a negative potential relative to the emitter, that is, the collector (A) exerts a repulsive force on the photoelectrons. If an electron is to travel from the emitter to the collector, the change in its potential energy e  V between emitter

photoelectric effect

Light from lamp illuminates… lamp

A

…a metal surface, the cathode, which emits electrons.

C

G

 V 

Electrons can reach a collecting electrode, the anode,… …and electron flow is measured by meter.

FIGURE 37.10 Schematic diagram of the apparatus for the investigation of the photoelectric effect.

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stopping potential

Quanta of Light

and collector must be less than or equal to its initial kinetic energy. If the change in potential energy exceeds the initial kinetic energy, then the electron will reverse its motion and return to the emitter. The critical potential Vstop that stops the flow of electrons from emitter to collector is called the stopping potential. The measured value of this stopping potential gives us the initial kinetic energy of the electrons: K  eVstop

work function 

Experimentally, it is found that the kinetic energy determined in this way increases directly with the frequency of the incident light. For example, Fig. 37.11 is a plot of kinetic energy vs. frequency of the light for photoelectrons ejected from sodium. Note that, according to this plot, if the frequency is below 4.4  1014 Hz, then the light is incapable of ejecting electrons. Einstein’s quantum theory of light accounts for these experimental observations as follows. The electrons in the illuminated electrode absorb photons from the light, one at a time. When an electron absorbs a photon, it acquires an energy hf. But before this electron can emerge from the electrode, it must overcome the restraining forces that bind it to the metal of the electrode. The energy required for this is called the work function of the metal, designated by f. The remaining energy of the electron is then hf  f, and this must be the kinetic energy of the emerging electron: K  hf  f

Einstein’s photoelectric equation

threshold frequency

(37.10)

(37.11)

Some electrons suffer extra energy losses in collisions within the metal before they emerge; thus, hf  f actually is the maximum possible kinetic energy with which electrons can emerge. Equation (37.11) is Einstein’s photoelectric equation. It shows that the kinetic energy does indeed increase directly with the frequency, in agreement with the data of Fig. 37.11. According to Eq. (37.11), a minimum frequency is required to achieve the ejection of an electron. This minimum frequency, called the threshold frequency, corresponds to the ejection of an electron of zero kinetic energy; such an electron is just barely ejected. The threshold frequency fthresh is given by 0  hfthresh  f or fthresh 

f h

(37.12)

K Kinetic energy increases linearly with frequency of light.

3 eV

2

1

FIGURE 37.11 Kinetic energies (in electron-volts) of photoelectrons ejected from sodium by light of different frequencies.

3

4

5

6

7

A minimum photon energy, the work function, is required to emit an electron.

8

9

10

11

121014 Hz

f

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37.3

Photons and the Photoelectric Effect

When an electron absorbs a photon of this frequency, all of the energy of the photon is used to overcome the restraining forces that bind the electron to the metal, and no kinetic energy is left over for the ejected electron.

According to Fig. 37.11, the threshold frequency for sodium is 4.4  1014 Hz (photons of a frequency below this are incapable of ejecting electrons). What is the work function for sodium? Express the answer in eV. EXAMPLE 5

SOLUTION: With fthresh  4.4  1014 Hz, Eq. (37.12) gives

f  hfthresh  6.63  1034 Js  4.4  1014 Hz  2.9  1019 J  2.9  1019 J 

1 eV 1.6  1019 J

 1.8 eV

The work function for platinum is 6.2 eV. If ultraviolet light of frequency 5.0  1015 Hz illuminates a platinum electrode, what is the maximum kinetic energy of the ejected electrons? What is the stopping potential? EXAMPLE 6

S O L U T I O N : With f  6.2 eV  6.2  1.6  1019 J  9.9  1019 J, we

find from Eq. (37.11) K  hf  f  (6.63  1034 Js  5.0  1015 Hz)  9.9  1019 J  2.3  1018 J The stopping potential is, from Eq. (37.10), Vstop 

2.3  1018 J K   14 V e 1.6  1019 C

(37.13)

Einstein’s photoelectric equation was verified in detail by a long series of meticulous experiments by R. A. Millikan (the data in Fig. 37.11 are due to him). In order to obtain reliable results, Millikan found it necessary to take extreme precautions to avoid contamination of the surface of the photosensitive electrode. Since the surfaces of metals exposed to air quickly accumulate a layer of oxide, he developed a technique for shaving the surfaces of his metals in a vacuum, by means of a magnetically operated knife. The results of these experiments gave strong support to the quantum theory of light. This success of Einstein’s theory was all the more remarkable in view of the failure of the classical wave theory of light to account for the features of the photoelectric effect. According to classical theory, an electromagnetic wave acts on the electron by means of its electric field, which exerts a force on the electron. Therefore the crucial parameter that determines the ejection of a photoelectron should be the intensity of light, since this determines the strength of the electric field in the wave. If an intense wave strikes an electron, it should be able to jolt it loose from the metal, regardless of the frequency of the wave. Furthermore, the kinetic energy of the ejected electron should depend on the intensity of the wave. The observational evidence contradicts these predictions of the classical theory: A wave with a frequency below the threshold frequency never ejects an electron, regardless of its intensity. And, furthermore, the

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PHYSICS IN PRACTICE

Quanta of Light

P H O T O M U LT I P L I E R

The photoelectric effect finds many practical applications in sensitive electronic devices for the detection of light. For instance, in a photomultiplier tube, an incident photon ejects an electron from an electrode at the faceplate of the tube (Fig. 1). To convert this single electron into a measurable pulse of current, electric fields within the tube accelerate this single photon electron

100 V

electron toward a second electrode (called a dynode; see the figure), where its impact ejects several secondary electrons. These, in turn, are accelerated toward a third electrode, where their impact ejects tertiary electrons, and so on. Thus, the single electron from the first electrode generates an avalanche of electrons. In a high-gain photomultiplier tube, a pulse of 109 electrons emerges from the last electrode, delivering a pulse of current to an external circuit. In this way, the photomultiplier tube can detect the arrival of individual photons. A similar solid-state device known as an avalanche photodiode is also used for photon counting. Some sensitive video cameras rely on the same multiplier principle to convert the arrival of a photon at a photosensitive faceplate into a pulse of current. This permits these cameras to take pictures in faint light, where there are few photons.

200 V dynode

300 V

400 V 500 V

FIGURE 1 Schematic diagram of a photomultiplier tube. The curved electrodes are called dynodes. For the purpose of this diagram, it has been assumed that each electron impact on a dynode releases two electrons. The arrows show an avalanche of electrons.

FIGURE 2 Photomultiplier tubes.

kinetic energy depends on the frequency [as specified by Eq. (37.11)] and not on the intensity. High-intensity light ejects more photoelectrons, but does not give the individual electrons more kinetic energy.

✔

Checkup 37.3

QUESTION 1: The colors of light range from red to violet. What color has the most ener-

getic photons? Platinum has a larger work function than sodium. Qualitatively, how does the K vs. f plot for platinum differ from that for sodium (Fig. 37.11)?

QUESTION 2:

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37.4

The Compton Effect

QUESTION 3: Figure 37.12 shows a plot of current vs. applied potential for the photoelectric current emitted by the surface of metal illuminated with light of a given wavelength. Qualitatively, why is the current zero if V Vstop? Why does the current level off for large positive V ? Why do the curves differ for different intensities of light? QUESTION 4: Sodium has a work function of 1.8 eV; for platinum, the value is 6.2 eV. Compared with the plot of the maximum kinetic energy of ejected electrons as a function of the frequency of the incident light for sodium (Fig. 37.11), a similar plot for platinum has (A) A larger slope (B) A smaller slope (C) The same slope

1269

For each intensity of light, current is zero for V Vstop…

…and levels off for high V.

I (a) (b) V Vstop Current is smaller for lower intensity of light.

37.4 THE COMPTON EFFECT Very clear experimental evidence for the particle-like behavior of photons was uncovered by Arthur Holly Compton in 1922. Compton had been investigating the scattering, or the deflection, of X rays by a target of graphite (see Fig. 37.13). According to Maxwell’s theory, X rays are merely light waves of extremely high frequency. But according to quantum theory, they ought to consist of photons; and since their frequency is much higher than that of ordinary light, the energy of the X-ray photons ought to be much larger than that of photons of ordinary light (X rays will be discussed in more detail in the next section). When Compton bombarded the graphite with X rays of one selected wavelength, he found that the scattered X rays had wavelengths somewhat larger than that of the incident X rays. Classical theory cannot explain such a change of wavelength. When a classical electromagnetic wave is incident on a graphite target, its oscillating electric fields exert forces on the electrons in the carbon atoms, and this causes the electrons to oscillate at the same frequency as the wave. The oscillating, accelerated electrons then radiate a new electromagnetic wave, which spreads outward in all directions. This radiated wave is the scattered wave. Its frequency is necessarily the same as that of the oscillating electrons, which is the same as that of the incident wave. Compton soon recognized that the change of wavelength could be understood in terms of collisions of photons with electrons, collisions in which the photons behave like particles. In such a collision the electron of a carbon atom can be regarded as free, because the force binding the electron to the atom is insignificant compared with the force exerted by the incident photon. When the photon bounces off the electron, the electron recoils and thereby picks up some of the photon’s energy—the deflected photon

detector A thin beam of X rays approaches target… incident X rays

target

…and is deflected (scattered) in several directions by target.



Detector measures X rays emerging at an angle .

scattered X rays

FIGURE 37.13 Scattering of X rays by a target of graphite.

FIGURE 37.12 I vs. V for two different values of the intensity of light: (a) high intensity; (b) low intensity.

ARTHUR HOLLY COMPTON (1892– 1962) American experimental physicist. For his discovery of the Compton effect, he received the Nobel Prize in 1927.

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is left with a reduced energy. Since the energy of the photon is E  hf  hcl, a reduction of the energy of the photon implies an increase of wavelength. Qualitatively, we expect that photons that suffer the most violent collisions will be deflected through the largest angles and lose the most energy; and therefore such photons should emerge with the longest wavelength. And this is just what Compton found in his experiments. For a quantitative discussion of the photon–electron collision, we need an expression for the momentum of the photon. We can derive this expression from the relativistic formula for the momentum of a particle. Since the speed of the photon is the speed of light, the photon must be regarded as an ultra-relativistic particle. For such a particle, Eq. (36.52) tells us that p

E c

(37.14)

[Note that according to Eq. (36.50), the energy of a particle with a speed equal to the speed of light can be finite only if the rest mass m is zero; thus, photons must be regarded as particles of zero rest mass.] Since the energy of the photon is E  hf  hcl, the expression for the momentum can be written as p

momentum of photon

h l

(37.15)

With the expression (37.15) for the momentum of the photon, Compton calculated the change of energy and the change of wavelength of a photon in an elastic collision with an electron, initially at rest. For a photon that emerges from the collision at an angle u (the deflection angle; see Fig. 37.14), he calculated from the laws of conservation of energy and momentum that the change of wavelength is h (1  cos u) mec

¢l 

Compton wavelength shift

(37.16)

This change of wavelength is called the Compton shift. The change of wavelength is always small; it is always less than 0.005 nm (see the next example). Such changes of wavelength are difficult to detect unless the wavelength of the X rays is itself quite small, so ¢l is an appreciable fraction of the wavelength.

Deflection angle  is angle between photon directions of propagation before and after collision.

Incident photon collides with initially stationary electron.

photon  photon electron Electron emerges from collision.

FIGURE 37.14 Directions of propagation, or directions of the momentum, of a photon before and after collision with an electron.

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37.4

EXAMPLE 7

The Compton Effect

Evaluate the Compton shift for X rays scattered at 90 and for X rays scattered at 180.

SOLUTION: If the photon emerges from the collision at an angle of 90, then cos u  cos 90  0, and the wavelength shift of Eq. (37.16) is

¢l 

6.63  1034 Js h   2.43  1012 m mec 9.11  1031 kg  3.00  108 m/s  0.002 43 nm

If the photon emerges at 180, in a direction opposite to that of its initial direction of motion, then cos u  cos 180  1, and the wavelength shift is ¢l 

2h h  [1  (1)]   4.85  1012 m  0.004 85 nm mec mec

This is the maximum wavelength shift that can be produced by the Compton effect.

The formula (37.16) for the wavelength shift can be deduced by examining the conservation of energy and momentum in the photon–electron collision. As a simple special case of such a collision, consider a photon that recoils back in the direction it came from [that is, u  180 in Eq. (37.16)] while the electron, initially at rest, acquires a motion in the forward direction. For such a straight-line motion of photon and electron, write down the laws of conservation of energy and momentum, and deduce the wavelength change of the photon.

EXAMPLE 8

SOLUTION: We will use the relativistic expressions for momentum and energy,

because, in a collision with a high-energy photon, the electron will emerge with an energy large enough to require such a relativistic treatment. In the notation familiar from Chapter 11, we designate the initial momentum of the photon by p1, the final momentum by p 1; likewise, we designate the initial momentum of the electron by p2 (which is zero, since the electron is initially at rest), the final momentum by p 2. Conservation of momentum then tells us p1  p 1  p 2

(37.17)

and conservation of energy tells us E1  mec 2  E 1  E 2

(37.18)

where mec 2 is the initial energy of the electron (the rest-mass energy). To solve this pair of equations for the final photon energy, we need to express the momenta in Eq. (37.17) in terms of the energies. For this purpose, we first rewrite Eq. (37.17) as p 2  p1  p 1 and square both sides: ( p 2)2  ( p1  p 1)2

(37.19)

Now we substitute the relativistic relation (p 2)2  (E 2c)2  m2ec 2 for the electron [compare Eq. (36.51)] and the relations p1  E1 c and p 1  E 1 c for the photon: a

E 2 2 E 1 2 E1 b  m2ec 2  a  b c c c

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By means of Eq. (37.18) we can now eliminate the energy of the electron and obtain an equation that relates the initial and final energies of the photon: a

E1 E 1 2 E 1 2 E1  mec  b  m2ec2  a  b c c c c

When we expand the squares in this equation, several terms cancel, leaving us with the expression 2mec a

E 1 E1 E 1 E1  b 4 c c c c

Multiplying both sides by c(2me E1 E 1) gives 1 1 2   m E 1 c E1 c ec

(37.20)

But E1 c  p1  hl from Eq. (37.15), so Eq. (37.20) is equivalent to l

l 2   mec h h that is, ¢l 

2h mec

(37.21)

This result agrees with our general formula (37.16) when we set u  180 and cos u  1.

✔

Checkup 37.4

QUESTION 1: The colors of light range from red to violet. What color has photons of the largest momentum? QUESTION 2: Does the photon gas in a cavity in a hot body exert a pressure on the surrounding walls? QUESTION 3: Consider X rays of wavelengths 0.2 nm and 0.6 nm. If these suffer Compton collisions with electrons and emerge at the same scattering angle, which have the larger wavelength shift? Which have the larger percentage change of wavelength? QUESTION 4: Can the Compton effect occur with visible light? Would it be observable? Q U E S T I O N 5 : For Compton scattering, it is impossible that the wavelength shift ¢l is (A) Smaller than the wavelength of the incident photon (B) Larger than the wavelength of the incident photon (C) Less than 0.001 nm (D) Greater than 0.005 nm (E) Larger for larger scattering angles

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37.5

X Rays

1273

3 7 . 5 X R AY S X rays were discovered in 1895 by W. K. Röntgen in experiments with beams of energetic electrons. Röntgen found that when such a beam of electrons moving in an evacuated glass tube struck the wall of the tube, some invisible, mysterious rays were emitted that caused a faint luminosity (fluorescence) on a nearby sheet of paper impregnated with chemicals. He also found that these rays could affect covered and wrapped photographic plates—the rays could penetrate through the wrapping. These rays proved capable of penetrating though thick layers of opaque materials, and Röntgen immediately recognized the possible medical applications of these rays for making images of the tissues inside the human body, especially bones (see Fig. 37.15). The distance that X rays can penetrate through a material depends on the density of the material. When X rays pass through atoms, they tend to be absorbed by the atomic electrons. Therefore materials of high density, such as lead, with a concomitant high density of electrons, strongly absorb and block X rays. In X-ray photographs of parts of the human body, bones throw sharp shadows because their density is higher than that of the surrounding tissues. Organs made of soft tissues of low density, such as the gastrointestinal tract, do not throw sharp shadows, and to enhance the contrast of the X-ray photograph it is advantageous to fill the organ with a barium solution, a high-density material that blocks the X rays (see Fig. 37.16). Röntgen had named his rays “X rays” because their nature was unknown. Although he and his contemporaries suspected that they might be electromagnetic waves of extremely short wavelength, the conclusive experimental proof of this conjecture was not obtained until 1912, when Max von Laue argued that if X rays are waves, they should display interference effects when passing through crystals. The distances between the rows of atoms in a crystal, such as rock salt, are of the same order of magnitude as the wavelengths of X rays, and von Laue proposed that the crystal therefore can play the role of a “grating” for X rays, analogous to a multiple-slit grating used for interference experiments with light. Figure 37.17 shows a photograph of the interference pattern produced by X rays incident on a crystal. In this photograph the interference maxima show up as dark spots. The wavelength of the X rays can be deduced from the angular positions of the interference maxima and the known size of the spacings in the crystal. The wavelengths of X rays range from about 0.001 nm to 10 nm. The spots in such photographs of X-ray interference patterns are called Laue spots. The beautiful symmetry of the pattern of Laue spots reflects the symmetry of the arrangement of the atoms in the crystal. In modern crystallography laboratories, the patterns of Laue spots produced by X rays incident on a crystal are often used to investigate the structure of the crystal and that of the molecules in it. For instance, such Xray interference experiments played a large role in the determination of the structure of DNA. Since X rays are electromagnetic waves, their generation by the impact of energetic electrons on some sort of obstacle can be understood by the familiar mechanism of the emission of radiation by acceleration of electric charges, which we discussed in Chapter 33. When the electrons in the beam collide with the atoms in the obstacle, they will suffer sudden decelerations and radiate intense electromagnetic waves of short wavelength. This kind of radiation is called Bremsstrahlung (German for braking radiation). Figure 37.18 shows an X-ray tube in which electrons emerging from a hot filament are accelerated through a potential difference of several kilovolts and then

FIGURE 37.15 One of the first X-ray photographs prepared by Röntgen. It shows the hands of his wife; note the sharp image of the ring.

FIGURE 37.16 X-ray photograph of a stomach with barium shadowing.

Positions of interference maxima (Laue spots) indicate symmetry and spacing of atoms in a crystal.

FIGURE 37.17 Interference pattern produced by X rays incident on a crystal.

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Smooth part of X-ray spectrum (Bremsstrahlung) is due to electron decelerations…



When electrons are accelerated through a potential difference and then strike a metal target…

Quanta of Light

intensity

…and discrete line (characteristic) spectrum is due to specific kind of target atoms.

target

X rays

electrons …the sudden deceleration of electrons produces intense electromagnetic waves of short wavelength. 

FIGURE 37.18 An X-ray tube.

0.04 0.036 nm

0.06

0.08 nm



Minimum wavelength (maximum energy) corresponds to an electron giving all its energy to one photon.

FIGURE 37.19 Distribution of energy in the spectrum of X rays produced by electrons of 35 kilovolts incident on a molybdenum target.

strike an obstacle, or target, made of a heavy metal, such as tungsten or molybdenum. The X rays produced in the decelerations of the electrons escape through the side of the tube. Such X-ray tubes are widely used in medical X-ray machines. Figure 37.19 is a plot of the distribution of energy of the X rays generated by electrons of 35 kilovolts striking a molybdenum target. Note that the X-ray energy is smoothly distributed over a wide range of wavelengths, but there also are two conspicuous spikes in the energy distribution. The smooth portion of the X-ray spectrum is due to Bremsstrahlung, whereas the discrete spikes are generated in the interior of the atoms of molybdenum, in much the same way as spectral lines of visible light are generated by the atoms (see Chapter 38). In the plot of the energy distribution, the shape and the location of the broad peak of the Bremsstrahlung portion of the spectrum depend on the energy of the incident electrons. But the locations of the spikes do not depend on the electron energy; instead they depend on the material of the target. The spikes are called the characteristic spectrum; each kind of target atom has its own distinctive characteristic spectrum, just as each kind of atom has its own distinctive spectrum of visible light (for more on the characteristic spectra of atoms, see the next chapter). Let us ignore the spikes for now, and concentrate on the smooth Bremsstrahlung spectrum. Note that below a certain wavelength—for instance, below about 0.03 nm in Fig. 37.19—there is no Bremsstrahlung. The minimum wavelength emitted by the electrons is called the cutoff wavelength. Experimentally, the cutoff wavelength is found to be inversely proportional to the kinetic energy of the electrons. This cutoff makes no sense from the point of view of classical theory—if an electron suffers a sufficiently violent collision with an atom, it ought to be able to radiate waves of arbitrarily short wavelength (although the intensity of short-wavelength waves is expected to be low). But the cutoff is readily explained by the quantum theory of light, accord-

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X Rays

1275

ing to which the decelerating electron emits photons. Obviously, the maximum energy that a decelerating electron can give to a photon is all its energy. In this case, the electron emits only one single photon, with a maximum energy and a maximum frequency, or cutoff frequency, fcutoff. The energy hfcutoff of this single photon of maximum energy then equals the kinetic energy K of the electron: hfcutoff  K

(37.22)

from which fcutoff 

K h

The wavelength is related to the frequency by l  cf, so the cutoff wavelength is lcutoff 

hc K

(37.23)

If an X-ray tube is operated at a potential of 35 kilovolts, the electron energy is 35 keV. What is the cutoff wavelength for the X rays generated by electrons of this energy?

EXAMPLE 9

SOLUTION: In joules, the electron energy is

K  35 keV  1.60  1019 J/eV  5.6  1015 J Equation (37.23) then gives lcutoff 

6.63  1034 Js  3.00  108 m/s hc  K 5.6  1015 J

 3.6  1011 m  0.036 nm This is in agreement with the cutoff wavelength indicated in Fig. 37.19.

✔

Checkup 37.5

QUESTION 1: A fine slit cut in a sheet of lead produces noticeable diffraction effects for light, but not for X rays. Why not? QUESTION 2: Figure 37.19 shows the energy distribution for X rays produced by electrons of 35 keV incident on a molybdenum target. If we reduce the electron energy to 17.5 keV, qualitatively, what will change in this figure? QUESTION 3: Sometimes an X-ray photograph of the intestines reveals bubbles of gas. Keeping in mind that X-ray photographs are usually negatives (they turn dark where the X rays strike, as in Fig. 37.15), would you expect a bubble of gas to look light or dark? QUESTION 4: When electrons are accelerated through an electrostatic potential V0 and strike a target, the frequencies f of emitted X rays obey the relation (A) f  V0 h (B) f V0 h (C) f  eV0 h

(D) f eV0 h

(E) f  eV0 hc

cut-off wavelength

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3 7 . 6 WAV E V S . PA RT I C L E

wavicle

Light beam is incident from left on a plate with a narrow slit. video camera P 

Sensitive camera detects arrival of individual photons.

FIGURE 37.20 A diffraction experiment.

From the photoelectric effect and the Compton effect, we learned that photons have particle properties. On the other hand, we know that light displays interference and diffraction phenomena, which prove that photons also have wave properties. Thus photons are neither classical particles nor classical waves. They are some new kind of object, unknown to classical physics, with a subtle combination of both wave and particle properties. Arthur Eddington coined the name wavicle for this new kind of object.2 It is difficult to achieve a clear understanding of the character of a wavicle because these objects are very remote from our everyday experience. We all have an intuitive grasp of the concept of classical particles and classical waves from our experience with, say, billiard balls and water waves, but we have no such experience with wavicles. We can gain some insight into the interplay between the particle behavior and the wave behavior of a photon by a new, detailed examination of a simple diffraction experiment. Figure 37.20 shows a light beam striking a plate with a narrow slit. In the usual diffraction experiments described in Chapter 35, we installed a screen or a photographic film at the far right, and with this we recorded the intensity of the light in the diffraction pattern. Instead, in our new arrangement we will install the faceplate of a very sensitive video camera in place of the customary screen. With this, we can detect the individual photons of the light in the diffraction pattern. If the incident light beam has a very low intensity, so there is only one photon passing through the slit at a time, we can watch the photons arriving one by one at the faceplate of our video camera. Figure 37.21a shows a typical pattern of impacts of 30 photons. The pattern seems quite random. If the photons behaved like classical particles, they would travel along a straight line and they would reach only those points on the faceplate that are within the geometric shadow of the slit. The widely scattered impacts prove that the photons are certainly not traveling along such straight lines. Figure 37.21b shows the pattern of accumulated impacts for 300 photons, and Fig. 37.21c shows it for 3000 photons. In these figures we can recognize a tendency of the photons to cluster in bandlike zones. These zones correspond to the maxima of the diffraction pattern predicted by the wave theory of light. Finally, Fig. 37.21d shows the pattern of accumulated impacts for a very large number of photons; this is simply the familiar intensity pattern of light diffracted by a slit (see also Fig. 35.33). (a)

FIGURE 37.21 Patterns of impacts of photons on the video camera in Fig. 37.20. (a) 30 photons; (b) 300 photons; (c) 3000 photons; (d) a very large number of photons. The first three pictures are computer simulations of accumulated photon impacts; the last picture is a diffraction pattern obtained with a laser beam.

(b)

(c)

For a few photon impacts, pattern is not a shadow of slit, but is spread out and seems random…

(d)

…but after many impacts, photon distribution resembles diffraction pattern of wave theory of light.

2 Other names that have been proposed are quanticle and quon. Most physicists prefer to avoid all such neologisms; if anything, they favor the descriptive phrase quantum-mechanical particle or wave-mechanical particle.

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Wave vs. Particle

1277

From this diffraction experiment we learn that photons display both a wave aspect and a particle aspect. They behave like waves while passing through the slit; but they behave like particles when they strike the faceplate of the video camera. Whether the wave aspect or the particle aspect predominates depends on the experimental equipment with which the photon is interacting. The slit brings out the wave aspect of the photon; the faceplate of the video camera brings out the particle aspect. The behavior of the photons is governed by a probabilistic law. The point of impact of an individual photon on the faceplate is unpredictable. Only the average distribution of impacts of a large number of photons is predictable: the distribution of photons matches the intensity distribution calculated from the wave theory of light. Thus, the probability that a photon arrives at a given point on the faceplate is proportional to the intensity of the wave at that point. Since the intensity of an electromagnetic wave is proportional to the square of its electric field, we can write the proportionality of probability and intensity as [probability for photon at a point] r E 2

(37.24)

Here we have a connection between the wave and the particle aspects of the photon: the intensity of the photon wave at some point determines the probability that there is a photon particle at that point. This probability interpretation of the intensity of the wave was discovered by Max Born. Our single-slit experiment can also teach us something about the limitations that quantum theory imposes on the ultimate precision of measurement of the position of a wavicle. Suppose we have a light wave consisting of one photon and we want to measure the position of this photon. The position has x, y, and z components; we will concentrate on the y component, perpendicular to the direction of propagation. Figure 37.22 shows the wave propagating in the horizontal direction; the y direction is vertical. To determine this vertical position of the photon, we use a narrow slit placed in the path of the wave. If the photon succeeds in passing through this slit, then we will have achieved a determination of the vertical position to within an uncertainty ¢y  a

probability interpretation of wave intensity

A photon incident from left on a slit…

y

p

a



x

…will emerge at some unknown angle  within the diffraction pattern.

FIGURE 37.22 Photon passes through a narrow slit and emerges at an angle u.

(37.25)

where a is the width of the slit. If the photon fails to pass through this slit, then our measurement is inconclusive and will have to be repeated. By making the slit very narrow, we can make the uncertainty of our determination of the y coordinate very small. But this has a surprising consequence for the y component of the momentum of the photon: if we make the uncertainty in the y coordinate small, we will make the uncertainty in the y component of the momentum large. To see how this comes about, let us recall that according to our preceding discussion of the single-slit experiment, the photon suffers diffraction by the slit and emerges at some angle u (Fig. 37.22). This angle u is unpredictable; all we can say about the photon after it emerges from the slit is that it will be heading toward some point within the diffraction pattern. Thus, the direction of motion of the photon is uncertain, and therefore the y component of its momentum is uncertain. If we make the slit very narrow, the diffraction pattern will become very wide, and the uncertainty in the direction of motion and the uncertainty in the y component of the momentum will become very large. As a rough quantitative measure of the magnitude of the uncertainty in direction, we can take the angular width of the central diffraction maximum (most of the intensity of the photon wave is gathered within the region of this central maximum, and

MAX BORN (1882–1970) German, and later British, theoretical physicist. He was awarded the Nobel Prize somewhat tardily in 1954 for his discovery of the probabilistic interpretation of quantum waves in 1926.

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hence the photon is most likely to be found in this region). This estimate of the uncertainty of the angle gives us ¢u  l a

(37.26)

The y component of the momentum is py  p sin u (see Fig. 37.22); since we are concerned with a small angle, we can use the approximation sin u  u and therefore py  p u. The uncertainty in py is then ¢py  p ¢u  pla

(37.27)

But, according to Eq. (37.15), p  hl, and therefore ¢py  ha

(37.28)

Thus, the uncertainty in the y component of the momentum is inversely proportional to the width of the slit. Comparing Eqs. (37.25) and (37.28), we find that the product of the uncertainty of position and the uncertainty of momentum is ¢y ¢py  h

(37.29)

This equation states that ¢y and ¢py cannot both be small; if one is small then the other must be large, so that their product equals Planck’s constant. Although we have obtained Eq. (37.29) by examining the special case of a position measurement by means of a slit, it turns out that this relation is actually of general validity for any kind of position measurement—the product of the uncertainty in the position and the uncertainty in the momentum always equals or exceeds Planck’s constant. A more rigorous derivation uses the rms uncertainty—the square root of the mean of 1 2 the squares of the deviations from the mean value; for example, ¢x  [(x  x)2] . Careful consideration of a variety of arrangements for the simultaneous measurement of position and momentum shows that the product of rms uncertainties rigorously obeys the inequality ¢y ¢py 

Heisenberg uncertainty relation

h 4p

(37.30)

Equation (37.30) is one example of a Heisenberg uncertainty relation. There are corresponding relations for the other components of position and momentum. The Heisenberg uncertainty relations tell us that there exist ultimate, insuperable limitations in the precision of our measurements. At the macroscopic level, the quantum uncertainties in our measurements can be neglected. But at the atomic level, these quantum uncertainties are often so large that it is completely meaningless to speak of the position or momentum of a wavicle.

Suppose we measure the vertical position of a photon by means of a narrow horizontal slit of width 1.0  105 m. With what uncertainty in vertical momentum does the photon emerge from this slit?

EXAMPLE 10

SOLUTION: Equation (37.28) tells us that for such a position measurement WERNER HEISENBERG (1901–1976) German theoretical physicist. He was one of the founders of the new quantum mechanics, and received the Nobel Prize in 1932.

¢py 

6.63  1034 Js h  6.6  1029 kgm/s  a 1.0  105 m

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Summary

✔

1279

Checkup 37.6

QUESTION 1: If photons were classical particles, what pattern of impact points would we find on the faceplate placed beyond the narrow slit in Fig. 37.22? QUESTION 2: Suppose that the amplitude of the electric field at some point in the diffraction pattern displayed in Fig. 37.21c is one-half the amplitude at the center. By what factor is the probability for a photon smaller than at the center? QUESTION 3: Suppose that an electron moving in a TV tube has an uncertainty of 106 m in position. Can this electron be free of uncertainty in momentum? In velocity? In energy? QUESTION 4: Given that an electron has zero uncertainty in its y position (¢y  0), can this electron have zero uncertainty in the x component of the momentum (¢px  0)? The z component of the momentum (¢pz  0)? QUESTION 5: Consider a double-slit experiment, with slit width and slit spacing several times the wavelength of the incident light. However, now the intensity is greatly reduced, so that only individual photons are incident on the slits. After a long time, the transmitted light recorded by a sensitive camera would show (A) A double-slit interference pattern (B) A single-slit diffraction pattern (C) A random pattern of uniform intensity (D) Two sharp shadows of the slits without diffraction effects

S U M M A RY PHYSICS IN PRACTICE

(page 1268)

Photomultiplier

PLANCK’S CONSTANT

ENERGY QUANTIZATION OF OSCILLATOR

WIEN’S DISPLACEMENT LAW

h  6.63  1034 Js E  n hf

(37.2)

n  0, 1, 2, …

(37.3)

lmax  (2.90  103 mK)T

(37.6)

energy flux (arbitrary units) classical 1450 K 1450 K

1250 K

STEFAN–BOLTZMANN CONSTANT

ENERGY AND MOMENTUM OF A PHOTON EINSTEIN’S PHOTOELECTRIC EQUATION

where  is the work function of the metal

1000 K

S  sT 4

where S is the energy flux radiated by a blackbody STEFAN–BOLTZMANN LAW

(37.7) 0

s  5.67  108 W/(m2K4 ) E  hf

and

p  hfc  hl

K  hf  f

2000

4000 wavelength

6000 nm

(37.8) (37.15) (37.11)

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COMPTON WAVELENGTH SHIFT OF PHOTON

¢l 

h (1  cos u) mec

photon

(37.16)

 photon electron

CUTOFF WAVELENGTH FOR X RAYS

cutoff 

hc K

intensity

(37.23)

Minimum wavelength (maximum energy) corresponds to an electron giving all its energy to one photon.

0.04

PROBABILITY INTERPRETATION OF WAVE

0.06

0.08 nm



[probability for presence of photon] r [intensity of wave] r E 2 (37.24)

INTENSITY HEISENBERG UNCERTAINTY RELATION FOR y AND p y

¢y ¢py 

h 4p

(37.30)

QUESTIONS FOR DISCUSSION 1. Is the light emitted by a neon tube thermal radiation? The light emitted by an ordinary incandescent lightbulb? 2. Does your body emit thermal radiation? 3. In Example 2 we calculated the radiative heat loss from the ground during a clear night. Qualitatively, how would a cover of snow on the ground affect the result? 4. The insulation used in the walls of homes consists of a thick blanket of fiberglass covered on one side by a thin aluminum foil. What is the purpose of these two layers? 5. Black velvet looks much blacker than black paint. Why? 6. For protection against the heat of sunlight, parts of the Lunar Lander (and some other spacecraft) were wrapped in shiny aluminum foil. Why is shiny foil useful for this purpose? 7. If you look into a kiln containing pottery heated to a temperature equal to that of the walls of the kiln, you can scarcely see the pottery. Explain. 8. According to Fig. 37.3, at what wavelength is the energy flux maximum for a body at 1450 K? At 1250 K? At 1000 K? Do these wavelengths satisfy Wien’s Law?

12. Fluorescent paints achieve their exceptionally bright orange or red color by converting short-wavelength photons into longwavelength (red) photons. Why can we not make such a paint in blue or violet colors? 13. When light of a given wavelength ejects photoelectrons from the surface of a metal, why is it that not all of the photoelectrons emerge with the same kinetic enegy? 14. Can a particle of mass zero ever be at rest? 15. According to Eq. (37.16), a photon suffers a maximum change of wavelength in a collision with an electron if it emerges at an angle u  180, and a minimum change of wavelength (no change) if it emerges at an angle u  0. Is this reasonable? 16. Suppose that a photon and an electron have the same momentum. Which has the larger energy, taking into account both the rest-mass energy and the kinetic energy? 17. Can the Compton effect occur with visible light? Would it be observable? 18. Photons of short wavelength are more particle-like than photons of long wavelength. Why?

9. The quantization of electric charge is consistent with classical physics, but the quantization of energy is not. Does this make sense?

19. Give an example of an experiment in which photons behave like waves. Give an example of an experiment in which they behave like particles.

10. Suppose that Planck’s constant were much larger than it is, say, 1034 times larger. What strange behavior would you notice in a simple harmonic oscillator consisting of a mass hanging on a spring?

20. What happens to the y momentum that a photon gains or loses in the diffraction experiment described in Fig. 37.22?

11. Why do we not notice the discrete quanta of light when we look at a lightbulb?

21. According to Eq. (37.24), we can predict only the probability that a photon will be found at some given point. Does this mean that quantum physics is not deterministic?

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Problems

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PROBLEMS 37.1 Blackbody Radiation 37.2 Energy Quanta

10. At what wavelength does your skin radiate a maximum flux of thermal radiation? Assume that the temperature of your skin is 33C.

1. Consider a seconds pendulum, that is, a pendulum that has a period of 2 seconds. What is the magnitude of one energy quantum for such a pendulum? Would you expect that quantum effects are noticeable in such a pendulum?

11. Nanomechanical oscillators with a frequency of 5.0 GHz and masses as small as 2.0  1020 kg can be fabricated. What is the amplitude of oscillation of such an oscillator if its energy of oscillation is one energy quantum?

2. In molecules, the atoms can vibrate about their equilibrium positions. For instance, in the H2 molecule, the hydrogen atoms vibrate about their equilibrium positions with a frequency of 1.31  1014 Hz. What is the magnitude of the energy quantum for this oscillating system? Note that although there are two masses in this system, they must be regarded as a single oscillator because the vibrational motions of the two masses are always equal.

12. Thermonuclear reactions near the core of the Sun maintain a temperature of 5.0  107 K. Assume that the distribution of this radiation is thermal. At what wavelength does the spectrum of this radiation have a maximum energy flux? What is this part of the electromagnetic spectrum called? What is the energy per second emitted per square meter near the Sun’s core due to this radiation?

3. In a solid, each atom is held in its position by elastic forces, which permit the atom to oscillate about its equilibrium point. The frequency of such oscillations of an aluminum atom in solid aluminum is 8.0  1011 Hz. The energy of this oscillatory motion is quantized. What is the magnitude of an energy quantum? 4. If we take Planck’s model of the walls of a blackbody cavity seriously, we will have to assume that the oscillating masses are electrons. Imagine an electron oscillating with a frequency of 2.0  1015 Hz under the influence of a springlike force. What is the amplitude of oscillation of this electron if its energy of oscillation is one energy quantum? Two energy quanta?

13. Low temperatures are commonly attained using liquid helium, which has a latent heat such that 1.0 watt will boil approximately 1.0 liter of liquid helium per hour. Consider a region with a surface area of 0.10 m2. What is the rate of energy transfer into the region due to thermal radiation if it is surrounded by a room-temperature (295 K) environment? If surrounded by a liquid nitrogen (77 K) environment? Which is preferable in order to conserve liquid helium? 14. The nichrome heater wire in a toaster has a radius of 0.20 mm and a total length of 1.0 m. Assume that the wire acts as a blackbody and emits 1200 W of thermal radiation. What is the surface temperature of the wire? At what wavelength is the thermal radiation maximum?

5. An oscillator of frequency 3.00  1013 Hz consists of a mass of 1.67  1027 kg attached to a spring. What is the amplitude of oscillation of this oscillator if its energy of oscillation is one energy quantum? Two energy quanta?

15. A black hole radiates with a thermal spectrum due to quantum effects. A black hole with a radius of 30 km radiates a total of 8.8  1031 W of such thermal radiation, or Hawking radiation. What is the temperature of this blackbody?

6. Suppose that two stars have the same size but the temperature of one is twice that of the other. By what factor will the thermal power radiated by the hotter star be larger than that radiated by the cooler star?

*16. The tungsten filament of a lightbulb is a wire of diameter 0.080 mm and length 5.0 cm. The filament is at a temperature of 3200 K. Calculate the power radiated by the filament. Assume the filament acts like a blackbody.

7. The theoretical expression for the constant s in the Stefan–Boltzmann Law is

*17. The temperature of the surface of the Sun is 5800 K and the Sun’s radius is 6.96  108 m. If the Sun radiates like a blackbody, what can you predict for the thermal energy flux it emits? What can you predict for the energy flux of sunlight arriving at the Earth, at a distance of 1.5  1011 m? Compare with the measured value, 1.35  103 W/m2.

s

2p5k4 15h3c 2

Verify that the numerical value implied by this theoretical expression agrees with Eq. (37.8). 8. The flux of thermal radiation from the star Procyon B is observed to have a maximum at a wavelength of 440 nm. Assuming the star radiates like a blackbody, what temperature can you deduce for the surface of this star? 9. Melting iron has a temperature of 1808 K. At what wavelength does the iron radiate a maximum flux? Assume that the iron acts like a blackbody.

*18. The star Procyon B is at a distance of 11 light-years from Earth. The flux of its starlight reaching us is 1.7  1012 W/m2, and the surface temperature of the star is 6600 K. Calculate the size of the star. *19. Consider the integral over wavelengths of Planck’s expression, Eq. (37.4), for the distribution of blackbody radiation

S



dS 



Sldl 



0

q

2phc2 l5  (e hclkT  1)

dl

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Show that the total intensity emitted by a blackbody is proportional to the fourth power of the absolute temperature, as required by the Stefan–Boltzmann Law. (Hint: You need not evaluate the integral; you need only use the substitution x  hclkT to make the integral dimensionless.) *20. At the Earth, the flux of sunlight per unit area facing the Sun is 1.35  103 W/m2. The Earth absorbs heat from the sunlight and reradiates heat as thermal infrared radiation. For equilibrium, the power arriving from the Sun must equal the average power radiated by the surface of the Earth. This permits us to make a rough prediction for the average temperature of the Earth. (a) Assume that the Earth absorbs all of the sunlight striking it. What is the power of the sunlight absorbed? (Hint: The relevant area is the cross-sectional area pR2 of the Earth.) (b) Assume that the surface of the Earth radiates like a blackbody. If the temperature of the surface is T, what is the expression for the power radiated? (Hint: The relevant area is the total surface area 4pR2.) If this power is to match the power calculated in (a), what must be the value of T ? **21. Deduce the surface temperature of Pluto by the method described in the preceding problem. Pluto is 39 times as far away from the Sun as the Earth, and hence the flux of sunlight at Pluto is (39)2 times smaller than the flux at the Earth. **22. Maximize Planck’s expression for the intensity of blackbody radiation per unit wavelength, Eq. (37.4), and thus obtain Wien’s Law, Eq. (37.5). (Hint: Once you obtain the condition for a maximum, consider the dimensionless variable x  hclkT. Find a solution for x iteratively, for example, using a calculator.) **23. In Problem 37 of Chapter 24, you will find a description of the Thomson model of the hydrogen atom. (a) What is the frequency of oscillation of the electron in this atom? (b) The electron can be regarded as an oscillator. Show that if the energy of the electron is one quantum, the amplitude of oscillation exceeds the radius (5  1011 m) of the atom.

37.3 Photons and the Photoelectric Effect 24. Photons of green light have a wavelength of 550 nm. What is the energy and what is the momentum of one of these photons? 25. You are lying on a beach, tanning in the sun. Roughly how many photons strike your skin in one hour? Assume that the energy flux of sunlight is as described in Example 4. 26. For each of the following kinds of electromagnetic waves, find the energy of a photon: FM radio wave of wavelength 3.0 m, infrared light of 1.0  105 m, visible light of 5.0  107 m, ultraviolet light of 1.0  107 m, X rays of 1.0  1010 m. 27. A radio transmitter radiates 10 kW at a frequency of 8.0  105 Hz. How many photons does the transmitter radiate per second? 28. The energy flux in the starlight reaching us from the bright star Capella is 1.2  108 W/m2. If you are looking at the

star, how many photons per second enter your eyes? The diameter of your pupil is 0.70 cm. Assume that the average wavelength of the light is 500 nm. 29. A laser emits a light beam with a power of 1.0 W. The wavelength of the light is 630 nm. How many photons per second does this laser emit? 30. The energy density in a radio wave of wavelength 3.0 m is 2.0  1013 J/m3. What is the corresponding density of photons? 31. Calculate the range of energies of photons of visible light, with wavelengths from 700 nm to 400 nm. Express your answers in units of electron-volts. 32. For light-sensing applications that do not require the extreme sensitivity of a photomultiplier (see Physics in Practice: Photomultiplier), photodiodes are often used. A typical photodiode can measure an incident photon flux provided the total power incident is at least 1.0  1014 W; otherwise, the signal is lost in the electrical noise of the photodiode. Laser light of wavelength 633 nm is to be detected. What is the minimum number of photons per second required? 33. Light of wavelength 486 nm is incident on a material with a work function of 2.26 eV. What is the maximum kinetic energy of ejected electrons? When light of wavelength 434 nm is used, what stopping potential is necessary? 34. Show that if we express the energy of a photon in keV and the wavelength in nanometers, then E  1.24>l 35. According to Fig. 37.11, what is the work function of sodium? Express your answer in electron-volts. 36. The work function of potassium is 2.26 eV. What is the threshold frequency for the photoelectric effect in potassium? 37. The work functions of K, Cr, Zn, and W are 2.26, 4.37, 4.24, and 4.49 eV, respectively. Which of these metals will emit photoelectrons when illuminated with red light (l  700 nm)? Blue light (l  400 nm)? Ultraviolet light (l  280 nm)? 38. The photons emitted by a sodium atom have a wavelength of 589 nm when at rest. If this atom is moving away from you at a speed of 2.0  106 m/s, what is the energy that you measure for one of these photons? *39. By inspection of Fig. 37.11, find the slope of the line in eV/Hz. Convert these units into J. s, and verify that the slope is the same as Planck’s constant. *40. The binding energy of an electron in a hydrogen atom is 13.6 eV. Suppose that a photon of wavelength 40 nm strikes the atom and gives up all of its energy to the electron. With what kinetic energy will the electron be ejected from the atom?

37.4 The Compton Effect 41. X rays emitted by molybdenum have a wavelength of 0.072 nm. What are the energy and the momentum of one of the photons of these X rays?

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42. In a collision with an initially stationary electron, a photon suffers a wavelength increase of 0.0022 nm. What must have been the deflection angle of the photon? 43. For an experiment on the Compton effect, you want the X rays emerging at 90 from the incident direction to suffer an increase of wavelength by a factor of 2. What wavelength do you need for your incident X rays? 44. A photon of wavelength 0.030 nm collides with a free electron at rest. Calculate the wavelength if the photon emerges from this collision with a deflection of 30. 45. X rays of wavelength 0.030 nm are incident on a graphite target. Calculate the wavelength of the X rays that emerge from this target with a deflection of 60. Calculate the wavelength of the X rays that emerge from this target with a deflection of 120. 46. X-ray photons of wavelength 0.154 nm are produced by a copper source. Suppose that 1.00  1018 of these photons are absorbed by a target each second. (a) What is the total momentum p transferred to the target each second? (b) What is the total energy E of the photons absorbed by the target each second? (c) For these values, verify that the force on the target is related to the rate of energy transfer by dpdt  (1c)(dEdt). 47. Calculate the percentage change in wavelength, (¢ll)  100%, when an X-ray photon of wavelength   0.0010 nm is Compton-scattered through an angle of (a) 10, (b) 30, and (c) 180. 48. Suppose that a photon is “Compton-scattered” from a proton instead of an electron. What is the maximum wavelength shift in this case? *49. In a Compton scattering experiment, X rays of wavelength l are incident. The maximum energy transferred to an electron is Ee. Find an expression for l in terms of Ee, h, me, and c. What is the value of l for Ee  25 keV? *50. Derive the general Compton-shift formula (37.16). [Hint: Repeat the calculation of Example 8, but replace Eq. (37.19) by the relation p21  2p1p 1 cos u  p 12  p 22, obtained from inspection of the diagram representing the vector sum of momentum vectors. Note that here the p’s represent magnitudes of momentum vectors and they are positive, whereas in Example 8 the p’s are positive or negative depending on the direction of motion.] *51. What is the maximum energy that a free electron (initially stationary) can acquire in a collision with a photon of energy 4.0  103 eV? *52. A photon of initial wavelength 0.040 nm suffers two successive collisions with two electrons. The deflection in the first collision is 90 and in the second collision it is 60. What is the final wavelength of the photon?

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37.5 X Rays 53. What is the energy of the photons in X rays of a wavelength of 0.050 nm? 54. An X-ray tube is being operated with electrons of energy 25 keV. What is the cutoff wavelength of the emitted X rays? 55. We want to use an X-ray tube to generate X rays of wavelength 1.00 nm. What is the minimum potential difference we must use to accelerate the electrons in the tube? 56. The tube in a medical X-ray machine can be operated at potential differences in the range from 25 kV to 150 kV. What is the cutoff wavelength of the X rays emitted when the tube is operated at 25 kV? At 150 kV? 57. For studies of the structure of materials, a characteristic spectral line of copper, with a wavelength of 0.1542 nm, is often used. For such X rays to be emitted, what is the minimum potential difference that could be used to accelerate electrons toward a copper target? 58. Tungsten targets are often used in X-ray tubes to produce an intense, broad spectrum of Bremsstrahlung. To study the spacing of different planes of atoms in a particular crystal, X-ray wavelengths throughout the range 0.10–40 nm are desired. What is the corresponding minimum accelerating potential for electrons?

3 7 . 6 Wa v e v s . P a r t i c l e 59. A photon passes through a horizontal slit of width 5.0  106 m. What uncertainty in the vertical position will this photon have as it emerges from the slit? What uncertainty in the vertical momentum? 60. A photon traveling in the z direction passes through a square hole 2.0  106 m on a side. What is the uncertainty in each of the x and y components of the momentum? Assuming the uncertainty in the z component of the momentum is negligible, what is the uncertainty in the total momentum? *61. A particular semiconductor laser emits photons with an uncertainty in position of 1.5 cm along the direction of propagation, known as the longitudinal coherence length. What is the corresponding uncertainty in the momentum of the photons? The photons have an average wavelength of 678 nm. What is the uncertainty in this wavelength? *62. Consider a radio wave in the form of a pulse lasting 0.0010 s. The pulse then has a length of 0.0010 s  c  3.0  105 m. Since an individual photon of this radio wave can be anywhere within this pulse, the uncertainty in the position of the photon is ¢x  3.0  105 m along the direction of propagation. (a) According to Heisenberg’s relation, what is the corresponding uncertainty in the momentum of the photon? (b) What is the uncertainty in the frequency of the photon?

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Quanta of Light

REVIEW PROBLEMS *63. Consider a mass of 1.0 g attached to a spring of spring constant 3.0  102 N/m. What is the frequency of oscillation of this system? What is the magnitude of an energy quantum? What is the amplitude of oscillation if the energy of the system equals one quantum?

68. In order for a photoelectron to be emitted from a nickel surface, the wavelength of the incident photon must be no longer than 247.5 nm. What is the work function for nickel? When a photon of wavelength 200 nm is used, what stopping potential is required?

*64. We know from Chapter 15 that at small amplitudes a pendulum behaves like a simple harmonic oscillator. Suppose that a pendulum consists of a mass of 0.10 kg attached to a (massless) string of length 1.0 m.

69. A photon loses 5.0% of its initial energy when a collision with an electron at rest deflects it by 90. What is the wavelength of the incident photon? With what speed does the electron emerge from the collision?

(a) Taking into account the quantization of energy, what is the least (nonzero) amount of energy that this pendulum can have? (b) What is the amplitude of oscillation of the pendulum with this least amount of energy? *65. Interplanetary and interstellar space is filled with thermal radiation of a temperature of 2.7 K left over from the Big Bang. (a) At what wavelength is the flux of this radiation maximum? What is this part of the electromagnetic spectrum called? (b) What is the power incident on the surface of the Earth due to this radiation? *66. If you stand naked in a room, your skin and the walls of the room will exchange heat by radiation. Suppose the temperature of your skin is 33C; the total area of your skin is 1.5 m2. The temperature of the walls is 15C. Assume your skin and the walls behave like blackbodies. (a) What is the rate at which your skin radiates heat? (b) What is the rate at which your skin absorbs heat? What is your net rate of loss of heat? 67. An incandescent bulb radiates 40 W of thermal radiation from a filament of temperature 3200 K. Estimate the number of photons radiated per second; assume that the photons have an average wavelength equal to the lmax given by Wien’s Law.

70. The energy density of starlight in intergalactic space is 1.0  1015 J/m3. What is the corresponding density of photons? Assume the average wavelength of the photons is 500 nm. *71. If you want to make a very faint light beam that delivers 1 photon per square meter per second, what must be the amplitude of the electric field in this light beam? The wavelength of the light is 500 nm. *72. A photon has an energy of 5.0 eV in the reference frame of the laboratory. What is the energy of this photon in the reference frame of a proton moving through the laboratory at a speed of 12 c in the same direction as the photon? (Hint: Use the Doppler-shift formula given in Section 36.3.) *73. In a collision with a free electron, a photon of energy 2.0  103 eV is deflected by 90. What energy does the electron acquire in this collision? *74. A photon of energy 1.6  108 eV collides elastically with a proton initially at rest. The photon is deflected by 45. What is its new energy? *75. Figure 37.19 shows two of the discrete spectral lines of molybdenum. According to this figure, what are the wavelengths of these spectral lines? Would these spectral lines appear if instead of using electrons of 35 keV we were to use electrons of 19 keV in the X-ray tube?

Answers to Checkups Checkup 37.1 1. If the stove were polished instead of black, it would not be as

good an emitter of thermal radiation, since a black surface radiates heat most efficiently, and a shiny surface does not. A good absorber is a good emitter, and a poor absorber is a poor emitter. 2. The roof color provides direct evidence of the light absorbed or

emitted: a roof of dark-colored shingles absorbs more sunlight, and a roof of light-colored shingles reflects more sunlight. 3. The shiny surface layer of the suit reflects sunlight, and thus

prevents overheating.

4. (D) Less intense at all wavelengths. As discussed in Section

37.1 and shown in Fig. 37.3, a blackbody with a peak at a longer wavelength also produces less radiation at all wavelengths (and also has a lower temperature).

Checkup 37.2 1. For the first peak, lmaxT  1450 K  2000 nm  2.9 

103 mK, in agreement with Wien’s Law; likewise for the other two peaks.

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Answers to Checkups

2. Wien’s Law asserts that the wavelength at maximum intensity

varies inversely with the absolute temperature, so doubling the temperature changes the wavelength of the maximum of the spectrum by a factor of 12. The Stefan–Boltzmann Law indicates that the total energy radiated varies as the fourth power of temperature, so doubling the temperature increases the total energy radiated from the surface by a factor of 16.

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4. Yes, visible photons scatter in the same way as X rays and

experience a Compton shift. The effect is extremely difficult to observe, since the change is so small, only a few picometers out of hundreds of nanometers. 5. (D) Greater than 0.005 nm. A wavelength shift greater than

0.005 nm is impossible; as in Example 7, the maximum possible wavelength shift is 2hmec  0.004 85 nm.

3. No to both—only the total energy of the oscillator is quan-

tized, that is, the sum of kinetic and potential energies is quantized. 4. Yes, the energies are quantized; but for frequencies around 1

Hz, the energy quanta are about E  hf  6.63  1034 J, and thus individual increments in the motion are much too small to be noticed.

5. For such a large value of Planck’s constant, the energy quanta

would be nearly one-tenth of one joule, and thus the amplitude of oscillation of a pendulum would jump by a noticeable amount when one quantum of energy is added or removed. 6. (D) Blue; red. By Wien’s Law, the hottest star emits more

strongly at shorter wavelengths (the blue) and the coolest at longer wavelengths (the red). See Fig. 33.23 for the relation between the colors of the spectrum and wavelength.

Checkup 37.5 1. X-ray wavelengths are much shorter than those of visible

light; diffraction effects increase with the ratio of the wavelength to the slit width, and so are much smaller for X rays. 2. Since lcutoff  hc  K [Eq. (37.23)], halving the kinetic energy

will double the cutoff wavelength to 2  0.036 nm (compare Example 9). The characteristic spikes will remain at the same wavelengths but disappear for molybdenum since lcutoff  0.072 nm is too large; their locations are determined by the target material.

3. A bubble would allow more X rays to pass than the surround-

ing tissue, and so would appear dark in a negative. 4. (D) f eV0  h. The maximum energy of the photon is the

kinetic energy of the electron, E  hf  K  eV0.

Checkup 37.3 1. Violet light has the shortest wavelength, and thus the highest

frequency and the most energetic photons. See Fig. 33.23 for the relation between the colors of the spectrum and wavelength. 2. For a larger work function, the threshold frequency is larger,

that is, the straight line intercepts the horizontal frequency axis at a larger value of f. 3. For repulsive voltages more negative than Vstop, electrons do

not have enough kinetic energy to reach the collector. The current levels off for sufficiently high positive voltages because then all ejected electrons have enough energy to reach the collector and, for a given intensity, no more are available. The current decreases for lower intensity since the number of photons, and thus the rate of ejection of electrons, is proportional to intensity. 4. (C) The same slope. By Eq. (37.11), the slope of all such plots

is Planck’s constant h.

Checkup 37.4 1. Violet light has the shortest wavelength and, since p  h>l,

the largest momentum. See Fig. 33.23 for the relation between the colors of the spectrum and wavelength. 2. Yes, when a photon bounces off the wall, its momentum

reverses, and the wall experiences an impulsive force. 3. At any given scattering angle, the wavelength shift ¢l is the

same for all incident wavelengths. Since the wavelength shift is fixed, the percentage change is largest for the smaller wavelength, 0.2 nm.

Checkup 37.6 1. We would find a narrow line of the same width as the narrow

slit; classically, the slit would merely create an ordinary shadow. 2. Since the probability that a photon arrives at a given point is

proportional to the intensity of the wave at that point, and since the intensity is proportional to the square of the electric field amplitude, the point where the amplitude is decreased by a factor of 2 has a photon arrival probability smaller by a factor of 4. 3. No; the uncertainty principle dictates that there must be

uncertainty in the momentum p. The electron is thus also not free from uncertainty in the velocity v  pm, or in the kinetic energy K  p2 2m. 4. Yes in both cases. The uncertainty relation applies to the same

component of the position and momentum. Additionally, Eq. (37.29) implies that if the uncertainty in, say, pz is zero, then the uncertainty in z must be infinite. 5. (A) A double-slit interference pattern. Like the case of the

single slit, the distribution of photons matches the intensity distribution calculated from the wave theory of light; that is, the probability that a photon arrives at a given point is proportional to the intensity of a classical wave at that point.

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38 H

Spectral Lines, Bohr’s Theor y, and Quantum Mechanics

410.3 nm

434.2 nm

486.3 nm

656.5 nm

H

H

H

H

350 nm

660 nm

CONCEPTS IN CONTEXT 38.1 Spectral Lines 38.2 Spectral Series of Hydrogen 38.3 The Nuclear Atom 38.4 Bohr’s Theory 38.5 Quantum Mechanics; the Schrödinger Equation

When a sample of a chemical element is heated, often by means of an electric discharge, it emits characteristic electromagnetic radiaton of discrete wavelengths. The visible light in such a discrete-line spectrum for hydrogen gas is shown in the upper frame of the figure above; the lower frame includes additional lines in the ultraviolet spectrum of hydrogen. In our study of the spectra of hydrogen, we will be able to ask:

? The spacings between these lines decrease systematically from right to left, from long wavelengths to short wavelengths. This systematic pattern suggests that the wavelengths form a series, described by a simple mathematical formula. What is this formula? (Section 38.2, page 1291)

? Are there other series of spectral lines in the spectrum of hydrogen? What are they, and what is the shortest wavelength in any of these series? (Section 38.2, and Example 1, page 1292) 1286

Concepts in Context

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Spectral Lines

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? How can we calculate the wavelengths of the series from the quantum processes within the hydrogen atom? (Example 2, page 1300)

? What series of wavelengths will hydrogen absorb, when illuminated with light containing a wide range of wavelengths? (Example 3, page 1301)

T

he photograph shown in Fig. 38.1 gives convincing visual evidence that solids, liquids, and gases are made of atoms, small grains of matter with a diameter of about 10 10 m. This photograph was prepared with a powerful electron microscope of a special design. Unfortunately, not even this microscope is sufficiently powerful to reveal the inside of the atom. For the exploration of the internal structure of the atom, we still have to rely on the technique developed by Ernest Rutherford and his associates around 1910: bombard the atom with a beam of particles and use this beam as a probe to “feel” the interior of the atom. By 1910, most physicists had come to believe that atoms are made of some combination of positive and negative electric charges, and that the attractions and repulsions between these electric charges are the basis for all the chemical and physical phenomena observed in solids, liquids, and gases. Since electrons were known to be present in all of these forms of matter, it seemed reasonable to suppose that each atom consists of a combination of electrons and positive charge. The vibrational motions of the electrons within the atom would then result in the radiation of electromagnetic waves; this was supposed to account for the emission of light by the atom. However, both the arrangement of the electric charges within the atom and the mechanism that accounts for the characteristic colors of the emitted light remained mysteries until Rutherford’s discovery of the nucleus and Niels Bohr’s discovery of the quantization of atomic states. In this chapter we will look at these two momentous discoveries. The exploration of the internal structure of the atom led to the inescapable conclusion that in the atomic realm Newton’s laws of motion are not valid. Electrons and other subatomic particles obey new equations of motion that are drastically different from the old equations of motion obeyed by planets, billiard balls, or tennis balls. The new equations of motion incorporate the quantization of energy and the wave– particle behaviors we discussed in the preceding chapter, and they extend quantization to the realm of the atom. The new theory of motion that rules the realm of the atom is called quantum mechanics. The discovery of an entirely new set of laws of motion was the greatest scientific revolution of the twentieth century.

Scanning tunneling microscopy image resolves individual atoms.

Surface of this crystal has a defect, a missing atom, known as a vacancy.

FIGURE 38.1 A scanning tunneling microsope image of the surface of a semiconductor, gallium arsenide (GaAs).

quantum mechanics

38.1 SPECTRAL LINES The earliest attempts at a theory of atomic structure ended in failure—these early theories were not able to explain the characteristic colors of the light emitted by atoms. These colors show up very distinctly when a small sample of gas is made to emit light by the application of heat or of an electric current. For instance, if we put a few grains of ordinary salt into a flame, the sodium vapor released by the salt will glow with a characteristic yellow color. If we put neon gas into an evacuated glass tube and connect the ends of the tube to a high-voltage generator (Fig. 38.2), the gas will glow with the familiar orange red color of neon signs (Fig. 38.3). The light emitted by an atom can be precisely analyzed with a prism (Fig. 38.4); this breaks the light up into its component colors. In the arrangement shown in Fig. 38.4, each discrete color generates a bright line, called a spectral line. Each kind of atom has its own discrete spectral lines. The color print on page 1289 shows the spectral lines of hydrogen, helium, lithium, mercury, and sodium; the numbers above

Online Concept Tutorial

43

spectral line

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spectrum

When terminals of a tube of low-pressure gas are connected to a high-voltage generator, …

4000 V

…an electric current flows through the gas and makes it glow.

FIGURE 38.2 An electric discharge tube.

FIGURE 38.3 A neon sign.

Spectral Lines, Bohr's Theory, and Quantum Mechanics

the visible spectral lines give the wavelengths in nanometers. Hydrogen has four spectral lines in the visible region (already mentioned in Chapter 34; see Fig. 34.29) and many ultraviolet and infrared lines not visible to the human eye. From the color print we see that the set of spectral lines, or spectrum, belonging to hydrogen is unmistakably different from the spectra of helium, mercury, lithium, and sodium—the spectrum of an atom can serve as a fingerprint for its identification. Spectroscopy provides us with a useful alternative to the traditional “wet” analysis familiar to students of chemistry. By heating a small sample of an unknown substance—preferably by means of an electric discharge—and inspecting the spectral lines, we can identify the kinds of atoms in the sample. It so happens that an atom capable of emitting light of a given wavelength is also capable of absorbing light of that wavelength. In spectroscopy laboratories, scientists often take advantage of this to perform the quantitative analysis of samples of atoms with absorption lines rather than emission lines. When we illuminate a sample of atoms with white light (a mixture containing all colors or wavelengths), the atoms will absorb light of their characteristic wavelength, and upon analyzing the remaining light with a prism, we find dark absorption lines in the continuous background generated by the white light. The last picture in the color print on page 1289 shows such an absorption spectrum for sodium vapor. The dark lines of this absorption spectrum coincide with the bright lines in the emission spectrum (see the next-to-last picture on page 1289). One advantage of spectroscopy over “wet” chemistry is that the analysis can be performed even on minuscule amounts of material. What is more, atoms can be identified at a distance. For example, we can identify the atoms on the surface of the Sun by careful analysis of the distribution of colors in sunlight— we do not need to pluck a sample of atoms from the Sun. The power of this technique is best illustrated by the story of the discovery of helium (the “Sun element”—helios is Greek for the Sun). In 1868, this gas was yet unknown to chemists when astronomers discovered it on the Sun by means of its spectral lines; 30 years later chemists finally found traces of helium in minerals on the Earth. By spectroscopic techniques, astronomers can identify atoms in remote stars, clouds of interstellar gas, galaxies, and quasars. For example, Fig. 38.5 shows the spectrum of the star Caph in the constellation Cassiopeia; the spectral absorption lines indicate the presence of hydrogen, calcium, iron, manganese, chromium, etc.; these lines are formed in the layers of gas on the surface of the star. Using spectroscopy, astronomers can perform a “chemical analysis” of the material on this star, even though it is many light-years away. Note that apart from these discrete spectral lines the bulk of the starlight is white light, a continuous and more or less uniform mixture of all colors. This is the thermal radiation emitted from the stellar surface. As we know from the preceding chapter, this

spectral lines

Prism disperses different colors of light in beam…

…and reveals a set of separate spectral lines on screen.

slit

Slit provides narrow beam of light from discharge tube. screen prism discharge tube

FIGURE 38.4 Analysis of light by means of a prism.

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38.1

400.0 nm

500.0 nm

Spectral Lines

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600.0 nm

700.0 nm

spectrum of white light G

F

E

D2 D1

C

B

spectrum of sunlight with main Fraunhofer lines 410.3 nm 434.2 nm

486.3 nm

656.5 nm

hydrogen 402.7 nm

438.9 nm 492.3 nm 447.3 nm 471.4 nm 501.7 nm

587.7 nm

668.0 nm

helium 413.3 nm

460.4 nm

lithium 404.8 nm 407.9 nm 436.0 nm

497.3 nm

491.7 nm

610.5 nm

546.2 nm

577.1 nm 579.2 nm

mercury 589.2 nm 589.8 nm

sodium

sodium, absorption spectrum (All wavelengths are measured in vacuum.)

671.0 nm

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Spectral Lines, Bohr's Theory, and Quantum Mechanics

Strong absorption lines on left are due to ionized calcium… 400

390

FIGURE 38.5 A portion of the spectrum of light from the star Caph ( Cassiopeiae), from 390 nm to 450 nm. All the spectral lines displayed here are absorption lines.

Fraunhofer lines

410

Ca Ca

H

420

430

…and other two strong lines are due to hydrogen.

440

450 nm

H

kind of light does not retain the fingerprint of the atoms that produced it. Light originating in the stellar interior cannot escape directly, but is first tossed back and forth (scattered) many times by the restless atoms of the hot stellar gas. The random motion of these atoms communicates random changes of wavelength to the light, and what finally emerges from the stellar interior is a continuous mixture of a wide range of wavelengths. Figure 38.6 shows a portion of the spectrum of the white light from our Sun. White light is a continuous and nearly uniform mixture of all colors. However, the high-resolution spectrum in Fig. 38.6 displays many dark lines in the spectrum of the Sun, caused by absorption in the layers of gas on the surface of the Sun. These dark lines are called the Fraunhofer lines.

Sun’s strong calcium and hydrogen absorption lines are similar to those from star in Fig. 38.5, … 400

390

Ca Ca

410

H

420

430

Ca

Fe

440

450 nm

H

…but there are also strong lines of iron, and many other somewhat weaker lines.

FIGURE 38.6 A portion of the spectrum of light from the Sun, from 390 nm to 450 nm.

✔

Checkup 38.1

If, instead of spectral lines, you want the prism in Fig. 38.4 to produce spectral dots (one dot for each color), how do you have to change the experimental arrangement?

QUESTION 1:

The flame of the gas burner in a kitchen stove is bluish. Is this color that of the continuum thermal radiation, or the color of one (or several) spectral lines? QUESTION 3: Sodium lamps are widely used to illuminate streets and parking lots. According to the information on the color print on page 1289, what is the wavelength of the light? QUESTION 2:

When the otherwise continuous spectrum of radiation from a small opening to a hot cavity exhibits reduced intensity at a few particular wavelengths, such “dark lines” are most likely due to (A) Shadows caused by the shape of the cavity opening. (B) Absorption of light by atoms in the cavity. (C) Destructive interference from multiple sources within the cavity. (D) Diffraction minima related to the size of the cavity opening.

QUESTION 4:

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Spectral Series of Hydrogen

1 1 1 1  a  2b l 91.176 nm 4 n

(38.1)

with n  3, 4, 5, etc. This infinite series of spectral lines is called the Balmer series. Note that if n approaches infinity, the wavelength approaches the ultimate value l  4 91.176 nm  364.70 nm. This value is called the series limit, because there are no spectral lines of shorter wavelength beyond this value. Balmer’s formula is usually written compactly as

1 1 1  R a 2  2b l 2 n

656.5 nm

Spacing of lines decreases systematically from right… 486.3 nm

43

434.2 nm

Online Concept Tutorial Careful examination of the spectral lines produced by an element reveals certain systematic regularities in the spacings of the lines. These regularities sometimes become especially striking if, instead of examining only the visible region of the spectrum, we examine both the visible region and the adjacent ultraviolet and infrared regions. Figure 38.7 again shows the spectrum of hydrogen in the visible and the near-ultraviolet regions. We can notice immediately that the spacings of the lines decrease systematically as we look at shorter and shorter wavelengths. The hydrogen lines in Fig. 38.7 are said to form a spectral series. In the spectra of other elements we find similar series; however, the spectra usually contain several overlapping series, and this makes it somewhat harder to perceive the regularities in the spacings. The systematic pattern in the spacing of the spectral lines of hydrogen shown in Fig. 38.7 suggests that the wavelengths of these lines should be described by some simple mathematical formula. Table 38.1 lists the wavelengths of many of these spectral lines; the spacing between the lines is smaller at shorter wavelengths, where very many spectral lines are crowded together. In 1855, Johann Balmer scrutinized the numbers in such a table and discovered that the wavelengths accurately fit the formula

38.2 SPECTRAL SERIES OF HYDROGEN

1291

397.1 nm 410.3 nm

38.2

H H H

H

H

…to left, and such lines form a spectral series.

FIGURE 38.7 Spectrum of hydrogen in the visible and near-ultraviolet regions. The spectral lines of this spectral series are consecutively labeled H, H, Hg, and so on. The numbers give the wavelengths of the spectral lines.

Concepts in Context

series limit

(38.2)

Balmer series

(38.3)

Rydberg constant

where R is called the Rydberg constant, R

1  1.096 78 107 m1 91.176 nm

Balmer’s formula was purely descriptive, or phenomenological; it did not explain the atomic mechanism responsible for the production of the spectral lines. Nevertheless, it proved very fruitful because it led to more general formulas describing other series

TA B L E 3 8 . 1 WAVELENGTH

656.47 nm a

a

THE BALMER SERIES IN THE HYDROGEN SPECTRUM l 486.27 nm

434.17 nm

Wavelengths are measured in vacuum.

410.29 nm

397.12 nm

389.02 nm

383.65 nm

379.90 nm

etc.

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of spectral lines. Balmer proposed that there might be other series in the hydrogen spectrum, with the 22 in the parentheses in Eq. (38.2) replaced by 12, or 32, or 42, etc. This yields the series of wavelengths

Lyman, Paschen, and Brackett series Concepts in Context

1 1 1  Ra 2  2b l 1 n

n  2, 3, 4, …

(38.4)

1 1 1  Ra 2  2b l 3 n

n  4, 5, 6, …

(38.5)

1 1 1  Ra 2  2b l 4 n

n  5, 6, 7, …

(38.6)

These series of spectral lines were actually discovered some years after Balmer proposed them; they are called, respectively, the Lyman, the Paschen, and the Brackett series. The first of these series lies in the ultraviolet region of the spectrum; the other series lie in the infrared. We can combine all the formulas for all the spectral series of hydrogen into a single general formula 1 1 1  Ra 2  2b l n2 n1

wavelengths of spectral series of hydrogen

(38.7)

where n1 and n2 are positive integers and n1 is larger than n2.

Concepts in Context

EXAMPLE 1

According to Eq. (38.7), what is the shortest wavelength of light that a hydrogen atom will emit or absorb?

SOLUTION: To find the shortest wavelength, we must chose n1 and n2 in Eq. (38.7) so as to obtain the largest possible value for the right side. This means that the positive term must be as large as possible, and the negative term as small as possible, which demands n2  1 and n1  q and gives

1 1 1  Ra 2  2b  R l 1 q

(38.8)

and l  1 R  91.176 nm Since n1  1, this spectral line belongs to the Lyman series—it is the series limit for the Lyman series.

✔

Checkup 38.2

QUESTION 1: Figure 38.7 shows that spectral lines of short wavelength in the Balmer series are tightly spaced. Is this also true for the other spectral series of hydrogen? QUESTION 2: Consider the series limits for the Lyman, Balmer, and Paschen series. Which has the longest wavelength? The shortest?

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38.3

The Nuclear Atom

1293

QUESTION 3: To find the longest wavelength in the Lyman series, what value of n1 must you select? In the Balmer series? In the Paschen series? In the Brackett series? QUESTION 4: In Example 1, we found that the shortest wavelength emitted by hydrogen, l  91.176 nm, is in the Lyman series. We previously saw that the Bolmer series limit is   364.70 nm. One of the spectral lines of hydrogen has l  121.57 nm. In which series is this line? (A) Lyman (B) Balmer (C) Paschen (D) Brackett

3 8 . 3 T H E N U C L E A R AT O M The regularity in the series of spectral lines of the atom must be due to an underlying regularity in the structure of the atom. We may think of an atom as analogous to a musical instrument, such as a flute. The atom can emit only a discrete set of spectral lines, just as the flute can emit only a discrete set of tones which make up a musical scale. The regularity in the spacing of tones in this musical scale is due to an underlying regularity in the structure of the flute—the tube of the instrument has regularly spaced toneholes that determine what kind of standing waves can build up within the tube and what kind of waves will be radiated. J. J.Thomson, the discoverer of the electron, made one of the first attempts at explaining the emission of light in terms of the structure of the atom. Having established that electrons are a ubiquitous component of matter, Thomson proposed the following picture: An atom consists of a number of electrons, say, Z electrons, embedded in a cloud of positive charge. The cloud is heavy, carrying almost all of the mass of the atom. The positive charge in the cloud is Ze, so it exactly neutralizes the negative charge Ze of the electrons. In an undisturbed atom, the electrons will sit at their equilibrium positions, where the attraction of the cloud on the electrons balances their mutual repulsion (Fig. 38.8). But if the electrons are disturbed by, say, a collision, then they will vibrate around their equilibrium positions, and this accelerated motion will cause the emission of electromagnetic radiation, that is, the emission of light.This model of the atom, called the “plumpudding” model, does yield frequencies of vibration of the same order of magnitude as the frequency of light, but it does not yield the observed spectral series; for instance, on the basis of this model, hydrogen should have only one single spectral line, in the far ultraviolet. And in 1910, experiments by Ernest Rutherford and his collaborators established conclusively that most of the mass of the atom is not spread out over a cloud—instead, the mass is concentrated in a small kernel, or nucleus, at the center of the atom. Rutherford had been studying the emission of alpha particles from radioactive substances. These alpha particles carry a positive charge 2e, and they have a mass of 6.64 1027 kg, about 4 times the mass of a proton (alpha particles are nuclei of helium atoms; that is, they are completely ionized helium atoms, with all electrons removed; see Section 40.3). Some radioactive substances, such as radioactive polonium and radioactive bismuth, spontaneously emit alpha particles with kinetic energies of several million electron-volts. These energetic alpha particles readily pass through thin foils of metal, thin sheets of glass, or other materials. Rutherford was much impressed by the penetrating power of these alpha particles, and it occurred to him that a beam of these particles can serve as a probe to “feel” the interior of the atom. When a beam of alpha particles strikes a foil of metal, the alpha particles penetrate the atoms and are deflected by collisions with the subatomic structures; the magnitude of these deflections gives a clue about the subatomic structures. For example, if the interior of the atom had the “plum-pudding”

In the (incorrect) “plum-pudding” model, an atom consists of a heavy cloud of positive charge…

…and electrons sitting at equilibrium positions.

FIGURE 38.8 The lithium atom according to the “plum-pudding” model.

“plum-pudding” model

nucleus

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SIR ERNEST RUTHERFORD (1871– 1937) British experimental physicist and director of the Cavendish Laboratory at Cambridge. Rutherford identified alpha and beta rays. He founded nuclear physics with his discoveries of the nucleus and of transmutation of elements by radioactive decay; he also produced the first artificial nuclear reaction. He was awarded the Nobel Prize in chemistry in 1908.

structure proposed by J. J. Thomson, then the alpha particles would suffer only very small deflections, since neither the electrons, with their small masses, nor the diffuse cloud of positive charge would be able to disturb the motion of a massive and energetic alpha particle. The crucial experiments were performed by H. Geiger and E. Marsden working under Rutherford’s direction. They used thin foils of gold and of silver as targets and bombarded these with a beam of alpha particles from a radioactive source. After the alpha particles passed through the foil, they were detected on a fluorescent screen, which registers the impact of each particle by a faint scintillation (Fig. 38.9). To Rutherford’s amazement, some of the alpha particles were deflected by such a large angle that they came out backward. In Rutherford’s own words, “It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.” Rutherford immediately recognized that the large deflection must be produced by a close encounter between the alpha particle and a very small but very massive kernel inside the atom. He therefore proposed the following picture: An atom consists of a small nucleus of charge Ze containing almost all of the mass of the atom; this nucleus is surrounded by a swarm of Z electrons, of charge Ze. Thus, the atom is like a solar system— the nucleus plays the role of Sun and the electrons play the role of planets. On the basis of this nuclear model of the atom, Rutherford calculated what fraction of the beam of alpha particles should be deflected through what angle. The deflections are caused by the repulsive electric force between the positive alpha particle (charge 2e) and the positive nucleus (charge Ze). If an alpha particle passes close to the nucleus, the electric force will be large, and it will be deflected by a large angle; if it passes farther from the nucleus, the electric force will be smaller, and it will be deflected by a smaller angle. Figure 38.10 shows the trajectories of several alpha particles approaching a nucleus; these trajectories are hyperbolas. The perpendicular distance between the nucleus and the original (undeflected) line of motion is called the impact parameter. In order to suffer a large deflection, the alpha particle must hit an atom with a very small impact parameter, 1013 m or less; since the alpha particles in the beam strike the foil of metal at random, only very few of them will score such a close hit, and only very few will be deflected by a large angle.

impact parameter

Alpha particles are emitted by a radioactive source… evacuated box

foil

microscope

Spectral Lines, Bohr's Theory, and Quantum Mechanics

Perpendicular distance between original line of motion and nucleus is the impact parameter b.

b

source of  particles

fluorescent screen

…and their deflection by atoms in a thin metal foil is detected upon impact with a screen.

FIGURE 38.9 Rutherford’s apparatus.

nucleus Particles with smaller impact parameters suffer larger deflections.

FIGURE 38.10 Four hyperbolic orbits of different impact parameters.

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38.4

✔

Bohr’s Theory

1295

Checkup 38.3

QUESTION 1: The hydrogen atom has one single electron. What does this atom look like according to the “plum-pudding” model? If the cloud of positive electric charge is spherically symmetric, what is the equilibrium position of the electron? QUESTION 2: If an alpha particle approaches a nucleus with an impact parameter of zero, what will happen to the alpha particle? QUESTION 3: Is the distance of closest approach between a nucleus and an alpha particle in a hyperbolic orbit (such as shown in Fig. 38.10) larger or smaller than the impact parameter? QUESTION 4: If the alpha particles in Fig. 38.10 had a negative electric charge 2e instead of their positive charge 2e, how would this affect the deflections illustrated in this figure? QUESTION 5: Rutherford’s experiments used foils of silver (atomic number 47) or gold (atomic number 79) as targets. For the same impact parameter, an alpha particle of a given energy incident on a silver target is deflected (A) Through a smaller angle than when incident on a gold target. (B) Through the same angle as when incident on a gold target. (C) Through a larger angle than when incident on a gold target.

3 8 . 4 B O H R ’ S T H E O RY Rutherford’s experiments did reveal the gross arrangement of the electrons in the atom, but not the details of their motion. Since the electrons make up the outer layers of an atom, their arrangement and motion determine the chemical bonds of the atom and the emission of light, that is, they determine all the chemical and spectroscopic properties. But when physicists tried to calculate the electron motion according to the laws of classical mechanics and electromagnetism, they immediately ran into trouble. To gain some insight into the source of this trouble, consider the case of the hydrogen atom. Suppose that the single electron of this atom is moving around the nucleus, according to the laws of classical mechanics, in a circular orbit of atomic size, that is, with a radius of about 1010 m. In such an orbit, the electron would have a centripetal acceleration due to the attractive electric force of the nucleus, which would be very large, about 1023 m/s2. Because of this acceleration, the electron would emit high-frequency electromagnetic radiation, that is, it would emit light (see Chapter 33). The energy carried away by the light would have to be supplied by the electron. Hence the emission process would have the same effect on the electron as a friction force—it would remove energy from the electron. This kind of friction would cause the electron to leave its circular orbit and gradually spiral in toward the nucleus, in just the way the residual atmospheric friction on an artificial satellite in a low-altitude orbit around the Earth causes it to spiral down toward the ground. A calculation using the laws of classical mechanics and electricity shows that the rate of emission of light by the orbiting electron in a hydrogen atom would be quite large. Correspondingly, the rate of energy loss of the electron would be large—the electron would spiral inward and collide with the nucleus within a time as short as 1010 s! Thus, our classical calculation leads us to the incongruous conclusion that hydrogen atoms, and other atoms, ought to be unstable—all the electrons ought to collapse into the nucleus almost instantaneously. Furthermore, the light that the electron emit-

Online Concept Tutorial

43

NIELS BOHR (1885–1962) Danish theoretical physicist. He worked under J. J. Thompson and Rutherford in England and then became director of the Institute of Theoretical Physics in Copenhagen, for the foundation of which he was largely responsible. After formulating the quantum theory of the atom, he played a leading role in the further development of the new quantum mechanics. He received the Nobel Prize in 1922.

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ted during the spiraling motion would have to be a wave of continually increasing amplitude and increasing frequency (in musical terminology, crescendo and glissando); this is so because when the electron came closer to the nucleus, the electric force would become larger, leading to a larger centripetal acceleration and a higher frequency of the orbital motion. Hydrogen atoms do not behave as this classical calculation predicts. Hydrogen atoms are stable, and when they do emit light, they emit discrete frequencies (spectral lines) instead of a continuum of frequencies. These irreconcilable disagreements between the observed properties of atoms and the calculated properties gave evidence of a serious breakdown of the classical mechanics of Newton and the classical theory of electromagnetism. Although these theories had proved very successful on a macroscopic scale, they were in need of some drastic modifications on an atomic scale. In 1913, Niels Bohr took a bold step toward resolving these difficulties. He made the radical proposal that, at the atomic level, the laws of classical mechanics and of classical electromagnetism must be replaced or supplemented by other laws. Bohr expressed these new laws of atomic mechanics in the form of several postulates: Bohr’s quantum postulates

stationary state

quantization of angular momentum

Electron moving in assumed circular orbit experiences a centripetal acceleration…

v

a

r …produced by attractive Coulomb force of nucleus.

FIGURE 38.11 Electron in circular orbit around a proton.

1. The orbits and the energies of the electrons in an atom are quantized, that is, only certain discrete orbits and energies are permitted. When an electron is in one of the quantized orbits, it does not emit any electromagnetic radiation; thus, the electron is said to be in a stationary state. The electron can make a discontinuous transition, or quantum jump, from one stationary state to another. During such a transition from one stationary state to another stationary state of lower energy, the electron does emit radiation. 2. The laws of classical mechanics apply to the orbital motion of the electrons in a stationary state, but these laws do not apply during the transition from one state to another. 3. When an electron makes a transition from one stationary state to another, the excess energy E is released as a single photon of frequency f  Eh. 4. The permitted orbits are characterized by quantized values of the orbital angular momentum L.This angular momentum is always an integer multiple of h2 : Ln

h 2p

n  1, 2, 3, …

(38.9)

The number n is called the angular-momentum quantum number. Let us now see how to calculate the stationary states and the spectrum of the hydrogen atom on the basis of these postulates. For the sake of simplicity, we will assume that the electron moves in a circular orbit around the proton, which remains at rest (Fig. 38.11). The centripetal acceleration of the electron is v2 r. This centripetal acceleration is produced by the force of attraction between the electron and the proton, that is, the Coulomb force e 2 (4p0r 2). Thus, the equation of motion mea  F for the electron is 1 e2 v2 (38.10) me  r 4p0 r 2 The orbital angular momentum for a circular orbit is L  mevr [see Eq. (13.34)]. According to Bohr’s postulate, this orbital angular momentum must be h 2p multiplied by an integer n: mevr  n

h 2p

(38.11)

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38.4

Bohr’s Theory

1297

It is convenient to rewrite this as mevr  nU

(38.12)

where U (pronounced “h-bar”) is Planck’s constant h divided by 2p:

U 

6.63 1034 J s h   1.05 1034 J s 2p 2p

(38.13)

h (h-bar)

From Eq. (38.12), we obtain nU mer

v

(38.14)

and when we substitute this into Eq. (38.10), we obtain an equation for the radius of the orbit, me a

1 e2 nU 2 1 b  mer r 4p0 r 2

(38.15)

Solving this for the radius, we find r

4p0n2 U 2

(38.16)

mee 2

From Eq. (38.16) we see that the radii of the permitted orbits are proportional to n2. The smallest value of n is n  1, and this leads to the radius of the smallest permitted orbit, which is called the Bohr radius and designated by a0: a0 

4p0 U 2 mee 2

 0.529 1010 m  0.0529 nm

(38.17)

The radii of the other permitted orbits are multiples of the Bohr radius: r

4p0n2 U 2 mee 2

 n2a0

(38.18)

For n  2, 3, 4, …, this gives r  4a0, 9a0, 16a0, …, respectively. Figure 38.12 shows the permitted circular orbits, drawn to scale.

Orbits permitted by Bohr theory are circular, …

n1

n2

n3 n4

…with radii r  n2a0 that are square-integer multiples of Bohr radius a0  0.0529 nm.

FIGURE 38.12 The possible Bohr orbits of an electron in the hydrogen atom.

0

2 4 6 8 10 1010 m

Bohr radius

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The energy of the electron in one of these orbits is a sum of kinetic and potential energies. The kinetic energy is K  12mev2, and the potential energy is the electrostatic potential energy U  (e) (1 4p0)e r  (1 4p0)e 2 r [see Eq. (25.11)]. Hence the total energy E  K  U is E

1 1 e2 mev2  2 4p0 r

(38.19)

Next, we substitute v2 from Eq. (38.14) and then r from Eq. (38.16), and we obtain E

 

1 n2 U 2 1 e2 me 2 2  2 me r 4p0 r

(38.20)

2 mee 2 mee 2 1 n2 U 2 e2 me 2 a b  a b 2 4p0 4p0n2 U 2 me 4p0n2 U 2

mee4

1

2(4p0)2 U 2 n2



mee4

1

(38.21)

(4p0)2 U 2 n2

or E

mee4

1

2(4p0)2 U 2 n2

(38.22)

This energy depends on the quantum number n, and as a reminder of this dependence, we label this energy with a subscript n: En  

mee4

1 2 2

2(4p0) U n2

(38.23)

According to this equation, the energies of all the stationary states are negative. The stationary state with the least energy, or the most negative energy, is the state with n  1, for which E1   

mee4 2(4p0)2 U 2 9.11 1031 kg (1.602 1019 C)4 2 (4p 8.85 1012 F/m)2 (1.055 1034 J s)2

 2.18 1018 J  13.6 eV

(38.24)

And, by Eq. (28.23), the energies of the other stationary states are fractions of this energy:

energies of stationary states

energy-level diagram

En  

13.6 eV n2

(38.25)

Thus, E1  13.6 eV, E2  (13.6 4) eV, E3  (13.6 9) eV, and so on. Figure 38.13 displays these quantized energies in an energy-level diagram. Each horizontal line represents one of the energies given by Eq. (38.25). According to Bohr’s

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38.4

E eV 0 –1 –2

Bohr’s Theory

1299

In an energy-level diagram, each horizontal line represents one of the allowed energies… n

6 5 4 3

Pfund series Brackett series Paschen series

–3 2 –4

Balmer series

–5 …and a colored arrow shows an electron’s possible quantum jump.

–6 –7 –8 –9 –10 –11

Lowest energy state is called the ground state.

–12 –13 –13.6 –14

1 Lyman series

FIGURE 38.13 Energy-level diagram for the hydrogen atom. Jumps (blue) that end in the ground state (n = 1) give rise to the Lyman series; jumps (red) that end in the first excited state (n  2) give rise to the Balmer series; etc.

assumptions, the electron emits a photon when it makes a transition, or a quantum jump, from one stationary state to a lower stationary state. Such quantum jumps have been indicated by arrows in Fig. 38.13. The stationary state of lowest energy (n  1) is called the ground state, the next state (n  2) is called the first excited state, the next state (n  3) is called the second excited state, and so on. Ordinarily, the electron of the hydrogen atom is in the ground state, that is, the circular orbit with radius a0 and energy 13.6 eV. This is the configuration of least energy, and it is the configuration into which the atom tends to settle when it is left undisturbed. As long as the atom remains in the ground state, it does not emit light. To bring about the emission of light, we must first kick the electron into one of the excited states, that is, a circular orbit of larger radius and higher energy. We can do this by heating a sample of atoms or by passing an electric current through the sample. Collisions between the atoms will then disturb the electronic motions and occasionally kick an electron into a larger orbit. From there, the electron will spontaneously jump into a smaller orbit, emitting a photon. Note that the quantum jumps indicated by colored arrows in Fig. 38.13 form several series: one series consists of all those jumps (indicated by blue arrows) that end in the ground state, another series consists of all those jumps (indicated by red arrows) that end in the first excited state, etc. These series of jumps give rise to the series of spectral lines: the Lyman series, the Balmer series, etc. From our formula for the energies of the states of the hydrogen atom we can calculate the frequency of the light emitted in a quantum jump from some initial state to a final state, as in the following example.

ground state and excited states

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Concepts in Context

Spectral Lines, Bohr's Theory, and Quantum Mechanics

Calculate the frequency and the wavelength of the light emitted by the electron in a quantum jump from the second excited state (n  3) to the first excited state (n  2). EXAMPLE 2

SOLUTION: The energy of the initial state is E3  (13.6 9) eV, and the energy

of the final state is E2  (13.6 4) eV. Hence the electron releases an energy ¢E  E3  E2  

13.6 13.6 eV  a eV b  1.89 eV 9 4

According to Bohr’s postulates, this energy is radiated as a single photon of frequency f

1.89 eV 1.60 1019 J/eV ¢E 1.89 eV    4.56 1014 Hz h h 6.63 1034 J s

and wavelength l

c 3.00 108 m/s   6.58 107 m  658 nm 14 f 4.56 10 Hz

This wavelength agrees with wavelength of the first spectral line of the Balmer series (see Table 38.1), except for a round-off error.

More generally, we can calculate the frequency of the light emitted in a quantum jump from some initial state of quantum number ni to some final state of quantum number nf . The initial energy of the electrons is Ei and the final energy is Ef . Thus, the electron releases an energy Ei  Ef , Ei  Ef 

mee4 2 2

2(4p0) U

a

1 n2f



1 n2i

b

(38.26)

This energy is radiated as a single photon of frequency f

frequency of emitted photon

Ei  Ef ¢E  h h

(38.27)

From the frequency of the light, we can calculate the wavelength. Since the wavelength is inversely proportional to the frequency, l  c f, Eq. (38.27) implies the following expression for the wavelength of the emitted light:

wavelength of emitted photon

Ei  Ef Ei  Ef 1   l hc 2pUc

(38.28)

mee4 1 1 1  a 2  2b 2 3 l 4p(4p0) U c nf ni

(38.29)

or, with Eq. (38.26),

Comparison of Eqs. (38.7) and (38.29) yields the following theoretical formula for the Rydberg constant:

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38.4

R

Bohr’s Theory

mee4

1301

(38.30)

4p(4p0)2 U 3c

Upon insertion of the accurate values of the fundamental constants given in Appendix 6, we obtain R

9.109 53 1031 kg (1.602 189 1019 C)4 4p(4p 8.854 178 1012 F/m)2 (1.054 589 1034 J s)3 2.997 925 108 m/s

 1.097 37 107 m1

(38.31)

This theoretical value of R agrees quite well with the experimental value quoted in Eq. (38.3).1

Suppose that the atoms in a sample of hydrogen gas are initially in the ground state. If we illuminate these atoms with light (from some kind of lamp), what frequencies will the atoms absorb? EXAMPLE 3

SOLUTION: Absorption of light is the reverse of emission. When an electron in

an atom absorbs a photon (supplied by the lamp), it jumps from the initial state to a state of higher energy. The energy of the photon must match the energy difference between the states. Thus, the frequencies of the photons that the electrons can absorb when jumping upward from the ground state are exactly those frequencies that they emit when jumping downward into the ground state, that is, the frequencies of the Lyman series (see Fig. 38.13).

What is the ionization energy of the hydrogen atom, that is, what energy must we supply to remove the electron from the atom when it is initially in the ground state?

EXAMPLE 4

SOLUTION: To remove the electron from the atom, we must lift it into an orbit

of infinite radius, that is, an orbit of quantum number n  q. The energy required to accomplish this transition from n  1 to n  q is E1  Eq  13.6 eV a

1 1

2



1 q2

b  13.6 eV

This is the ionization energy.

With his theory Bohr attained the goal of explaining the regularities in the spectrum of hydrogen in terms of the regularities of the structure of the atom. By showing that this structure is based on a simple numerical sequence, he fulfilled the ancient dream of Pythagoras of a universe based on simple numerical ratios, a dream that arose from an analogy with musical instruments. Bohr’s theory tells us how the atom plays its tune.

1

The small disagreement between the theoretical value of R given in Eq. (38.31) and the experimental value given in Eq. (38.3) is due to the motion of the nucleus of the hydrogen atom, which we have neglected in our calculation. A careful calculation that takes into account the motion of electron and proton about their common center of mass requires the substitution of the so-called reduced mass m  memp (me  mp) for the electron mass, and eliminates the disagreement. See also Problem 31.

Concepts in Context

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✔

Spectral Lines, Bohr's Theory, and Quantum Mechanics

Checkup 38.4

Why is Bohr’s postulate of stationary states in direct contradiction with classical mechanics and electromagnetism? QUESTION 2: If an electron in a hydrogen atom makes a transition from some state to a lower state, does its kinetic energy increase or decrease? Its potential energy? Its orbital angular momentum? QUESTION 3: Suppose that an electron is initially in the n  5 state. It then makes three quantum jumps in succession, such that the first jump produces a photon in the Paschen series, the second a photon in the Balmer series, and the third a photon in the Lyman series. In the energy-level diagram of Fig. 38.13, find the arrows that indicate these transitions. QUESTION 4: In the energy-level diagram of Fig. 38.13, what transition would produce the series limit of the Balmer series? (A) n  2 to n  1 (B) n  3 to n  2 (C) n  q to n  1 (D) n  q to n  2 (E) n  11 to n  2 QUESTION 1:

38.5 QUANTUM MECHANICS; .. T H E S C H R O D I N G E R E Q U AT I O N Bohr’s theory is a hybrid. It relies on some basic classical features (orbits) and grafts onto these some quantum features (quantum jumps, quanta of light). In the 1920s the cooperative efforts of several brilliant physicists—L. de Broglie, E. Schrödinger, W. Heisenberg, M. Born, P. Jordan, P. A. M. Dirac—established that the remaining classical features had to be eradicated from the theory of the atom. Bohr’s semiclassical theory had to be replaced by a new quantum mechanics with an entirely different equation of motion. The basis of the new quantum mechanics was laid by the discovery that electrons— as well as protons, neutrons, and all the other “particles” found in nature—have not only particle properties but also wave properties. When a beam of electrons is made to pass through an extremely narrow slit, the electrons exhibit diffraction. This means that electrons are neither classical particles nor classical waves. Electrons, just like photons, are a new kind of object with a subtle combination of particle and wave properties. Electrons are wavicles. As in the case of photons [see Eq. (37.15)], the wavelength associated with an electron or some other wavicle is inversely proportional to its momentum: l

de Broglie wavelength

h p

(38.32)

This formula was postulated by de Broglie, and it is called the de Broglie wavelength. This wavelength is quite small, even for electrons of the lowest energies attainable in experiments with beams of electrons.

What is the de Broglie wavelength of an electron of kinetic energy 1.0 eV, which is about the lowest energy that can be attained in experiments with beams of electrons?

EXAMPLE 5

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Quantum Mechanics; the Schrödinger Equation

1303

SOLUTION: In joules, a kinetic energy of 1.0 eV is 1.6 1019 J. This kinetic

energy is related to the momentum as follows: E

p2 2me

Hence p  22me E  22 9.11 1031 kg 1.6 1019 J  5.4 1025 kg m/s According to the de Broglie equation, the wavelength of the electron is then l

6.63 1034 J s h  p 5.4 1025 kg m/s

 1.2 109 m  1.2 nm

The wave properties of electrons were confirmed experimentally by C. J. Davisson and L. Germer, who observed interference effects with beams of electrons scattered by a crystal. In these experiments the crystal acts as a grating for electron waves, in much the same way as it acts as a grating for X rays in X-ray interference experiments (see the discussion of X-ray interference and the production of Laue spots in Section 37.5). Davisson and Germer found that constructive interference of the electron waves scattered by the rows of atoms in the crystal produced strong interference maxima in selected directions, and they were able to confirm that these directions agreed with calculations based on the de Broglie formula (38.32). In the new quantum mechanics, or wave mechanics, the motion of an electron is described by a wave equation, the Schrödinger equation. This Schrödinger equation plays the same role for electrons as the wave equation derived from the Maxwell equations plays for photons [see Eq. (33.33)]. As in the case of photons, the behavior of an electron is governed by a probabilistic law. The electron is represented by an electron wavefunction c (Greek letter psi) calculated from the Schrödinger equation, and the intensity of this electron wave at some point determines the probability that there is an electron particle at that point [compare Eq. (37.24)]: [probability for an electron at a point] r c2

(38.33)

Furthermore, as a consequence of their wave properties, electrons obey the Heisenberg uncertainty relations for position and momentum [see Eq. (37.30)], ¢y ¢py 

U 2

(38.34)

These quantum uncertainties are of crucial importance for the behavior of an electron inside an atom. For such an electron, the uncertainty in the position is very large— about as large as the size of the atom. This implies that the electron follows no definite orbit. It is therefore not surprising that the Bohr theory should have failed in all attempts at calculating the electron motions in the helium atom and in other atoms with several electrons; what is surprising is that this theory should have succeeded as well as it did in the case of the hydrogen atom.

LOUIS VICTOR, PRINCE DE BROGLIE (DE BROY) (1885–1962) French theoretical physicist. He discovered the de Broglie wavelength by reasoning that if waves have particle properties, then maybe particles have wave properties. For his discovery of the wave properties of matter he was awarded the Nobel Prize in 1929, after the existence of these wave properties was confirmed experimentally.

Schrödinger equation

probability interpretation of wavefunction

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Spectral Lines, Bohr's Theory, and Quantum Mechanics

EXAMPLE 6

Consider an electron in the ground state of hydrogen. Show that a well-defined orbit is inconsistent with the Heisenberg uncertainty relations.

SOLUTION: If the electron is to follow a well-defined orbit, the uncertainty in its

momentum (in any direction) must be much smaller than the magnitude of the momentum. According to Eq. (38.14), the speed of the electron in the smallest circular orbit (n  1) is v  U mer, and the magnitude of the momentum is p  mev  U r. For a well-defined orbit, the uncertainty of the momentum must be much smaller than this magnitude of the momentum: ¢py V

ERWIN SCHRÖDINGER (1887–1961) Austrian theoretical physicist. Another of the founders of the new quantum mechanics, he received the Nobel Prize in 1933 for the discovery of his wave equation.

U r

(38.35)

Furthermore, for a well-defined orbit, we must require that the uncertainty in the position must be much smaller than the size of the orbit: ¢y V r

(38.36)

In view of the inequalities (38.35) and (38.36), the product y py must be smaller than U : ¢y ¢py V U

(for a well-defined orbit)

(38.37)

This is inconsistent with the Heisenberg uncertainty relation (38.34).

In wave mechanics, the quantization of the energy in the hydrogen atom and other atoms is an automatic consequence of the wave properties of the electron. The attractive electric force of the nucleus confines the electron wave to some region near the nucleus and causes the wave to reflect back and forth across the region, forming a standing wave. The different stationary states of the atom correspond to different standing-wave modes. As in the case of standing waves on a string, the standing electron waves in the atom have a discrete set of eigenfrequencies. For electrons and other particles, just as for photons, the energy is related to the frequency by energy of stationary state in terms of frequency

E  hf

(38.38)

Thus a discrete set of frequencies for electron standing waves corresponds to a discrete set of energies. The ground state of the atom is analogous to the fundamental mode of the string, the first excited state is analogous to the first overtone, and so on. However, whereas the determination of the eigenfrequencies of standing waves on a string is a quite trivial mathematical exercise, the determination of the eigenfrequencies of the electron waves in an atom is a formidable mathematical problem, which requires an investigation of the solutions of the Schrödinger wave equation. Although we will not deal here with the mathematical complexities of the Schrödinger wave equation in three dimensions, we can gain some insight into how electron waves determine the discrete energies in the hydrogen atom by means of the following simple calculation. Let us assume that the electron travels around the nucleus along an orbit of radius r, but instead of thinking of the electron as a particle, as in the Bohr theory, let us think of it as a wave. Figure 38.14 shows a “snapshot” of such an electron wave at one instant of time. If the wave is to have a well-defined amplitude at all points, it must repeat whenever we go once around the circumference—if it did

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Quantum Mechanics; the Schrödinger Equation

not, then the wave amplitude would have two different values at a single point, which makes no sense. Hence, the wave is subject to the condition that some integer number of wavelengths must fit around the circumference: 2pr  nl

1305

If we think of an electron orbiting the nucleus as a wave…

(38.39)

Although this equation looks like the condition for a standing wave on a string of length 2pr, we are here dealing with a traveling wave, for which the entire wave pattern in Fig. 38.14 rotates rigidly around the center. With the de Broglie relation l  h p, Eq. (38.39) becomes 2pr 

nh p

or rp 

nh 2p

(38.40)

Since rp is the orbital angular momentum, this equation coincides with Bohr’s quantization condition for the angular momentum, Eq. (38.9). Thus, the wave picture of the electron implies the quantization of the angular momentum and, therefore, the quantization of the energy. But we must not take this calculation too seriously—its legitimacy is questionable, since it relies in part on the wave picture and in part on the particle picture. Furthermore, analysis of the Schrödinger equation shows that it is not enough to consider the behavior of the wave around the circumference; we must also consider the behavior of the wave outward along each radius. Thus, this simple calculation provides no more than a crude qualitative sketch of the role of the wave properties of electrons in the atom. There are some applications where a simple one-dimensional calculation gives accurate and meaningful results, such as electron reflection from a barrier, electron transmission through a barrier (known as “tunneling”; see Physics in Practice: Ultramicroscopes), and confinement of electrons in one direction in some modern layered semiconductor devices. In one dimension, each allowed stationary state is described by an electron wavefunction c  c(x) that is a solution to the time-independent Schrödinger equation:



U2 d2 c(x)  U(x) c(x)  E c(x) 2me dx2

(38.41)

When the potential energy U(x) in this equation is some known function of position, then solutions may be found for the allowed values of the total energy E and the corresponding allowed position-dependent wavefunctions c  c(x). Since the terms in Eq. (38.41) with U and E involve the potential and total energies, respectively, it is not surprising that the first term represents the kinetic energy K  p2 2me. Thus the Schrödinger equation incorporates a wave-mechanical version of the the energy relation K  U  E of classical mechanics.

Consider an electron in one dimension. The electron is confined to a region of length L, say, to the interval 0  x  L (see Fig. 38.15). Inside this region, the potential energy U (x)  0, and outside U (x)  q. This is sometimes referred to as a “particle in a box” or the “infinite potential EXAMPLE 7

…then some integer number of wavelengths must fit around the circumference.

FIGURE 38.14 In this example, six wavelengths of the electron wave fit around the circumference of the orbit.

time-independent Schrödinger equation

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CHAPTER 38

well.” What electron wavefunctions are solutions to the time-independent onedimensional Schrödinger equation? What are the corresponding allowed energies?

Since potential energy U(x) is infinite for x  0 and x  L, …

SOLUTION: Since the potential energy U(x) is infinite outside the box, the only possible solution for the Schrödinger equation (38.41) outside the box is c(x)  0. We thus know that c(x)  0 both at x  0 and at x  L. Inside the box, we have U(x)  0, and the Schrödinger equation becomes

U

Spectral Lines, Bohr's Theory, and Quantum Mechanics

 U0

U 2 d2 c(x)  E c(x) 2me dx2

We can rearrange this in the more familar form 0

x

L

position

d2

…electron is confined to this region, where U(x)  0.

FIGURE 38.15 The one-dimensional infinite potential well. The electron is confined to the region 0  x  L.

2

dx

c(x)  

2meE U2

c(x)

(38.42)

We encountered similar equations, where the second derivative of a function is proportional to the negative of the function, in our study of oscillations in Chapter 15. By analogy with Eq. (15.15), we know solutions to Eq. (38.42) are of the form c(x)  A sin (kx  d) where the wave vector k  2p l of the sinusoidal function is given by

3(x)

k2  x L

0

2(x)

Wave functions resemble standing waves on a string: an integer number of half wavelengths must fit into length L.

2me E

(38.43)

U2

Since we know that c(0)  0, we require c(0)  A sin (0  d)  A sin d  0

(38.44)

The amplitude A must be nonzero, since the probability of finding the electron cannot be zero everywhere. Equation (38.44) thus determines that d  0. Hence our wavefunctions are of the form c(x)  A sin (kx)

(38.45)

x

Similarly, since we know that c(L)  0, we require

L

0

c(L)  A sin (kL)  0 which has solutions when the argument of the sine function is an integer multiple of p. Hence the solutions correspond to discrete values of the wave vector k, which we label with a quantum number n:

1(x)

x 0

L

kn  n

p L

for

n  1, 2, 3, …

(38.46)

Substituting these values into Eq. (38.45), we see that the corresponding allowed wavefunctions are FIGURE 38.16 Wavefunction (x) for the first few states of the infinite potential well.

cn(x)  A sin a n

px b L

(38.47)

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Quantum Mechanics; the Schrödinger Equation

The wavefunctions corresponding to n  1, 2, and 3 are shown in Fig. 38.16; in all cases, an integer number of half wavelengths fits into the length L of the box. This condition on the wavelength is analogous to the condition for standing waves on a string. From Eq. (38.43) we can express the energy in terms of the wave vector k, E

U 2k2 2me

U2 p 2 an b 2me L

(38.49)

The allowed energies are thus square-integer multiples of the ground-state energy E1, En  n2E1

for

Allowed energies are squareinteger multiples of the groundstate energy, En = n2E1. E 25E1

n5

16E1

n4

9E1

n3

4E1

n2

U 2 2 = E 1 2meL2

n1

(38.48)

or, inserting the allowed values of kn, we obtain the allowed energies En 

1307

n  1, 2, 3, …

(38.50)

FIGURE 38.17 Energy-level diagram for the one-dimensional infinite potential well.

where the ground-state energy is E1 

U 2p2 2meL2

(38.51)

The energies of the first five wavefunctions are shown in the energy-level diagram of Fig. 38.17. COMMENTS: Since U (x)  0 inside the box, the energy of an electron confined

to a region of length L is all kinetic energy. Note that in the n  1 ground state, this energy is nonzero, and is known as zero-point energy; even at absolute zero temperature, the energy of the electron cannot be reduced below this value. This zero-point energy of the infinite potential well is inversely proportional to the square of the length L of the confinement region. Thus confinement costs energy. Note also that the higher kinetic energy of an excited state corresponds to a shorter wavelength. This is also evident from the de Broglie relation, since K  p2 2m  h2 2ml2.

zero-point energy

The probabilistic information contained in the wavefunction c was mentioned in Eq. (38.33): the probability of finding the electron at a given point in space is proportional to the square of the magnitude of the wavefunction at that point.2 The probability of finding an electron at a single point is infinitesimal, so the quantity c2 in Eq. (38.33) represents the probability per unit volume for finding an electron in a small volume around some point. Similarly, in one dimension the quantity c(x)2 represents a probability per unit length, that is, c(x)2 is the probability that the electron is in a region of width dx centered at the point x.Thus in practice, the probabilistic relation (38.33) means that the probability P of finding the electron in the region between two points a and b is given by b

P

 c(x) dx 2

(38.52)

a

2

We use the square of the magnitude of c to allow for cases where c is a complex function. Although in practice the wavefunction is often complex, in this text we will only encounter examples where c is a real function.

probability of finding electron in a region

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Spectral Lines, Bohr's Theory, and Quantum Mechanics

Since the sum of the probabilities of all possible locations of the electron must be 1, that is, the electron must be somewhere, we must also require that the wavefunction satisfy



normalization condition

q

c(x)2dx  1

(38.53)

q

This requirement is known as the normalization condition. In addition to satisfying the Schrödinger equation and being continuous and smooth at boundaries, a meaningful wavefunction must satisfy the normalization condition. In classical mechanics, an electron with energy E in a onedimensional box of length L simply bounces back and forth between the walls at constant speed. A classical electron thus has equal probability of being anywhere in the box; the probability per unit length for an electron to be at any point is then Pclassical  1 L. For example, classical mechanics predicts that an electron would be found in the middle half of the box with a probability of exactly 12, or 50% of the time. What are the normalized wavefunctions for an electron in the one-dimensional box of Example 7? In the quantum-mechanical ground state, what is the probability per unit length that the electron is at the center of the box? What is the probability of finding the electron in the middle half of the box? EXAMPLE 8

SOLUTION: In order to use the wavefunctions cn(x)  A sin (npx L) from Eq.

(38.47) to calculate probabilities, we must determine the overall multiplicative constant A from the normalization condition (38.53). Since c(x)  0 outside the box, we need only consider the region inside the box. For any quantum number n, inserting the solutions (38.47) into the normalization condition (38.53) gives 1



0

L

L

cn(x)2dx 

 A sin 2

2

0

an

px b dx  12 LA 2 L

(38.54)

where we have used the fact that for any integer number of half wavelengths, the average value of the square of the sine function is 12, and so the integral over the length L is 12L. Solving for A, we obtain A

2 AL

(38.55)

independent of the quantum number n. So the normalized wavefunctions are cn(x) 

2 npx sin a b AL L

(38.56)

For the ground state, we use the n  1 wavefunction. The probability per unit length that the electron is at the center of the box is obtained by evaluating c1(x)2 at x  L 2: L 2 2 pL 2 2 2 p 2 ` c1 a b `  ` sin a b `  sin2 a b  2 BL L L 2 L Thus, compared with classical mechanics, where Pclassical  1 L, quantum mechanics predicts that the electron is twice as likely to be at the center of the box.

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Quantum Mechanics; the Schrödinger Equation

3L 4



c1(x)2dx 

L4

3L 4



L4

2 L

2 px b dx sin2 a L L

x L

0

With a change of variable u  px L, this becomes 2

sin u du

p4

—

2(x) 2

3p 4



Probability of finding an electron is zero at edges and nodes…

—

2 L P Lp

—

P

3(x) 2

—

The middle half of the box extends from x  14L to x  34L, so the probability of finding the electron between these points is

1309

2 L

From our table of integrals (Appendix 4),  sin2 u du  12 u  14 sin 2u. Hence

x L

0

P

3p 4 2 1 3 1 2 1 1 1 a u  sin 2u b `  e c p  a b d  a p  b f p 2 p 4 8 4 8 4 p 4

…and has the same maximum value for any n(x).



1 1   0.818 2 p

Thus quantum mechanics indicates that the electron in the ground state is much more likely to be in the middle half of the box than classical physics would predict: approximately 82% compared with 50%. COMMENT: The probability distribution depends on the state. For the ground

state, the electron is most likely to be found at the center of the box, but for the first excited state, it is least likely to be found there (see Fig. 38.18).

The above examples of an electron in a box demonstrate that we cannot be certain about the position of the electron. Since the electron may be traveling to the right or to the left, its momentum is also uncertain. Within our small confinement region or, similarly, within the atom, electrons always behave very much like waves. But outside the atom, they will sometimes behave pretty much like classical particles. Roughly, we can say that classical mechanics will be a good approximation whenever the quantum uncertainties are small compared with the relevant magnitudes of positions and momenta. For instance, for the electrons in the beam of a TV tube, the quantum uncertainty in the momentum is negligible compared with the magnitude of the momentum. Under these conditions, classical mechanics gives an adequate description of the motion of the electrons. What we have said about the wave mechanics of electrons also applies to other particles, or wavicles, found in nature—they all have wave properties and they all have quantum uncertainties in their position and momentum. Strictly speaking, even large macroscopic bodies have wave properties. For example, an automobile is a wavicle and it has some quantum uncertainty in its position. However, it turns out that the quantum uncertainties are very small whenever the mass of the body is large compared with atomic masses—the quantum uncertainty in the position of an automobile is typically no more than about 1018 m, a number that can be ignored for all practical purposes. Hence, for automobiles and other macroscopic bodies, quantum effects are completely insignificant and classical mechanics gives an excellent description of the motion of these bodies.

—

—

1(x) 2

2 L

x 0

L

FIGURE 38.18 Probability per unit length, c(x)2, as a function of position for the first few states of the infinite potential well.

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PHYSICS IN PRACTICE

Spectral Lines, Bohr's Theory, and Quantum Mechanics

U LT R A M I C R O S C O P E S

The wave properties of electrons find practical application in the transmission electron microscope (TEM), which employs electron waves to form a highly magnified image of an object in the same way as an ordinary microscope employs light waves. The maximum magnification attainable by ordinary microscopes is limited by the wavelength of light. If we attempt to observe an object as small as or smaller than the wavelength of light, the image becomes indistinct, because the light waves suffer pronounced diffraction when passing through and around such a small object, and the resulting diffraction fringes blur the image. The electron waves used in typical electron microscopes have wavelengths of 0.05 nm, which is 10 000 times shorter than the wavelength of light; the electron waves are therefore much less susceptible to diffraction than light. The main “optical” elements in a transmission electron microscope are the same as in an ordinary microscope, an objective lens and an ocular, or projector, lens (see Fig. 1). The “lenses” are not made of glass, but of magnetic fields, carefully shaped so as to provide deflections similar to those experienced by light in a glass lens. The electron rays emerging from the projector lens are intercepted by a detector array, which records the image digitally for viewing and analysis. Another kind of electron microscope is the scanning electron microscope (SEM). The principle of operation of this microscope is quite different, and bears no resemblance to the operation of ordinary light microscopes. The scanning electron microscope relies on the particle properties of electrons as well as their wave properties. Instead of forming an

image by means of refracted electron waves, the scanning electron microscope uses a diffraction-limited fine beam of electrons to bombard the object. The beam “scans” across the object line by line in a sweep pattern, like the sweep pattern of the electron beam in an ordinary TV tube. A detector picks up the current of electrons that recoil from the object and of secondary electrons knocked out of the object by the incident primary electrons (Fig. 2). This detected current is amplified and fed into a video monitor, which displays a picture. Scanning electron microscopes produce very crisp pictures with an exceptional depth of field and strong shadows that give a vivid three-dimensional impression, but they do not attain the extremely high magnifications of transmission electron microscopes. (a)

electron gun condenser lens

objective lens scanning coil

electron gun

detector

specimen amplifier to vacuum pump video monitor

(b)

(a)

electron beam

(b)

specimen objective lens projector lens

to vacuum pump

detector array

FIGURE 1 (a) Schematic diagram of a transmission electron microscope (TEM). (b) Picture of a human chromosome prepared with a TEM.

FIGURE 2 (a) Schematic diagram of a scanning electron microscope (SEM). (b) Picture of human chromosomes prepared with an SEM. Figure 1b is a close-up view of one such chromosome.

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Quantum Mechanics; the Schrödinger Equation

A newer and more powerful kind of electron microscope is the scanning tunneling microscope (STM). This microscope scans the surface of the object with a fine tungsten needle whose motion is precisely controlled by a delicate suspension system (see Fig. 3a). A potential difference is applied between the needle and the surface of the object. The needle is not actually in contact with the surface; it merely sweeps across the surface line by line. The gap between the needle and the surface is effectively an insulator, which tends to block the motion of the electrons from the needle to the surface of the object. However, the wave properties of electrons permit them to spread for some distance into this gap, and leak across to the surface. This kind of leakage of an electron across a gap where its motion is classically forbidden is called tunneling. The probability for an electron to succeed in tunneling across the gap depends drastically on the size of the gap.

1311

(b)

Whenever the needle, during its sweep, comes near any peak in the surface—such as a bulging atom—the tunneling probability starts to increase drastically, and so does the current from the needle to the surface. This incipient increase of the current is detected by the electronic circuit connected to the needle, and an amplified feedback signal is sent to the suspension system to lift the needle, so as to keep it at constant height from the peak. Thus, the needle skims over the peaks and valleys of the surface, maintaining a relatively constant height above the “terrain.” The amplified signal used to control the suspension is also sent to a computer, where it is processed and then fed into a video monitor for display. Modern scanning tunneling microscopes attain magnifications of up to 108, and they permit us to “see” individual atoms (Fig. 3b). This technique for constructing a picture of the surface by sweeping a needle across it is also exploited in the atomicforce microscope (AFM). But the needle of this kind of microscope is placed in direct contact with the surface—it is pressed against the surface with a force of, say 108 N, and it is lifted or lowered by the suspension system so as to keep the force constant during the sweep. Thus, the needle actually explores the shape of the surface by directly feeling it, just as you might explore the shape of a surface by feeling it with the tip of your finger. An alternative AFM technique is “tapping mode” (see Example 2 of Chapter 15), where the needle oscillates and taps the surface. While tapping the surface during scanning, the needle is lifted or lowered to keep the oscillation amplitude constant. Magnifications attained with atomicforce microscopes are nearly the same as those attained with scanning tunneling microscopes. The picture displayed in Fig. 4 was prepared with an atomic-force microscope.

FIGURE 3 (a) Schematic diagram of a scanning tunneling electron microscope (STM). (b) Picture of iron atoms prepared with an STM. The 48 iron atoms arranged in a circle on the surface of a copper crystal form a confining potential known as a “quantum corral,” and the ripples in the image show the wave nature of the surface copper electrons.

FIGURE 4 Picture of a carbon nanotube lying across ultra thin stripes of gold, prepared with an atomic-force microscope (AFM).

(a) tunneling current

preamp

piezoelectric scanner

horizontal sweep motion control specimen stage

vertical motion control specimen needle stepper motor

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✔

Spectral Lines, Bohr's Theory, and Quantum Mechanics

Checkup 38.5

QUESTION 1: One electron has a kinetic energy of 1 eV; another has a kinetic energy of 100 eV. Which has the shorter de Broglie wavelength? By what factor? QUESTION 2: Suppose that an electron has a de Broglie wavelength of 20 nm. If we increase the speed of this electron by a factor of 4, what will be its new de Broglie wavelength? QUESTION 3: An electron and a proton have the same kinetic energy. Which has the longer de Broglie wavelength? Q U E S T I O N 4 : What is the probability per unit length that an electron in the first excited state (n  2) of a one-dimensional box is at the center of the box? An electron in the second excited state (n  3)? (See Fig. 38.18.) QUESTION 5: Bohr’s theory of the hydrogen atom assumes that the electron remains in a given orbital plane, say, the x–y plane, and that it has no motion in the z direction. What are ¢z and ¢pz in this case? Is this consistent with the uncertainty principle? QUESTION 6: Take the proton-to-electron mass ratio to be mp me  2000. For the ground states of the respective one-dimensional boxes, what is the ratio of the energy of a proton in a region of width 1015 m to that of an electron in a region of width 1010 m?

(A) 0.02

(D) 5 106

(B) 50 (C) 2000

(E) 2 1013

S U M M A RY PHYSICS IN PRACTICE

SPECTRAL SERIES OF HYDROGEN

QUANTIZATION OF ANGULAR MOMENTUM

PLANCK’S CONSTANT

(page 1310)

Ultramicroscopes

(“h-bar”)

BOHR RADIUS

ENERGY OF STATIONARY STATES OF HYDROGEN

1 1 1  Ra 2  2 b l n2 n1

R  1.096 78 107 m1

(38. 7)

n  1, 2, 3, …

(38.9)

h  1.05 1034 J s 2p

(38.13)

L  nU U 

a0 

En  

4p0 U 2 mee 2

 0.0529 nm

mee4

1 2 2

2

2(4p0) U n



13.6 eV n2

(38.17)

(38.23, 38.25)

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Questions for Discussion

FREQUENCY AND WAVELENGTH OF PHOTON

f

EMITTED IN TRANSITION

Ei  Ef h

DE BROGLIE WAVELENGTH OF WAVICLE

and

1313

Ei  Ef f 1   c l hc

(38.27)

l  h p

(38.32)

E  hf

(38.38)

ENERGY OF STATIONARY STATE IN TERMS OF FREQUENCY TIME-INDEPENDENT SCHRÖDINGER EQUATION



One-dimension:

U2 d2 c(x)  U(x) c(x)  E c(x) 2me dx2

(38.41)

CONDITIONS FOR WAVEFUNCTION

One-dimension: c(x) is a solution to the Schrödinger equation



c(x) is everywhere continuous

q

dc(x) dx is everywhere continuous [except if U(x) is infinite]

c(x)2dx  1 (normalization)

q

PROBABILITY FOR FINDING ELECTRON IN A

b

P

REGION a  x  b

 c(x) dx 2

(38.52)

a

ELECTRON IN A BOX

Wavefunctions:

Energies:

One-dimension, 0  x  L: cn(x) 

U

2 npx sin a b BL L

En  n2E1  n2

U 2p2

∂ n  1, 2, 3, …

U0

2

2me L

0

position

L

x

(38.50, 38.51)

QUESTIONS FOR DISCUSSION 1. Do the spectral lines seen in a stellar spectrum (for example, Fig. 38.5) tell us anything about the chemical composition of the stellar interior? 2. The lower spectrum of hydrogen shown in the chapter photo displays all of the spectral lines simultaneously. Since a hydrogen atom emits only one spectral line at a time, how can all the lines be visible simultaneously? 3. The target used in Rutherford’s scattering experiment was a very thin foil of metal. What is the advantage of a thin foil over a thick foil in this experiment?

4. How can Rutherford’s experiment tell us something about the size of the nucleus? 5. At low temperatures, the absorption spectrum of hydrogen displays only the spectral lines of the Lyman series. At higher temperatures, it also displays other series. Explain. 6. The planets move around the Sun in circular orbits. Is their orbital angular momentum quantized? 7. In a muonic atom, a muon orbits around the nucleus. The mass of the muon is 207 times the mass of the electron. What is the Bohr radius for a muonic atom with a hydrogen nucleus?

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Spectral Lines, Bohr's Theory, and Quantum Mechanics

8. Given that the orbital angular momentum of an atom is quantized, can we conclude that the orbital magnetic moment is also quantized?

11. Show that photons obey the de Broglie relation.

9. Bohr’s theory of the hydrogen atom can be adapted to the singly ionized helium atom, that is, the helium atom with one missing electron. To what other ionized atoms can Bohr’s theory be adapted?

13. According to the de Broglie relation, the wavelength of an electron of very small momentum is very large. Could we take advantage of this to design an experiment that makes the wave properties of the electron obvious?

10. According to the Complementarity Principle, formulated by Bohr, a wavicle has both wave properties and particle properties, but these properties are never exhibited simultaneously; if the wavicle exhibits wave properties in an experiment, then it will not exhibit particle properties, and conversely. Give some examples of experiments in which wave or particle properties (but not both simultaneously) are exhibited.

12. If the de Broglie wavelengths of two electrons differ by a factor of 2, by what factor must the kinetic energies differ?

14. Describe the interference pattern expected for an electron wave incident on a plate with two very narrow parallel slits separated by a small distance. 15. Electron microscopes achieve high resolution because they use electron waves of very short wavelength, usually less than 0.05 nm. Why can we not build a microscope that uses photons of equally short wavelength?

PROBLEMS †

38.2 Spectral Series of Hydrogen 1. Use Eq. (38.4) to calculate the wavelengths of the first four lines of the Lyman series. 2. Show that the spectral lines of the Balmer series all have a higher frequency than the spectral lines of the Paschen series. Do the spectral lines of the Paschen series all have a higher frequency than those of the Brackett series? 3. A hydrogen atom at rest emits a photon. By Eq. (37.15), p  h l, the photon of highest momentum has the shortest wavelength, corresponding to the Lyman series limit. What is the recoil velocity of the hydrogen atom when this photon is emitted? 4. Which of the spectral lines of the Brackett series is closest in wavelength to the first spectral line (n2  5; n1  6) of the Pfund series? By how much do the wavelengths differ? *5. When astronomers examine the light of a distant galaxy, they find that all the wavelengths of the spectral lines of the atoms are longer than those of the atoms here on Earth by a common multiplicative factor. This is the red shift of light; it is a Doppler shift caused by the motion of recession of the galaxy, away from Earth. In the light of a galaxy beyond the constellation Virgo, astronomers find spectral lines of wavelengths 411.7 nm and 435.7 nm. (a) Assume that these are two spectral lines of hydrogen, with the wavelengths multiplied by some factor. Identify these lines. What is the factor by which these wavelengths are longer than the normal wavelengths of the two spectral lines? What is the factor by which the frequencies are lower?

†

For help, see Online Concept Tutorial 43 at www.wwnorton.com/physics

(b) If the speed of recession is low compared with the speed of light, the Doppler shift of light obeys a formula similar to that for the Doppler shift of sound [see Eq. (17.17)]. Calculate the speed of recession of the galaxy. *6. One of the series of spectral lines of the lithium atom is the principal series, with the following wavelengths, measured in vacuum: 617.0 nm, 323.4 nm, 274.2 nm, 256.3 nm, 247.6 nm. These wavelengths approximately fit the formula 1 1 1  Rc  d l (1  s)2 (n  p)2

n  2, 3, 4, …

where R  1.097 29 107 m1 is the Rydberg constant for lithium, and where s and p are constants characteristic of the series. Find the values of these constants. (Hint: Write the equations for 1 for two different wavelengths, eliminate the unknown s by subtracting these formulas, and then solve for p by trial and error.) *7. Another of the spectral series of the lithium atom is the diffuse series, with the following wavelengths, measured in vacuum: 610.5 nm, 460.4 nm, 413.4 nm, 391.6 nm, 379.6 nm. These wavelengths approximately fit the formula 1 1 1  Rc  d l (2  p)2 (n  d ) 2

n  3, 4, 5, …

where, as in the preceding problem, R  1.097 29 107 m1 and where p and d are constants. (a) Find the values of these constants. (b) The principal series (see Problem 6) and the diffuse series of lithium are analogous to two spectral series of hydrogen. Which two series? (Hint: See the preceding problem.)

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Problems

38.3 The Nuclear Atom *8. What is the distance of closest approach for a 5.5-MeV alpha particle in a head-on collision with a gold nucleus? With an aluminum nucleus? *9. The nucleus of platinum has a radius of 6.96 1015 m and an electric charge of 78e. What must be the minimum energy of an alpha particle in a head-on collision if it is to just barely reach the nuclear surface? Assume the alpha particle is pointlike. *10. Consider a model of the hydrogen atom with a pointlike nucleus, but assume that the electron charge is uniformly distributed over a sphere of radius R  0.053 nm. The atom is in equilibrium when the nucleus is at the center of the electron charge distribution. Find the frequency of oscillation when the electron sphere is displaced from equilibrium by a distance x  R. What is the wavelength of a photon with this frequency? *11. The probability for a scattering event to occur is equal to the fraction of the area of the target for which that event will occur. The area per nucleus for an event is the scattering cross section s. This area is related to the impact parameter b for that event by s  pb2. The impact parameter for a deflection of more than 90 (backscattering) is 1.0 1013 m for alpha particles of energy 5.7 MeV scattering from gold nuclei. What is the scattering cross section for backscattering? Consider an incident beam of alpha particles with a beam radius of 2.0 millimeters and a target with 5.0 1017 gold nuclei in the beam area. What is the probability for backscattering? *12. An alpha particle of energy 5.5 MeV is incident on a silver nucleus with an impact parameter 8.0 1015 m. The distance of closest approach of the particle is 2.7 1014 m. Find the speed at the point of closest approach. (Hint: The angular momentum is conserved.) 5

**13. A foil of gold, 2.1 10 cm thick, is being bombarded by alpha particles of energy 7.7 MeV. The particles strike at random over an area of 1.0 cm2 of the foil of gold. (a) How many atoms are there within the volume 1.0 cm2 2.1 105 cm under bombardment? The density of gold is 19.3 g/cm3, and the mass of one atom is 3.27 1025 kg. (b) It can be shown that to suffer a deflection of more than 30, an alpha particle must strike within 5.5 1014 m of the center of a gold nucleus. What is the probability for this to happen? (Hint: See Problem 11.) (c) If 1.0 1010 alpha particles impact on the foil, how many will suffer deflections of more than 30? †

3 8 . 4 B o h r ’s T h e o r y 14. If you wanted to give an apple an angular momentum of U, at what rate would you have to spin it about its axis? Treat the apple as a uniform sphere of mass 0.20 kg and radius of 4.0 cm.

†

For help, see Online Concept Tutorial 43 at www.wwnorton.com/physics

1315

15. What is the angular momentum of a vinyl record (a uniform disk) rotating at 3313 revolutions per minute? The moment of inertia of the record is 1.3 102 kg m2. Express the answer as a multiple of U . 16. What are the angular momentum, the kinetic energy, the potential energy, and the net energy of an electron in the smallest (n  1) Bohr orbit in the hydrogen atom? Express your answers in SI units. 17. What is the speed of an electron in the smallest (n  1) Bohr orbit? Express your answer as a fraction of the speed of light. Is it justified to ignore relativistic effects in the Bohr model? 18. What is the centripetal acceleration of an electron in the smallest (n  1) Bohr orbit in the hydrogen atom? 19. Hydrogen atoms in highly excited states with a quantum number as large as n  732 have been detected in interstellar space by radio astronomers. What is the orbital radius of the electron in such an atom? What is the energy of the electron? 20. If a hydrogen atom is initially in the first excited state, what is the longest wavelength of light it will absorb? What is the shortest wavelength of light it will absorb? 21. If you bombard hydrogen atoms in their ground state with a beam of particles, the collisions will (sometimes) kick atoms into one of their excited states. What must be the minimum kinetic energy of the bombarding particles if they are to achieve such an excitation? 22. Suppose that the electron in a hydrogen atom is initially in the second excited state (n  3). What wavelength will the atom emit if the electron jumps directly to the ground state? What two wavelengths will the atom emit if the electron jumps to the first excited state and then to the ground state? 23. A hydrogen atom emits a photon of wavelength 102.6 nm. From what stationary state to what lower stationary state did the electron jump? *24. Show that the speed of an electron in a Bohr orbit with quantum number n is given by v

1 e2 n 4p0 U

Also, find an expression for the centripetal acceleration of an electron in the nth Bohr orbit. *25. The quantity U (mec) is called the Compton wavelength. The quantity e2 (4p0mec 2) is called the “classical electron radius.” Show that the Bohr radius, the Compton wavelength, and the classical electron radius are in the ratios 1::2, where   e2 (4p0 Uc). The quantity  is called the fine-structure constant. What is the numerical value of this constant? *26. A hydrogen atom is initially in the ground state. In a collision with an argon atom, the electron of the hydrogen atom absorbs an energy of 15.0 eV. With what speed will the electron be ejected from the hydrogen atom? *27. Suppose that a sample of hydrogen atoms, initially all in the ground state, is under bombardment by a beam of electrons of

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kinetic energy 12.2 eV. In an inelastic collision between a hydrogen atom and one of the incident electrons, the hydrogen atom will occasionally absorb all, or almost all, the kinetic energy of the electron and make a transition from the ground state to an excited state. If so, what excited state will the hydrogen atom attain? What are the possible spectral lines the atom can emit subsequently? *28. The singly ionized helium atom (usually designated HeII or He) has one electron in orbit around a nucleus of charge 2e. (a) Apply Bohr’s theory to this atom and find the energies of the stationary states. What is the value of the ionization energy, that is, the energy that you must supply to remove the electron from the atom when it is in the ground state? Express the answer in electron-volts. (b) Show that for every spectral line of the hydrogen atom, the ionized helium atom has a spectral line of identical wavelength. *29. Assume that, as proposed by J. J. Thomson, the hydrogen atom consists of a cloud of positive charge e, uniformly distributed over a sphere of radius R. However, instead of placing the electron in static equilibrium at the center of the sphere, assume that the electron orbits around the center with uniform circular motion under the influence of the electric centripetal force (e2 4p0)(r R3). If the angular momentum of this orbiting electron is quantized according to Bohr’s theory (so that L  n U ), what are the radii and the energies of the quantized orbits? What are the frequencies of the photons emitted in transitions from one quantized orbit to another? What must be the value of R if at least two orbits are to fit inside this atom? **30. In principle, Bohr’s theory also applies to the motion of the Earth around the Sun. The Earth plays the role of the electron, the Sun that of the nucleus, and the gravitational force that of the electric force.

(Hint: The electron and the proton move in circles of radii re  r

mp mp  me

and

rp  r

me mp  me

where r is the distance between the electron and the proton. According to Bohr’s theory, the net angular momentum of this system of two particles is quantized, L  nU.) **32. The atom of positronium consists of an electron and a positron (or antielectron) orbiting about their common center of mass. According to Bohr’s theory, the net angular momentum of this system is quantized, L  n U. What is the radius of the smallest possible circular orbit of this system? What is the wavelength of the photon released in the transition from n  2 to n  1? (Hint: the electron and the positron have the same mass. See also Problem 31.)

38.5 Quantum Mechanics; the Schrödinger Equation 33. What must be the energy of an electron if its wavelength is to equal the wavelength of visible light, about 550 nm? 34. Find the de Broglie wavelength for each of the following electrons with the specified kinetic energy: electron of 20 keV in a TV tube, conduction electron of 5.4 eV in a metal, orbiting electron of 13.6 eV in a hydrogen atom, orbiting electron of 91 keV in a lead atom. 35. What is the de Broglie wavelength of an electron in the ground state of hydrogen? In the first excited state? 36. What is the de Broglie wavelength of a tennis ball of mass 0.060 kg moving at a speed of 1.0 m/s? 37. An electron microscope operates with electrons of kinetic energy 40 keV. What is the wavelength of such electrons? By what factor is this wavelength smaller than that of visible light?

(a) Find a formula analogous to Eq. (38.18) for the radii of the permitted circular orbits of the Earth around the Sun.

38. A photon and an electron each have an energy of 6.0 103 eV. What are their wavelengths?

(b) The actual radius of the Earth’s orbit is 1.50 1011 m. What value of the quantum number n does this correspond to?

39. If the de Broglie wavelengths of two electrons differ by a factor of 4, by what factor must their kinetic energies differ?

(c) What is the radial distance between the Earth’s actual orbit and the next larger orbit? **31. In our calculation of the energies of the stationary states of hydrogen we pretended that the proton remains at rest. Actually, both the electron and the proton orbit about their common center of mass. Show that the energies of the stationary states, taking into account this motion of the proton, are given by En  

me4

1

2(4p0)2 U 2 n2

where m

memp me  mp

40. The “thermal” neutrons in a nuclear reactor typically have a kinetic energy of about 0.050 eV. What is the de Broglie wavelength of such a neutron? 41. Suppose that the velocity of an electron has been measured to within an uncertainty of — 1.0 cm/s. What minimum uncertainty in the position of the electron does this imply? 42. The nucleus of the aluminum atom has a diameter of 7.2 1015 m. Consider one of the protons in this nucleus. The uncertainty in the position of this proton is necessarily less than 7.2 1015 m. What is the minimum uncertainty in its momentum and velocity? 43. If the position of a parked automobile of mass 2.0 103 kg is uncertain by _1.0 1018 m, what is the corresponding uncertainty in its velocity?

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Review Problems

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44. For an electron confined to a one-dimensional box of length L  0.10 nm, what is the kinetic energy in the ground state? What is the corresponding speed? Is it justified to ignore relativistic effects for such an electron?

*53. Show that the de Broglie wavelength of an electron can be calculated from the formula

45. Suppose that a particle confined to a one-dimensional box of length L is in its first excited state. What is the probability per unit length that the particle is in a small interval near x  14 L? Near x  12 L? Near x  34 L?

where the wavelength l is expressed in nm and the kinetic energy K is expressed in eV.

46. Consider the possibility that a nucleus might contain an electron. What would be the energy of an electron in the ground state of a one-dimensional box of nuclear size, say 1.0 1014 m? Since nuclear energies are typically only on the order of 1 MeV, do you expect that an electron might be confined within the nucleus? 47. For a particle in the state of quantum number n of a onedimensional box of length L, at how many points within the box is the probability per unit length of finding the particle zero? At how many points is it maximum? In each case, describe the locations of those points. 48. An electron in a one-dimensional box of length L  0.30 nm makes a transition from the first excited state to the ground state. What is the wavelength of the emitted photon? 49. Use symmetry and the result of Example 8 to determine the probability that an electron in the ground state of a onedimensional box is in the region 0  x  14 L. *50. Neutrinos of energy 10 MeV are emitted by the sun. What is the de Broglie wavelength of such a neutrino? (Hint: Since the neutrino is a particle of very small mass, its momentum at such high kinetic energy must be calculated according to the formula for the momentum of an ultra-relativistic particle.)

l  1.23 1K

*54. Consider the solutions to the Schrödinger equation for a particle in a one-dimensional box. For n large, show that the probability per unit length, averaged over a small but suitable interval, yields the classical result Pclassical  1 L. *55. Obtain the energies [Eqs. (38.50–38.51)] of an electron confined to an interval from x  0 to x  L along the x axis by a simple de Broglie wavelength calculation: (a) What are the de Broglie wavelengths of standing electron waves in this interval? Assume that, as for a standing wave on a string, the amplitude of oscillation is zero at x  0 and x  L, so that an integer number of half wavelengths fit into the length L. (b) What are the momenta corresponding to these wavelengths? (c) What are the energies of the stationary states? (d) Evaluate the energies of the ground state and the first three excited states numerically for L  0.10 nm. **56. The harmonic oscillator potential is U(x)  12mv20x2; a particle of mass m in this potential oscillates with frequency 0. The ground-state wavefunction for a particle in the harmonic oscillator potential has the form 2

c(x)  Aeax

*51. What is the de Broglie wavelength of an electron if its kinetic energy is equal to its rest-mass energy? (Hint: For a relativistic particle the momentum must be calculated according to the relativistic formula.)

(a) By substituting U(x) and c(x) into the one-dimensional, time-independent Schrödinger equation [Eq. (38.41)], find expressions for the ground-state energy E and the constant a in terms of m, U, and v0.

*52. What is the de Broglie wavelength of the Earth in its motion around the Sun?

(b) Apply the normalization condition to determine the constant A in terms of m, U, and v0.

REVIEW PROBLEMS 57. The series limit for the Balmer series is 364.7 nm. What are the series limits for the Lyman, Paschen, and Brackett series? 58. Are the wavelengths of all the spectral lines of the Lyman series shorter than the series limit of the Balmer series? Are the wavelengths of all the spectral lines of the Balmer series shorter than the series limit of the Paschen series? Are the wavelengths of all the spectral lines of the Paschen series shorter than the series limit of the Pfund (n2  5) series? 59. If a hydrogen atom is in the ground state, what is the longest wavelength it will absorb?

60. What is the frequency of the orbital motion for an electron in the smallest (n  1) Bohr orbit? In the next (n  2) Bohr orbit? Do either of these frequencies coincide with the frequency of the light emitted during the transition from n  2 to n  1? 61. Find the orbital radius, the speed, the angular momentum, and the centripetal acceleration for an electron in the n  2 Bohr orbit of hydrogen. 62. When a hydrogen atom is initially in the ground state, the ionization energy (or the energy for removal of the electron) is

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Spectral Lines, Bohr's Theory, and Quantum Mechanics

13.6 eV. What is the ionization energy for a hydrogen atom initially in the first excited state? 63. For an electron in the first excited state of a one-dimensional box, what is the probability for finding the electron in the interval 0  x  18 L? Compare this with the classical result, P  18. 64. An electron makes a transition from the n  3 state to the n  1 state of a one-dimensional box, emitting a photon of wavelength 450 nm. What is the length of the box? *65. An alpha particle of energy 5.5 MeV is incident on a copper nucleus with an impact parameter of 5.0 1015 m. Find the distance of closest approach of the particle, and find the speed at the point of closest approach. (Hint: Both the energy and the angular momentum of the alpha particle are conserved.) *66. The muon (or mu meson) is a particle somewhat similar to an electron; it has a charge e and a mass 206.8 times as large as the mass of the electron. When this particle orbits around a proton, the two particles form a muonic hydrogen atom, similar to an ordinary hydrogen atom, but with the muon playing the role of the electron. Calculate the Bohr radius of this muonic atom and calculate the energies of the stationary states. What is the energy of the photon emitted when the muon makes a transition from the n  2 to the n  1 state? *67. An electron in a hydrogen atom is initially in the ground state. The electron absorbs a photon from an external light source and thereby makes a transition to the n  4 state. What must have been the energy and the wavelength of the absorbed photon? If the electron now jumps spontaneously to the n  3 state, what are the energy and the wavelength of the emitted photon?

*68. The doubly ionized lithium atom (usually designated LiIII or Li2 ) has one electron in orbit around a nucleus of charge 3e. What is the radius of the smallest Bohr orbit in doubly ionized lithium? What is the energy of this orbit? *69. Consider a helium atom in interstellar space in a circular orbit around a meteoroid of mass 4.0 kg and radius 10 cm under the influence of the gravitational force. We can apply Bohr’s theory to this system; the meteoroid plays the role of the nucleus, the atom that of the electron, and the gravitational force that of the electric force. Find formulas analogous to Eqs. (38.18) and (38.23) for the orbital radii and the energies of the permitted circular orbits. Because of the finite size of the meteoroid, the smallest feasible orbit has a radius of 10 cm. What is the quantum number and what is the energy (in eV) of a helium atom in this orbit? *70. What is the de Broglie wavelength of a nitrogen molecule (N2) in air at room temperature (20C)? Assume that the molecule is moving with the rms speed of molecules at this temperature. *71. Interferometric methods permit us to measure the position of a macroscopic body to within ; 1.0 1012 m. Suppose we perform a position measurement of such a precision on a body of mass 0.050 kg. What uncertainty in momentum is implied by the Heisenberg relation? What uncertainty in velocity? *72. Consider an electron in a circular orbit of quantum number n in a hydrogen atom. The orbit is well defined provided that ¢p V p and ¢y V r. Show that if n W 1, these requirements are not in conflict with the Heisenberg uncertainty relations. Thus, large orbits in the hydrogen atom are well defined.

Answers to Checkups Checkup 38.1 1. The original spectra are in the form of lines because of the

presence of the slit between the light source and the prism; to produce spectral dots, you would have to replace the slit by a small hole. 2. For the moderately hot temperature of a normal flame, we saw

in Chapter 37 that the continuum thermal radiation is peaked in the infrared, not the blue; thus the bluish color is due to spectral lines. 3. According to the color print on page 1289, this yellow light is

due to two nearly equal wavelengths (a “doublet”) with values 589.2 nm and 589.8 nm.

4. (B) Absorption of light by atoms in the cavity. Before emis-

sion from the opening, thermal radiation can strike atoms in the cavity; the atoms will absorb light at their characteristic wavelengths, resulting in dark absorption lines.

Checkup 38.2 1. Yes. In any series, the separation of successive spectral lines is

given by the difference of the inverse square of successive integers, which approaches zero as the integers approach infinity. 2. Similar to the result of Example 1, the series limits are given

by 1 l  R n22. For the series given, the largest wavelength corresponds to n2  3 for the Paschen series, and the shortest wavelength to n2  1 for the Lyman series.

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Answers to Checkups

3. In each case, the longest wavelength is given by the smallest

difference in the two terms of Eq. (38.7); this corresponds to choosing n1  n2  1. Since n2  1, 2, 3, and 4 for the Lyman, Balmer, Paschen, and Brackett series, respectively, the corresponding values of n1 are 2, 3, 4, and 5. 4. (A) Lyman. Similar to Example 1, the shortest wavelength of

each series is given by l  n22 R  n22 91.176 nm. Thus the series with n2 = 2, 3, and 4 have wavelengths too long to include the given wavelength, so the given wavelength must belong to the n2  1, or Lyman series.

1319

decreases as n decreases. According to Bohr’s postulate [see Eq. (38.9)], the orbital angular momentum is proportional to n and thus decreases when the hydrogen atom makes a transition to a lower state. 3. Transitions of the Paschen series all end at n  3, so the first

transition is from n  5 to n  3. Transitions of the Balmer series all end at n  2, so the second transition is from n  3 to n  2; the final transition must then be from n  2 to n  1. Transitions that end at n  1 are in the Lyman series. 4. (D) n  q to n  2. Transitions of the Balmer series all end

at n  2; the transition that produces the series limit has the largest energy difference, and so begins at n  q.

Checkup 38.3 1. According to the “plum-pudding” model, the positive charge 10

m, is spread over a cloud of atomic size, approximately 10 and the electron is a pointlike particle in this cloud. If the cloud of positive charge is spherically symmetric, the electron is in equilibrium when at rest at the center of the cloud, where the electric field is zero. 2. An impact parameter of zero means that the alpha particle

approaches the nucleus head-on; thus, the alpha particle bounces back, along its initial line of motion (or if it has sufficient energy, it penetrates into the nucleus). 3. Since the impact parameter is the perpendicular distance

between the nucleus and the undeflected original line of motion, the distance of closest approach in the repulsively deflected hyperbolic orbit is larger than the impact parameter. 4. If the incident particles were negatively charged, the deflections

would be attractive; thus, the orbits would go partially around the nucleus, and the alpha particle would emerge in a somewhat downward direction, instead of being deflected upward. 5. (A) Through a smaller angle than when incident on a gold

target. For the smaller charge Ze on the silver nucleus, the repulsive electric force is smaller, and so the deflection is smaller.

Checkup 38.4 1. According to classical mechanics and electromagnetism, an

electron in a circular orbit is accelerating and thus must radiate, lose energy, and spiral inward toward the nucleus. 2. The kinetic energy is the first term in Eqs. (38.19)–(38.21),

which is positive and inversely proportional to the square of the quantum number n. For a transition to a lower state, n decreases, so the kinetic energy increases. The potential energy, the second term in the same equations, is also inversely proportional to the square of n, but is negative and so

Checkup 38.5 1. Higher energy corresponds to shorter wavelength. Since the

kinetic energy is proportional to the square of the momentum (that is, K  12 mv2  p2 2m) and the de Broglie wavelength is l  h p, the wavelength varies inversely with the square root of the energy. So the 100-eV electron has a factor of 10 shorter wavelength than the 1-eV electron. 2. The de Broglie wavelength is inversely proportional to the

momentum and, since p  mv, also to the speed. A factor of 4 increase in speed thus results in a factor of 4 decrease in wavelength to 5 nm. 3. Since the de Broglie wavelength is p  h l, the kinetic

energy is K  p2 2m  h2 2ml2. Equality of energies thus implies that the wavelength must be longer for the electron, because of its smaller mass.

4. Since the n  2 state has a node at the center of the box

[c(L 2)  0], the probability per unit length that an electron is at the center is zero. The n  3 state has a maximum at the center. In Example 8, we found that all the stationary states of the one-dimensional box have the same wavefunction amplitude, 12 L, and thus the probability per unit length at the maximum is 2L.

5. If an electron remains within the x–y plane, then it has

¢z  0. If there is no motion in the z direction, then the electron has ¢pz  0. These are inconsistent with the uncertainty relation, which requires ¢z ¢pz  U 2. 6

6. (D) 5 10 . The energy varies inversely with the mass and

inversely with the square of the length, so compared with the electron, the proton energy will be both decreased by a factor of 2000 and increased by a factor of 1010, for an overall increase by a factor of 5 106. The given lengths correspond to the sizes of nuclear and atomic regions; atomic energies are typically of order eV, and nuclear energies, MeV.

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CHAPTER

39

Quantum Structure of Atoms, Molecules, and Solids

CONCEPTS IN CONTEXT

39.4 Energy Bands in Solids

This scanning tunneling microscope (STM) image depicts the electron densities associated with individual atoms on the surface of gallium arsenide, a modern semiconductor preferred in some applications over silicon, the material predominantly used in the fabrication of transistors and integrated circuits. The atoms are arranged in a periodic structure, known as a lattice. As we learn about the quantum structure of materials, we will consider such questions as:

39.5 Semiconductor Devices

? The nearly free electrons in a metal interact with the atoms in a lat-

39.1 Principal, Orbital, and Magnetic Quantum Numbers; Spin 39.2 The Exclusion Principle and the Structure of Atoms 39.3 Energy Levels in Molecules

tice, and this restricts the permitted energies of the electrons. How do these restrictions arise? (Section 39.4, page 1336)

? The bright spot in the photo of the crystal lattice shows an indium atom, an impurity. Such impurities are intentionally added to a 1320

Concepts in Context

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39.1

Principal, Orbital, and Magnetic Quantum Numbers; Spin

1321

semiconductor to modify the conductivity. How does the presence of impurity atoms affect the conductivity? (Section 39.4, page 1339)

? How are regions with different impurities combined to make useful devices? (Section 39.5, page 1340)

T

he physical and chemical properties of atoms and of molecules depend on the quantum behavior of their electrons. These electrons occupy most of the volume of the atom, and their arrangement in different quantum states determines the size and the shape of the atom, the chemical bonds that the atom forms with other atoms, the energy required for ionization, the spectrum of light emitted and absorbed, and so on. Likewise, the physical properties of a solid—such as diamond, silicon, silver, copper—depend on the quantum behavior of the electrons in the solid. The spacing of the crystal lattice of the solid, the electric and thermal conductivities, the magnetic properties, and the mechanical properties (such as elasticity and hardness) all hinge on the arrangement of the electrons. For instance, we will see in Section 39.4 how the arrangement of the electrons determines the ability of the solid to conduct an electric current, and we will see how the differences among conductors, semiconductors, and insulators depend on what stationary quantum states are available in the solid and which of these are occupied by electrons. The arrangement of the electrons in the stationary states of an atom or a solid is subject to an important restriction: no more than two electrons can occupy the same orbital state. This is called the Pauli Exclusion Principle. In Section 39.1, we will become acquainted with the spin, or the intrinsic angular momentum, of the electron, and in Section 39.2, we will see that the Exclusion Principle is intimately linked to the spin. In later sections we will examine the implications of the Exclusion Principle for the arrangement of the electrons in atoms and in solids.

3 9 . 1 P R I N C I PA L , O R B I TA L , A N D MAGNETIC QUANTUM NUMBERS; SPIN In the preceding chapter, we saw that Bohr’s theory characterizes the stationary states of the hydrogen atom and their energies by a single quantum number n. However, Bohr’s simple theory deals only with circular orbits. We know from the study of planetary orbits (Chapter 9) that the general orbit of a particle moving under the influence of an inverse-square force is an ellipse with one focus at the center of attraction. For an electron in such an elliptical orbit around a nucleus, the quantum number n, now called the principal quantum number, characterizes the overall size of the ellipse, that is, its major axis; this quantum number also characterizes the energy of the electron [the formula for the quantized energies of the elliptical orbits is the same as that for the circular orbits, Eq. (38.25)]. But for the complete definition of the ellipse we need two extra quantum numbers l and m that characterize the elongation of the ellipse and the inclination of the ellipse, that is, its orientation in space. These two extra quantum numbers l and m are called, respectively, the orbital quantum number and the magnetic quantum number. For a classical orbit, the elongation of the ellipse is closely related to the magnitude of the angular momentum of the electron. Among ellipses of equal sizes (equal major axes), the most elongated has the least angular momentum. Thus, the orbital quantum number l characterizes both the elongation of the ellipse and the magnitude of its angular momentum.

Online Concept Tutorial

44

principal quantum number

orbital quantum number magnetic quantum number

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CHAPTER 39

Angular momentum vector L is normal to plane of orbit.

The inclination of the ellipse is closely related to the direction of the angular momentum. To understand this relationship, we have to think of the angular momentum as a vector, with a direction perpendicular to the plane of the orbit (see Fig. 39.1). The orientation of the orbit in space is then specified by the direction of the angularmomentum vector; for instance, if the orbit is inclined at an angle of 35 relative to the x–y plane, as in Fig. 39.1, then the angular momentum vector is inclined at an angle of 35 relative to the z axis. As we will see below, the magnetic quantum number m specifies the z component of the angular-momentum vector, and it thereby characterizes both the direction of the angular-momentum vector and the inclination of the orbit. A large value of the magnetic quantum number m corresponds to a large value of this z component and a small inclination of the orbit; a small value of the magnetic quantum number m corresponds to a small value of this z component and a large inclination of the orbit, near 90. (The quantum number m is called “magnetic” because it acquires a special significance when the atom is placed in a magnetic field; the energy of the orbit then depends not only on the size of the orbit and on the principal quantum number n, but also on the orientation of the orbit and, therefore, on the “magnetic” quantum number m.) Although the two extra quantum numbers l and m were originally introduced on the basis of semiclassical considerations involving elliptical orbits, a later, more rigorous analysis based on wave mechanics and the Schrödinger equation confirmed that these quantum numbers are, indeed, required for the description of the stationary quantum states. In wave mechanics a stationary state corresponds to a three-dimensional standing wave, and the quantum numbers n, l, and m characterize the “shape” of this standing wave, and they also characterize the energy, the magnitude of the angular momentum, and the z component of the angular momentum. According to wave mechanics, the magnitude of the angular momentum is quantized. However, this quantization differs from that of the simple Bohr theory in that the quantum number for angular momentum is the orbital quantum number l, not the principal quantum number n. Furthermore, the formula for the quantization condition is somewhat more complicated. From the study of the mathematical properties of the Schrödinger equation, it can be demonstrated that the magnitude L of the angular momentum obeys the quantization condition

z Since plane of orbit is tilted 35 from x –y plane…

Lz L

35

…L is tilted 35 from the z axis.

y 35 x

r

p

FIGURE 39.1 This elliptical orbit is inclined at an angle of 35 relative to the x–y plane. The angular-momentum vector L is perpendicular to the plane of the orbit; this vector is therefore inclined at an angle of 35 relative to the z axis. The z component of the angular-momentum vector is Lz.

Quantum Structure of Atoms, Molecules, and Solids

L  1l (l  1)U

magnitude of angular momentum

(39.1)

where the orbital quantum number l is restricted to integer values from 0 to n  1, that is, l  0, 1, 2, # # #, n  1

(39.2)

For example, if n  3, then the possible values of the orbital quantum number are l  0, l  1, and l  2; and the corresponding magnitudes of the angular momentum are L  10  (0  1)U  0 L  11  (1  1)U  12U and L  12  (2  1)U  16U Note that the smallest possible magnitude of the angular momentum is actually zero, in contrast to what was claimed by the Bohr theory, where the smallest magnitude of the angular momentum was U.

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Principal, Orbital, and Magnetic Quantum Numbers; Spin

1323

Since the angular momentum is a vector quantity, it has not only magnitude, but also a direction. However, according to wave mechanics, the direction of the angularmomentum vector is only partially determined. The direction of the angular-momentum vector always has substantial quantum uncertainties, and, correspondingly, the x, y, and z components of this vector have substantial uncertainties. It can be demonstrated that at most one of the components is well determined—the other two components are completely uncertain. If we take the well-determined component to be the z component, then the quantization condition for this component is Lz  mU

(39.3)

z component of angular momentum

Here, the magnetic quantum number m is restricted to integer values from l to  l ; that is, m  l , l  1, ..., 0, ..., l  l , l

(39.4)

Note that, for a given value of l, there are 2l  1 possible values of m. Each of these possible values of m corresponds to a possible direction of the angular-momentum vector relative to the z axis. For instance, if l  2, then there are five possible values of m, namely, m  2, m  1, m  0, m  1, and m  2; thus, there are five possible choices for the direction of the angular-momentum vector relative to the z axis. Figure 39.2 shows these possible directions. From Eq. (39.3) we see that the z component of the angular momentum has a maximum possible value of lU; this corresponds to the best attainable alignment of the angular momentum with the z axis. If we compare this maximum possible value of the z component with the magnitude of a nonzero angular momentum, we see that the former is always smaller than the latter, lU  1l (l  1)U. This means that the angular-momentum vector is never perfectly aligned with the z axis. The uppermost vector drawn in Fig. 39.2 indicates the best attainable alignment for l  2.

z component of angular momentum for a given l may range from l… z

m  2, Lz  2 L m  1, Lz  

␪

m  0, Lz  0

y

x m  1, L z   m  2, L z  2

EXAMPLE 1

For the angular-momentum vector corresponding to a quantum number l  2, calculate the angle u between the vector and the z axis for the cases m  2, m  1, and m  0.

…to l in integer multiples of . z component is always smaller than magnitude so nonzero L is always tilted away from z axis.

SOLUTION: The directions of the angular-momentum vector for these cases are

illustrated in Fig. 39.2. The magnitude of the angular-momentum vector is L  1l (l  1)U  12  (2  1)U  16U. In the case m  2, the z component is Lz  mU  2U, and the angle u is given by cos u 

Lz L



2 2U   0.82 16U 16

FIGURE 39.2 Possible directions of the angular-momentum vector for the case l  2. In this case, there are five possible values of Lz, characterized by m  2, 1, 0, 1, and 2.

With a calculator, we find that the angle with this cosine is 35. Likewise, in the case m  1, the z component is Lz  U , and cos u 

U  0.41 16U

which gives us an angle of 66. Finally, in the case m  0, the z component is Lz  0, which gives us an angle of 90. This angular-momentum vector lies in the x –y plane.

Each possible set of values of n, l, and m corresponds to one kind of three-dimensional standing wave, often called an orbital. The simplest of these waves is that for the

orbital

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CHAPTER 39

spin quantum number ms spin, or intrinsic angular momentum

Quantum Structure of Atoms, Molecules, and Solids

ground state, with n  1, l  0, and m  0. Figure 39.3a shows a picture of the intensity of this wave, denoted c2, the square of the amplitude of the wavefunction. In this picture, generated from the theoretical formulas of wave mechanics by a computer, the density of the dots indicates the intensity of the wave; thus, the wave is strongest near the center (near the nucleus of the atom), and gradually fades with increasing distance from the center. The intensity of the wave at any point is proportional to the probability for finding the electron at this point; thus, in the ground state of the hydrogen atom, the most probable position for the electron is right at the nucleus. Because of this probabilistic interpretation, the pictures of wave intensity shown in Fig. 39.3 are often called probability clouds. Waves with other values of n, l, and m correspond to the excited states; all these waves are more complicated than the wave for the ground state, especially if n, l, and m are large. Figures 39.3b–d give some examples of waves corresponding to some excited states. In addition to these quantum numbers n, l, and m, one more quantum number is needed for the complete characterization of the stationary states of an electron in the hydrogen atom. This is the spin quantum number ms that characterizes the spin, or intrinsic angular momentum, of the electron. The quantum number ms was originally proposed by Wolfgang Pauli in an attempt to describe the “hyperfine” structure of the spectral lines: when the spectral lines of hydrogen and of other atoms are examined with a spectroscope of high resolving power, they are often found to consist of pairs, or doublets, of very closely spaced lines. This implies that one or both of the atomic energy levels involved in the transition must be a closely spaced pair of energy levels, and it implies that, besides n, l, and m, there must be another quantum number that distinguishes between the two energy levels in the pair. Pauli assigned the values ms  12 and ms  12 to the two energy levels in the pair, but offered no explanation of the physical significance of this new quantum number.

For l = 0, probability of finding electron is spherically symmetric and is highest at nucleus.

For many excited states, probability is zero at nucleus and has complicated patterns.

(a)

(b)

(c)

(d)

Ground state, n  1, l  0, m  0.

First excited state, n  2, l  1, m  0.

First excited state, n  2, l  1, m  1 or 1.

Second excited state, n  3, l  2, m  1 or 1.

FIGURE 39.3 Intensities of possible standing waves, or orbitals, in the hydrogen atom. The nucleus is at the center of each picture. The density of the dots is proportional to the intensity of the wave, and thus to the probability of finding the electron, c2.

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39.1

Principal, Orbital, and Magnetic Quantum Numbers; Spin

(a)

1325

(b)

In a simple picture, we imagine electron spinning. spin up

Sense of rotation… spin down

…determines direction of intrinsic spin angular momentum.

FIGURE 39.4 Simple picture of the electron as a ball of charge spinning about its axis. (a) The electron can spin in the counterclockwise direction, seen from above, so the direction of the angular momentum is up. (b) Alternatively, the electron can spin in the clockwise direction, so the direction of the angular momentum is down.

Shortly thereafter, S. Goudsmit and G. Uhlenbeck suggested that the electron has an intrinsic spin angular momentum, and that the two values of ms correspond to different orientations of the axis of spin (see Fig. 39.4). They imagined the electron as a small ball of charge spinning about its axis, like the Earth spinning about its axis, with an angular momentum of magnitude 212 (12  1)U, which means that the intrinsic angular momentum quantum number, or the spin quantum number, is 12. If we use the same rule for the possible directions of the spin as for the possible directions of the orbital angular momentum, we find that there are 2  12  1  2 possible directions for the spin. One of these directions is characterized by a magnetic quantum number ms  12, and the other by a magnetic quantum number ms  12; these two possible spin directions are called spin up and spin down, respectively. Goudsmit and Uhlenbeck’s simple picture of the spin as due to a rotation of the electron about its axis proved untenable. A consistent, relativistic picture of the spin was later contrived by P. A. M. Dirac. Modern wave mechanics tells us that the spin is an angular momentum generated by a circulating energy flow in the electron wave. The orbital angular momentum is also generated by such a circulating energy flow in the electron wave, but the energy flow that generates the spin is distinguished in that it persists even in the electron wave of an isolated, free electron at rest (that is, an electron outside of the hydrogen atom). Although the simple picture of the electron as a rotating ball of charge is false, it can sometimes serve as a convenient crutch for our imagination. According to this simple picture, we expect that the electron has a magnetic moment, since a piece of charge rotating about an axis amounts to a current loop (see Fig. 39.5). The energy difference between the two energy levels that gives rise to closely-spaced pairs of spectral lines (“doublets”, such as seen in the sodium and mercury spectra in the color print on page 1289) is due to this magnetic interaction of the magnetic moment with the magnetic field generated by the motion of the nuclear charge seen in the reference frame of the electron. As already mentioned in Section 30.4, the magnetic moment of the electron also plays an important role in the behavior of ferromagnetic materials.

WOLFGANG PAULI (1900 –1958) Austrian and later Swiss theoretical physicist. For his discovery of the Exclusion Principle, he was awarded the Nobel Prize in 1945. Pauli made an important contribution to the theory of beta decay by proposing that the emission of the beta particle is always accompanied by the emission of a neutrino (see Chapter 40).

For a negative charge, rotation around axis in one direction… rotation spin

…is equivalent to loops of current around axis in opposite direction.

currents magnetic moment

Current loops imply electron has a magnetic moment.

FIGURE 39.5 A ball of negative charge spinning about an axis.

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CHAPTER 39

Quantum Structure of Atoms, Molecules, and Solids

Many other elementary particles besides electrons have spin. For example, the proton and the neutron have spin angular momenta of magnitude 212 (12  1)U (spin quantum number 12), the photon has a spin angular momentum of magnitude 11(1  1)U (spin quantum number 1), etc. The quantum numbers n, l, m, and ms provide a complete characterization of the stationary states of the hydrogen atom. The energies of the stationary state depend mainly on the principal quantum number n. According to Eq. (38.25), the energies are E

PAUL ADRIEN MAURICE DIRAC (1902–1984) British theoretical physicist, professor at Cambridge. He formulated the relativistic wave equation for the electron, incorporating spin, and he used this equation to predict the existence of antielectrons. He received the Nobel Prize in 1933, when antielectrons were confirmed experimentally.

TA B L E 3 9 . 1

QUANTUM NUMBER

13.6 eV

(39.5)

n2

Thus, stationary states of the same value of n have (almost) the same energy, regardless of the values of the other quantum numbers. Classically, this means that the energy depends on the overall size of the elliptical orbit, but not on its elongation, or on its orientation in space. However, a wave-mechanics calculation of the energies of the stationary states of the hydrogen atom shows that the energies depend slightly on the orbital quantum number l and on the orientation of the spin relative to the orbital angular momentum. This means that Eq. (39.5) for the energies of the stationary states of the hydrogen atom is not quite accurate. But the deviations from Eq. (39.5) are very small, and we can often ignore them. Table 39.1 lists the permitted quantum numbers for the stationary states of the hydrogen atom.1 These quantum numbers can also be used to characterize the stationary states of atoms other than hydrogen, but the energies of other atoms are not given by the simple formula (39.5).

QUANTUM NUMBERS OF ELECTRONIC STATES

SYMBOL

VALUES

PHYSICAL QUANTITY

principal

n

1, 2, 3, p

energy

orbital

l

0, 1, 2, p , n  1

magnitude of orbital angular momentum, L  1l (l  1)U

magnetic

m

l, l  1, p , 0, p , l  1, l

z component of orbital angular momentum, Lz  m U

spin

 12,

ms

12

EXAMPLE 2

z component of spin angular momentum, Sz  ; 12 U

(a) What are the permitted quantum numbers for the case n  1? (b) What are the possible quantum numbers for the case n  2?

SOLUTION: (a) For n  1, Table 39.1 tells us that the only permitted value of l

is 0. Furthermore, if l  0, then the only permitted value of m is also 0. The

1

Strictly, these quantum numbers are appropriate for a hydrogen atom placed in a magnetic field. For a hydrogen atom by itself, the appropriate quantum numbers are n and some intricate combinations of l, m, and ms. For the sake of simplicity, we will ignore this complication.

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39.1

Principal, Orbital, and Magnetic Quantum Numbers; Spin

permitted values of ms are 12 and 12, regardless of the values of the other quantum numbers. Hence, for n  1, the list of permitted quantum numbers is n  1;

l  0;

m  0;

ms  12

or n  1;

l  0;

m  0;

ms  12

(b) For n  2, Table 39.1 tells us that the permitted values of l are 0 and 1. If l  0, then the only permitted value of m is 0, and we find the same list as in part (a): n  2;

l  0;

m  0;

ms  12

or n  2;

l  0;

m  0;

ms  12

However, if l  1, then the permitted values of m are 1, 0, and 1, and we find the following list: n  2;

l  1;

m  1;

ms  12

n  2;

l  1;

m  1;

ms  12

n  2;

l  1;

m  0;

ms  12

n  2;

l  1;

m  0;

ms  12

n  2;

l  1;

m  1;

ms  12

or n  2;

l  1;

m  1;

ms  12

Each alternative listed here represents a possible electronic state.

✔

Checkup 39.1

Consider the ground state of the hydrogen atom. According to wave mechanics, what is the magnitude of the orbital angular momentum? According to Bohr’s theory? According to wave mechanics, what is the value of the quantum number n? According to Bohr’s theory? QUESTION 2: If the orbital quantum number is l  3, what is the magnitude of the angular momentum? QUESTION 3: If the principal quantum number is n  3, what are the possible values of the orbital angular-momentum quantum number? QUESTION 4: Figures 39.3b, c, and d give the probability clouds for some excited states of the hydrogen atom. According to these figures, what are the probabilities for finding the electron at the center (at the nucleus)? QUESTION 5: If the orbital angular-momentum quantum number is l  1, what are all the possible values of the magnetic quantum number m? QUESTION 1:

(A) 0 (B) 0 or 1 (C) 0, 1, or 2 (D) 12 or  12

(E) 1, 0, or 1

1327

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Online Concept Tutorial

39.2 THE EXCLUSION PRINCIPLE AND T H E S T R U C T U R E O F AT O M S

44

atomic number Z

periodic table

Pauli Exclusion Principle

Quantum Structure of Atoms, Molecules, and Solids

The paramount question in atomic structure is the determination of the detailed arrangement of the electrons in their orbital states around the nucleus. An atom of atomic number Z has Z electrons in orbitals around a nucleus of positive charge Ze. The electron arrangement, or configuration, determines all the physical and chemical properties of the atom—if the electron configuration is known, all the properties of the atom can be deduced by theoretical considerations. For instance, the observed similarities of chemical properties among select groups of elements are due to similarities in their electronic configurations. Chemists list similar elements in columns in the periodic table of the elements (see Table 39.2). Thus, the elements helium, neon, argon, krypton, etc., are listed in the last column; these are the noble gases, or inert gases, which are practically unable to react chemically with anything at all. The elements fluorine, chlorine, bromine, iodine, etc., are listed in the column adjacent to the noble gases; these are the halogens, irritating, corrosive substances, all with quite noticeable and distinctive colors (pale yellow, greenish yellow, red, and blue violet, respectively). And the elements lithium, sodium, potassium, rubidium, etc., are listed in the first column; these are the alkalis, silvery white metals, which are extremely reactive. The pattern of the elements displayed in the periodic table can be explained by a study of the electron configurations. For the case of the hydrogen atom, the determination of the electron configuration is trivial: the single electron of this atom is in one or another of the stationary states characterized by the quantum numbers n, l, m, and ms. If the atom is in the ground state, the values of the quantum numbers of the electron configuration are n  1, l  0, m  0, and ms  ; 12; thus, everything is fixed, except the direction of the spin, which can be up or down. But for atoms with several electrons, the determination of the electron configuration is not so trivial. It might be tempting to suppose that the ground state of the atom (the state of least energy) is attained by placing all the electrons in the lowest stationary state, with n  1, l  0, m  0, ms  ; 12, as for the hydrogen atom. But this would imply that all the atoms ought to have a spectral series similar to that of hydrogen, and it would also imply that atoms with a large number of electrons, or with a large atomic number Z, ought to be very small, since the Bohr radius for an atom of nuclear charge Ze is a0>Z [if the nuclear charge is Ze instead of e, then in the denominator of the formula (38.17) for the Bohr radius a0 of hydrogen, we must replace one of the factors of e by Ze]. These conclusions are in stark conflict with the observed properties of atoms: the spectra of most atoms are quite different, and the sizes of atoms of large Z—such as lead or bismuth—are considerably larger than hydrogen, which is the smallest of all atoms. The rule that governs the configuration of the electrons in an atom is the Pauli Exclusion Principle, which was discovered by Wolfgang Pauli and is often known simply as the Exclusion Principle. For atoms, the Exclusion Principle asserts Each stationary state of quantum numbers n, l, m, and ms can be occupied by no more than one electron. Since for each stationary orbital state of quantum numbers n, l, and m there are two possible spin states (ms  ; 12), we can also rephrase the Exclusion Principle as follows: each stationary orbital state of quantum numbers n, l, and m can be occupied by no more than two electrons.

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The Exclusion Principle and the Structure of Atoms

1329

THE PERIODIC TABLE OF THE CHEMICAL ELEMENTS a

TA B L E 3 9 . 2

IA (1) 1

2

Periods

3

1 H

VIIIA (18)

1.00794

IIA (2)

3 Li

4 Be

6.941

9.012182

11 Na

12 Mg

22.98977 24.3050

4

19 K 39.0983

5

6

20 Ca

Group designation Atomic number Symbol for element Atomic mass VIIIB IIIB (3)

IVB (4)

VB (5)

VIB (6)

VIIB (7)

(8)

(9)

21 Sc

22 Ti

23 V

24 Cr

25 Mn

26 Fe

27 Co

40.078 44.955910 47.867

51.9961 54.938049 55.845

VIA (16)

VIIA (17)

4.002602

5 B

6 C

7 N

8 O

9 F

10 Ne

10.811

12.0107

14.0067

13 Al

14 Si

15 P

72.64

49 In

50 Sn

114.818

118.710

121.760

81 Tl

82 Pb

83 Bi

204.3833

207.2

91.224

92.90638

95.94

98.9072

72 Hf

73 Ta

74 W

75 Re

76 Os

77 Ir

178.49

180.9479

183.84

186.207

190.23

192.217

104 Rf

105 Db

106 Sg

107 Bh

108 Hs

109 Mt

110 Ds

111 Uuu

112 Uub

114 Uuq

262.12

265.1306

(268)

(271)

(272)

(285)

(289)

60 Nd

61 Pm

62 Sm

63 Eu

64 Gd

65 Tb

144.9127

150.36

151.964

93 Np

94 Pu

95 Am

†Actinides

54 Xe

69.723

48 Cd

57 *La

92 U

83.798

53 I

65.409

47 Ag

88.90585

91 Pa

79.904

52 Te

63.546

56 Ba

90 Th

78.96

51 Sb

33 As

87.62

140.116 140.90765 144.24

74.92160

32 Ge

55 Cs

59 Pr

36 Kr

31 Ga

46 Pd

26.98154 28.0855 30.97376

85.4678

58 Ce

39.948

35 Br

30 Zn

43 Tc

*Lanthanides

35.453

34 Se

29 Cu

42 Mo

223.0197 226.0277 227.0277 261.1089 262.1144 263.118

32.065

28 Ni

41 Nb

89 †Ac

18 Ar

(10)

40 Zr

88 Ra

17 Cl

IIB (12)

39 Y

101.07 102.90550 106.42

78 Pt

107.8682 112.411

79 Au

80 Hg

195.078 196.96654 200.59

66 Dy

67 Ho

15.9994 18.99840 20.1797

16 S

IB (11)

58.93320 58.6934

45 Rh

VA (15)

38 Sr

87 Fr

44 Ru

IVA (14)

37 Rb

132.90545 137.327 138.9055

7

50.9415

127.60 126.90447 131.293

84 Po

85 At

97 Bk

98 Cf

99 Es

86 Rn

208.98037 208.9824 209.9871 222.0176

68 Er

69 Tm

70 Yb

157.25 158.92534 162.50 164.93032 167.26 168.93421 173.04

96 Cm

2 He

IIIA (13)

100 Fm

101 Md

102 No

71 Lu 174.967

103 Lr

232.0381 231.0359 238.0289 237.0482 244.0642 243.0614 247.07003 247.0703 251.0796 252.083 257.0951 258.0984 259.1011 262.110

a In each box, the upper number is the atomic number. The lower number is the atomic mass, that is, the mass (in grams) of one mole or, equivalently, the mass (in atomic mass units) of one atom. Numbers in parentheses denote the atomic mass of the most stable or best-known isotope of the element; all other numbers represent the average mass of a mixture of several isotopes as found in naturally occuring samples of the element.

Pauli originally proposed this principle as an empirical rule, based on the observed features of atomic spectra. It was later established that the Exclusion Principle is intimately linked to the value of the spin of the electron; the Exclusion Principle can be shown to be a necessary consequence of the quantum theory of particles of half-integer spin. Thus, protons and neutrons also obey the Exclusion Principle, a fact of great importance for the configuration of these particles in the interior of the nucleus (see the next chapter). In contrast, particles of integer spin, such as photons, do not obey the Exclusion Principle. There is no limit to the number of such particles that can be packed into a given stationary state, for instance, one of the standing-wave states in a cavity filled with blackbody radiation. For our investigation of the electron configuration of atoms, we will find it convenient to start with a list of all the available states, in order of increasing energy (see Example 2). The states of lowest energy have n  1; there are two such states:

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TA B L E 3 9 . 3

K shell (states with n  1)

n1

l0

m0

ms  12

n1

l0

m0

ms  12

Next, consider n  2; there are eight available states: TA B L E 3 9 . 4

L shell (states with n  2)

n2

l0

m0

ms  12

n2

l0

m0

ms  12

n2

l1

m  1

ms  12

n2

l1

m  1

ms  12

n2

l1

m0

ms  12

n2

l1

m0

ms  12

n2

l1

m  1

ms  12

n2

l1

m  1

ms  12

Likewise, for n  3, there are 18 available states, and so on. The groups of states of a given value of n are called shells, and they are conventionally labeled with the letters K, L, M, etc., from the innermost to the outermost shell. Thus, the two states with n  1 form the K shell, the eight states with n  2 form the L shell, the eighteen states with n  3 form the M shell, etc. In the simple Bohr theory, these groups of states were called shells because the sizes of the orbits in each group are similar; thus, these groups of orbits form layers around the nucleus, like the layers of an onion. But in wave mechanics there is no such simple way to visualize the shells. According to the Exclusion Principle, each of the states listed above can accommodate one, and only one, electron. Thus, if an atom with Z electrons is in its ground state, the electrons will occupy the first Z of the states in the above list. We can therefore build up the configurations for all the atoms in the periodic table of elements by beginning with hydrogen and adding electrons one by one, sequentially filling the states in our list. The second element in the periodic table is helium, which has two electrons. To obtain its electron configuration, we must add one electron to the hydrogen configuration; since the single electron of hydrogen occupies one of the states of the K shell, we can place the second electron in the other available state in the K shell. Helium therefore has a full K shell. The third element is lithium, with three electrons. When we add one electron to the helium configuration, we must place this third electron in the L shell, with n  2. Detailed calculations show that in multielectron atoms, the l  0 states are slightly lower in energy than the l  1 states, so this third electron is in one of the two states with quantum numbers n  2, l  0, and m  0. The next element is beryllium, with four electrons. Thus, we must add one more electron in the L shell, in the other available state with quantum numbers n  2, l  0, and m  0.

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39.2

The Exclusion Principle and the Structure of Atoms

We can continue in this way, filling up the states in our list one by one. With the tenth element, neon, we will have filled the L shell. And with the eleventh element, sodium, we must place one electron in the M shell, and so on. This simple procedure for building up the electron configurations of the atoms provides us with an immediate explanation of the similarities of the elements in a column of the periodic table. For instance, the similarity in the chemical behavior and the similarity in the spectra of helium and neon can be traced to a similarity of their electron configurations: both these atoms have full shells of electrons. The full-shell electron configuration is quite stable, tending neither to capture nor to lose an electron; these and other noble gases are chemically inert. Likewise, the chemical and spectroscopic similarities of hydrogen, lithium, and sodium can be traced to a similarity of their electron configurations: they all have a single electron outside of a full shell of electrons. This single, outer electron tends to come off the atom fairly easily, and in chemical reactions these atoms all tend to lose an electron. In contrast, fluorine and chlorine are one electron short of a full shell, and in chemical reactions they tend to capture an electron to complete their shell. Thus, the Exclusion Principle in conjunction with a simple counting procedure for the available stationary states is sufficient to explain the broad, qualitative features of the periodic table of elements. Detailed calculations, based on wave mechanics, provide quantitative theoretical results for ionization energies, spectral lines, atomic sizes, etc., in agreement with the observed atomic properties. For atoms other than hydrogen there is no simple formula for the energies of the stationary states. However, in an atom of fairly large atomic number, say, Z 20, the dominant force on the innermost electrons is the attractive Coulomb force exerted by the positive charge Ze of the nucleus, and the repulsive forces exerted by the other electrons can mostly be neglected. Thus, these innermost electrons have hydrogenlike orbitals. The energy of a single electron in such an orbital is given approximately by Eq. (38.23), with one modification: the product e  e of the electron and the proton charge must be replaced by the product e  Ze of the electron and the nuclear charge; hence e4 in Eq. (38.23) must be replaced by e4Z2, which leads to the following approximate formula for the energy: En  

meZ2e4 2 2

1 2

2(4p0) U n



Z2  13.6 eV n2

(39.6)

From this formula, we can estimate the frequency and the wavelength of light emitted during a quantum jump from some initial state to a final state. However, such a quantum jump between the innermost orbitals of the atom is not possible if the atom has its full complement of electrons—all the orbitals are then occupied by electrons, and the Exclusion Principle forbids quantum jumps into an already occupied orbital.Thus, a jump is possible only if some external disturbance first removes one of the electrons from the atom, leaving a gap into which some other electron can jump. Such a process occurs when the target atoms in an X-ray tube are subjected to the impact of the electron beam. The target atoms are disturbed by this impact, and sometimes an electron in one of the innermost orbitals is ejected, leaving a gap into which another electron can jump. The photon emitted when an electron jumps into the vacant innermost orbital has a very short wavelength; it is an X ray. Thus, the quantum jumps of the innermost electrons of atoms give rise to the characteristic spectrum of X rays mentioned in Section 37.5. These characteristic X rays correspond to the two sharp peaks in Fig. 37.19. When one of the electrons in the innermost orbital is removed, the one remaining n  1 electron “shields” a part of the nuclear charge from the view of the electron making the quantum jump. For a nearby electron in an n  2 state jumping into a

1331

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single available n  1 state, the effect of the shielding electron is to reduce the effective nuclear charge by one elementary charge. Thus, we can obtain somewhat better estimates of the energies of these states by replacing the nuclear charge Z in Eq. (39.6) by the shielded charge Z  1: En  

me(Z  1)2e4 1 2(4p0)2 U 2 n2



(Z  1)2  13.6 eV n2

(39.7)

Henry Moseley used this predictable behavior of the transitions to the K shell to correct the positions of several heavy metals in the periodic table; these were previously positioned by mass, not by the atomic number Z.

Suppose that in an atom of copper (Z  29) in the target of an X-ray tube, one of the electrons in the n  1 state is ejected during the impact of the electron beam on the target. Suppose that subsequently one of the other electrons in the atom jumps from the n  2 state into this available empty n  1 state. What are the energy and the wavelength of the photon emitted during this quantum jump? EXAMPLE 3

HENRY MOSELEY (1887–1915) English physicist, lecturer at Manchester, where he worked under Rutherford. Moseley was a skillful experimenter, and his brilliant investigations of the characteristics of spectral lines of X rays led to firm assignments of atomic numbers for chemical elements. He was killed in action in the futile Gallipoli campaign, at age 28.

SOLUTION: The initial energy of the electron is

E2  

(Z  1)2  13.6 eV 22



(29  1)2  13.6 eV   2.67  103 eV 4

and the final energy is E1  

(Z  1)2  13.6 eV 12



(29  1)2  13.6 eV   1.066  104 eV 1

Hence the energy of the photon is E2  E1  8.00  103 eV  1.28  1015 J The frequency of the photon is f 

1.28  1015 J 1.28  1015 J   1.93  1018 Hz 34 h 6.63  10 J s

and the wavelength is l

c 3.00  108 m/s   1.55  1010 m  0.155 nm f 1.93  1018 Hz

COMMENTS: This agrees well with the experimentally determined wavelength,

l  0.154 18 nm. This characteristic X ray is known as copper K radiation, where the  refers to the dominant transition, here to the K shell from the n  2 state. In engineering and materials research, diffractometers often use X rays of this wavelength, somewhat smaller than atomic spacings, to determine crystal structures.

✔

Checkup 39.2

Consider the K, L, and M shells. Which of these has the largest number of states? Which the least?

QUESTION 1:

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39.3

Energy Levels in Molecules

1333

When a shell is full, does it always have equal numbers of electrons of spin up and spin down? QUESTION 3: The ionization energy for the hydrogen atom is 13.6 eV. What is the energy required for the second ionization of a helium atom (Z  2), that is, the energy for the removal of the second electron (if the first electron has already been removed previously)? (A) 1  13.6 eV (B) 2  13.6 eV (C) 3  13.6 eV (D) 4  13.6 eV (E) 9  13.6 eV QUESTION 2:

39.3 ENERGY LEVELS IN MOLECULES The chemical bonds that bind two or more atoms together in a molecule, such as O2 or NaCl, arise from a rearrangement of the outer electrons, or valence electrons, of the atoms. In some molecules (e.g., O2), the outer electrons are shared between the atoms, a sharing that produces an attractive force (covalent bond). In some other molecules (e.g., NaCl), one atom loses an electron to the other atom, and the atom with the missing electron is then electrically attracted by the atom with the extra electron (ionic bond). Hydrogen atoms are especially susceptible to losing electrons to neighboring atoms. When a hydrogen atom is located between two other atoms, it often loses its electron to these neighbors, and the residual proton of the hydrogen atom then electrically attracts the neighboring atoms, holding them together (hydrogen bond). The chemical bonds are elastic—they behave rather like springs tying the atoms together. The spring holds the atoms at an average equilibrium distance, but permits the atoms to oscillate back and forth about this average distance, with some kinetic and potential energy. This means that the energy of the molecule is the sum of the electronic energy of the atoms and the vibrational energy of the motion of the atoms in relation to each other. We will first focus on the vibrational energy of a molecule and examine its quantization. The mass of the atom is concentrated in its center, in the nucleus, which is much smaller than the interatomic distances in a molecule. We can therefore schematically represent a molecule—for instance, a diatomic molecule—as a system of pointlike masses connected by a massless spring, which represents the bond between the two atoms (see Fig. 39.6). The atoms oscillate in unison relative to the center of mass, which we can regard as fixed. Thus, the system is an oscillator, and the energy of this oscillator is subject to Planck’s quantization condition, Eq. (37.3). If the frequency of oscillation is f, the energy of the vibration is quantized according to 2 E  n hf

n  0, 1, 2, p

(39.8)

The corresponding energy-level diagram is shown in Fig. 39.7. The molecule will emit a photon if it makes a transition from an upper level to a lower level. The vibrational transitions in a molecule are restricted by a selection rule: the transition must proceed from one level to the next, that is, transitions spanning two or more levels in one jump are forbidden.

2

More precisely, the energies are E  (n  12 )hf. However, the additional 12 hf, the zero-point energy, while important for the calculation of some quantities, does not affect any of the transitions or other properties discussed here.

In a diatomic molecule, we represent nuclei as two pointlike masses…

…and the chemical bond as a massless spring.

FIGURE 39.6 Representation of an oscillating diatomic molecule.

vibrational energy of molecule

selection rule

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This selection rule can be deduced from wave mechanics. The arrows in Fig. 39.7 indicate the permitted transitions. The frequencies of the spectral lines emitted during all these transitions are therefore the same,

For vibrational frequency f, energy levels are evenly spaced with E  hf… E 4hf

n4

3hf

n3

2hf

n2

hf

n1 n0

…and only transitions between adjacent levels are allowed, so all emitted photons have f light f.

FIGURE 39.7 Energy-level diagram for the oscillating molecule. The arrows indicate the possible transitions.

rotational energy of molecule

flight 

hf ¢E  f h h

(39.9)

Thus, the frequency of the spectral lines equals the frequency of vibration of the molecule. Typically, the frequencies of vibration of molecules are of the order of 1013 Hz, and the wavelengths of the emitted spectral lines lie in the infrared. Besides the vibrational motion, the molecule can also perform rotational motion. For the purposes of this rotational motion, we can regard the molecule as two pointlike masses linked by a massless rod, that is, a dumbbell (see Fig. 39.8). If the moment of inertia of the dumbbell about a perpendicular axis through the center of mass is I, then the kinetic energy of rotation is E  12I2

(39.10)

where  is the angular frequency of the rotation. Let us express this in terms of the angular momentum. The magnitude of the angular momentum is I, and this is quantized in the usual way [see Eq. (39.1)], so we can write Iv  1J ( J  1)U

J  0, 1, 2, p

(39.11)

where J is called the rotational quantum number. Solving Eq. (39.11) for  and substituting this into Eq. (39.10) implies that the energy of the rotational motion is quantized, that is, E

J ( J  1)U 2 2I

J  0, 1, 2, p

(39.12)

Figure 39.9 displays the energy-level diagram for the rotational states of a molecule. The transitions are, again, subject to the selection rule that they must proceed from one level to the next. Such transitions are indicated by the arrows in Fig. 39.9. Note that here we considered only rotation about a perpendicular axis through the center of mass. For rotation about the axis through the atoms, the moment of inertia of a diatomic moelcule is many orders of magnitude smaller, since the mass is con-

Axis of rotation is through center of mass…

…and is perpendicular to line joining two atoms.

FIGURE 39.8 A rotating diatomic molecule can be regarded as two pointlike masses linked by a massless rigid rod.

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39.3

Energy Levels in Molecules

centrated in the nuclei. Equation (39.12) tells us that for such small values of I, the energy levels will be very high, too high for practical consequences.

The moment of inertia of the HCl molecule about its axis of rotation is 2.66  1047 kg m2. What is the energy of the first excited rotational state of this molecule? The second excited rotational state? EXAMPLE 4

1335

Spacing between adjacent rotational energy levels varies in proportion to J of upper level. E 102 I

J4

62 I

J3

32 I

J2

2 I

J1

SOLUTION: For the first excited state, J  1 and Eq. (39.12) gives

E

(1.05  1034 J s)2 2U 2   4.14  1022 J 47 2 2I 2.66  10 kg m

For the second excited state, J  2 and E

3  (1.05  1034 J s)2 2  3  U2   1.24  1021 J 2I 2.66  1047 kg m2

The rotational transitions in a molecule involve much smaller energies than the electronic transitions in an atom, such as the electronic transitions in a hydrogen atom. This means that the photons emitted in purely rotational molecular transitions are of rather low energy, and their wavelengths lie in the far infrared region of the spectrum. However, rotational molecular transitions are often observed in conjunction with a simultaneous electronic transition in one of the atoms of the molecule or with a vibrational transition of the molecule. This increases the energy of the transition and reduces the wavelength of the spectral line. By inspection of Fig. 39.9 we see that successive rotational transitions have slightly different energies and wavelengths; thus, they give rise to a group, or sequence, of adjacent spectral lines. This is called a spectral band. Figure 39.10 is a photograph of several spectral bands in the spectrum of the N2 (nitrogen) molecule.

Successive rotational transitions give rise to sequences of spectral lines, or spectral bands.

FIGURE 39.10 Several bands of spectral lines emitted by the N2 molecule.

✔

Checkup 39.3

For a vibrational transition in a molecule, the frequency of the radiation equals the frequency of vibration of the molecule. Is it likewise true that for a rotational transition in a molecule, the frequency of the radiation equals the frequency of the rotational motion? QUESTION 2: Suppose that a molecule is initially in the J  4 state in Fig. 39.9. If it makes the sequence of downward transitions indicated by the arrows in this figure, how many photons does it emit? Which of these photons has the most energy? The least energy? Which has the longest wavelength? The shortest? QUESTION 1:

J0

0

Only transitions between adjacent levels are allowed.

FIGURE 39.9 Energy-level diagram for the rotating molecule. The arrows indicate the possible transitions.

spectral band

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QUESTION 3: The moment of inertia of the deuterium molecule (D2) is twice as large

as that of the hydrogen molecule (H2). Which has a larger energy difference between the ground state and the first excited rotational state? By what factor? Q U E S T I O N 4 : The frequencies of the photons emitted during the four transitions shown in Fig. 39.9 are (in units of U I ) (A) 1, 2, 3, 4 (B) 0, 1, 3, 6 (C) 1, 3, 6, 10 (D) 4, 7, 9, 10

39.4 ENERGY BANDS IN SOLIDS

Near each positive ion…

U

+

+

+

+

x

…potential energy of electron dips, indicating an attractive force.

FIGURE 39.11 Potential energy of an electron in a crystal. The plus signs mark the positions of the atoms along a row in the crystal.

As discussed in Section 22.5 in a metal the outermost, or valence, electrons of the atoms are detached from their atoms, and they are free to wander all over the volume of the metal. However, whenever such a “free” electron passes near one of the positively charged atoms of the metal, it experiences an attractive force. Figure 39.11 is a rough sketch of the potential energy for an electron moving along a row of atoms in the crystal lattice of a metal; the dips in the potential energy indicate the attractive force. While the electron moves along this row of atoms, the attractive force acts repetitively, each time the electron passes near an atom. Under suitable conditions, such a repetitive action of a force can lead to a large cumulative effect. According to wave mechanics, we have to think of the electron as a wave, and an encounter with an atom scatters the wave: some fraction of the wave proceeds in its original direction of motion and some fraction is reflected. The repetitive scatterings at the atoms in the row will build up a large reflected wave by constructive interference if the scattering by one atom produces a reflected wave that is in phase with the reflected wave produced by the next atom. This will happen if the extra distance for a round trip from one atom to the next and back is equal to one de Broglie wavelength or an integer multiple of one wavelength. Designating the distance between the atoms by a, we can express the condition for constructive interference of all the reflected waves as 2a  l, 2l, 3l, p

(39.13)

Since the de Broglie wavelength is related to the momentum of the electron by l  h p [see Eq. (38.32)], we can express the condition for total reflection in terms of the momentum of the electron: h 2h 3h , , p 2a  , p p p

(39.14)

h 2h 3h , , , p 2a 2a 2a

(39.15)

or p

Concepts in Context

If this condition is satisfied, the reflected wave will gain more and more strength at each reflection at each atom, and finally match the strength of the incident wave. The result is a standing wave, which travels neither right nor left. The energy of such a standing wave is strongly affected by the interaction of the electron with the atoms of the lattice. The energy of the standing wave will be low if the peaks of intensity of the wave coincide with the locations of the atoms (see Fig. 39.12a), because the electron then has a large probability for being found near an atom, where the potential energy is low. The energy of the standing wave will be high if the peaks of intensity of the wave fall between the locations of the atoms (see Fig. 39.12b), because the electron then has a large probability of being found between one atom and the next, where the potential energy is high. Thus, the wavefunction

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39.4

Energy Bands in Solids

depicted in Fig. 39.12a corresponds to a low-energy state of the electron, and that in Fig. 39.12b corresponds to a high-energy state. The transition from one of these states to the other involves a quantum jump of the electron. To appreciate the implications of such a quantum jump, suppose we take an electron of low initial momentum and we gradually increase its energy and momentum. At first, the energy of the electron will increase smoothly with the momentum. But at the critical value p  h 2a given by Eq. (39.15), the energy has to increase by a step that represents the energy difference between the low-energy standing wave and the highenergy standing wave. Beyond this critical value, the energy again increases smoothly with the momentum, until we reach the second critical value p  2h2a, where there is another step in the energy, etc. From this discussion we see that the permitted energies of a free electron in a crystal lattice occur in several permitted intervals, within which the energy increases smoothly with momentum. These permitted intervals are separated by forbidden intervals, where the electron energy increases by a step. Figure 39.13 shows such permitted and forbidden energy intervals on an energy-level diagram. The permitted intervals, shown with lines in the diagram, are called energy bands; the forbidden intervals are called energy gaps. The precise widths of the permitted energy bands and the forbidden energy gaps depend on the details of the crystal lattice, but all crystals with “free” electrons have some kind of band pattern in their energy-level diagram. As in the case of the electron configuration of atoms, we can deduce the electron configuration of crystals by means of the Exclusion Principle. In a crystal in its ground state, the electrons settle in the available states of lowest energy. To discover the electron configuration, we proceed as before: we take all the “free” electrons and pack them, one by one, into the available energy bands. The lowest energy bands will then be completely filled, but the upper energy band will be either filled or partially filled, depending on the number of “free” electrons and the number of available states. The differences among the electric properties of conductors, semiconductors, and insulators arise from the partial or complete filling of the energy bands. A conductor, such as copper or silver, has a band partially filled with electrons (see Fig. 39.14). When the electrons in this partially filled band are subjected to an electric field, they absorb energy from the field, and they make transitions to some of the slightly higher, empty states of the band. Thus, the electrons respond to the electric field, and they begin to carry an electric current. Permitted energies of an electron in a crystal occur throughout intervals called energy bands…

1337

A high probability of finding an electron near positive ions intensity implies a low electron energy…

(a)

x

0 (b)

intensity

x

0

…and a high probability of finding an electron between atoms implies a high electron energy.

FIGURE 39.12 Standing electron waves in a crystal lattice. The diagrams show the intensity of the wave, which is proportional to the probability of finding the electron, c2. (a) This standing wave has its peaks of intensity at the locations of the atoms (red dots). (b) This standing wave has its peaks of intensity midway between the locations of the atoms.

In a conductor, empty states are only very slightly higher in energy… E

E …that are separated by forbidden intervals called energy gaps. gap

gap

partially filled band …than full states, so electrons can easily make transitions and carry current.

gap

FIGURE 39.13 Energy-level diagram for an electron moving in a crystal.

FIGURE 39.14 In a conductor, an upper energy band is only partially filled with electrons. The portion of the band filled with electrons is shown in blue.

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E

conduction band In an insulator, empty states are separated by a large energy gap…

 6 eV

FIGURE 39.15 In an insulator, the lower energy bands are completely filled with electrons. The gap between the uppermost filled band and the next, empty band is fairly large.

semiconductor

valence band conduction band

n-type semiconductor

p-type semiconductor

valence band

…from full states, so electrons cannot readily make transitions, and cannot carry current.

In an insulator, such as diamond, the lower bands are completely filled with electrons (see Fig. 39.15). When the electrons in a full band are subjected to an electric field, they cannot make transitions into other states in the band, because all of these states are already full, and the Exclusion Principle forbids transitions into already full states. This implies that the electrons in the full band cannot respond to the electric field; they therefore cannot accelerate in the direction of the electric force, and they cannot begin to carry a current. The only way the electrons in the full band could respond to the electric field is by making transitions to a higher, empty energy band; but to do this without external help requires that an electron acquire a large amount of energy from random thermal fluctuations, which is difficult, since the typical energy of random thermal fluctuations is small compared with the energy gap at room temperature. In a semiconductor, such as silicon, germanium, or gallium arsenide, the bands are completely filled with electrons, as in an insulator. However, the energy gap between the last full band and the next, empty band is much smaller than in an insulator (see Fig. 39.16); typically, the width of the energy gap separating the full band from the next, empty band is less than or approximately equal to 1 eV, whereas in an insulator the gap is often 5 eV or more. In a semiconductor at room temperature, the random thermal fluctuations of the energy will permit many of the electrons at the top of the full band to make a transition to the next, empty band. This means these electrons have nearby empty states; they can respond to the electric field, and they can carry a current. The uppermost full band in an insulator or a semiconductor is called the valence band, and the empty band above it is called the conduction band (in a metal, the conduction band is the valence band, and this band is only partially filled). The values of the resistivities of semiconductors are between those of conductors and insulators. The resistivities of semiconductors vary over a wide range; the resistivities may be 104 to 1015 times as large as the resistivities of conductors. Semiconductors fall into two categories: n type and p type. In an n-type semiconductor, the carriers of current are free electrons that have reached the conduction band. Thus, the mechanism for conduction is the same as in a metallic conductor. However, the resistance of a semiconductor is higher than that of a metallic conductor, because the semiconductor has fewer free electrons in its conduction band than a metal in its partially filled upper band. Also, a (pure) semiconductor differs from a metal in that the resistivity decreases as the temperature increases (see also Fig. 27.8). This curious behavior is due to an increase in the number of free electrons—as the temperature increases, more electrons are excited into the conduction band by random thermal fluctuations, and these extra free electrons more than compensate for the extra friction experienced by each at the higher temperature. In a p-type semiconductor, the carriers of current are “holes” of positive charge. This type of semiconductor has a valence band that is almost, but not quite, filled with elec-

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39.4

Energy Bands in Solids

E

1339

When electron is pulled out of right end… (a)

– In a semiconductor, empty states of conduction band are separated by only a small gap… …from full states of valence band…

–

–

–

–

–

–

conduction band …there is a missing electron, or “hole.” (b)

–

 1 eV valence band

FIGURE 39.16 In a semiconductor, the lower energy bands are almost completely filled with electrons, but some electrons reach the conduction band because the gap between the uppermost filled band and the next, empty band is small.

–

–

–

–

–

Next electron jumps into hole… (c)

…so some electrons can be thermally excited, and can then carry current.

trons. Thus, there are missing electrons, or “holes,” in the electron distribution, and if these holes move, they will transport charge. To see how such a transport of charge comes about, consider Fig. 39.17, showing an array of electrons and positive ions. In Fig. 39.17a, these electrons and ions form neutral atoms. Suppose that the right end of this array is connected to the positive pole of a battery (not shown) and the left end to the negative pole. If the battery pulls an electron out of the right end, it will leave the array with a hole, or missing electron, at the position of the last atom (Fig. 39.17b). The electrons will then play a game of musical chairs: the electron from the next-tolast atom will jump into this hole, leaving a hole at the position of the next-to-last atom (Fig. 39.17c); and then the electron from the next atom will jump, etc. The collective motion of the electrons from left to right can be conveniently described as the motion of a hole from right to left. The hole virtually carries positive charge from the right to the left. In essence, this is the mechanism for conduction in a p-type semiconductor. Instead of free electrons, this type of semiconductor has free holes. A flow of current is then a flow of holes, and the direction of the current is the same as the direction of motion of the holes. Semiconductors usually contain both free electrons and free holes. Whether a semiconductor is n type or p type depends on which kind of charge carrier dominates. The concentration of free electrons and of free holes is largely determined by the impurities that are present in the material. Donor impurities consist of atoms that release a valence electron when placed in the semiconductor, and they increase the number of free electrons. Acceptor impurities consist of atoms that trap electrons when placed in the semiconductor, and they thereby generate holes. Hence, a semiconductor with donor impurities will be n type and one with acceptor impurities will be p type. For instance, silicon with arsenic impurities, which have one more valence electron than silicon, is an n-type semiconductor, and silicon with boron impurities, which have one less valence electron than silicon, is a p-type semiconductor. Even though the deliberately added impurity atoms may amount to only a few parts per million, they completely change the conductivity because the semiconductor has so few current carriers to start with. For typical devices at room temperature, such “doping” of silicon with impurities increases the density of free charge carriers, and thus the conductivity, by a factor of 103 to 106 compared with pure silicon.

–

–

–

–

–

–

…and each electron, in turn, moves to right. (d)

–

–

–

–

–

–

Motion of many electrons to right is equivalent to one hole moving to left.

FIGURE 39.17 A row of positive ions (red balls marked ) and electrons (blue dots marked ).

Concepts in Context

donor impurities

acceptor impurities

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✔

Quantum Structure of Atoms, Molecules, and Solids

Checkup 39.4

QUESTION 1: Suppose that a crystalline solid has an energy band that is completely filled and all bands above this filled band are empty. Is this solid a conductor, an insulator, or a semiconductor? What extra information do you need to decide? QUESTION 2: Must conductors always be n type? Can they be p type? QUESTION 3: Suppose we cool a semiconductor to very low temperature, so all electrons settle into the lowest possible states. Will the semiconductor conduct? QUESTION 4: Equation (39.14) says that the forbidden values of the momentum are evenly spaced. Are the forbidden values of the energy (the gaps) also evenly spaced? QUESTION 5: How does p-type silicon differ from n-type silicon? (A) p-type has acceptor atoms and free electrons; n-type has donor atoms and holes. (B) p-type has acceptor atoms and holes; n-type has donor atoms and free electrons. (C) p-type has donor atoms and free electrons; n-type has acceptor atoms and holes. (D) p-type has donor atoms and holes; n-type has acceptor atoms and free electrons.

Concepts in Context

39.5 SEMICONDUCTOR DEVICES The manipulation of the resistivity of semiconductor materials by intentional contamination with carefully selected impurities plays a crucial role in the manufacture of semiconductor devices, such as diodes, transistors, and integrated circuits. It is a characteristic feature of semiconductor materials that the addition of impurities to the material has a drastic effect on the resistivity. For instance, the silicon used in electronic devices is usually contaminated, or “doped,” with small amounts of arsenic or boron; the addition of just one part per million of arsenic will decrease the resistivity of silicon by a factor of more than 105. Pure semiconductor materials are hardly ever used in practical applications. In most cases, the presence of impurities is what gives the semiconductor materials their interesting and useful electric properties.

Rectifier (Diode) A semiconductor rectifier consists of a piece of n-type and a piece of p-type semiconductor joined together. The n-type semiconductor has free electrons, and the ptype semiconductor has free holes; when they are joined, some of the free electrons will wander from the n region into the p region, and some of the holes will wander from the p region into the n region. Wherever the electrons and the holes meet, they annihilate each other—the electron falls into the hole and fills it, which means that both the electron and the hole disappear. This recombination of some electrons and holes leaves residual positive and negative ions near the interface of the two regions, and the electric charges of these ions generate an electric field across the interface (see Fig. 39.18). This electric field opposes any further wandering of holes or electrons from one region into the other. Because of this lack of mobile charge carriers, the interface is known as the depletion region. When such a p–n junction is connected to a battery or some other source of emf, it will permit the flow of current from the p region into the n region, but not in the opposite direction. Figure 39.19 shows the p–n junction connected to the source of emf so the p region is at high potential and the n region at low potential, a configuration

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39.5

Semiconductor Devices

1341

Very near junction, free electrons in n region can fall into holes in p region… negative ion

free electron p type

free hole

n type

positive ion

…leaving a region depleted of free charges.

Without free charges, (fixed) ions produce an electric field.

FIGURE 39.18 Pieces of p-type and n-type semiconductor in contact. The plus and minus signs represent the ions of the lattice. The blue dots represent electrons and the red dots holes. The arrows indicate the electric field generated by the ions at the interface.

called “forward bias.” The source of emf pumps a steady flow of electrons into the n region and it removes electrons from the p region, which is equivalent to pumping holes into the p region. The electrons and the holes meet at the junction and they recombine. This process can continue indefinitely, and therefore the source of emf can continue to pump current around the circuit indefinitely. Figure 39.20 is a plot of the current vs. the voltage applied to the p–n junction. The current increases steeply with the voltage, because the electric field associated with the applied voltage tends to cancel the internal electric field at the p–n junction, and this makes it easier for the electrons and holes to meet at the center. Evidently, the current is not simply proportional to the voltage, and the p–n junction does not obey Ohm’s Law. Now, consider what happens if the p–n junction is connected to the source of emf so the p region is at low potential and the n region is at high potential, as shown in Fig. 39.21. This configuration is called “reverse bias.” Many of the free electrons in the n region then flow away through the wire on the right, and many of the holes in the p region flow away through the wire on the left. Consequently, the region depleted of charge carriers near the interface becomes wider, and the flow of current stops almost immediately—the p–n junction blocks the current. Current increases steeply…

Under “forward bias,” source of emf pumps holes into p region and electrons into n region… p

Under “reverse bias,” source of emf pulls holes from p region and electrons from n region…

I

p

n

n

I





…which meet and annihilate at junction, so a steady current can flow around circuit.

FIGURE 39.19 A p–n junction connected to a source of emf. The n region is at low potential and the p region at high potential (forward bias).

 0.6 V

V

…with increasing forward bias voltage.

FIGURE 39.20 Plot of current vs. voltage for the p-n junction.



…which widens depletion region at junction, and stops the current.

FIGURE 39.21 A p–n junction connected to a source of emf. The n region is at high potential and the p region at low potential (reverse bias).

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The p–n junction is called a rectifier because it can be used to convert an alternating current into a direct current. If the junction is connected to a source of alternating emf, it will pass current only during the “forward” part of the cycle. The alternating positive and negative emf then yields a periodic sequence of positive current pulses (see Fig. 39.22). Such solid-state rectifiers find many practical applications; for instance, they are used in the “alternators” that generate DC power in the electrical systems of automobiles, and they are found in AC-to-DC power adapters for portable electronics.

When alternating emf is applied to diode, current flows during forward-biased part of cycle… I

0

2␲ ␻

4␲ ␻

Quantum Structure of Atoms, Molecules, and Solids

t

Tr a n s i s t o r ( B i p o l a r Tr a n s i s t o r )

…and current is blocked during reverse-biased part.

FIGURE 39.22 Current passed by the rectifier vs. time. The negative portions of the alternating current are blocked by the rectifier, and only the positive portions remain.

A bipolar transistor consists of a thin piece of semiconductor of one type sandwiched between two pieces of semiconductor of the other type. Figure 39.23 illustrates an n–p–n junction transistor. The thin piece in the middle is called the base, and the pieces at the ends are called the emitter and the collector, respectively. The transistor has three terminals, which are connected to two sources of emf, VB and VC , so the emitter-base junction has a forward bias and the base-collector junction has a reverse bias. In this configuration, the emitter-base junction acts as a diode with forward bias, and it permits the flow of electrons from the emitter into the base. However, the electrons that enter the p region fail to recombine with holes, because this base region is quite thin and contains only a low density of holes, and the electrons pass through it before they have a chance to meet with a hole. The electrons wander to the base-collector junction, and the electric field across this junction (indicated by the longer red arrows in Fig. 39.23) pulls the electrons into the collector. They leave the collector via the terminal connected to its end, and they continue around the external circuit, forming the external collector current IC. Of the electrons that enter the base, a small fraction wander to the terminal connected to the base, and they leave via the wire connected there, forming the external base current IB. The use of the transistor as an amplifier for currents and voltages hinges on the relationship between the collector and the base currents: a small change in the base current IB or the base voltage VB leads to a quite large change in the collector current IC. As in the case of the diode with forward bias, if we increase the base potential VB, we cause a drastic increase in the flow of electrons entering the base from the emitter. But most of these electrons flow straight through to the collector, and only a small fraction flow to the terminal connected to the base. Thus, the result of an increase of base potential is a large increase of collector current, but only a small increase of base current. For a typical transistor, the ratio of the collector current increment and the base current increment is of the order of 100 or 200, and this ratio has a fixed value,

Electrons are injected into base at this forward-biased junction…

FIGURE 39.23 An n–p–n junction transistor. Two sources of emf VB and VC are connected to the base and the collector, respectively.

n

p

n

emitter

base

collector



IB

 VB



 VC

…but most wander into this junction, where electric field pulls them to collector…

IC

…so that IC is much larger than IB.

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39.5

Semiconductor Devices

1343

FIGURE 39.24 Some transistors. FIGURE 39.25 Integrated circuit with many miniature circuit elements.

over a wide range of currents. The ratio of these current increments is called the current gain factor,3 [current gain factor] 

¢IC ¢IB

(39.16)

This means that whenever we change the base current IB by some amount (by adjusting the voltage VB), we will change the collector current by an amount a hundred or so times larger. The transistor amplifies the current—a small input current at the base results in a much larger output current at the collector. For instance, in the “amplifier” in a radio receiver, the weak current picked up by the radio antenna is amplified by sending it into a transistor, as input current at the base. Further amplification can be achieved by connecting several transistors in tandem, so the output of each serves as input for the next. Transistors are used in a wide variety of electronic circuits, to amplify and control currents. Figure 39.24 shows some ordinary transistors. In an integrated circuit, such as shown in Fig. 39.25, many transistors and other circuit elements of extremely small size are built up on a single crystal of silicon. The small transistors are not manufactured by sticking together separate pieces of n- and p-type material, but by diffusing suitable concentrations of acceptor and of donor impurities into different adjoining layers of the silicon crystal.

In forward-biased LED, electron and hole meet at junction and recombine, producing a photon.

Light-Emitting Diode (LED) In principle, a light-emitting diode is simply a p–n junction operated with forward bias. At such a junction, electrons arriving from the n region meet holes arriving from the p region, and they recombine, that is, the electrons fall into the holes. But this jump is a transition of the electron from a state of high energy in the conduction band to a state of lower energy in the valence band, a transition that releases energy. In gallium arsenide and some other semiconducting materials, the released energy takes the form of a photon of visible light (see Fig. 39.26). Thus, the p–n junction emits light when an electric current passes through it. Such light-emitting diodes have many practical applications in luminous displays in the dials of measuring instruments, watches, electronic calculators, clocks, automobile speedometers, and so on. 3

In electronics catalogs and transistor circuits, the symbol  or hfe is usually used for the current gain factor.

p

n

I





FIGURE 39.26 A p–n junction used as a light-emitting diode.

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CHAPTER 39

Solar Cell (Photodiode)

Photons strike junction of solar cell, producing free electrons and holes, …

p

Quantum Structure of Atoms, Molecules, and Solids

n

I

…generating current in external circuit.

FIGURE 39.27 A p–n junction used as a solar cell.

sunlight Top layer is made thin so light can reach junction…

3 ␮m

A solar cell is simply a light-emitting diode operating in reverse. When sunlight is absorbed at the p–n junction, it excites an electron from the valence band to the conduction band, creating a free electron and a free hole (see Fig. 39.27). The electric field at the junction then pulls the electron toward the n region and the hole toward the p region. This means negative charge flows into the n region and from there into the external wire connected on the right; while positive charge flows into the p region and from there into the external wire connected on the left. Thus, sunlight striking the junction generates an electric current in the external circuit. Solar cells and other photodiodes are commonly manufactured out of p-type silicon and n-type silicon. The emf of such a silicon solar cell is only about 0.6 V, and the current it delivers is fairly small; most photodiodes produce a few tenths of an ampere of current per watt of incident light. For the solar cell device geometry shown in Fig. 39.28, with an area of about 4 cm2, the current delivered in full sunlight is about 0.1 A. Many calculators are powered by photodiodes, eliminating the need for batteries. Solar cells are routinely used to generate electric power on communications satellites and other satellites in orbit around the Earth (see Fig. 39.29). Solar cells have been used to generate power to drive experimental vehicles (see Fig. 39.30). Arrays of solar cells are commercially available (Fig. 39.31). Some attempts have also been made to use them to generate fairly large amounts of electric power for residential and industrial use (see Fig. 39.32). For instance, a solar power station at Pocking, Germany generates about 10 megawatts.

n-type silicon

1 mm

 p-type silicon



2 cm …and area of junction is made large so more light can be absorbed.

FIGURE 39.28 Device geometry for solar cell (side view).

FIGURE 39.29 Solar panels on the International Space Station.

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Summary

FIGURE 39.30 A race between experimental vehicles powered by solar cells covering their bodies.

✔

1345

FIGURE 39.31 Panel of solar cells.

Checkup 39.5

If the n region of a p–n junction is at a potential of 3 V and the p region at 5 V, is the bias forward or reverse? QUESTION 2: In the rectifier, the light-emitting diode, and the solar cell, electrons and holes are created or recombined at the p–n junction. Which devices involve creation, which recombination? QUESTION 3: You place a piece of n-type silicon in contact with p-type silicon. What is the direction of the electric field at the interface? (A) Electric field points from n-type region to p-type region. (B) Electric field points from p-type region to n-type region. (C) Electric field points parallel to the plane of the interface. (D) There is no electric field without an external bias. QUESTION 1:

FIGURE 39.32 Solar power station at Carrisa Plains in California.

S U M M A RY QUANTUM NUMBERS OF ATOMIC STATES

n, l, m, and ms (principal, orbital, magnetic, and spin quantum numbers)

n  1, 2, 3, # # # l  0, 1, 2, # # # , n  1 m  l,  l  1, # # # , 0, # # # , l  1, l ms   12,  12

MAGNITUDE OF ANGULAR MOMENTUM

L  1l (l  1)U

(39.1)

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Quantum Structure of Atoms, Molecules, and Solids

Lz  mU

z COMPONENT OF ANGULAR MOMENTUM

z

(39.3)

m  2, Lz  2 L m  1, Lz  

␪

m  0, Lz  0

y

x m  1, L z   m  2, L z  2

ELECTRON SPIN (INTRINSIC ANGULAR MOMENTUM)

212 (12  1)U

Magnitude: z component:

ms U

Spin up:

ms  12

Spin down:

ms  12

PAULI EXCLUSION PRINCIPLE Each stationary state of quantum numbers n, l, m, and ms can be occupied by no more than one electron.

E  n hf

VIBRATIONAL ENERGIES OF MOLECULE

n  0, 1, 2, p

(39.8)

E 4hf

n4

3hf

n3

2hf

n2

hf

n1 n0

ROTATIONAL ENERGIES OF MOLECULE

E

J( J  1)U 2 2I

J  0, 1, 2, p

E 102 I

J4

62 I

J3

32 I

J2

2 I

J1 J0

0

CONDUCTOR

Partially filled conduction band.

E

E

E

INSULATOR Full valence band, empty conduction band, large gap between.

(39.12)

conduction band partially filled band

conduction band  6 eV  1 eV valence band

Full valence band, empty conduction band, small gap between. SEMICONDUCTOR

conductor

insulator

valence band

semiconductor

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Problems

1347

QUESTIONS FOR DISCUSSION 1. According to classical mechanics, what kind of orbit would have zero angular momentum? 2. If there were no Exclusion Principle, what would be the electron configuration of lithium?

9. A spring-loaded butterfly valve in a large water pipe is controlled by a stream of water from a separate pipe (see Fig. 39.33). Is this a reasonable hydraulic analog of a transistor? How could you use such a device to amplify a water current?

3. The bond in the NaCl molecule is similar to the bond in the KBr molecule. Explain this similarity.

IB

4. Rotational transitions in a molecule give a band spectrum, but vibrational transitions do not. Explain. 5. How does n-type silicon differ from p-type?

IC

6. If you dope silicon with phosphorus impurities, will the silicon be n type or p type? 7. What kind of valve in a hydraulic circuit is analogous to a diode rectifier? Draw a picture of a water pipe with such a valve. 8. Do the current IC and the voltage VC in a transistor obey Ohm’s Law?

FIGURE 39.33 The butterfly valve in the large pipe is hinged at the center. The stream of water from the small pipe strikes the upper portion of the butterfly valve and pushes it open.

PROBLEMS †

39.1 Principal, Orbital, and Magnetic Quantum Numbers; Spin 1. According to wave mechanics, what are the possible values of the orbital quantum number l if the principal quantum number is n  1? If n  2? If n  3? 2. Suppose that a state in the hydrogen atom has orbital quantum number l  5. What are the possible values of the magnetic quantum number m? 3. Suppose that the magnitude of the orbital angular momentum vector is 120 U . What are the permitted values of the z component of the angular momentum? 4. Pretend that the electron is a small sphere of uniform density of radius 2.8  1015 m. (a) What is the moment of inertia of the electron according to this model? (b) What angular velocity of rotation is required to give the sphere an angular momentum of magnitude 234 U? What is the corresponding speed of rotation of a point on the equator of the sphere? Does this model make any sense? 5. An electron in an atom is in a state of quantum numbers l  2, m  2, and ms  12. What is the z component of the total angular momentum of the electron? What is the z com-

†

For help, see Online Concept Tutorial 44 at www.wwnorton.com/physics

ponent of the total magnetic moment of the electron? [Hint: Eq. (30.23) relates the z component of the spin magnetic moment, B  z,spin , to the z component of the spin angular momentum, 12 U . The proportionality of the orbital magnetic moment to the orbital angular momentum is half as large.] 6. What are the possible values of the orbital angular momentum of the hydrogen atom in its first excited state? Taking into account the spin, what is the maximum possible value of the z component of the total angular momentum? What is the corresponding value of the z component of the magnetic moment of the atom? (Hint: see Problem 5.) 7. Within a shell of an atom, the groups of states of the same l are called a subshell. For instance, in the L shell, the two states with l  0 form one subshell, and the six states with l  1 form another subshell. (a) How many subshells are there within the M shell? How many states are there in each of these subshells? (b) Neon has a complete l  0 subshell and a complete l  1 subshell in the L shell. What atom has similar subshells in the M shell? 8. The circular orbits of orbital angular momentum nU in Bohr’s theory roughly correspond to the wave-mechanical states of maximum orbital quantum number, that is, l  n  1. If n  2, compare the magnitude of the angular momentum given by Bohr’s theory with the magnitude given by wave mechanics. Repeat for n  4, n  10, and n  500.

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Quantum Structure of Atoms, Molecules, and Solids

9. The r meson is a particle of spin quantum number 1. What is the magnitude of the spin angular momentum of the r meson? What are the possible values of ms for this particle?

according to Bohr theory and the energy calculated according to the improved formula. (b) For n  2, l  1, find the difference, in eV, between the energies of the states of spin parallel and antiparallel to the orbital angular momentum calculated according to the improved formula. Which of these states has the lower energy?

10. In many atoms, the nucleus of the atom has a spin. For example, the nucleus of one of the isotopes of magnesium (25Mg) has spin, with a spin quantum number of 5 2. What is the magnitude of the spin angular momentum of this nucleus? What are the possible values of ms for this nucleus?

*17. For an electron in a state of given orbital (l) and magnetic (m) quantum numbers, the magnitude of the angular momentum and the z component of the angular momentum are well defined, but the x and y components are completely uncertain. Show that nevertheless the sum of the squares of the x and y components is well defined according to the formula

11. Consider the angular-momentum vector with quantum number l  3. What is the smallest possible angle that this angular momentum can make with the z axis? 12. The spin angular momentum of the electron has a magnitude 212 (12  1)U and a magnetic quantum number ms  12 or ms  12. For each of these two values of ms , calculate the angle between the direction of the spin angular momentum vector and the z axis.

(L2x  L2y )  l (l  1)U 2  m2 U 2 With the additional assumption that, on the average, L2x and L2y are equal, show that the rms values of the x and y components of the angular momentum are

13. The electron orbital angular momentum gives rise to a magnetic moment with z component mz,orbital  mBm, where mB  eU 2me is the Bohr magneton and m is the magnetic quantum number. In Chapter 30, we saw that the electron spin angular momentum gives rise to a magnetic moment with z component mz,spin  ; mB. Write an expression that relates mz,spin to the spin quantum number ms. Is this relation different from that for mz,orbital? 14. Consider the angular-momentum quantum numbers for a macroscopic object, such as a toy top with moment of inertia 2.0  104 kg m2 spinning at 25 revolutions per second. The axis of the top makes an angle of 25 with the vertical (with the Earth’s gravitational field). What are approximate values of the angular-momentum quantum numbers l and m?

2L2x  2L2y 

(This is similar to what happens in the case of the velocity components of a molecule of gas in a container. The, say, x component of the velocity is equally likely to be positive or negative, and therefore is completely unpredictable; nevertheless, the rms value 2v2x is well defined.) †

39.2 The Exclusion Principle and the Structure of Atoms 18. How many possible electron states are there in the M shell (n  3) of a hydrogen atom? Make a list of these states, like the lists in Tables 39.3 and 39.4.

15. Similar to the Bohr magneton [see Problem 13 or Eq. (30.23)], a nuclear magneton is defined by mN  eU 2mp, where mp is the mass of a proton. Unlike the electron, however, the z component of the nuclear spin magnetic moment is not related to the nuclear magneton in a simple way (due to the internal structure of nuclei). For example, the z component of the magnetic moment of a proton is 1.41  1026 J T, and that of a 13C nucleus (an isotope of carbon) is 3.55  1027 J/T. Express each of these magnetic moments as a mutiple of a nuclear magneton.

19. List the quantum numbers of all the electrons of a boron atom in its ground state. 20. List the quantum numbers of all the electrons of a carbon atom in its ground state. 21. List the quantum numbers of all the electrons of an Na ion in its ground state. 22. What are the quantum numbers n and l for the outermost electron of the Li atom? The Na atom? The K atom? In what ways are these quantum numbers of these different atoms similar?

*16. As stated in Section 39.1, the energies of the stationary states of the hydrogen atom depend slightly on the orbital angularmomentum quantum number l. An improved formula for the energy of the state of quantum numbers n and l for nonzero l is En,l 

mee4 2 2 2

2(4p0) U n

c1

e4 2 2 2

(4p0) U c n

a

1 l  12 ;

1 2



23. Suppose that the spin quantum number of the electron were 32 instead of 12. How many possible spin directions would the electrons have in this case? What would be the possible permitted quantum numbers for the case n  1 (K shell)? For the case n  2 (L shell)? How many electrons could be placed in the K shell? The L shell? Compare the resulting periodic table of elements with the familiar periodic table in Section 39.2.

3 bd 4n

; 12

where the term corresponds to the spin parallel and antiparallel, respectively, to the orbital angular momentum. (a) For the case of the first excited state, n  2, l  1, and the spin antiparallel to the orbital angular momentum, find the difference, in eV, between the energy calculated

l (l  1)  m2 U B 2

*24. What energy is required to eject one of the electrons from the n  1 state in molybdenum (Z  42) out of the atom? Express your answer in eV. †

For help, see Online Concept Tutorial 44 at www.wwnorton.com/physics

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Problems

*25. Calculate the energy and the wavelength of the X ray emitted during a quantum jump of an electron from the n  3 to the n  1 state in copper (Z  29). *26. An X-ray tube has a tungsten target (Z  74). Calculate the energies and the wavelengths of the X rays emitted in the quantum jumps from the n  2 to the n  1 state and from the n  3 to the n  1 state. *27. For the molybdenum atom (Z  42), the wavelength of the characteristic X ray emitted in the quantum jump from the n  2 to the n  1 state is approximately 0.071 nm. For what atom is the wavelength of the corresponding characteristic X ray twice as large? Half as large? *28. In addition to X rays from copper and molybdenum targets, commercial X-ray diffractometers often provide X rays of wavelengths 0.229 nm, 0.194 nm, or 0.179 nm. Each of these is due to a transition from the n  2 to the n  1 state of a different element. What are the three elements? *29. For transitions of electrons to a vacant L-shell state (n  2) from the higher-energy states of a many-electron atom with nuclear charge Z, the nuclear charge is shielded by both Kshell and L-shell electrons. The effective charge seen by an electron making a transition to the L shell is given by ZL  Z  7.4. What is the wavelength of the X ray emitted when an electron in a gold atom (Z  79) makes a transition from the n  3 state to the n  2 state? *30. After H. G. J. Moseley measured the energy of characteristic K X-ray photons, he plotted the atomic number Z as a function of the inverse of the square root of the photon wavelength, 1 1l. Such a plot is linear. Find (a) the slope and (b) the y intercept (Z-axis intercept) of such a plot. *31. A sample of an unknown element is being used as the target in an X-ray tube. It is found that the characteristic X-ray spectrum displays a strong spectral line at l  0.0228 nm. Assume that this spectral line results from the quantum jump of an electron from the n  2 to the n  1 state. Can you identify the unknown element?

39.3 Energy Levels in Molecules 32. The frequency of vibration of the H2 molecule is 1.31  1014 Hz. What are the energies of the vibrational states? What is the frequency of the emitted radiation? The wavelength? 33. The atoms of deuterium (D) and of hydrogen (H) have the same electron configuration (one electron), but the deuterium is a heavier atom than hydrogen, because it has more mass in its nucleus. The mass of the deuterium atom is 2.014 u, whereas the mass of the hydrogen atom is 1.008 u. Given that the frequency of vibration of the H2 molecule is 1.31  1014 Hz, deduce the frequency of vibration of the D2 molecule. 34. The photon emitted in a vibrational transition of the hydrogen bromide (HBr) molecule has frequency 7.7  1013 Hz.

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What is the spring constant of the chemical bond holding this molecule together? You may assume, for simplicity, that the bromine atom remains essentially at rest, and only the hydrogen atom moves. *35. In the O2 molecule, the distance between the two oxygen nuclei is 0.20 nm. (a) What is the moment of inertia of the molecule for rotation about the perpendicular axis through the center of mass? (b) What are the energies of the first, second, and third excited rotational states? Express these energies in eV. *36. The distance between the K and the Br nuclei in the KBr molecule is 0.282 nm, and the center of mass is at a distance of 0.093 nm from the Br nucleus. (a) What is the moment of inertia of a KBr molecule rotating about its center of mass? (b) What are the energies of the first, second, and third excited rotational states? Express these energies in eV. *37. Consider the HD molecule, where one of the atoms of the hydrogen molecule H2 has been replaced by a deuterium atom. Deuterium and hydrogen have the same electronic structure, but the mass of a deuterium atom is 2.014 u, and that of a hydrogen atom is 1.008 u. Given that the frequency of vibration of the H2 molecule is 1.31  1014 Hz, deduce the frequency of vibration of the HD molecule. [Hint: You may use the result of Problem 41 of Chapter 15, where such an oscillation about the center of mass was shown to have an angular frequency v  1k m, where m is the reduced mass m  m1m2 (m1  m2).] *38. In a vibrational transition, a nitric oxide (NO) molecule emits a photon of frequency 5.6  1013 Hz. What is the spring constant of the chemical bond holding this molecule together? (Hint: See Problem 37.) *39. In Section 19.4, we asserted that quantum mechanics enables us to neglect rotations about an axis through the atoms of a diatomic or other linear molecule. Show this explicitly by calculating the energy of the first rotational excited state of H2 for (a) rotation about the axis through the atoms and (b) rotation about a perpendicular axis. Compare these two energies. For (a), assume the proton is a uniform sphere with radius 1.0  1015 m and the atomic electron is a uniform sphere with radius 5.0  1011 m; for (b), use the interatomic distance of 7.4  1011 m. *40. What are the ratios of the frequencies emitted in rotational transitions in a molecule from the first excited state to the ground state, from the second excited state to the first, from the third to the second, from the nth to the (n  1)th? **41. According to spectroscopic measurements, the energy difference between the first and the second excited rotational states of the N2 molecule is 2.38  1022 J. Deduce the moment of inertia of the molecule. Deduce the center-to-center distance between the N atoms.

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CHAPTER 39

Quantum Structure of Atoms, Molecules, and Solids

39.4 Energy Bands in Solids

n

p

n

42. Consider a crystal with a spacing of 0.10 nm between one atom and the next. What are the de Broglie wavelengths and the energies at which the electron wave cannot propagate through this crystal? Express the energies in electron-volts. 43. In silicon, the energy gap between the valence and the conduction bands is 1.1 eV. If we want to excite an electron from the top of the valence band to the conduction band by means of a photon, what is the maximum permitted wavelength for the photon? *44. The spacing of atoms in the crystal lattice of a metal, and thus the wavelength for maximum repetitive scattering of electron waves by the atoms, varies with the direction of electron wave propagation in the crystal. Consider a simple cubic arrangement of atoms as in Fig. 39.34, where the length of the cube edge is a  2.0  1010 m. Find the two longest de Broglie wavelengths and corresponding energies in eV at which the electron wave cannot propagate (a) along a cube edge, (b) along the diagonal of a square face of a cube, and (c) along the body diagonal of a cube.

a



RB

 

VB

RC

 VC

Vout

FIGURE 39.35 A transistor acting as a voltage amplifier. The source of emf VB provides the input signal, and the voltage across the two free terminals constitutes the output signal.

*47. We want to connect two transistors in tandem, so that the net current amplification of the combination is the product of the individual current amplifications of the two transistors. Design a circuit that will accomplish this. *48. A solar cell delivers 0.10 A at 0.60 V. How many such solar cells do you need, and how must you connect them, to obtain 2.0 A at 6.0 V for charging a battery? *49. A solar cell of area 5.0 cm2 facing the Sun delivers 0.10 A at 0.60 V. What is the power delivered by the solar cell? Compare with the incident power of sunlight (1.0 kW m2) and deduce the efficiency for the conversion of energy of light into electric energy.

FIGURE 39.34 Structure of a simple cubic crystal.

*50. In practical transistor circuits, a single source of emf is used to provide the two emfs VB and VC of Fig. 39.23. A typical circuit with a 9.0-V emf is shown in Fig. 39.36. The transistor has a current gain factor of 100 and it is desired to operate with IC  1.0 mA with the two 1.0-k resistors in place. Assume that the potential across the forward-biased base–emitter junction is 0.60 V. The resistors R1 and R2 are known as bias resistors. Use Kirchhoff ’s rules to find the required values of R2 and R1. n

39.5 Semiconductor Devices

p

n

1.0 k

1.0 k

45. Figure 39.23 shows a circuit diagram for an n–p–n transistor. Draw the analogous diagram for the p–n–p transistor, and explain how a small current IB leads to a large current IC . 46. When a transistor is connected to a circuit as shown in Fig. 39.35, it serves as an amplifier of voltage. The voltage gain factor is defined as the ratio of the output voltage (measured across the resistor RC ) to the input voltage VB . Evaluate this ratio if RB  3000 and RC  6000 . Assume that the current gain factor for this transistor is 100, and that the internal resistance of the emitter–base junction (a diode with forward bias) is negligible.

IC  1.0 mA

IB R1





R2

9.0 V

FIGURE 39.36 A transistor circuit with a single source of emf.

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Review Problems

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REVIEW PROBLEMS 51. (a) How many of the stationary states (counting spin states) of the hydrogen atom have energy 13.6 eV? Energy 3.4 eV? (b) How many of the stationary states (counting spin states) of the hydrogen atom have energy 3.4 eV and l  0? Energy 3.4 eV and l  1? Energy 3.4 eV and l  3? 52. Suppose that all you know about a state of the hydrogen atom is that the magnetic quantum number is m  3. What conclusions can you draw about the value of the orbital quantum number? The principal quantum number? The energy of the state? (Hint: Is l  0 compatible with the given value of m? Is l  1? Is l  2?) *53. Consider the possible directions of the orbital angularmomentum vector for an electron with orbital quantum number l  1. What are the possible values of the magnetic quantum number m? For each value of m, calculate the angle between the angular-momentum vector and the z axis. Draw a diagram showing the possible orientations of the angularmomentum vector for all the different values of m. *54. Suppose that we regard the proton as a sphere of uniform density with a radius of 1.0  1015 m rotating rigidly about its axis. According to classical mechanics, if the spin angular momentum of this sphere is to have a magnitude of 13 4U, what must be the angular velocity of rotation? What must be the speed of a point on the equator? 55. List the quantum numbers of all the electrons of a magnesium atom in its ground state. 56. What are the quantum numbers n and l for the two outermost electrons of a Be atom? The Mg atom? The Ca atom? In what way are these quantum numbers of these different atoms similar? *57. The conventional range of wavelengths for X rays extends from 10 nm to 0.01 nm. Suppose you want to generate characteristic X rays within this range of wavelengths by means of the transition of an internal electron in an atom, from the first excited state to the ground state. What atomic numbers are suitable? Which atoms do these correspond to? *58. Calculate the wavelengths of the characteristic X rays emitted by the inner electrons in molybdenum atoms in transitions from the second and the third excited states to the ground state.

(a) Calculate the frequency of vibration of the molecule. For the sake of simplicity, assume that the fluorine atom remains at rest and only the hydrogen atom moves. (b) Calculate the energy of the first excited vibrational state of the molecule. 61. Consider the HCl molecule described in Example 4. Suppose that this molecule is initially in the J  3 rotational state. If the molecule sequentially makes purely rotational transitions to the J  2 state, then to the J  1 state, and finally to the J  0 state, what are the energies and the wavelengths of the photons emitted in each of these transitions? 62. The distance between the two nuclei in the H2 molecule is 0.074 nm. (a) What is the moment of inertia of an H2 molecule rotating about its center of mass? What are the energies of the first, second, and third excited rotational states? Express these energies in eV. (b) Suppose we replace one of the hydrogen atoms by a deuterium atom (D, or 2H), whose mass is twice that of the hydrogen atom. Where is the new center of mass? What is the moment of inertia of the HD molecule about its center of mass? What are the energies of the first, second, and third excited rotational states? By what factor do these energies differ from those of the H2 molecule? 63. (a) The binding energy of the NaCl crystal (salt) is 765 kJ/mole; this is the energy required to dissociate the crystal into separate ions. This crystal is cubic, as is illustrated in Fig. 39.37. Pretend that each ion forms a bond with only the six nearest ions. What is the energy per Na–Cl bond? Express the answer in electron-volts. (b) The distance between each ion and the nearest ion is 0.281 nm. What is the Coulomb energy of a pair of ions separated by this distance? Express the answer in electronvolts. Explain why the answers obtained in (a) and (b) are of the same order of magnitude, although not exactly equal. Na Cl Na

*59. When a block of metal serving as target in an X-ray tube is bombarded with a beam of fast electrons, it emits not only Bremsstrahlung but also characteristic X rays of wavelengths 0.167 nm, 0.141 nm, and 0.133 nm. Assuming that these X rays arise in transitions from excited states into the ground state, identify the metal. 60. In the HF molecule, the chemical bonds holding the two atoms together behave like a massless spring of a spring constant k  9.7  102 N/m.

FIGURE 39.37 Structure of NaCl crystal. The yellow balls are sodium ions and the green balls are chlorine ions.

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CHAPTER 39

Quantum Structure of Atoms, Molecules, and Solids

64. Figure 39.38 shows a full-wave rectifier consisting of four diodes connected together. This rectifier not only blocks the negative portion of an entering alternating current, but also reverses this portion, so the current is positive at all times. Describe the flow of current through the four diodes when the entering alternating current is positive and when it is negative.

(b) Iin

t

(a) Iout

Iin p

n

p n Iout n

p

FIGURE 39.38 (a) Four diodes connected to form a full-wave rectifier. (b) The current going into the rectifier, and the current coming out.

n

p

t

Answers to Checkups Checkup 39.1 1. According to wave mechanics, the ground-state orbital angular momentum is L  0; according to Bohr theory, L  U. For both theories, the quantum number n  1 in the ground state. 2. The magnitude of the orbital angular momentum is given by L  1l (l  1)U ; for l  3, this is L  13  (3  1)U  213U. 3. The orbital angular-momentum quantum number may take values from 0 to n  1, so for n  3, the possible values of l are l  0, l  1, and l  2. 4. The intensity drops to zero at the center of each plot, so the probability of finding the electron at the nucleus is zero in each of the electronic states corresponding to Figs. 39.3b, c, and d. 5. (E) 1, 0, and 1. For a given value of l, the magnetic quantum number m may take on integer values from l to l.

Checkup 39.2 1. For any principal quantum number n, the values of the orbital angular momentum quantum number l range from 0 to n  1, and for each l there are 2l  1 values of the magnetic quantum number m. There are two values of spin for each state. For the K shell, we have the least number of states, two: n  1,

l  0, m  0, and ms  ; 21. For the L shell, we have two similar states but with n  2, and six states with l  1 [2  (2l  1)], for a total of eight. For the M shell, we have the largest number: eight states similar to the L-shell states but with n  3, plus ten states with l  2 [2  (2l  1)], for a total of eighteen. 2. Yes, for each orbital state there are a spin up and a spin down state, and if all states are full, for every spin up electron there is a spin down electron. 3. (D) 4  13.6 eV. Neutral helium (Z  2) has two electrons. After one is removed, the single remaining electron is in a purely hydrogenic orbit, but with nuclear charge Z  2. Equation (39.6) gives the energies of such orbits as En  Z2  (13.6 eV) n2. To remove the electron, we take it from the n  1 state to n  q, which requires an energy of removal ¢E  Z2  (13.6 eV)  4  13.6 eV.

Checkup 39.3 1. No. From Eq. (39.9), we recall that the energy of rotation is proportional to the square of the frequency of rotation. We know that the frequency of radiation is proportional to the difference of the energies of two states, and so is proportional to the difference of two frequencies squared, and thus is not equal to any single rotational frequency.

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Answers to Checkups

2. The molecule emits four photons, corresponding to the four transitions shown. From the spacings in Fig. 39.9, we see that the photon emitted in the first (top) transition has the largest energy, and the photon emitted in the last (bottom) transition has the least energy. Since E  hf  hc l, the first photon has the shortest wavelength, and the last photon the longest. 3. The rotational energy of any state is inversely proportional to the moment of inertia [see Eq. (39.12)], so the lighter H2 molecule has the larger energy difference, by a factor of 2. 4. (A) 1, 2, 3, 4. We can obtain the energy difference ¢E between two adjacent states directly from the corresponding adjacent values on the vertical axis in Fig. 39.9. The photon frequency is photon  ¢E U , so starting with the bottom transition we have photon  1, 2, 3 and 4 times U I. This sequence continues: for a transition from a state with any J to the next lower (J  1) state, ¢E  ( U 2 2I )  [ J ( J  1)  ( J  1 ) J ]  (U 2 I )  J. with J  1, 2, 3, 4, etc.

Checkup 39.4 1. A crystalline solid with a full band and only empty bands above it is an insulator or a semiconductor. To decide, we would need to know the magnitude of the energy gap. 2. Most conductors are n type, but a conductor can be p type. A p-type conductor must have holes that behave like moving positive charges (as illustrated in Fig. 39.17); this happens when a band is almost filled. Such behavior occurs in conductors when different bands overlap, so electrons that would otherwise be near the top of one band spill over into the lower-energy states of an overlapping band. 3. No. At sufficiently low temperature, there are, in essence, no electrons with enough thermal energy to cross the energy gap. 4. No. When the lattice potential is not too strong, the evenly spaced values of the momentum and the dominantly kinetic

1353

energy (E  p2 2m), imply that the forbidden values of the energy will be spaced roughly in proportion to the square of the corresponding momentum values. (When such behavior is generalized to crystal structures in three dimensions, or when the lattice potential is strong, then more complicated sequences of forbidden values of energy occur.) 5. (B) p-type has acceptor atoms and holes; n-type has donor atoms and free electrons. Acceptor atoms remove electrons from the valence band and leave behind positively charged (p-type) holes; donor atoms contribute electrons (negatively charged, n-type) to the conduction band.

Checkup 39.5 1. Forward. A diode is forward-biased when the p-type region is at a higher potential than the n-type region. 2. The forward-biased rectifier involves recombination, since the external source of emf pumps many electrons into the n region and many holes into the p region, and they recombine at the junction. (A reverse-biased diode involves the creation of an extremely small number of thermally generated electron–hole pairs.) The light-emitting diode is a forward-biased rectifier and so also involves recombination. The solar cell involves creation, since the absorption of photons creates electron–hole pairs. 3. (A) Electric field points from n-type region to p-type region. In the interface region, the mobile carriers are depleted, and since these carriers came from neutral impurity atoms, the impurity ions left behind are opposite in sign to the mobile carriers of each region. The electric field thus points from the positive donor ions of the n-type region to the negative acceptor ions in the p-type region (see Fig. 39.18).

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CHAPTER

40

Nuclei

CONCEPTS IN CONTEXT

40.4 The Law of Radioactive Decay

This nuclear reactor, immersed in a pool of water, releases energy by the fission, or splitting, of nuclei of uranium. Such nuclear reactions produce “penetrating radiations,” that is, radiations capable of penetrating through cloth, paper, and skin. For protection, nuclear reactors are shielded by thick layers of concrete, or, as in the photo, by a thick layer of water. With the concepts of this chapter, we can consider such questions as:

40.5 Fission

? Why does fission release energy, and what is the amount of energy

40.1 Isotopes 40.2 The Strong Force and the Nuclear Binding Energy 40.3 Radioactivity

40.6 Nuclear Bombs and Nuclear Reactors 40.7 Fusion

released per fission? (Section 40.5, pages 1377 and 1376)

? What are the penetrating radiations produced in nuclear reactions, and how are they produced? (Section 40.3, page 1365; and Example 5, page 1371)

? What is the energy of the penetrating radiation released in the fission of uranium? (Example 3, page 1367) 1354

Concepts in Context

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40.1

Isotopes

1355

? A typical reactor releases a power of 1200 MW. How much uranium does this consume per year? (Example 7, page 1382)

R

utherford’s first experiments on the bombardment of atoms with a beam of alpha particles established that the nucleus of the atom is very small, but contains most of the mass of the atom. The nucleus is therefore very dense, and it must be made of massive particles packed very tightly together. In later experiments, Rutherford proceeded to explore the structure of the nucleus, again using a beam of alpha particles as a probe. He found that if the projectiles were energetic enough to penetrate the nucleus, they would often split it into two pieces, two smaller nuclei. The smallest such piece that could be split off was a nucleus of hydrogen, or a proton, and Rutherford therefore conjectured that all other nuclei also contain protons. However, all these other nuclei have more mass and less charge than expected if they contained nothing but protons—there must be some neutral particles in the nucleus or, alternatively, some electrically neutral combination of particles of opposite charges. The mystery of the neutral constituent of the nucleus was not solved until 1932, when J. Chadwick discovered the neutron, a particle of about the same mass as the proton but of zero electric charge. This discovery led to the modern view of the nucleus as a tightly packed conglomerate of protons and neutrons (see Fig. 40.1). Since the average distance between the protons in the nucleus is quite short, the repulsive electric force among the nuclear protons is very large. This force would burst the nucleus apart if there were not an extra, even larger, attractive force holding the protons and the neutrons together. This extra force is the nuclear force, or the “strong” force. Acting on two adjacent protons in a nucleus, this attractive force is about 100 times as large as the repulsive electric force. Thus, the strong force completely overwhelms the electric force. However, in heavy nuclei—such as uranium—with a large number of protons and a large total electric charge, the electric repulsion becomes important. The fission of uranium, as manifested in the explosion of a nuclear bomb, provides a spectacular demonstration of the electric force overpowering the strong force.

Nuclei are made of protons and neutrons. Generically, these two kinds of constituents of the nucleus are called nucleons. Table 40.1 lists the main properties of protons and neutrons. The values of the masses listed in this table are expressed in atomic mass units, where 1 u  1.660 54  1027 kg. The value of 12 listed for the spin is the spin quantum number. According to the usual rule for the magnitude of an angular momentum [see Eq. (39.1)], the magnitude of the spin is actually 212(12  1)U; but physi-

NUCLEON

neutron

8.210–15 m

Protons and neutrons are tightly packed in nucleus.

FIGURE 40.1 The nucleus of the argon atom, consisting of 18 protons (red) and 22 neutrons (gray).

40.1 ISOTOPES

TA B L E 4 0 . 1

proton

THE NUCLEONS

MASS

SPIN

RADIUS

proton

mp  1.007 28 u

1 2

1  1015 m

neutron

mn  1.008 66 u

1 2

1  1015 m

nucleon atomic mass unit u

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CHAPTER 40

isotopes

mass number A

Nuclei

cists often list the values of the spins of particles by giving the spin quantum number, rather than the magnitude of the spin angular momentum. It is instructive to compare these particles with the electron. Both the proton and the neutron have masses about 1840 times as large as that of the electron.1 Their spin quantum number is 12, the same as that of the electron. In contrast to the electron, which is a pointlike particle of no discernible size, both the proton and the neutron are small spheres, of a radius of about 1015 m. The number of protons in the nucleus of a (neutral) atom of a given element matches the number of its electrons, that is, it matches the atomic number of the element. For example, the carbon atom has six electrons and it has six protons in its nucleus. All the atoms of a given chemical element, such as carbon, have exactly the same chemical properties, because they all have exactly the same number of electrons and the same electron configuration. However, the atoms of a chemical element can differ in mass, because their nuclei can have different numbers of neutrons. Thus, all carbon atoms have six protons in their nuclei, but some have six neutrons, some have seven, some have eight, and so on. Atoms with the same number of protons in their nuclei but different numbers of neutrons are called isotopes. Carbon has eleven known isotopes, designated 8C, 9C, 10C, 11C, 12C, 13C, 14C, 15C, 16C, 17C, and 18C, (see Fig. 40.2). The superscript before the chemical symbol (for instance, the superscript “12” in “12C”) indicates the sum of the number of protons and the number of neutrons; this sum is called the mass number. If we designate the mass number by the symbol A, then ANZ

atomic number Z

(40.1)

where N is the number of neutrons and Z is the number of protons, or the atomic number. Since the mass of each proton and each neutron is approximately one atomic mass unit, the mass number is approximately equal to the mass of the nucleus in atomic mass units. Natural samples of atoms of carbon or any other chemical element contain characteristic percentages of different isotopes. Natural carbon, as found in coal, is a mixture of 98.90% of the isotope 12C and 1.10% of the isotope 13C. Carbon dioxide, as found in air, contains not only the isotopes 12C and 13C but also a very small amount (about 2.4  1010 %) of the isotope 14C. The other isotopes of carbon do not occur

All nuclei of a given element have the same number of protons…

…but isotopes of an element have different numbers of neutrons.

11C

12C

6 protons 5 neutrons

6 protons 6 neutrons

13C

14C

6 protons 7 neutrons

6 protons 8 neutrons

Expressed in atomic mass units, the electron mass is 5.49  104 u.

1

FIGURE 40.2 Some of the isotopes of carbon.

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40.1

Isotopes

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naturally; they can be produced only artificially in transmutation of elements in a nuclear reactor or in a particle accelerator. All chemical elements have several isotopes (see the excerpt from the chart of isotopes in Table 40.2). Hydrogen has three isotopes (1H, or ordinary hydrogen; 2H, or deuterium; 3H, or tritium). Helium has five isotopes, lithium has six, and so on. Some of these isotopes occur in nature, others can be produced only by artificial means. Most of the isotopes listed in the chart of isotopes are unstable; they decay by a spontaneous nuclear reaction and transmute themselves into another element. The decay is accompanied by the emission of alpha rays, beta rays, or gamma rays. The alpha rays are high-speed alpha particles (that is, helium-4 nuclei), the beta rays are highspeed electrons or antielectrons, and the gamma rays are high-energy photons. We will examine these decay processes in Section 40.3.

TA B L E 4 0 . 2

EXCERPT FROM THE CHART OF ISOTOPES a

Ne

Z

17 Ne 18 Ne 19 Ne 20 Ne 21 Ne 22 Ne 23 Ne 24 Ne 25 Ne 26 Ne 27 Ne 0.109 s +

10 20.179

F

9 O

13 O

8 11 N

14.0067

C

11.0267

9C

0.127 s +

6 12.011

B

8B

10.811

9.012 18

Li

15 N

16 N

17 N

11 C

12 C

13 C

14 C

15 C

16 C

10 B

11 B

12 B

13 B

14 B

6.941

10 Be

5 Li –21

6 Li

7 Li

8 Li

9 Li

4 He

1H

99.985%

0.00013%

~100%

2H

3H

0.015%

1n

5 He

0.85 s –

6 He

0.017 s – 

11 Be

12 Be

1.008 665

1

2

00.63 s – 

24 F

25 F

17

2.2 s – 

24.0093

25.0138

22 O

23 O

16

13.6 s – 

3.4 s – 

23.0101

23.0193

19 N

20 N

21 N

15

0.42 s

19.0176

20.0238

21.0289

17 C

18 C

19 C

14

17.0226

18.0267

19.0370

17 B

13

– 

17.0986

14 Be

0.011 s –

11

12

14.0440

11 Li

0.17 s –

8 He

0.122 s – 

5

3

10.6 min –

0

0.020 s – 

27.0072

23 F

4.0 s –

0.009 s –

9

10

6.015 123 7.016 005

12.33 yr – no  1.007 825 2.014 102 3.016 049

0

a

92.5%

2  10–21 s 0.802 s n,  – no  3.016 029 4.002 603 5.0122

4.002 60

1.0079

9 Be

5.0125

3 He

2

80.2%

0.74 s –

26.0005

22 F

21 O

99.756% 0.037% 0.204% 26.9 s 122 s + – no   15.003 065 15.994 915 16.999 131 17.999 159

4.16 s 9.97 min 99.63% 0.37% 7.11 s + – – no    1.007 94 14.003 074 15.000 109 16.006 100

0.61 s – 

20 O

18 N

14 N

3.38 min – 

 

19 O

13 N

21 F

37.6 s – 

4.36 s – 

18 O

12 N

19.8%

9.22%

11.1 s – 

17 O

8 Be

7.5%

18.998 403

20 F

16 O

7 Be

s

100%

16.011

2.45 s 20.4 min 98.89% 1.11% 5730 yr + – – no   EC 11.011 433 12.000 00 13.003 355 14.003 242 s

19 F

15 O

6 Be

~10 p, 

3

109.8 min + EC

14 O

–21 –16  106 yr 13.8 s ~ 3  10 5 s 53.3 day ~1  10 s 100% 1.6  2 – – p, , Li EC  no   6.019 73 7.016 930 8.005 305 9.012 183

4

H

9 B–19

18 F

66.0 s + no 

15.0180

9.013 33 10.012 938 11.009 305

Be

1

19.4 s + 

0.774 s ~8  10 + p

5

He

10 C

0.27%

19.992 439 20.993 847 21.991 384

17 F

70.5 s + 

0.011 s + 

7

16 F

~10–19 s p

90.5%

0.0089 s +

15.9994

N

1.74 s +

17.0177

15 F

18.9984

1.67 s +

7

8

6

4 N

The number Z, increasing vertically along the chart, is the number of protons in the isotope; it coincides with the atomic number. The number N, increasing horizontally, is the number of neutrons. In each box, the number directly below the symbol for the isotope gives the abundance in percent for naturally occurring isotopes, or else the half-life for unstable, artificially produced isotopes (the half-life is the time required for one-half of a sample of unstable isotope to decay). The Greek letters indicate the emissions that accompany the decay:  rays (helium nuclei),  rays (electrons),  rays (antielectrons), or  rays. The bottom number gives the mass of the neutral atom (nucleus plus Z electrons) in atomic mass units. The bottom number in the shaded boxes gives the atomic mass averaged in proportion to the abundance of naturally occuring isotopes.

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Experiments on the bombardment of nuclei with alpha particles and other particles indicate that the size of the nucleus is proportional to the cube root of the mass number. Specifically, these experiments indicate that the radius R of a nucleus of mass number A is R  (1.2  1015 m)  A 13

radius of nucleus

(40.2)

For example, the carbon-12 nucleus has a radius R  1.2  1015 m  (12)13  2.7  1015 m whereas the uranium-238 nucleus has a radius R  1.2  1015 m  (238)13  7.4  1015 m We can gain some feeling for how small these radii are by comparing them with the radius of an atom, typically about 1010 m. The comparison tells us that the radius of the nucleus is less than 1/10000 of the radius of the atom, so the volume of the nucleus is only 11012, or one-trillionth, of the volume of the atom. The proportionality between R and A 13 implies that the number of nucleons per unit volume is the same for all nuclei. Since the volume of a sphere is 43pR3, the number of nucleons per unit volume is A (4p3)R3



A (4p 3)(1.2  1015A 13)3 m3

 1.4  1044 nucleons/m3

(40.3)

The mass of each nucleon is about 1.67  1027 kg, and the mass density of the nuclear material is therefore 1.67  1027 kg 

1.4  1044 3

m

 2.3  1017 kg/m3

(40.4)

This means that one cubic centimeter, or 106 m3, of nuclear material would have a mass of 230 million tons! According to Eq. (40.3), the volume per nucleon is 1(1.4  1044) m3. We can think of this volume as a cube enclosing the nucleon; the edge of the cube or, equivalently, the distance from one nucleon to its nearest neighbor is therefore the cube root of the volume, 1 (1.4  1044)13 m  2  1015 m. By comparing this with the radius of a proton or neutron, about 1  1015 m, we see that inside the nucleus the nucleons are so tightly packed together that they touch or almost touch (see Figs. 40.1 and 40.2).

✔

Checkup 40.1

QUESTION 1: How many protons, how many neutrons, and how many nucleons are there in the nuclei of each of the following carbon isotopes: 8C, 9C, and 10C? QUESTION 2: How many protons, how many neutrons, and how many nucleons are there in the nuclei of each of the following hydrogen isotopes: 1H, 2H, and 3H? QUESTION 3: Which of the isotopes in Table 40.2 has the largest number of neutrons? The largest number of protons? The largest mass number?

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40.2

The Strong Force and the Nuclear Binding Energy

1359

QUESTION 4: Which of the following isotopes has the largest ratio of the number of neutrons to the number of protons? (A) 2 H (B) 3 H (C) 4 He (D) 6 Li (E) 12C QUESTION 5: One nucleus has twice the radius of a second nucleus. What is the ratio of the number of nucleons in the first nucleus to the number in the second? (A) 2 (B) 3 (C) 4 (D) 6 (E) 8

40.2 THE STRONG FORCE AND THE NUCLEAR BINDING ENERGY Since the protons within a nucleus are at such short distances from one another, they exert very large repulsive electric forces on one another. For two neighboring protons, separated by a center-to-center distance of r  2.0  1015 m, Coulomb’s Law gives an electric repulsive force of F

(1.6  1019 C)2 1 e2 1   4p0 r 2 4p0 (2.0  1015 m)2

(40.5)

 58 N This is, roughly, four times the weight of this book; acting on a mass of only 1027 kg, the magnitude of this force is colossal. Obviously, some extra force must be present in the nucleus to prevent it from instantaneously bursting apart under the influence of the mutual electric repulsion of the protons. This extra force is the strong force, already mentioned in Section 6.4. This force acts equally between any two nucleons, regardless of whether they are protons or neutrons (the force is “charge-independent”). Figure 40.3 is a plot of the potential energy associated with the strong nucleon–nucleon force, calculated from experimental data on nuclear collisions. From Chapter 8, we know that a potential energy that increases as the nucleons separate corresponds to an attractive force. A decreasing potential energy means that the force does positive work when the nucleons separate; this corresponds to a repulsive force. We therefore see from the plot of the potential energy that the strong force is attractive over a range of internucleon distances from  0.7  1015 m to  2  1015 m. In this range of distances, the strong force is much larger than the electric force, as much as 100 times larger. The strong force is repulsive for internucleon distances less than  0.7  1015 m; this means that the nucleons have a hard core that resists interpenetration. For distances larger than  2  1015 m, the strong force decreases drastically and finally vanishes. Thus, in contrast to the electric force, which fades only gradually and reaches out to large distances, the strong force cuts off sharply and has only a short range. In order to feel the strong force, the nucleons must be touching or almost touching; that is, the force acts only between nearest neighbors in the nucleus. In consequence of the short-range character of the strong force, a nucleon deep inside the nucleus does not experience any net force–the nucleon interacts only with its nearest neighbors, and since these pull it with equal forces in almost all directions, the net force on the nucleon is zero or nearly zero (see Fig. 40.4). However, a nucleon at the nuclear surface has neighbors only on the side that is toward the interior, and hence these will exert a net force pulling the nucleon inward (see Fig. 40.4). Altogether this means that nucleons are more or less free to wander about the interior of the nucleus,

strong force U

200 MeV

Strong force is attractive for internucleon distances of ≈ 0.7–2  10–15 m, …

100

110–15 210–15 m

r

–100 …resists nucleon interpenetration at shorter distances, … …and cuts off sharply, having only a short range.

FIGURE 40.3 Potential energy for the strong force acting between two nucleons.

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but whenever they approach the nuclear surface, strong forces pull them back and prevent their escape. This suggests that the nucleons in the nucleus behave somewhat like the molecules in a drop of water; such molecules are free to wander throughout the volume of the drop, but when they approach the water surface, intermolecular forces hold them back. This similarity between nuclei and drops of water rests on a similarity of the laws of force. The intermolecular force has general features rather similar to those displayed in Fig. 40.3; the force is attractive over a short range and then becomes strongly repulsive when the molecules begin to interpenetrate. The hard repulsive core makes the water nearly incompressible, whereas the short-range attraction provides a cohe…while net strong force on nucleons sive force that prevents water droplets from falling apart. The balance of attraction inside nucleus is essentially zero. and repulsion encourages water molecules to stay at a particular distance from one FIGURE 40.4 Strong forces on a nucleon another, and this gives water a particular, uniform density. at the nuclear surface, and strong forces on a Because of the similarities between a liquid and the nuclear material, the nucleus can nucleon in the nuclear interior. be crudely regarded as a droplet of incompressible “nuclear fluid” of uniform density. The fluid is, of course, made of nucleons, but for some purposes we can ignore the individual nucleons, and we can calculate the properties of nuclei in terms of the gross properties of a liquid. For example, the spherical shape adopted by most nuclei can easily be understood as follows: Any nucleon located on the surface of a globule of nuclear fluid experiences an inward force pulling it back into the volume, and consequently the fluid tends to shrink its exposed surface to the smallest value compatible with its (fixed) volume. Since a sphere has the least surface area for a given volume, the globule of fluid will take the shape of a spherical droplet. In a stable nucleus, the repulsive electric forces among the protons are held in check by the attractive strong forces. To achieve this balance of forces, the presence of neutrons is an advantage: a nucleus with more neutrons will have a larger size and, therefore, a large average distance between pairs of protons—the neutrons in the nucleus dilute the repulsive effect of the electric force. Consequently, all stable nuclei, with the exception of hydrogen and one isotope of helium, contain at least as A 150 many neutrons as protons; heavy nuclei, such as uranium, contain sub Heavy nuclei contain 22 0 stantially more neutrons than protons. 140 many more neutrons 20 than protons. Figure 40.5 is a plot of the number of neutrons vs. the number of 0 130 protons (N vs. Z). On this plot, blue dots indicate the stable nuclei and 18 0 120 red dots indicate unstable nuclei, that is, radioactive isotopes. Note that 16 there is no stable nucleus beyond bismuth (Z  83). However, several 0 110 elements beyond bismuth have some isotopes with very long lifetimes; 14 0 100 these are therefore almost stable, and they occur naturally. 12 90 The energy stored in a nucleus is a sum of the potential energies 0 contributed by the electric and the strong forces and the kinetic energies 80 N of the nucleons. The potential energy is negative, and its magnitude is 10 70 larger than that of the kinetic energy. Thus, the “stored” energy is neg0 Z  ative, and energy is released when the nucleus is assembled out of its con60 80 N stituent nucleons. Conversely, energy must be supplied to take the nucleus 50 60 apart into its constituent nucleons. The energy that must be supplied to 40 4 take the nucleus apart, or the energy released during the assembly of the 0 Net strong force on nucleons near surface is directed inward…

30

A

20



For Z  2, stable nuclei (blue) contain at least as many neutrons as protons.

20

10 0

0

10

20

30

40

50 Z

60

70

80

90

100

FIGURE 40.5 Number of neutrons (N ) vs. number of protons (Z) for stable nuclei (blue dots) and unstable nuclei (red dots).

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40.2

B.E./A MeV Binding energy is about 8 MeV per nucleon for almost all nuclei.

20Ne

Fe

Zn Kr

Mo

Te

Sm

16O

7 11B

1361

24Mg

Ca

8 12C

The Strong Force and the Nuclear Binding Energy

Lu

Hg

4He

Ra

6 Binding energy per nucleon increases with mass number for light nuclei,…

5

…and decreases with mass number for heavy nuclei.

6Li

4 3

3He

2 1

2H

50

100 150 mass number, A

200

FIGURE 40.6 Average binding energy per nucleon vs. mass number.

nucleus out of its constituent nucleons, is called the binding energy (B.E.). Figure 40.6 is a plot of B.E./A, the binding energy divided by the number of nucleons, or the average binding energy per nucleon. The curve plotted in Fig. 40.6 is called the curve of binding energy. The energy unit used in this plot is the MeV, where 1 MeV  106 eV  1.6022  1013 J

curve of binding energy

(40.6)

This unit is widely used in nuclear physics. The binding energy of a typical nucleus is a rather large amount of energy. As may be seen from Fig. 40.6, the average binding energy per nucleon is in the vicinity of 8 MeV for almost all nuclei; thus a nucleus with a mass number A typically has a binding energy of about A  8 MeV. To put this number in perspective, let us compare it with the rest-mass energy of the nucleons. Each nucleon has a mass of about one atomic mass unit. The energy corresponding to one atomic mass unit is 1 u  c 2  1.6605  1027 kg  (2.9979  108 m/s)2  1.4924  1010 J or, in MeV units [see Eq. (40.6)],2 1 u  c 2  1.4924  1010 J 

1 MeV 1.6022  1013 J

 931.5 MeV

(40.7)

Thus, each nucleon has a rest-mass energy of about 930 MeV, and the A nucleons in the nucleus have a rest-mass energy of about A  930 MeV. The ratio of binding energy to rest-mass energy is then about 8930  0.009, which means the binding energy is nearly 1% of the rest-mass energy! By Einstein’s mass-energy formula, the mass associated with the binding energy is B.E./c 2. This mass is carried away by the energy released during the assembly of the 2

We have retained four significant figures in this result because we will be needing a precise value of the equivalence between energy and mass later on.

energy equivalent of atomic mass unit

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Nuclei

nucleus from its constituent protons and neutrons. The mass of a typical nucleus is therefore about 1% less than the sum of the masses of these protons and neutrons. The mass difference is called the mass defect,

mass defect

B.E.  [mass defect]  [mass of Z protons and N neutrons] c2  [mass of nucleus]

(40.8)

Experimental values of the binding energy of a nucleus usually are obtained via the mass defect. A precise measurement of the mass of the nucleus is compared with the sums of the masses of the constituent protons and neutrons; the difference, or mass defect, then gives the binding energy according to Eq. (40.8). This is how the experimental points in the plot of the curve of binding energy were obtained. What is the nuclear binding energy of the isotope 238U? Express the energy in MeV. The mass of one atom of this isotope is 238.0508 u.

EXAMPLE 1

SOLUTION: The atomic number of uranium is Z  92 and uranium-238 has

mass number A  238, so this isotope has 92 protons and N  A  Z  146 neutrons. According to Eq. (40.8), B.E. c2

 [92mp  146mn]  [mass of nucleus]

(40.9)

We want to re-express this equation in terms of the mass of the atom, since we know this mass. The mass of the uranium nucleus is the mass of the uranium atom minus the mass of the 92 electrons of this atom. Hence B.E. c2

 [92mp  146mn]  ([mass of

238

U atom]  92me)

But 92 proton masses plus 92 electron masses equals 92 hydrogen masses, because a hydrogen atom consists of one proton and one electron. We therefore obtain the following convenient formula for the binding energy: B.E. c2

 92[mass of 1H atom]  146mn  [mass of

238

U atom]

With [mass of 1 H atom]  1.007 825 u (from Table 40.2, page 1357) and mn  1.008 66 u, this yields B.E. c2

 92  1.007 825 u  146  1.008 66 u  238.0508 u  1.934 u

The binding energy is therefore B.E.  1.934 u  c 2 Since u  c 2  931.5 MeV [see Eq. (40.7)], the result is B.E.  1.934  931.5 MeV  1802 MeV

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40.2

The Strong Force and the Nuclear Binding Energy

Besides its role as nuclear binding force, the strong force also plays a crucial role in nuclear reactions that occur when the nuclei in some target material are bombarded by a beam of protons, neutrons, or alpha particles. When such a projectile strikes the nuclear surface, the strong force pulls the projectile into the nucleus. The projectile then either remains bound within the nucleus, forming a heavier nucleus; or else the projectile disrupts the internal structure of the nucleus to such an extent that the nucleus ejects one or several fragments, that is, one or several protons, neutrons, alpha particles, or even other nuclei. Such nuclear reactions involve transmutation of elements: the original nucleus is changed into a new nucleus of different mass number and atomic number. The first such transmutation of elements was discovered in 1919 by Rutherford when he bombarded nitrogen with alpha particles. He found that this led to the transmutation of nitrogen into oxygen, according to the reaction3 

1363

transmutation of elements

14

N S 17O  p

As we will see in Section 40.3, the alpha particle is actually the nucleus of a helium4 atom; hence the reaction can also be written3 4

He  14N S 17O  1H

(40.10)

In his early experiments, Rutherford used a naturally radioactive material as the source of his beam of alpha particles. But in the 1930s, physicists began to build machines for the artificial acceleration of beams of charged particles. The first of these accelerators were electrostatic; they accelerated particles by means of strong static electric fields produced by a large amount of electric charge accumulated on a spherical capacitor (see Fig. 40.7). Many of the later accelerators use a combination of electric and magnetic fields. Thus, in the cyclotron (see Section 30.1), a uniform magnetic field holds protons in a circular orbit, while an electric field acts on them periodically, gradually increasing their energy step by step. The investigation of nuclear reactions initiated by projectiles from accelerators led to the discovery of a multitude of new isotopes, most of them short-lived and highly radioactive. These investigations also led to a better understanding of the details of the strong force. The energy of the projectile required to initiate a nuclear reaction or the energy released in a nuclear reaction can be calculated from the rest-mass energies of the isotopes that participate in the reaction. The energy released in the reaction is called its Q value. For a reaction with initial masses m1, m2, p and final masses m 1, m 2, p , the Q value is the difference between the initial and the final sums of rest-mass energies: Q  (m1c 2  m2c 2  )  (m 1c 2  m 2c 2  )

(40.11)

If the rest-mass energy after the reaction is smaller than that before, the Q value is positive, and energy is released in the reaction. Such a reaction can proceed even if the energy of the projectile is very low (nearly zero). If the rest-mass energy after the reaction is larger than that before, the Q value is negative, and energy must be supplied to make the reaction proceed; that is, the reaction absorbs energy, instead of releasing energy. The energy required to make the reaction proceed must come from the kinetic energy of the incident projectile; thus, the 3

These reactions as written include only nucleons, and do not account for the atomic electrons of the atoms or ions involved.

FIGURE 40.7 The Cockcroft–Walton accelerator built in 1932.

Q value of reaction

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projectile needs a minimum energy to initiate the reaction. The kinetic energy of the projectile is then (partially) converted into rest-mass energy of the reaction products.

Calculate the Q value for the reaction (40.10). What can you say about the minimum kinetic energy of the alpha particle required to initiate this reaction? EXAMPLE 2

SOLUTION: The Q value is the difference between the total rest-mass energy before the reaction and the total rest-mass energy after the reaction. The total restmass energy before the reaction (40.10) is the rest-mass energy of the alpha particle (or 4He nucleus) plus the rest-mass energy of the 14N nucleus. The total rest-mass energy after the reaction is the sum of the rest-mass energies of the 17O nucleus and the proton. Thus,

Q  [mass of 4He nucleus]c 2  [mass of 14N nucleus]c 2  [mass of 17O nucleus]c 2  mpc2

(40.12)

Since the chart of isotopes lists the masses of the atoms, rather than the masses of nuclei, we want to express Eq. (40.12) in terms of atomic masses. For this purpose, we add the rest-mass energy of 18 electrons to the first two terms on the right side of Eq. (40.12), and we subtract the rest-mass energy of 18 electrons from the last two terms: Q  [mass of 4He nucleus  4me]c 2  [mass of 14N nucleus  14me]c 2  [mass of 17O nucleus  17me]c 2  [mp  me]c 2 Each of the terms in brackets is now the mass of a complete atom: Q  [mass of 4He atom]c 2  [mass of 14N atom]c 2  [mass of 17O atom]c 2  [mass of 1H atom]c 2 Substituting the values of the masses listed on the chart of isotopes (Table 40.2, page 1357), we then find Q  4.002 603 u  c 2  14.003 074 u  c 2  16.999 131 u  c 2  1.007 825 u  c 2  0.001 279 u  c 2 Since u  c 2  931.5 MeV, we can express our final result as Q  0.001 279  931.5 MeV  1.191 MeV This negative Q value indicates that the sum of rest-mass energies after the reaction is larger than before, that is, the reaction absorbs energy. The absorbed energy is 1.191 MeV, and this energy must be supplied by the alpha particle. The minimum kinetic energy of the alpha particle must therefore be at least 1.191 MeV. COMMENT: Although the minimum energy required for the reaction is 1.191

MeV, the reaction can proceed only with alpha particles of somewhat higher kinetic energy, because momentum conservation dictates that some kinetic energy must be given to the reaction products.

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40.3

✔

Radioactivity

1365

Checkup 40.2

QUESTION 1: By inspection of Fig. 40.3, determine where the strong force has its largest attractive value. 238 QUESTION 2: In Example 1 we found that the binding energy of U is 1802 MeV. Accordingly, what is the value of B.E./A for this isotope? QUESTION 3: Can a stable nucleus have a negative mass defect, that is, a mass excess? 17 QUESTION 4: Consider the reverse of the reaction (40.10), that is, bombardment of O by protons to form helium and nitrogen, 1H  17O S 4He  14N. What is the Q value of this reaction? What can you say about the minimum kinetic energy of the proton required to initiate this reaction? Ignore the recoil of the oxygen nucleus. QUESTION 5: According to Fig. 40.6, which of the nuclei named in this figure has the largest binding energy per nucleon (largest B.E./A)? Which has the largest binding energy (largest B.E.)?

(A) 2H, Fe

(B) 2H, 4He

(C) 4He, Ra

(D) Fe, Ra

(E) Fe, 4He

40.3 RADIOACTIVITY Radioactivity was discovered in 1896 by Henri Becquerel.Through an accident, he noticed that samples of uranium minerals emitted invisible rays which could penetrate through sheets of opaque materials and make an imprint on a photographic plate. Subsequent investigations established that uranium, and many other radioactive substances, emit three kinds of rays: alpha rays (),beta rays (), and gamma rays (). Of these, the alpha rays are the least penetrating; they can be stopped by a thick piece of paper. The beta rays are more penetrating; they can pass through a foil of lead or a plate of aluminum. The gamma rays are the most penetrating; they can pass through a thick wall of concrete. When these three kinds of rays are aimed into a magnetic field, the alpha and beta rays are deflected in opposite directions, whereas the gamma rays proceed without deflection (see Fig. 40.8).This simple experiment demonstrates that alpha rays and beta rays are electrically charged, with charges of opposite signs, whereas gamma rays are neutral. By more detailed experiments. Becquerel identified the beta rays as high-speed electrons. Some years later, Rutherford investigated the nature of the alpha rays and demonstrated that they are identical to nuclei of helium. Thus, electrons and helium nuclei are, somehow, manufactured within the sample of uranium or other radioactive substance, and they are ejected at high speed. Gamma rays are high-energy photons emitted by the radioactive substance; the energy of these gamma-ray photons is typically a thousand times as large as that of X-ray photons. The “manufacture” and ejection of these high-speed electrons and helium nuclei occurs in the nuclei of the radioactive substance, by nuclear decay reactions. In the following, we will discuss the broad features of these nuclear reactions.

Online Concept Tutorial

45

Concepts in Context

alpha rays (), beta rays (), and gamma rays ()

When rays enter magnetic field,  and  rays are deflected in opposite directions,… 

 

…and  rays are undeflected.

Alpha Decay The alpha particle consists of two protons and two neutrons; it has the same structure as the helium-4 nucleus. When a radioactive nucleus ejects an alpha particle consisting of two protons and two neutrons, the radioactive nucleus does not create these protons

FIGURE 40.8 Alpha (), beta (), and gamma () rays emitted by a radioactive source. A magnetic field is perpendicular to the plane of the page.

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and neutrons out of nothing—the nucleus merely takes two of its own protons and neutrons and spits them out. Since the nucleus loses two protons and neutrons, both the atomic number and the mass number of the nucleus decrease. The atomic number decreases by 2, and the mass number by 4. Thus, the original isotope becomes a different isotope of a different chemical element. The following examples of alpha decays illustrate such transmutations of chemical elements:

parent nucleus and daughter nucleus fission

ANTOINE HENRI BECQUEREL (1852– 1908) French physicist. He was awarded the 1903 Nobel Prize for his discovery of radioactivity.

238

U S 234 Th  

(40.13)

226

Ra S 222Rn  

(40.14)

In the first of these alpha-decay reactions, uranium is transmuted into thorium, and in the second, radium into radon. The original nucleus in a decay reaction is called the parent, and the resulting nucleus is called the daughter. The alpha-decay reaction can be regarded as a fission, or splitting, of the nucleus into two smaller nuclei. The, say, uranium nucleus fissions into a thorium nucleus and a helium nucleus. The total number of protons and of neutrons is unchanged in this fission (see Fig. 40.9). The fission occurs spontaneously, because of an instability in the original nucleus. Large nuclei, such as uranium or radium, contain many protons, which exert repulsive electric forces on each other. Although the repulsive electric force is balanced by the attractive strong force, the balance of these forces is rather precarious because the electric force easily reaches from one end of the nucleus to the other, whereas the strong force acts only between adjacent nucleons and therefore, in a large nucleus, cannot reach directly from one end to the other. Thus, any accidental, spontaneous elongation of the nucleus can shift the balance in favor of the electric force— the nucleus then elongates more and more, and ultimately bursts apart into two fragments. In true fission, the two fragments are of approximately equal size. In alpha decay, we are dealing with an extreme case of fission, with two fragments of very unequal size. The ejection of an alpha particle is strongly favored over ejection of, say, a hydrogen nucleus or a lithium nucleus, because the helium nucleus is an exceptionally tightly bound nucleus. In consequence of this large binding energy, the formation of an alpha particle, just before its ejection, makes more energy available for driving the fission reaction. All the large, heavy nuclei beyond bismuth are unstable; they all are subject to alpha decay or other forms of spontaneous fission. Besides, many isotopes of somewhat smaller nuclei suffer from the same instability. 4He 238U

Alpha decay of an unstable nucleus is often favored…

2 protons 2 neutrons

…since 4He nucleus is tightly bound, availing more energy for fission.

234Th

92 protons 146 neutrons Atomic number of daughter nucleus has decreased by two, and mass number by four.

FIGURE 40.9 Fission of uranium into thorium and helium.

90 protons 144 neutrons

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40.3

Radioactivity

1367

The kinetic energy of the alpha particles ejected in reactions such as (40.13) and (40.14) can be calculated from the masses of the participating nuclei and the mass of the alpha particle.

Calculate the energy released in the alpha decay of 238U, and calculate the kinetic energy of the alpha particle ejected in this reaction. The atomic masses of the isotopes of uranium, thorium, and helium in this reaction are 238.0508 u, 234.0436 u, and 4.0026 u, respectively.

EXAMPLE 3

Concepts in Context

SOLUTION: The energy released in the reaction (40.13) is simply the difference between the total rest-mass energy before the reaction and the (smaller) total restmass energy after the reaction. The total rest-mass energy before the reaction is the rest-mass energy of the uranium nucleus, [mass of U nucleus]c 2. The total restmass energy after the reaction is the sum of the rest-mass energies of the thorium nucleus and the alpha particle, [mass of Th nucleus]c 2  [mass of He nucleus]c 2. The energy released in the reaction is therefore

Q  [mass of

238

U nucleus]c 2  [mass of

234

Th nucleus]c2

 [mass of 4He nucleus]c 2

(40.15)

To express this in terms of the masses of the atoms rather than the masses of the nuclei, we add the rest-mass energy of 92 electrons to the first term on the right side of Eq. (40.15) and we subtract the same amount from the other two terms: Q  [mass of

238

U nucleus  92me]c 2  [mass of

234

Th nucleus  90me]c 2

 [mass of 4He nucleus  2me]c 2

(40.16)

Now each of the terms in square brackets is the mass of a complete atom: Q  [mass of

238

U atom]c 2  [mass of

234

Th atom]c 2  [mass of 4He atom]c 2 (40.17)

We know that these masses are 238.0508 u, 234.0436 u, and 4.0026 u, respectively; hence, Q  238.0508 u  c 2  234.0436 u  c 2  4.0026 u  c 2  0.0046 u  c 2 or, with u  c 2  931.5 MeV, Q  0.0046  931.5 MeV  4.3 MeV Since the thorium nucleus is much heavier than the alpha particle, it suffers almost no recoil, and the alpha particle carries away almost all of the energy released in the reaction. Thus, the kinetic energy of the alpha particle will be 4.3 MeV.

In many cases of alpha decay, the daughter nucleus is also unstable, and decays some time after its formation, either by alpha decay or by beta decay. The daughter of the daughter then decays, and so on. The sequence of daughters descending from the original parent is called a radioactive series. The series ends when it reaches a stable iso-

radioactive series

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tope. For instance, the decay of uranium initiates a radioactive series, which ultimately ends with a stable isotope of lead.

Beta Decay The simplest beta-decay reaction is the decay of the neutron. The free neutron is unstable, and it decays into a proton, an electron, and an antineutrino: n S p  e   neutrinos () and antineutrinos (  )

In this equation, the symbol  (nu-bar) represents the antineutrino. Neutrinos () and antineutrinos (  ) are particles of very small mass and spin quantum number 12, which travel at nearly the speed of light. Thus, they are somewhat similar to photons. However, in contrast to photons, which interact with electric charges, neutrinos and antineutrinos do not interact directly with electric charges—in fact, they hardly interact with anything at all, and they pass through the entire bulk of the Earth with little hindrance. As their names indicate, neutrinos and antineutrinos are antiparticles of each other; they can annihilate each other, producing a flash of light (gamma rays). For a free neutron, the decay reaction (40.18), on the average, takes a time of 15 minutes.4 For a neutron in a nucleus, the reaction may proceed at a faster rate or at a slower rate, depending on whether the nucleus promotes the reaction by supplying extra energy, or inhibits the reaction by withdrawing energy. When the reaction occurs in a nucleus, the electron and antineutrino are ejected, and the net effect is the conversion of a neutron into a proton, which increases the atomic number of the nucleus by 1, while leaving the mass number unchanged. Thus, the original isotope is transmuted into an isotope of the next chemical element. Two examples of beta decays are the following: Co S 60 Ni  e  

(40.19)

C S 14 N  e  

(40.20)

60

14

In negative beta decay, electron and antineutrino are ejected. 60Co

e–

—

27 protons 33 neutrons

60Ni

28 protons 32 neutrons Atomic number of daughter nucleus has increased by one, transmuting cobalt into nickel. Mass number is unchanged.

FIGURE 40.10 Beta decay of 60Co into 60 Ni. The number of neutrons is 33 before the decay and 32 after; the number of protons is 27 before the decay and 28 after.

(40.18)

In each of these reactions, the number of neutrons (N ) decreases by 1, and the number of protons (Z) increases by 1 (see Figs. 40.10 and 40.11). The antineutrino emitted in these reactions is almost impossible to detect. Its existence, however, can be inferred from the conservation of energy. In the beta decay of, say, radioactive cobalt-60 [Eq. (40.19)], sometimes the electron emerges with one energy, and sometimes with another.This is because the energy of the decay is shared between the electron and the neutrino, and sometimes one of these particles carries off most of the energy, sometimes the other. Hence the energy of the electron is unpredictable—it can be anything from zero up to a maximum amount that corresponds to the electron carrying away all the energy of the decay. In fact, it was on the basis of this variability of the energy of the ejected electron that Pauli first proposed the existence of the neutrino, since this was the only way to preserve the Law of Conservation of Energy. When Pauli made this proposal in 1931, there was no direct evidence for the existence of an extra particle; neutrinos and antineutrinos were detected only much later, in experiments with nuclear reactors. Beta-decay reactions of the type (40.19) and (40.20) involve the conversion of a neutron into a proton and the ejection of an electron and an antineutrino and occur for many nuclei that have more neutrons than the stable isotopes of a given element. In nuclei that have fewer neutrons than the stable isotopes, there are often beta-decay

4

The average lifetime of free neutrons is 15.0 min, but the half-life (see next section) is 10.6 min.

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40.3

e–

e

22Na

—



14N

6 protons 8 neutrons

11 protons 11 neutrons

…carbon is transmuted into nitrogen.

1369

In positive beta decay, positron (antielectron) and neutrino are ejected.

In this negative beta decay… 14C

Radioactivity

7 protons 7 neutrons

FIGURE 40.11 Beta decay of 14C into 14 N with ejection of an electron and an antineutrino. The number of neutrons is 8 before the decay and 7 after; the number of protons is 6 before the decay and 7 after.

22Ne

10 protons 12 neutrons Atomic number of daughter nucleus has decreased by one, transmuting sodium into neon. Mass number is unchanged.

FIGURE 40.12 Beta decay of 22 Na into 22 Ne, with ejection of an antielectron and a neutrino. The number of neutrons is 11 before the decay and 12 after; the number of protons is 11 before the decay and 10 after.

reactions involving the conversion of a proton into a neutron and the ejection of an antielectron, or positron, and a neutrino. The positron is the antiparticle of the electron; it has the same mass and spin as the electron, but the opposite electric charge. An example of a beta-decay reaction with ejection of a positron is 22

Na S 22 Ne  e  

positron (e)

(40.21)



Here the symbol e represents the positron. After the positron is ejected in this decay, it sooner or later collides with one of the abundant atomic electrons in the surrounding environment and annihilates with it, producing gamma rays. In the reaction (40.21), the number of protons (Z) decreases by 1, and the number of neutrons (N ) increases by 1 (see Fig. 40.12). All the beta-decay reactions are instigated by a new kind of force, the “weak” force. This is one of the four fundamental forces found in matter (see Sections 6.4 and 41.3). Whereas the electromagnetic and the strong forces play an important role in determining the structure of atoms and of nuclei, the weak force plays no role in this. Its only effect is to engender the beta-decay reactions (40.19)–(40.21) and other similar reactions.

What is the maximum kinetic energy of the beta rays emitted in the beta decay of 14C? The atomic masses of 14C and of 14N are 14.003 24 u and 14.003 07 u, respectively.

EXAMPLE 4

SOLUTION: The energy released in the reaction (40.20) is, again, the difference

between the total rest-mass energy before the reaction and the (smaller) total restmass energy after the reaction. The total rest-mass energy before the reaction is

“weak” force

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the rest-mass energy of the carbon nucleus, [mass of 14C nucleus]c 2. The total rest-mass energy after the reaction is the sum of the rest-mass energies of the nitrogen nucleus and the electron, [mass of 14N nucleus]c 2  mec 2 (the rest-mass energy of the neutrino is negligible). The energy released in the reaction is therefore Q  [mass of

14

C nucleus]c 2  [mass of

14

N nucleus]c 2  mec 2

If we add and subtract the rest-mass energy of 6 electrons on the right side of this equation, we obtain Q  [mass of

14

C nucleus  6me]c 2  [mass of

 [mass of

14

C nucleus  6me]c 2  [mass of

14

N nucleus  6me]c 2  mec 2

14

N nucleus  7me]c 2

Here the two terms in brackets are the masses of the complete atoms, so 14

C atom]c 2  [mass of

Q  [mass of

14

N atom]c 2

Substituting these masses and substituting u  c 2  931.5 MeV, we find Q  14.003 24 u  c 2  14.003 07 u  c 2  0.000 17 u  c 2  0.000 17  931.5 MeV  0.16 MeV The nitrogen nucleus, being much heavier than the electron, remains at rest, or nearly at rest. The decay energy is therefore shared between the electron and the antineutrino. If the antineutrino carries away almost none of the energy, then the electron will acquire nearly all the energy. Thus, the maximum possible electron kinetic energy is 0.16 MeV.

Gamma Emission Gamma rays are high-energy photons emitted by nuclei when nucleons make transitions from one stationary nuclear state to another. The emission of a gamma ray by a transition of a nucleon is similar to the emission of visible photons or of X rays by a transition of an atomic electron. When a nucleus suffers alpha decay or beta decay, it is often left in an excited state, and it then eliminates the excitation energy in the form of a gamma ray. Thus, gamma emission is usually one step in a two-step process, with an alpha or beta decay preceding the gamma emission. For example, the beta decay of 60Co leads to subsequent gamma emission according to the sequence of reactions Co S 60Ni  e  

60

R 60

Ni  g

(40.22)

The isotope 60Co is commonly employed in high-intensity industrial irradiation cells, called cobalt cells. But, as shown in Eq. (40.22), the emitter of the gamma rays is nickel, not cobalt. As mentioned above, gamma rays are also generated when an electron annihilates with an antielectron. But this is a secondary process, which does not directly involve the nucleus.

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40.3

P R O B L E M - S O LV I N G T E C H N I Q U E S

Radioactivity

1371

NUCLEAR REACTIONS

As illustrated in Examples 2, 3, and 4, the Q value of a nuclear reaction is best calculated by adding electron masses to the nuclear masses, so as to form masses of atoms. The atomic masses can then be copied from charts or tables of isotopes (which always list atomic masses, rather than nuclear masses). However, it is necessary to take some extra care with this technique in the case of beta decays with antielectrons, such as the beta decay in Eq. (40.21). On the left side of Eq. (40.21), the nucleus is sodium, which requires 11 electrons to be a complete sodium atom. On the right side the nucleus is neon,

which requires only 10 electrons to become a complete neon atom. Hence, if we add 11 electron masses to the nuclear masses on each side, we will be left with one surplus electron mass on the right side. Furthermore, the right side has an antielectron mass (equal to the electron mass). Thus, the Q value for the reaction (40.21) is Q  [mass of 22Na atom]c 2  [mass of 22Ne atom]c 2  2mec 2

Figure 40.13 shows the energy-level diagram for the 60Ni nucleus. When this nucleus is formed in the beta decay of 60Co it is initially in the second excited state (2.50 MeV above the ground state). It makes a transition to the first excited state (1.33 MeV above the ground state), and then a transition to the ground state. What are the energies of the gamma rays emitted in these transitions?

EXAMPLE 5

Concepts in Context

SOLUTION: The first transition is from the energy level E2  2.50 MeV to the

energy level E1  1.33 MeV. The energy of the gamma ray emitted in this transition is E  E2  E1  2.50 MeV  1.33 MeV  1.17 MeV The second transition is from the energy level E 1  1.33 MeV to the ground state, at E0  0. The energy of the gamma ray emitted in this transition is E  E1  E0  1.33 MeV E … 60Ni is in its second excited state…

After beta decay of 60Co, … 60Co

60Ni

– …and emits one gamma ray in transition to first excited state, and another in transition to ground state.

2.50 MeV  1.33 MeV  0 MeV

FIGURE 40.13 Energy-level diagram for the 60 Ni nucleus formed in the beta decay of 60Co.

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Checkup 40.3

QUESTION 1: Why do alpha and beta emissions involve transmutation of elements whereas gamma emission does not? QUESTION 2: Figure 40.8 shows the deflections of alpha and beta rays in a magnetic field. If instead of consisting of an electron, the beta ray consisted of an antielectron, what would be its direction of deflection? QUESTION 3: Can a nuclear decay reaction have a negative Q value? QUESTION 4: An electron is emitted during the beta decay of a cesium-136 nucleus (136Cs). Which of the following nuclei results? (A) 135 Xe (B) 136 Xe (C) 137Cs (D) 136Ba (E) 137Ba

4 0 . 4 T H E L AW O F R A D I O A C T I V E D E C AY

Online Concept Tutorial

45

half-life t 1/2

In radioactive alpha or beta decay, the original isotope, or parent, becomes transmuted into another isotope, or daughter. If we have some given initial amount of parent material, say, 1 gram of radioactive strontium, some nuclei decay at one time, some at another, and the parent material disappears only gradually. In the case of radioactive strontium, one-half of the original amount disappears in 29 years, one-half of the remainder in the next 29 years, one-half of the new remainder in the next 29 years, and so on. Hence the amounts left at times 0, 29, 58, 87 years, etc., are 1, 12, 14, 18 gram, etc. The time interval of 29 years is called the half-life, or t12, of strontium. Mathematically, we can represent the amount of strontium at different times by the formula 1 t  t1  2 n  n0 a b 2

(40.23)

Here n is the number of strontium nuclei at time t and n0 is the number at the initial time, t  0. With t12  29 years, we can easily verify that at t  0, 29 years, 58 years, etc., the formula (40.23) yields the expected results: 1 0 n  n0 a b  n0  1  n0 2

at t  0

1 29 years29 years 1 1 1 n  n0 a b  n0 a b  n0  2 2 2

at t  29 years

1 58 years29 years 1 2 1 n  n0 a b  n0 a b  n0  2 2 4

at t  58 years

and so on. However, the formula (40.23) is valid not only at these times, but also at intermediate times. For instance, at t  45 years, 1 45 years29 years 1 4529 1 1.55 n  n0 a b  n0 a b  n0 a b 2 2 2  n0  0.34 Figure 40.14 is a plot of the number of remaining strontium nuclei vs. time.

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40.4

The Law of Radioactive Decay

1373

amount of radioactive strontium 1g

Amount of radioactive strontium decreases exponentially, to half of initial value after one half-life, …

1/2

…and to e–1 ≈ 0.368 of initial value after one mean lifetime .

0.368

MARIE SKLODOWSKA CURIE (1867– 1934) and PIERRE CURIE (1859–1906) French physicists and chemists. For their discovery of the radioactive elements radium and polonium, they shared the 1903 Nobel Prize with Henri Becquerel. On the death of Pierre, Marie succeeded him in his professorship at the Sorbonne. She received a second Nobel Prize, in chemistry, for further work on radium. Her daughter Irène Joliot-Curie, shared the 1935 Nobel Prize with Frédéric Joliot-Curie for their work on production of radioactive substances by bombardment with alpha particles.

1/4 1/8 1/16 1/32 0

29  58

87

116

145

174 yr

time

FIGURE 40.14 Amount of remaining radioactive strontium vs. time. In this plot, the amount of strontium is measured in grams. The initial amount is 1 gram, which corresponds to 6.70  1021 nuclei. After 29 years the remaining amount is 12 gram, which corresponds to 3.35  1021 nuclei.

We can conveniently express Eq. (40.23) in terms of the exponential function. Since 12  eln 12  eln 2 (see Math Help: The Exponential Function, page 910 in Chapter 28), we can rewrite Eq. (40.23) as n  n0(eln 2)tt12  n0et (ln 2)t12 If we adopt the notation t

t1  2 ln 2



t12 0.693

(40.24)

this becomes n  n0ett

(40.25)

Equation (40.25) is called the Law of Radioactive Decay, and t is called the mean lifetime, or the time constant. The inverse of t is often called the decay constant l  1t. It can be shown that t is indeed the average lifetime of all the radioactive nuclei in the initial amount of parent material. Note that the mean lifetime is longer than the half-life; for instance, for strontium, the half-life is 29 years, but the mean lifetime is t  (29 years)0.693  42 years. After one time constant, when t  t, Eq. (40.25) tells us that the number of strontium nuclei has decreased to a factor of e1  0.368 times the original number (see Fig. 40.14).

Law of Radioactive Decay mean lifetime  decay constant   1/

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TA B L E 4 0 . 3

RADIOISOTOPE

decay rate

SOME RADIOISOTOPES

RADIOACTIVITY

HALF-LIFE

14

C



5730 years

22

Na

, g

2.6 years

60

Co

, g

5.27 years

90

Sr

, g

28.8 years

131

I

, g

8.04 days

226

Ra

, g

1600 years

238

U

, g

4.5  109 years

With appropriate values of the half-life—and thus, by Eq. (40.24), of the decay time constant—the Law of Radioactive Decay (40.25) is, of course, applicable to the decay of any radioactive isotope. The lifetimes of different radioisotopes vary over a wide range. Some have half-lives of several billions of years; others have half-lives as short as a fraction of a second. Table 40.3 lists the half-lives of some radioisotopes. The chart of isotopes in Table 40.2 lists some more half-lives. Since each decay produces one alpha or beta ray, the rate of emission of rays by the parent material is equal to the decay rate, or the number of parent nuclei that decay per second. To find the instantaneous decay rate we differentiate n with respect to time, which gives us the instantaneous rate of change of the number of nuclei: n0 dn n d  n0ett   ett   t t dt dt

(40.26)

The negative sign of this rate of change indicates that the number of nuclei is decreasing. The number of nuclei that decay per second is the negative of the rate of change dndt: 

decay rate in terms of mean lifetime

dn n  t dt

(40.27)

Thus, the decay rate is directly proportional to the (instantaneous) number of parent nuclei and inversely proportional to the mean lifetime. This means that the decay rate is large if we have a large sample of radioactive material, and it is large if the mean lifetime is short. Equation (40.27) expresses a characteristic property of many random decay processes: the negative of the rate of change of a quantity is proportional to the quantity. It is straightforward to show by direct integration of (40.27) that for such a proportionality, the exponential decay of Eq. (40.25) follows. By substituting the definition (40.24) into Eq. (40.27), we can express the decay rate in terms of the half-life: decay rate in terms of half-life



dn n  0.693 dt t12

(40.28)

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40.4

The Law of Radioactive Decay

Since tables of radioactive isotopes list half-lives, the form (40.28) is convenient for calculations. The radioactive decay rate of a sample, or the number of disintegrations per second, is usually called the activity of the sample. The SI unit for the radioactive decay rate, or the activity, is the becquerel (Bq). 1 becquerel  1 Bq  1 disintegration/s

(40.29)

1375

activity

becquerel (Bq)

In practice, a larger unit is traditionally preferred; this larger unit is the curie 1 curie  1 Ci  3.7  1010 disintegrations/s

(40.30)

curie (Ci)

What is the decay rate of 1.00 gram of radioactive strontium, 90 Sr? The atomic mass of this strontium isotope is 89.9 g.

EXAMPLE 6

SOLUTION: Since the atomic mass of this strontium isotope is 89.9 g, the number

of moles in 1.00 gram of strontium is 1 89.9 mole, or (189.9)  6.02  10 23 atoms  6.70  1021 atoms. With t 12  28.8 years from Table 40.3, Eq. (40.28) then tells us that the decay rate is 

dn n 6.70  1021  0.693  0.693  dt t12 28.8 years  0.693 

6.70  1021 28.8  3.16  107 s

 5.10  1012 s1  5.10  1012 Bq Thus, one gram of strontium-90 emits about 5 trillion beta rays per second! The effects of penetrating radiation on a material depend on the amount of energy absorbed over time. The amount of absorbed radiation energy per kilogram of material is called the absorbed dose. The SI unit of absorbed dose is the gray (Gy), where 1 Gy  1 J/kg. A unit in common use is the rad, 1 rad  0.01 J/kg  0.01 Gy The damage that radiation inflicts on biological tissue depends on the kind of radiation; for example, an alpha particle damages many molecules in each cell in its path, whereas gamma rays damage fewer molecules per cell. To obtain a measure of the biological damage, the absorbed dose is multiplied by a correction factor, the relative biological effectiveness (RBE). The RBE for alpha particles is in the range 10–20, for beta particles it is 1–2, and for gamma rays and X rays it is 1. The measure of the biologically equivalent absorbed dose is the rem5, where [equivalent dose in rem]  [dose in rad]  RBE The SI unit of biological equivalent dose is the sievert (Sv), where 1 Sv  100 rem5. 5 An older measure of ionizing radiation is the roentgen, defined as the amount of radiation that would ionize a charge of 3.34  1010 C in 1 cm3 of air at standard temperature and pressure. The rem is an acronym for roentgen equivalent in man.

absorbed dose

gray (Gy) and rad

relative biological effectiveness (RBE)

equivalent dose in rem

sievert (Sv)

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PHYSICS IN PRACTICE

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R A D I O A C T I V E D AT I N G

Atmospheric carbon dioxide contains the stable isotopes 12C and 13C, and also the radioactive isotope 14C, with a half-life of 5730 years (the 14C isotope is produced by cosmic rays striking the upper atmosphere). While a plant or animal lives, it absorbs carbon dioxide from the air and therefore maintains the same relative abundance of the isotopes 12C and 14C as in air. When the organism dies, it ceases to absorb carbon, and then the 14C in its body decreases due to beta decay [Eq. (40.25)] according to the Law of Radioactive Decay. Thus, fresh wood or bones have the same abundance of 14C relative to 12C as air, but ancient wood or bones have less 14C. Archeologists take advantage of this decrease of 14C for the radioactive dating of materials found in ancient tombs or other sites. They determine the abundance of carbon (all isotopes) in a sample of material by chemical analysis, and they determine the abundance of 14C by measuring the beta activity of the sample. From this, the age can be calculated. For instance, samples of bone taken from the cadaver of the “iceman” (see Fig. 1) discovered in 1991 in the Alps near the Italian-Austrian border have about half the abundance of 14C relative to 12C as air. We can then conclude that about one half of the “normal” amount of 14C has decayed, and that the age of the sample must be about one half-life, that is, about

✔

5700 years (accurate analysis of the 14C abundance gives 5300 years). This means that the iceman died in the late Stone Age. Radioactive dating by carbon can be employed to establish reliable ages for samples as old as about 60 000 years. In older samples the residue of 14C is too small for accurate measurements of beta activity.

FIGURE 1 This “iceman” was preserved for thousands of years by the ice in which he was buried. An exceptionally warm summer melted the ice and revealed his cadaver.

Checkup 40.4 QUESTION 1: Table 40.3 lists some half-lives of radioactive isotopes. Suppose

you have 1.00 mg of each of these isotopes. Which has the highest decay rate? Which the lowest? 131 I, how long does it take for this to decay QUESTION 2: If you initially have 1 mg of 1 1 1 to 2 mg? To 4 mg? To 8 mg? QUESTION 3: According to the Law of Radioactive Decay, at time t  t, what fraction of the initial amount of parent material remains? For radioactive strontium, t  42 years; according to Fig. 40.14, how many grams of strontium remain at this time? 226 Ra (see Table 40.3) is currently Q U E S T I O N 4 : The decay rate of a sample of 9 8.0  10 disintegrations per second. What will be the number of disintegrations per second after 3200 years? (A) 4.0  109 (B) 2.0  109 (C) 1.0  109 (D) 5.0  108 (E) 2.5  105 14 226 QUESTION 5: Suppose you have 1.0 mg each of C, Ra, and 238U. Which has the larger decay rate, that is, which has the larger number of disintegrations per second? (See Table 40.3). (A) 14C (B) 226Ra (C) 238U

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40.5

Fission

1377

40.5 FISSION In heavy, large nuclei—such as uranium—the balance between the attractive strong force and the repulsive electric force is quite precarious. As already mentioned in our discussion of alpha decay, in such a nucleus the electric force becomes dominant if the nucleus suffers some extra elongation, either spontaneous or caused by an external disturbance, such as the impact of a neutron on the nucleus. The repulsive electric force then overpowers the attractive strong force, and the nucleus splits into two fragments. Such a splitting of a nucleus is called fission. Figure 40.15 illustrates the fission of a nucleus after it suffers a disturbance and elongation when hit by a neutron. Fission was first discovered by Otto Hahn, Lise Meitner, and Fritz Strassmann, who bombarded a sample of uranium with a beam of neutrons and found that the uranium fissions into barium and krypton, according to the reaction

Concepts in Context

fission

neutron  238U S 145Ba  94Kr More generally, it was found that uranium can fission in several different ways, with the formation of a variety of fission fragments, each with a mass of about one-half of the mass of the original uranium nucleus. The fission fragments are usually radioactive, and they often emit one or several neutrons. Thus, the net reaction is actually neutron 

238

U S fission fragments  2 or 3 neutrons

(40.31)

While the fission fragments fly apart, the repulsive electric force does positive work on the fragments, giving them a large kinetic energy. This kinetic energy represents the energy released in the reaction. We can calculate the amount of energy released in the reaction in the usual way, by comparing the mass of the uranium nucleus with the (smaller) sum of the masses of the fission fragments. But even without such a calculation we can see from the curve of binding energy (Fig. 40.6) that the fission of a large nucleus leads to a release of energy. By inspecting this figure we recognize that a large nucleus has a somewhat lower amount of binding energy per nucleon than a medium-sized nucleus (the curve has a maximum at A  56, corresponding to 56Fe). Hence the fission of a large nucleus into two medium-sized nuclei results in an increase of the net binding energy, that is, it results in a release of energy.

145Ba

Impact of a neutron on a uranium nucleus…

238U

n 56 protons 89 neutrons 94Kr

92 protons 146 neutrons 36 protons 58 neutrons

FIGURE 40.15 Fission of uranium triggered by the impact of a neutron.

…can result in fission fragments, such as these nuclei of barium and krypton.

LISE MEITNER (1878–1968) Austrian physicist. She studied under Boltzmann, and became the first woman to earn the Ph.D. in physics at the University of Vienna. In Berlin she and Otto Hahn collaborated for several decades, and she became the first female full profesor in Germany. She fled Nazi Germany to Sweden in 1938, just before Hahn discovered the splitting of uranium from chemical evidence. With her nephew, Otto Frisch, she coined the term nuclear fission and described its physical mechanism, but only Hahn was awarded the Nobel Prize in Chemistry for nuclear fission in 1944. Hahn, Meitner, and Hahn’s assistant Fritz Strassman shared the Enrico Fermi Award for their discovery in 1966. Meitnerium (Mt), atomic number 109, was named in her honor.

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Concepts in Context

Nuclei

On the average, the energy released in the fission of a uranium nucleus is about 200 MeV. We can gain a better appreciation of the magnitude of this energy by looking at the following numbers. The total energy released by the complete fission of 1 kg of uranium (a lump slightly larger than a golf ball) is 200 MeV times the number of atoms in 1 kg of uranium. Since a mole of uranium is 238 g, 1.00 kg of uranium is 1000 238  4.20 moles, or 6.02  1023  4.20 atoms  2.53  1024 atoms, and the energy released by the fission of the nuclei of these atoms is 200

MeV  2.53  1024 nuclei  5.1  1026 MeV nucleus  8.1  1013 J

(40.32)

This is equivalent to the energy released in the burning of 2 million liters of gasoline. It is also equivalent to the energy released in the explosion of about 20 000 tons of TNT. To exploit the energy released by the fission of uranium we rely on the neutrons that emerge in the reaction (40.31). If one initial fission occurs in a lump of uranium, the 2 or 3 neutrons that emerge from this reaction can strike other uranium nuclei and trigger their fission, and the neutrons that emerge from these fissions can trigger further fissions, and so on. Thus, one initial neutron can initiate an avalanche of neutrons and an avalanche of fission reactions. Such an avalanche is called a chain reaction (see Fig. 40.16). If no neutrons, or very few neutrons, are lost from the chain reaction, the rate of fission and the rate of release of energy grow drastically with time. For instance, if on the average two of the neutrons released in each fission succeed in generating further fission reactions and further neutrons, then the numbers of fission reactions in successive steps of the chain will be 2, 4, 8, 16, . . .. If this multiplicative growth continues unchecked, the rate of release of energy will become explosive.

chain reaction

95

Y

Three neutrons emerge from one initial fission… 235U

138

I

88Kr

…and can strike other uranium nuclei, triggering their fission… neutron

235U

92Kr

235U

141Ba 235

U 92Sr

145Ba

FIGURE 40.16 A chain reaction. A 235U nucleus absorbs a neutron and fissions, emitting several neutrons, which are absorbed by other 235U nuclei.

…and then more neutrons can trigger more fissions, resulting in a chain reaction.

140

Xe

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In the case of 238U, conditions are not favorable for sustaining a chain reaction. Uranium-238 is a fairly stable nucleus, which does not readily fission when a neutron strikes it. Instead, the nucleus often merely absorbs the incident neutron, removing it from the chain reaction. This loss of neutrons blocks the chain reaction. However, naturally occurring uranium also contains a small amount (0.72%) of the isotope 235U, which is unstable and much more susceptible to fission. This isotope is widely used as the nuclear fuel in applications of chain reactions in nuclear bombs and nuclear reactors. Another isotope used in such applications is 239Pu, an isotope of plutonium. But this isotope does not occur naturally; it must be manufactured by nuclear transmutation reactions involving uranium, in a nuclear reactor.

✔

Checkup 40.5

Does the energy released in fission of a large, heavy nucleus originate from the initial “strong” potential energy or the electric potential energy? QUESTION 2: Why are small, light nuclei incapable of fission? QUESTION 3: Is a chain reaction possible if, on the average, each fission of a nucleus releases less than 1 neutron? Exactly 1 neutron? More than 1 neutron? QUESTION 1:

When a neutron strikes a 235U nucleus resulting in fission, which of the following is true? (A) The rest mass of the products is greater than the sum of the masses of 235U and the neutron. (B) The rest mass of the products is less than the sum of the masses of 235U and the neutron. (C) The products always include Ba (barium) and Kr (krypton). (D) The total number of nucleons in the products is less than 236. (E) The total number of nucleons in the products is greater than 236.

QUESTION 4:

40.6 NUCLEAR BOMBS AND NUCLEAR REACTORS In a given mass of 235U or 239Pu, neutrons produced by spontaneous fission or stray neutrons coming from elsewhere can initiate the first step in the chain reaction. Whether the reaction keeps going depends on how many neutrons are lost from the chain, by absorption without fission (as in 238U) or by escape beyond the boundary of the mass. If the mass is large, few neutrons will reach its boundary before they are intercepted by a nucleus; thus a large mass inhibits escape of the neutrons and favors the chain reaction. The mass is said to be critical if the number of neutrons lost from the chain reaction (by escape or by absorption) equals the number of neutrons released by the fissions. In this case the chain reaction merely proceeds at a constant rate—as in a nuclear reactor. The mass is said to be supercritical if the number of neutrons lost from the chain is smaller than the number of neutrons released in fission reactions. In this case the chain reaction proceeds at an ever-increasing, runaway rate leading to an explosion—as in a nuclear bomb. For pure 235U arranged in a spherical configuration, the critical mass is about 50 kg. The simplest fission bomb, commonly known as an atomic bomb, or A-bomb, consists of two pieces of 235U whose separate masses are each less than the critical

critical and supercritical mass

atomic bomb

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To detonate the first nuclear bomb, a gun propelled one piece of uranium…

propellant explosive

neutron reflector

uranium

gun tube

FIGURE 40.17 A fission bomb. ENRICO FERMI (1901–1954) Italian and later American physicist. He worked on experimental and theoretical investigations of beta decay, the artificial production of isotopes by neutron bombardment, for which he received the 1938 Nobel Prize, and the fission of uranium. Fermi was one of the leaders of the Manhattan Project, and he provided the first experimental demonstration of a chain reaction.

hydrogen bomb

nuclear reactor

neutron reflector …toward another piece, exceeding critical mass and causing a chain reaction.

mass but whose combined mass is more than the critical mass. To detonate such a bomb, the two pieces of uranium, initially at a safe distance from each other, must be suddenly brought close together. In the first such bomb (see Fig. 40.17), the device used for the assembly of the two pieces of uranium consisted of a gun which propelled one piece of uranium toward the other at high speed. More sophisticated fission bombs consist of a (barely) subcritical mass of 239Pu; if this mass is suddenly compressed to higher than normal density, it will become supercritical. The sudden compression is achieved by the preliminary explosion of a chemical high explosive arranged in a layer around the mass of plutonium. The energy released in the explosion of an A-bomb is typically equivalent to that released in the explosion of about 20 kilotons of TNT (see Fig. 40.18). A much higher explosive yield is achieved by a hydrogen bomb, or H-bomb, in which an A-bomb is used to trigger fusion reactions similar to the fusion reactions that power the Sun (see Section 40.7). The energy released in the explosion of an H-bomb is typically equivalent to that in one or several megatons of TNT. Such an explosion would level an entire city, with complete devastation and complete destruction of life by incineration and crushing out to a radius of about 16 km from the center of the explosion. For the peaceful exploitation of nuclear fission in a nuclear reactor, we must keep the chain reaction under control, so it releases energy at a steady rate. This means that the configuration of the uranium or other nuclear fuel must be critical rather than super-

FIGURE 40.18 Explosion of a fission bomb.

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Nuclear Bombs and Nuclear Reactors

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critical. The most common type of reactor operates with “enriched” uranium consisting of a few percent 235U mixed with more than 95 percent of 238U. Such a uranium mixture cannot by itself maintain a chain reaction—the 238U soaks up too many of the neutrons. However, if the uranium is enmeshed in a substance capable of slowing down the neutrons released in fissions, a chain reaction becomes viable. The substance that slows the neutrons is called the moderator. The role of the moderator in a fission reaction is analogous to that of a catalyst in a chemical reaction. The moderator enhances the chain reaction because slow neutrons are more efficient at producing fissions in 235 U than fast neutrons, and they are also less likely to be absorbed by 238U. Inside the reactor, the uranium is usually placed in long fuel rods, and these are immersed in the bulk of the moderator (see Fig. 40.19). Fast neutrons released by fissions travel from the fuel rods into the moderator; there they lose their kinetic energy by collisions with the moderator nuclei; and then they wander back into one or another of the fuel rods and trigger further fissions. The three best moderators are ordinary water (H2O), heavy water (D2O), and graphite (pure carbon). The configuration of the reactor—the size, number, and location of the fuel rods and the shape of the moderator—must be designed so the reactor is nearly critical. Fine adjustments in the number of neutrons and the reaction rate are made by means of control rods of boron or cadmium. These substances greedily soak up neutrons, and by pushing the control rods in or pulling them out, the reaction rate can be decreased or increased. The main application of reactors is for the generation of electric power. In the United States, most of the reactors used for this purpose have water-filled cores. The water acts simultaneously as moderator and as coolant. The water circulates through the core and removes the heat energy released by the fission reactions (see Fig. 40.20).

To suppress the reaction, control rods are inserted to absorb neutrons from uranium fission in fuel rods.

moderator

fuel rods

control rods

Water moderator is also coolant, transferring heat energy to steam, …

…and steam drives turbines that generate electricity.

steam generator

turbine

control rod

pump cooling water hot water outlet

condenser reactor pump

FIGURE 40.20 Schematic diagram of nuclear power plant. reactor vessel

fuel rod moderator (water)

cool water intake

To enhance reaction, collisions in moderator slow neutrons, increasing fission in fuel rods.

FIGURE 40.19 Nuclear reactor.

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The heat is transferred from the water to steam, which drives a steam turbine connected to an electric generator. Thus, the nuclear reactor plays the role of the furnace of a conventional steam engine and uranium replaces coal or oil as fuel.

Concepts in Context

EXAMPLE 7

A nuclear reactor in a power plant produces 1200 MW of heat. How many kilograms of 235U does this reactor consume per year?

SOLUTION: The energy that must be supplied by fission reactions in one year is

the product of the power and the time, E  P ¢t  1200 MW  1 yr  1200  106 W  3.16  107 s  3.8  1016 J From Eq. (40.32) we know that the energy released in the fission of 1.0 kg of uranium is 8.1  1013 J. Hence, the number of kilograms required per year is 3.8  1016 J 

1.0 kg 8.1  1013 J

 470 kg

Nuclear power plants could meet all our energy requirements for the next several hundred years, or even the next several thousand years, if we exploit low-grade ores containing nuclear fuels. Unfortunately, nuclear fission yields rather dirty energy— the fission reactions generate dangerous radioactive residues. Nuclear power plants must be carefully designed to hold these residues in confinement. The reactor core is enclosed in a massive reactor vessel, and as an extra precaution this vessel and the attached pumps and pipes are enclosed in a strong containment shell (see Fig. 40.21). The elaborate safety features that are incorporated in the design of a nuclear power plant make the construction and the maintenance extremely expensive. Furthermore, when the load of fuel of the reactor has been spent, the residual radioactive wastes must be removed to a safe place to be held in storage for thousands of years until their radioactivity has died away. If some of the radioactive material contained in a nuclear reactor is released in the form of smoke or dust in an accidental explosion or fire, it can be carried away by winds, and it can then descend to the ground as lethal radioactive fallout. After the

FIGURE 40.21 The containment shells of the two nuclear reactors at the San Onofre nuclear generating station in California.

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40.6

Nuclear Bombs and Nuclear Reactors

disastrous nuclear accident that happened at Chernobyl, near Kiev in the former U.S.S.R., in 1986, a plume of fallout spread north and west into Europe and produced hazardous contamination thousands of kilometers away. An area of 30-km radius around the reactor was so heavily contaminated it had to be permanently evacuated. The immediate cause of the Chernobyl disaster was operator error—through a series of almost incredible blunders, the operators lost control of the reactor, and its power level surged to 100 times normal values, which overheated the core, blew it apart, and started a fire in the graphite moderator. The water-moderated reactors used in the United States are thought to be much safer than the Chernobyl reactor. But some experts question whether they are safe enough. Scenarios for conceivable reactor accidents have been analyzed in great detail. The worst that might happen is a loss-of-coolant accident, that is, the loss of the water that normally circulates through the core of the reactors serving both as moderator and as coolant. Although such a loss of water would shut off the fission chain, the reactor contains a large amount of radioactive residues, and the heat from the radioactivity can by itself overheat the reactor vessel beyond safe limits. As a precaution, reactors are equipped with emergency cooling systems, and they are also encased in a containment shell. Nuclear power reactors in the United States all use 235U as fuel. Unfortunately, the supply of this nuclear fuel is rather limited, and it is expected to become exhausted in this century. There are, however, several other nuclear fuels in larger supply. One of these is 238U. Although this isotope is incapable of supporting chain reactions, it can be converted into 239Pu, which supports chain reactions. The manufacture of 239Pu is an automatic side effect of the operation of nuclear reactors now in operation; in all of these reactors, the fuel rods contain a mixture of 235U and 238U, and fission neutrons striking the 238U gradually convert it into 239Pu. A reactor fueled with 239Pu not only makes good use of a material that would otherwise go to waste, but, if the reactor is surrounded by a blanket of 238U, it can also manufacture extra 239Pu. The number of neutrons released in the fission of 239Pu is so large that in an efficiently designed reactor neutrons can be diverted to the 238U without hindering the chain reaction. Such a reactor can produce more 239Pu (from 238U) than it consumes (from its original supply). A reactor of this kind is called a breeder reactor. Because of worries over the safety of these reactors, they have not been adopted in the United States, but dozens are operating in Europe.

✔

Checkup 40.6

QUESTION 1: If each fission of a uranium nucleus produces three neutrons, how many of these neutrons can you afford to lose, if you want to maintain a chain reaction? QUESTION 2: Why is it impossible to maintain a chain reaction in a very small mass of uranium, say, 100 g? 238 QUESTION 3: Why is the presence of U detrimental to the maintenance of a chain reaction in a mass of 235U? QUESTION 4: A breeder reactor produces more nuclear fuel than it consumes. Is this in conflict with conservation of energy? QUESTION 5: In a nuclear reactor, the moderator (A) Coordinates the activities of the technicians operating the reactor. (B) Controls the steam that drives the turbines of the electrical generator. (C) Absorbs neutrons before they decay into protons and electrons. (D) Slows neutrons so that they can be absorbed by 235U in fuel rods later. (E) Slows the reaction by permitting more neutrons to be absorbed by 238U.

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breeder reactor

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40.7 FUSION fusion

Fusion is the merging of two small nuclei to form a larger nucleus. The fusion of two small nuclei, such as the nuclei of hydrogen, deuterium, or tritium, releases energy. For instance, the heat of the Sun is generated by a fusion reaction called hydrogen burning, in which hydrogen nuclei fuse together to make helium nuclei. By inspection of the curve of binding energy, we can immediately recognize that the fusion of small nuclei leads to a release of energy. Figure 40.6 shows us that the smallest nuclei have exceptionally low binding energies per nucleon. Hence, the fusion of two such nuclei into a larger nucleus of higher binding energy results in release of energy. Note that the process of fusion is the reverse of fission: small nuclei (such as hydrogen) release energy when they fuse; large nuclei (such as uranium) release energy when they split. The hydrogen burning in the Sun proceeds in three steps; first hydrogen nuclei fuse to make deuterium (2H), then deuterium fuses with hydrogen to make helium-3 (3He), and finally helium-3 fuses with helium-3 to make helium-4 (4He): 1

HANS ALBRECHT BETHE (1906–2005) German and later American Physicist. During World War II he worked on the Manhattan Project as director of the Theoretical Physics Division. He received the Nobel Prize in 1967 for his investigations of nuclear reactions in stars.

thermonuclear reaction

H  1H S 2H  e   1

(40.33)

H  2H S 3He  g

(40.34)

He  3He S 4He  1H  1H

(40.35)

3

Each of the first two reactions must occur twice for the last reaction to occur once. Each of these reactions releases energy; when the first two reactions occur twice and the last occurs once (with the formation of one nucleus of 4He), the net energy released is 24.7 MeV. Four nuclei of 1H are consumed in this process (6 are consumed when the first two reactions occur twice, but 2 are regenerated when the last reaction occurs); hence the amount of energy released per nucleon of “fuel” is 24.7 MeV per 4 nucleons, or 6.2 MeV per nucleon. This number is to be compared with the energy released in the fission of uranium, about 200 MeV per 235 nucleons of “fuel,” or 0.85 MeV per nucleon. Thus, fusion of a given mass of 1H in the Sun releases about 7 times as much energy as the fission of an equal mass of 235U. The fusion reactions (40.33)–(40.35) are called thermonuclear, because they can proceed only at extremely high temperatures and pressures, such as the temperature of about 1.5  107 K in the core of the Sun. The high temperature is needed to overcome the electric repulsion that the hydrogen nuclei experience whenever they come close together. At high temperatures, the hydrogen nuclei have high speeds, and their collisions are sufficiently violent to overcome the electric repulsion and bring the nuclei into the intimate contact required for fusion. The thermonuclear reaction based on hydrogen fuel cannot be duplicated on Earth, because we cannot attain the pressure found in the core of the Sun. However, there are some fusion reactions based on deuterium (2H) and tritium (3H) that proceed at lower pressures, for example, 2

H  3H S 4He  neutron

(40.36)

Although the reaction (40.36) can proceed at attainable pressures, it requires a temperature of about 108 K, even higher than in the center of the Sun. Such temperatures have been attained in the explosions of H-bombs, where the preliminary explosion of a fission bomb heats and compresses a mixture of deuterium and tritium and thereby initiates fusion.

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40.7

Fusion

Experimental attempts at a peaceful exploitation of fusion power also rely on a deuterium–tritium fuel mixture, which is heated to the required temperature of about 108 K either by an intense electric current, or by a particle beam, or by a laser beam. At such a temperature, the deuterium and tritium will be in the form of a plasma, that is, a totally ionized gas, consisting of a mixture of independent nuclei and electrons. The plasma cannot be contained by the walls of a conventional reactor vessel, since any contact with the wall of the vessel would cool the plasma, and would also melt the wall. Instead, the plasma must be held suspended in the middle of the vessel, away from contact with the walls. One scheme for controlled fusion attempts to suspend the plasma by means of magnetic fields, that is, by magnetic confinement. Figure 40.22 shows the Princeton Tokamak Test Reactor, in which the plasma is confined inside a large toroidal solenoid (a solenoid shaped like a donut). The plasma is heated by a combination of electric currents induced in the plasma and particle beams aimed into it. A larger test reactor, a 500 MW fusion power generator known as ITER (International Thermonuclear Experimental Reactor), is under construction at Cadarache, France. Another scheme for fusion attempts to extract energy by exploding small pellets of a deuterium–tritium mixture in a combustion chamber by hitting them with intense laser beams. This scheme is called inertial confinement because the inertia of the pellet holds it together long enough for the reaction to occur. The beams heat the pellet so suddenly that it has no time to disperse before fusion begins. The pellet then explodes like a miniature hydrogen bomb. After the thermal energy has been extracted from the combustion chamber, the next pellet is placed in the chamber, and so on. Figure 40.23 shows the combustion chamber of the National Ignition Facility (NIF) at Lawrence Livermore National Laboratory, where 192 high-power laser beams converge to heat a deuterium–tritium pellet (see also Chapter 26). Although both magnetic and inertial confinement schemes have been successful in initiating fusion, the amounts of energy released in fusion have remained far below the amount of energy that had to be fed into the reactor chambers to heat the plasma. Nuclear fusion is an attractive source of energy because it bypasses many of the safety problems associated with nuclear fission, especially the production of heavy, long-lived radioactive nuclei. However, tritium is radioactive and thus raises safety concerns, both when used as a fuel and when produced in the liquid lithium coolant of proposed fusion reactors by absorption of fast neutrons. Since the half-life of tritium is fairly short (12.3 years) and extraneous tritium could be recycled for fuel,

FIGURE 40.22 The Princeton Tokamak Test Reactor.

1385

magnetic confinement

inertial confinement

FIGURE 40.23 Combustion chamber at the National Ignition Facility in Livermore, California.

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fusion is considered a potential source of much cleaner energy than fission. Besides, we have available an enormous supply of deuterium. There is enough heavy water mixed with the ordinary water in the oceans of the Earth to satisfy our energy needs for millions of years.

✔

Checkup 40.7

Why are large, heavy nuclei incapable of fusion? QUESTION 2: What is the net result of the sequence of fusion reactions (40.33)–(40.35), that is, what nuclei are consumed, and what nuclei are produced? QUESTION 3: Fusion requires high temperature. Is this also true for fission? QUESTION 4: Do the fusion reactions (40.33)–(40.35) produce any radioactive nuclei? QUESTION 5: Tritium is produced in the reaction QUESTION 1:

2

H  2H S 3H  1H

What is the net result of this reaction followed by reaction (40.36)? (A) 1H  2H  3H S 4He  1H  neutron. (B) 1H  2H  2H S 3He  1H  neutron. (C) 2H  2H  3H S 4He  1H  2 neutron. (D) 2H  2H  2H S 4He  1H  neutron. (E) 2H  2H  2H S 3He  1H  2 neutron.

S U M M A RY PROBLEM-SOLVING TECHNIQUES PHYSICS IN PRACTICE

(page 1371)

Nuclear Reactions

(page 1376)

Radioactive Dating

Atoms with the same number of protons in their nuclei, but different numbers of neutrons.

ISOTOPES

RADIUS OF NUCLEUS

(A is mass number.)

ATOMIC MASS UNIT u ENERGY EQUIVALENT OF ATOMIC MASS UNIT MASS DEFECT AND BINDING ENERGY (B.E.) Q VALUE OF REACTION

Initial masses m1, m2, p and final masses m 1, m 2, p

11C

12C

6 protons 5 neutrons

6 protons 6 neutrons

R  (1.2  1015 m)  A 13

(40.2)

1 u  1.660 54  1027 kg 1 u  c2  931.5 MeV [mass of N neutrons and Z protons]  [mass of nucleus]  B.E./c 2 Q  (m1c 2  m2c 2  )  (m 1c 2  m 2c 2  )

(40.7) (40.8) (40.11)

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Questions For Discussion

alpha decay:  particle (4He nucleus) beta decay:  particle (electron or antielectron) gamma decay:  ray (high-energy photon)

KINDS OF RADIOACTIVITY

LAW OF RADIOACTIVE DECAY

where t 

t1  2



1387

n  n0 ett

t1  2

, t is the half-life, ln 2 0.693 12 and t is the mean lifetime, or time constant, of the decay.

or

1 tt12 n  n0 a b 2



 

(40.23) (40.25)

DECAY RATE OR ACTIVITY



n dn n   0.693 t dt t12

(40.28)

1 Bq  1 disintegration/s

(40.29)

CURIE

1 Ci  3.7  1010 disintegrations/s

(40.30)

SI UNIT OF ABSORBED DOSE

1 gray  1 Gy  1 J/kg  100 rad

BECQUEREL

10–20 for alpha 1–2 for beta 1 for gamma

RELATIVE BIOLOGICAL EFFECTIVENESS (RBE)

EQUIVALENT ABSORBED DOSE

[equivalent dose in rem]  [dose in rad]  RBE

SI UNIT OF EQUIVALENT ABSORBED DOSE

1 sievert  1 Sv  100 rem

The splitting of a nucleus into two fragments, usually with the emission of one or more neutrons.

FISSION

The merging of two small nuclei to form a larger nucleus.

FUSION

QUESTIONS FOR DISCUSSION 1. What is the average overall density of a typical atom, such as iron? Why is this much smaller than the density of a nucleus, is given by Eq. (40.4)? 2. Naturally occurring magnesium has an atomic mass of 24.305. On the basis of this information, can you conclude that natural magnesium contains a mixture of several isotopes? Can you guess which isotopes?

3. According to Fig. 40.6, which isotope has the largest binding energy? 4. Why do alpha and beta emissions involve transmutation of elements whereas gamma emission does not? 5. If you irradiate a sample of material with alpha, beta, or gamma rays, is the sample likely to become radioactive? What if you irradiate it with neutrons?

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6. Radiation is sometimes used to sterilize food or surgical supplies. What are the advantages or disadvantages of this procedure over sterilization by heating? 7. Can radioactive dating with 14C be used to determine the age of a book printed during the last century? The age of deposits of mineral oil? The age of ancient Egyptian gold jewelry? Ancient Egyptian furniture? 8. Why is an absorbed dose of alpha radiation more likely to kill a biological cell than the same absorbed dose of gamma radiation? 9. Tritium (3H) is a radioactive isotope of hydrogen which occurs naturally in small concentrations in the ordinary water found in the environment. The half-life of this isotope is 12 years. Describe how you could take advantage of this isotope to determine the age of a bottle full of wine that your wine merchant claims is 25 years old. 10. Alpha particles are more massive than beta particles, yet for equal kinetic energies, the alpha particles are stopped by a thinner layer of material than the beta particles. Can you explain why an alpha particle loses its energy more quickly? (Hint: Think of the electric forces that act on the particles in a collision with an atom.) 11. All the best moderators for reactors consist of fairly lightweight nuclei. Why is a material with heavy nuclei, such as lead, not a good moderator?

12. Why is it difficult to separate the isotopes of 238U and 235U? 13. Some artificial satellites carry small nuclear reactors as power supplies. What danger does this pose for people on Earth? 14. Nuclear fission bombs produce a long-lived radioactive isotope of strontium (90Sr), an element that is chemically similar to calcium. Explain why strontium poses a severe hazard to humans and other vertebrates. 15. Hydrogen bombs operate with tritium, a radioactive isotope of hydrogen with a half-life of 12.3 years. When a hydrogen bomb is held in storage, its tritium gradually decays, and must be replenished with fresh tritium every few years. According to a recent proposal, arms control could be achieved by halting production of tritium. Every few years, the superpowers would then have to take some of their bombs out of service, extract the remaining tritium, and use it to replenish their other bombs. Suppose the United States and Russia each have 2000 hydrogen bombs now. Without fresh tritium, how many bombs will each superpower have in 25 years? In 50 years? 16. Why is it more difficult to achieve controlled fusion in a reactor than controlled fission? 17. A fusion reactor would produce a large amount of tritium (3H), a radioactive isotope of hydrogen. If the tritium were accidentally released into the environment, it would be likely to contaminate the water. Explain.

PROBLEMS 40.1 Isotopes 1. How many protons and neutrons are there in the nucleus of the isotope 16O? 56Fe? 238U? 2. What isotope has 82 protons and 122 neutrons in its nucleus? 3. What are the number of protons and the number of neutrons in each of the following isotopes: 24Na, 27Al, 52Cr, 52Mn, 63 Cu, 63Zn, 124Xe, 138La?

7. In any nuclear reaction, the nuclear electric charge must be conserved and the mass number must be conserved. Are the following reactions in accord with these conservation laws? Explain. 2

H  12C S 4He  9B

4

He  10B S 13C  n

n  238U S 121Ag  118Pd

4. What isotope has 17 protons and 18 neutrons in its nucleus? What isotope has 18 protons and 17 neutrons? Such isotopes in which the numbers of protons and neutrons are exchanged are called mirror nuclei. Find another example of mirror nuclei in the chart of isotopes in Table 40.2.

8. Naturally occurring boron is a mixture of 80.2% 11B and 19.8% 10B. From the masses of these isotopes (listed in the chart of isotopes in Table 40.2) calculate the atomic mass of naturally occurring boron.

5. Use the chart of isotopes in Table 40.2 to make a list of all the isotopes of oxygen. What are the number of protons and the number of neutrons in the nucleus of each of these isotopes?

9. According to Eq. (40.2), what is the nuclear radius of the smallest of the isotopes of carbon? The largest of these isotopes?

6. By inspection of Fig. 40.5, estimate the ratio of neutrons to protons in stable light nuclei (Z  20) and in stable heavy nuclei (Z  80).

10. The largest known nucleus is that of the isotope 266 of element 109, which was recently named meitnerium (Mt). What is the radius of the 266Mt nucleus?

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Problems

11. In some reference books, isotopes are listed with their atomic number as a left subscript; for example, deuterium (2H) would be listed as 12 H, and carbon-14 would be 146 C. Determine which 31 44 207 of the following are incorrectly written: 15 P, 22Sc, 118 50 Sn, 83 Pb. Since the subscript is unique for each element, we could consider using a notation without the atomic symbol of the ele89 ment. What isotope would be meant by 39 ? By 197 79 ? 12. Many stable nuclei have an even mass number A, and all but a few stable nuclei have an even number N of neutrons or an even atomic number Z. Certain even numbers for N and Z correspond to very stable nuclei; these “magic nuclei” have (Z or N )  2, 8, 14, 20, 28, 50, 82, or 126. These suggest a nuclear shell structure analogous to atomic shells, with independent shells for protons and neutrons. In the chart of isotopes in Table 40.2 for what value of N are there the most stable isotopes (that is, for what value of N are there the most naturally occurring isotopes, as indicated by any listed values for the percent abundance)? For what value of Z? What do the stable isotopes 36S, 37Cl, 38Ar, 39K, and 40Ca have in common? The most abundant stable isotopes of barium, lanthanum, cerium, praseodymium, and neodymium are 138Ba, 139La, 140 Ce, 141Pr and 142Nd. What do these have in common? 13. What stable isotope has one-half the radius of a 104Ru nucleus? One-half the radius of a 128Xe nucleus? See the chart of isotopes in Table 40.2. 14. Chlorine has two stable isotopes, 35Cl and 37Cl, with atomic masses 34.968 85 u and 36.965 90 u, respectively. In the periodic table, the average atomic mass of chlorine is 35.453 u. From these data, calculate the percent abundance of the two stable chlorine isotopes. 15. Nuclei with the same mass number but different atomic numbers are called isobars. Find the number of isobars for each mass number from 1 to 17. Which nuclei have no isobars? Use the chart of isotopes in Table 40.2. *16. Neutron stars are made (almost) entirely of neutrons, and they have approximately the same density as that of a nucleus. What is the radius of a neutron star of mass 0.50 times the mass of the Sun? *17. What fraction of the volume of your body is filled with nuclear material? The average density of your body is about 1000 kg/m3. *18. Suppose that an alpha particle of energy 4.4 MeV collides head-on with a stationary gold nucleus. What is the distance of closest approach? Does the alpha particle make contact with the surface of the nucleus? Since the gold nucleus is so heavy, assume for simplicity that it gains no energy in the collision.

40.2 The Strong Force and the Nuclear Binding Energy 19. The binding energy of the electron in the ground state of the hydrogen atom is 13.6 eV. Calculate the corresponding mass defect of the hydrogen atom; express this in atomic mass units. (Since the mass defects associated with atomic binding energies are small, they are usually ignored in nuclear physics.)

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20. According to Fig. 40.6, what is the binding energy of 56Fe? From this binding energy, calculate the mass of the isotope (in atomic mass units). 21. What are the mass defect (in atomic mass units) and the nuclear binding energy (in MeV) of the isotope 4He? The mass of 4He is given in the chart of isotopes in Table 40.2. 22. What are the mass defect (in atomic mass units) and the nuclear binding energy (in MeV) of the isotope 12C? 23. Nuclei with 2, 8, 14, 20, 28, 50, 82, or 126 protons or neutrons are called magic nuclei because they have exceptionally large binding energies and they are exceptionally stable (see also Problem 12). Compare the binding energy of the magic nucleus 16O with that of the nonmagic nucleus 16F. The masses of these isotopes are given in Table 40.2. 24. How much energy (in MeV) is required to remove one neutron from the nucleus of the isotope 14N? (Hint: Compare the binding energies of 14N and 13N; the masses of these isotopes are given in the chart of isotopes in Table 40.2. 25. Nuclei with “magic numbers” of protons or neutrons are exceptionally stable (see also Problems 12 and 23), analogous to a closed shell in atomic physics. When a nucleus has only one or a few nucleons outside a closed shell, such nucleons are often relatively weakly bound. Consider cerium-140, which has a magic number of 82 neutrons. Compare 140Ce and 142Ce to find the average binding energy for each of the two extra neutrons in 142 Ce. Compare your result with the average binding energy per nucleon in 140Ce. The atomic masses of 140Ce, 142Ce, and the neutron are 139.9054 u, 141.9092 u, and 1.0087 u. 26. Two nuclei with the same mass number but different atomic numbers are called isobars. Calculate the binding energy per nucleon for the stable isobars germanium-74 and selenium74. Which nucleus is more tightly bound? The atomic masses of 74Ge and 74Se are 73.9218 u and 73.9225 u, respectively. 27. Verify that in the reaction (40.10), both sides have the same number of protons and neutrons. *28. When boron is bombarded with alpha particles, the following reaction is observed: 4

He  10B S 13C  1H

How much energy is released in this reaction? *29. When 7Li is bombarded with protons, the following reaction occurs: 1

H  7Li S 7Be  n

What is the minimum kinetic energy required for the proton to initiate this reaction? Neglect the recoil energy of the 7Be nucleus. *30. Consider the reaction 1

H  3H S 3He  n

(a) What is the energy absorbed in this reaction? (b) We can initiate this reaction either by bombarding a tritium target with protons, or by bombarding a hydrogen

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target with tritium ions. To initiate this reaction, the kinetic energy of the bombarding particles must be somewhat larger than the energy absorbed in the reaction, since the recoil of the reaction products retains some of the energy supplied by the bombarding particle. If tritium ions are used as bombarding particles, the required kinetic energy is larger than if the protons are used as bombarding particles. Explain the difference.

absorbs one of its own atomic electrons. A neutrino is created in this process. What is the energy of the emitted neutrino when radioactive 37Ar transmutes via electron capture into the stable nucleus 37Cl? The atomic masses of 37Ar and 37Cl are 36.966 78 u and 36.965 90 u, respectively. 38. When 129I beta-decays into 129Xe, what is the maximum possible kinetic energy of the emitted electron? The maximum possible energy of the emitted neutrino? What are the minimum energies? The atomic masses of 129I and 129Xe are 128.904 98 u and 128.904 78 u, respectively.

*31. Consider the spontaneous fission of uranium into barium and krypton, according to the reaction

39. The nucleus 57Co is often used as a a source of gamma rays for experiments in magnetism, since these gamma rays are readily absorbed by iron nuclei. Suppose an atom of 57Co at rest emits a 0.707-MeV gamma ray. What is the recoil velocity of the cobalt atom?

238

U S 145Ba  93Kr

The masses of the isotopes 238U, 145Ba, and 93Kr are 238.050 79 u, 144.926 87 u, and 92.931 12 u, respectively. (a) Find the Q value for this reaction. (b) The uranium nucleus is initially at rest. Find the final speeds of the barium and krypton nuclei after the reaction, and find their individual energies. *32. For large nuclei, the total electrostatic potential energy can be approximated by that of a sphere of uniform positive charge density r. This energy can be calculated by building up spherical shells of charge. (a) Show that the potential at the surface of a small sphere of charge q and radius r is V

*40. The alpha decay of 210Po results in 206Pb. Calculate the energy of the emitted alpha particle. The masses of these two isotopes are 209.9829 u and 205.9745 u, respectively. *41. What is the maximum kinetic energy of the beta ray emitted in the beta decay of a neutron, according to the reaction (40.18)? Ignore the recoil of the proton. *42. If neutrons had a somewhat smaller mass, then the (slow) electrons in atoms could combine with protons in the nucleus according to the reaction e  p S n  

rr 2 1 q  4p0 r 30

How much smaller would the mass of the neutron have to be to make this reaction viable? What consequences would this reaction have for the existence of atoms and the existence of life?

2

(b) Adding a shell of charge dq  r d (volume)  r 4pr dr requires an energy dU  V dq Integrate such contributions from r  0 to r  R to obtain the total electrostatic energy of a sphere of radius R and total charge Q. (c) Use Eq. (40.2) and the result from (b) to estimate the average electrostatic energy per proton for 238U (Z  92). †

216

Po? 239Pu?

34. What isotope is formed by the negative beta decay of

85

Kr? 63Ni?

35. What isotope is formed by the positive beta decay, or beta plus decay, of 22Na? 64Cu? 36. The alpha decay of uranium [see Eq. (40.13)] is the first step in a radioactive series of decays. The next four steps are a negative beta decay, another negative beta decay, another alpha decay, and another alpha decay. What are the daughter nuclei produced in these four steps? 37. For some radioactive nuclei, the inverse of beta decay is favorable, a process known as electron capture; typically, the nucleus †

†

40.4 The Law of Radioactive Decay 44. According to Fig. 40.14, at what time will the remaining amount of radioactive strontium have fallen to 1 10 of the initial amount?

40.3 Radioactivity 33. What isotope is formed by the alpha decay of

*43. What isotope is formed in the negative beta decay of 14C? Calculate the maximum energy of the beta rays emitted in this decay. The masses of the relevant isotopes are given in the chart of isotopes in Table 40.2.

For help, see Online Concept Tutorial 45 at www.wwnorton.com/physics

45. According to the best available data, the half-life of 14C is believed to be 5730 years. However, according to previous data, the half-life was thought to be 5570 years, and age determinations based on this value of the half-life were in error. What percentage error in age determination does the error in the half-life introduce? For a sample from the year 3000 B.C., what is the error in years? 46. The isotope 238U found on the Earth was originally synthesized in nuclear reactions in supernovas that exploded in our Galaxy about 6.8  109 years ago and scattered debris through the Galaxy, some of which became trapped in the Earth during its formation, about 4.6  109 years ago. What fraction of the original amount of uranium synthesized in the

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Galaxy is still in existence? What fraction of the amount of uranium trapped in the Earth during its formation is still in existence? The half-life of 238U is 4.5  109 years. 47. The activity of an industrial 60Co irradiation cell is 1.0  1016 Bq when the cell is new. What will be the activity after the cell has been in use for 2.0 years? The half-life of 60 Co is 5.3 years. 48. What is the activity of 1.00 gram of pure radium, 226Ra? The half-life of this isotope is listed in Table 40.3. 49. What is the activity of 1.0 microgram of 22Na? The half-life of this isotope is listed in Table 40.3. 50. Some modern theories of elementary particles propose that the proton might be unstable with a very long half-life. Suppose the half-life for proton decay is 1.0  1033 years. What would be the decay rate for the protons in an experiment using 5.0  104 liters of water? What is the inverse of this decay rate, that is, how long on average would you have to wait for one disintegration? 51. The activity of a sample of 57Co is initially 12.2 mCi (millicuries). After 2.00 years, the activity has dropped to 1.87 mCi. From these data, what is the half-life of 57Co? If the initial sample was pure 57Co (atomic mass 56.9), what was the initial mass of this sample? 52. 1.00 mole of naturally occurring rubidium emits beta rays at a decay rate of 7.75  104 Bq due to the radioactive isotope 87 Rb, which has a natural isotopic abundance of 27.8%. From these data, calculate the half-life of 87Rb. 53. A 75-kg worker receives an accidental whole-body dose of 15 rads of beta radiation with an RBE of 1.7. What energy is absorbed? What is the absorbed dose in grays? What is the equivalent absorbed dose in rems? In sieverts? 54. What absorbed dose in rads of alpha radiation with an RBE of 20 causes the same amount of biological damage as 10 rads of gamma radiation? 55. During a radiation treatment of a 75-g tumor, a patient absorbs 5.0 J of gamma radiation. The tumor absorbs 30% of the radiation; by rotating the gamma-ray beam, the remaining energy is distributed throughout 1.5 kg of surrounding tissue. What is the radiation dose absorbed by the tumor? By the surrounding tissue?

40.5 Fission 40.6 Nuclear Bombs and Nuclear Reactors 56. According to the curve of binding energy in Fig. 40.6, what is the binding energy of a nucleus with mass number A  200? What is the binding energy of a nucleus with mass number A  100? What is the energy released in the fission of the A  200 nucleus into two A  100 nuclei? 57. The bomb exploded at Hiroshima had an explosive yield of about 20 kilotons of TNT, or 8.4  1013 J. How many kilo-

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grams of uranium actually underwent fission to release this amount of energy? 58. Use the curve of binding energy to find the nucleus of smallest mass number A for which fission into two nuclei of mass number A2 is energetically viable. (Hint: For fission to be viable, the height of the curve at A2 must be above the height at A. Why?) 59. Naturally occurring uranium contains 99.28% of the isotope 238 U and 0.72% of the isotope 235U. How many kilograms of natural uranium must we process to extract the 50 kg of 235U required for a bomb? Pretend that all of the 235U can be extracted. 60. Some artificial space probes that travel far from the Sun are powered by nuclear energy, typically the alpha decay of plutonium-238. Suppose such a satellite uses 2.0 kg of 238Pu. At what rate is energy generated from alpha decay? The atomic masses of 238Pu and 234U are 238.0496 u and 234.0409 u, respectively, and the half-life of 238Pu is 88 years. 61. How much energy is released in the following fission reaction? n  235U S 87Rb  137Cs  12 n The atomic masses of rubidium-87 and cesium-137 are 86.9092 u and 136.9071 u, respectively. *62. Consider the fission reaction n  235U S 144Ba  92Kr Given that the masses of the isotopes 235U, 144Ba, and 92Kr are 235.043 94 u, 143.922 85 u, and 91.926 27 u, respectively, calculate the energy released in this reaction.

40.7 Fusion 63. Some designs for fusion reactors include a liquid lithium blanket for the absorption of fast neutrons and energy. The lithium also generates more tritium for fuel according to the reaction n  6Li S 3H  4He How much energy is generated in this reaction? Use the chart of isotopes in Table 40.2. 64. To achieve fusion, a sufficient number density n of nuclei must be confined for a long enough time t to provide a sustained reaction. For deuterium–tritium fusion the condition is n t  1.0  1020 s/m3, known as Lawson’s criterion. (a) In magnetic confinement test reactors, confinement times of about 0.50 s have been achieved. What is the corresponding required density of nuclei? How does this compare with the density of an ideal gas at standard temperature and pressure? (b) For inertial confinement, suppose the fuel is in a liquid pellet with the same number density as that of protons in water. The laser pulse lasts only 1.0  1010 s; take this to be the confinement time. By what factor must the pulse compress the volume of the liquid to achieve Lawson’s criterion?

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*65. For deuterium and tritium to fuse, they must get close enough to touch. (a) Use Eq. (40.2) to estimate the radii of deuterium and tritium and thus obtain their center-to-center distance when barely touching. (b) Calculate the electrostatic potential energy of the two nuclei when touching. (c) If the nuclei approach head-on, each with kinetic energy equal to about half the amount obtained in (b), they can fuse. At a temperature T, there will be a broad distribution of thermal velocities. Suppose that at a temperature T, enough nuclei with kinetic energy equal to 10 times the average thermal kinetic energy 32 kT are present for fusion. From these considerations, estimate the temperature needed for fusion. *66. Consider the three fusion reactions (40.33)–(40.35) that occur in the core of the Sun. From the masses of the nuclei that participate in these reactions (see the chart of isotopes in Table 40.2), calculate the energy released in each reaction. What is the net energy if the first two reactions occur twice and the last reaction occurs once? *67. (a) The rate at which the Sun radiates heat and light is 3.9  10 26 W. If the energy for all of this radiation comes from the fusion reactions (40.33)–(40.35), what is the rate at which the Sun consumes hydrogen?

(b) The amount of hydrogen in the Sun is about 1.5  1030 kg. How long will this fuel last if the Sun goes on radiating at a steady rate? *68. Whenever the first two of the fusion reactions (40.33)–(40.35) proceed twice and the last proceeds once, 24.7 MeV is released, and so are two neutrinos. Hence, the number of neutrinos released is 1 per 12.3 MeV. Assume that the fusion reactions (40.33)–(40.35) account for all of the 3.9  1026 W of heat and light radiated by the Sun. Calculate the rate at which the Sun releases neutrinos. **69. A sequence of reactions that can be used in controlled fusion is the burning of deuterium “fuel,” as follows: 2 2

H  2H S 3H  1H

2 2

H  2H S 3He  n

H  3H S 4He  n

H  3He S 4He  1H

The net result of this sequence of reactions is the transmutation of six nuclei of deuterium (2H) into two nuclei of 4He, two protons (1H), and two neutrons. Calculate the energy released in each of these reactions. What is the net energy released per nucleon of fuel?

REVIEW PROBLEMS 70. How many protons and neutrons are there in the nucleus of the isotope 18Na? 39K? 231Pu? *71. Suppose you bombard a target of magnesium with alpha particles. If, in a head-on collision with a stationary magnesium nucleus, an alpha particle is to reach the nuclear surface just barely before being halted by the repulsive electric force, what must be the energy of the alpha particle? Express your answer in MeV. 72. A nucleus of radius 7.4  1015 m fissions into two equal spherical pieces. What is the radius of each piece? 73. Chadwick discovered the neutron when he bombarded boron with alpha particles, which resulted in the reaction 4

He  11B S 14N  n

Calculate the energy released in this reaction. 74. Consider the following nuclear reaction: 2

12

4

deuterons (nuclei of deuterium) of 1.2 MeV? Neglect the recoil of the carbon nucleus. 75. Bombardment of 13C with protons produces an isotope of nitrogen, according to the reaction 1

H  13C S 13N  n

What is the minimum energy that the proton must have to initiate this reaction? Neglect the recoil of the carbon nucleus. *76. Nuclear physicists sometimes deduce the mass of an isotope from the energy of the beta rays emitted by the isotope. For instance, the maximum energy of the beta rays emitted in the decay of 27Mg into 27Al is 2.610 MeV. Given that the mass of 27 Al is 26.9815 41 u, what mass can you deduce for 27Mg? *77. Find the maximum kinetic energy of the beta rays emitted in the decay 16

N S 16O  e  

10

H  C S He  B

What is the energy absorbed in this reaction? Can this reaction be initiated by bombarding carbon with a beam of

78. One of the dangerous radioactive isotopes in the radioactive waste produced by nuclear reactors is 85Kr, with a half-life of 10.8 yr. How long must we hold the radioactive waste in stor-

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age to reduce the activity of the 85Kr material to 0.10% of its initial value? 79. Bananas contain potassium, in the form of 39K (93.258%), 40K (0.0117%), and 41K (6.7302%). The isotope 40K is radioactive, with a half-life of 1.28  109 yr. The total amount of potassium in a banana is 0.50 g. Calculate the activity of a banana. 80. For diagnostic medical imaging, the isotope iodine-123 is absorbed by healthy thyroid tissue and emits 1.4-MeV gamma rays. Similarly, technetium-99 is absorbed by abnormal brain tissue and emits 0.14-MeV gamma rays. Thus, the absence or presence of gamma emission from a region of the thyroid or brain, respectively, is indicative of abnormal tissue. Assume that a compound containing 0.10 g of each isotope is injected into the bloodstream of a 25-kg child, and that 10% of the radiation is absorbed throughout the body. What is the absorbed dose in each case? *81. The partial eradication of the thyroid in patients suffering from hyperthyroidism can be accomplished by injecting a compound containing radioactive iodine 131I into the body; the iodine then concentrates in the thyroid and kills its cells. If the thyroid is to be subjected to an activity of 0.10 Ci, how many grams of 131I should be injected? The half-life of 131I is 8.04 days. 82. In a mass spectrometer, ions are accelerated through a potential and enter a region of uniform magnetic field, where they travel in a semicircular arc before reaching a detector (see Fig. 30.26). Atoms from an unknown substance are singly ionized and analyzed using a magnetic field of 0.425 T. The detector is fixed for an arc radius of 6.50 cm. When the accelerating voltage is varied, a large number of ions are detected at 5249 V, and a small number at 6123 V. From the chart of isotopes in Table 40.2, what is the unknown substance?

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83. When a neutron combines with a proton to form a deuteron (2H), a gamma ray with energy 2.223 MeV is released. The atomic masses of 1H and 2H are 1.007 825 u and 2.014 102 u, respectively. From these data, calculate the mass of the neutron. 84. Consider the fission of a 238U nucleus into two 119Pd nuclei. In the first instant after this fission, the two 119Pd nuclei are just barely in contact, and they then accelerate away from each other, impelled by their mutual electrostatic repulsion. If the center-to-center distance between the 119Pd nuclei is twice their radius (see Fig. 40.24). what is the electric potential energy? What is the force? What is the acceleration of each fragment? 2r

FIGURE 40.24 Two palladium nuclei.

85. The total worldwide supply of 235U in high-grade ores is estimated at 2.6  108 kg. The total worldwide demand for energy is about 5.0  1020 J/yr. Suppose that the entire energy demand were supplied by 1000-MW nuclear reactors fueled with 235U. (a) How many reactors would we need? (b) How long would the high-grade fuel last?

Answers to Checkups Checkup 40.1 1. The number of protons is 6 for all three isotopes; all isotopes

of a given chemical element have the same atomic number. The number of nucleons, or mass number, is given by the superscript before the element symbol, and so is 8, 9, and 10 for the respective isotopes. The number of neutrons is the number of nucleons minus the number of protons, and so is 2, 3, and 4, respectively. 2. Each hydrogen isotope has one proton (all isotopes of a given

element have the same atomic number), and the three isotopes have 0, 1, and 2 neutrons, and thus 1, 2, and 3 nucleons, respectively.

3. The rightmost column, containing only

27

Ne, corresponds to the largest number of neutrons (N  17). The top row contains isotopes of Ne; these have the most protons (Z  10). The isotope 27Ne has the largest mass number in the chart. 3

3

4. (B) H. Tritium, or H, has two neutrons and one proton;

each of the other isotope choices has an equal number of neutrons and protons. 5. (E) 8. Since Eq. (40.2) asserts that the radius is proportional

to the cube root of the atomic number, the atomic number is thus proportional to the cube of the radius, so the ratio is 23  8.

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Checkup 40.2

2. As discussed before and after Eq. (40.23), the number of

1. Since the force is the negative of the gradient of the potential,

the largest attractive (negative) force occurs where the slope of the potential vs. distance is most positive. From the plot of Fig. 40.3, this is at about 0.91.0  1015 m. 2. The binding energy per nucleon for

238

U is 1802 MeV/(238 nucleons)  7.6 MeV/nucleon, at the heavy-nuclei (low) end of the  8 MeV values common to all but the lightest nuclei.

radioactive nuclei is 12 the initial number after one half-life, or 8.04 days for 131I (see Table 40.3); the number is 41 the initial number after two half-lives, or 16.08 days, and is 18 the initial number after three half-lives, or 24.12 days. 3. The fraction of radioactive nuclei remaining at t  t is,

according to Eq. (40.25), e1  0.368. According to Fig. 40.14, the initial amount is 1.00 g, and so the amount remaining after t  42 years is 0.37 g. 9

3. No; this would mean the nucleus has more energy than the

4. (B) 2.0  10 . From Table 40.3, 3200 years is two half-lives,

separate protons and neutrons, and the nucleus could and would fly apart.

so the number of radioactive nuclei and the decay rate will decrease by a factor of (12)2  14 to the value 2.0  109.

4. The Q value of the reverse reaction is the negative of that of

the forward reaction, and so is 1.191 MeV. The energy required would be nearly zero, although for the reaction to occur, some energy is required to overcome the electric repulsion of the nucleus. 5. (D) Fe; Ra. The largest binding energy per nucleon occurs at

the peak of the curve in Fig. 40.6, or, of the nuclei named, for Fe (iron). Since the curve per nucleon is almost constant, decreasing very slowly for large mass numbers, the largest binding energy occurs at the far right end of the curve, or for the named nucleus with the largest mass number, Ra (radium).

Checkup 40.3 1. Because alpha and beta decays involve the emission of charged

particles, they result in a change of the atomic number of the nucleus; gamma decay involves the emission of a photon, which carries no charge, and so does not result in transmutation. 2. The direction of deflection for a positron would be to the left,

in the same direction as the  ray, since the antielectron carries an elementary positive charge. 3. No. To decay, that is, to spontaneously overcome the binding

energy and emit particles, the initial energy of the nucleus must be greater than the final energy of the system, and so the Q value (initial minus final rest-mass energy) must be positive. 4. (D)

136

Ba. The negative beta decay results in a unit increase in atomic number Z, transmuting cesium into barium. Beta decay does not change the mass number A  136, so the decay product is 136Ba.

C. 226Ra has a half-life 16005730  0.3 times that of C. However, 1.0 mg of 14C has more nuclei by a factor of 22614  16 than 1.0 mg of 226Ra, and thus has a higher decay rate by a factor of about 0.3  16  5.

5. (A)

14

14

Checkup 40.5 1. It is the stored electric potential energy of protons held close

together that is released when the nuclear fragments separate. 2. From the curve of binding energy, Fig. 40.6, we can see that

the fission of a light nucleus would result in a smaller binding energy per nucleon; that is, it would require a net increase of energy. 3. A release of less than one neutron per fission would result in

less than one additional fission, and so would not provide a chain reaction; similarly, exactly one neutron could in principle provide a steady reaction, but since some neutrons would be lost, it would not provide a true chain reaction. A release of even slightly more than one neutron per fission could indeed result in a chain reaction, since one or more additional fissions could be caused by each fission. 4. (B) The rest mass of the products is less than the sum of the

rest masses of 235U and the neutron. Since energy is released in the reaction, the rest mass of the system decreases. The other choices are not valid, because a variety of products are possible, and because the number of nucleons, 236, remains the same.

Checkup 40.6 1. To ensure a chain reaction, you could lose up to two; that is, to

Checkup 40.4 1. The decay rate is proportional to the number of nuclei and

inversely proportional to the half-life [Eq. (40.28)]. Since the atomic masses of the isotopes in Table 40.3 differ by less than a factor of 20, so do the number of nuclei in 1.0 mg. The extreme values of half-life in the table differ from the nearest values by much more than a factor of 20, so the smallest and largest half-life determine the highest and lowest decay rate, respectively. Thus, 131I has the highest decay rate and 238U, the lowest.

continue a steady reaction, each fission must produce one neutron that is later absorbed. 2. A mass of 100 g is too small in size to ensure that enough

neutrons are reabsorbed to support a chain reaction; that is, too many neutrons escape from the mass. 3. The

238

U nuclei absorb too many neutrons in nonfission events, preventing those neutrons from causing the fission of 235U.

4. No, the new nuclear fuel is produced from extra uranium

placed around the reactor, not from fuel being consumed inside the reactor.

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5. (D) Slows neutrons so that they are more likely to be absorbed

by 235U in fuel rods later. Thus, the moderator enhances the reaction.

Checkup 40.7 1. Heavy nuclei contain so many protons that electrostatic repul-

sive forces push the nuclei apart, so that fusion is not energetically favorable. 2. The reaction (40.35) requires that each of the reactions

(40.33) and (40.34) occur twice. The net result is that four 1H nuclei are consumed and one 4He nuclei is produced (along with two positrons, two neutrinos, and two gamma rays).

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3. No; fission can proceed spontaneously or with the absorption

of slow (low-temperature) neutrons. 4. No; all the nuclear products of reactions (40.33)–(40.35) are

stable nuclei. However, penetrating radiation is produced, including the gamma ray produced in reaction (40.34) and the gamma rays produced when the positron of reaction (40.33) annihilates an electron. 2

2

2

4

1

5. (D) H  H  H S He  H  neutron. While all the

net reactions listed are in principle possible, only (D) represents the sum of the given reaction (production of tritium from two deuterium nuclei) and the reaction (40.36) (production of 4He from the tritium product of the given reaction and a third deuterium nucleus).

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CHAPTER

41

Elementar y Particles and Cosmology

CONCEPTS IN CONTEXT 41.1 The Tools of High-Energy Physics 41.2 The Multitude of Particles 41.3 Interactions and Conservation Laws 41.4 Fields and Quanta 41.5 Quarks 41.6 Cosmology

Trails of small bubbles reveal the passage of high-energy particles in a bubble chamber filled with liquid hydrogen. A high-energy proton entered the chamber and struck a proton in a hydrogen nucleus, and the collision produced a spray of several kinds of new particles. While learning about elementary particles in this chapter, we will consider such questions as:

? How do charged particles make trails of bubbles in a bubble chamber? (Section 41.1, page 1399)

? The several tracks emerging from the collision shown in Fig. 41.5 reveal the creation of new particles in the collision. What reaction created these particles? (Section 41.1, page 1400)

? How can we calculate the momentum of a particle from the radius of curvature of the track? (Example 2, page 1401) 1396

Concepts in Context

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The Tools of High-Energy Physics

? Besides energy and momentum, what quantities are conserved in a reaction that creates particles, such as the reaction that created the new particles in the photos? (Section 41.3, page 1406)

I

n this final chapter we briefly explore two fascinating questions: What is the structure of the physical world on the smallest scale? And what is the structure of the physical world on the largest scale? To answer the first of these questions we must turn to particle physics and the search for the elementary, indivisible building blocks of matter. And to answer the second question we must turn to cosmology and the investigation of the dynamics of the Universe on a large scale. We will see that there are profound links between the physics of elementary particles and the evolution of the Universe. The study of these links relies on a wide range of tools—from high-energy accelerators used to re-create the kinds of particles that populated the Universe immediately after the Big Bang, to large optical and radio telescopes needed to observe distant regions of the Universe. Early in the last century physicists discovered that the atom is not an elementary, indivisible unit—each atom consists of electrons orbiting around a nucleus. The electron is a truly elementary particle, the first elementary particle to be discovered. But the nucleus is not an elementary, indivisible unit—each nucleus consists of protons and neutrons packed tightly together. In the 1930s, physicists began to build accelerating machines producing beams of energetic protons or electrons that could serve as projectiles; with these atom smashers physicists could split the nucleus. In the 1950s, they built much larger and more powerful accelerating machines; with these new machines they attempted to split the proton and neutron. But the result of these attempts was chaos: when struck by very energetic projectiles, the proton and neutron do not split into any simple subprotonic pieces. Instead, such violent collisions generate a multitude of new, exotic particles by the conversion of kinetic energy into mass. For want of a better name, these new particles were called “elementary particles.” However, most of these new particles are more massive and more complicated than protons and neutrons—they do not appear to be truly elementary, indivisible units. After much effort, theoretical physicists imposed some order on this chaos. They found convincing evidence that protons, neutrons, and many other “elementary particles” are made of very small, compact subunits. The subprotonic building blocks are called quarks.

41.1 THE TOOLS OF HIGH-ENERGY PHYSICS Protons and neutrons are much “harder” than atoms or nuclei. A projectile of a bombarding energy of a few eV can shatter an atom, and a projectile of a bombarding energy of a few MeV can shatter a nucleus. But to make a dent in a proton, we need a bombarding energy of a few hundred or thousand MeV. Elementary-particle physicists like to measure the energies of their projectiles in billion electron-volts (GeV) or in trillion electron-volts (TeV). Expressed in joules, these energy units are 1 GeV  109 eV  1.60  1010 J 1 TeV  1012 eV  1.60  107 J

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CHAPTER 41

accelerator

FIGURE 41.1 Panoramic view of Fermilab.

Elementary Particles and Cosmology

The acceleration of a projectile to such a high energy requires large, sophisticated accelerator machines. High energy—one GeV to thousands of GeV—characterizes the realm of particle physics. Collisions between particles at these high energies are capable of creating a large variety of new particles by the conversion of kinetic energy into mass. The two largest accelerators are the proton synchrotrons at the Fermi National Accelerator Laboratory (Fermilab, near Chicago) and at the Conseil Européen pour la Recherche Nucléaire (CERN, on the Swiss–French border near Geneva). Figure 41.1 shows an overall view of Fermilab; the accelerator is buried underground in a circular tunnel 6 km in circumference. The CERN accelerator tunnel is even larger, about 27 km in circumference. The large hadron collider (LHC) under construction at CERN will provide proton collisions with an energy of 14 TeV. The Fermilab Tevatron accelerator provides the largest energy now available, producing a beam of protons with an energy of 1 TeV and a speed of 99.9995% of the speed of light. The protons travel in an evacuated circular beam pipe (Fig. 41.2). Large magnets placed along this pipe exert forces on the protons, preventing their escape—the protons move as if on a circular racetrack. At regular intervals the pipe is joined to cavities connected to high-voltage oscillators; in each of these cavities, oscillating electric fields impel the protons to higher energy. After several hundred thousand circuits around the racetrack, the protons reach their final energy of 1 TeV. Before the protons are allowed to enter the main accelerator ring, they must pass through several smaller preliminary accelerators. The protons generated by a proton gun first pass through an electrostatic generator, then through a linear accelerator, and then through a “small” circular accelerator. Only after the protons have passed through these preliminary stages do they enter the main ring. Once the particles have been given their maximum energy, they are guided out of the accelerator by steering magnets and made to crash against a target consisting of a block of metal or a tankful of liquid. Within the material of the target, the high-energy particles collide violently with the protons or neutrons of the nuclei. The reactions that take place in the collisions create new particles by conversion of energy into mass. Unfortunately, not all of the kinetic energy of the incident particles can participate in these reactions. As we saw in Section 11.3, the velocity of the center of mass remains constant during the collision, and therefore the particles must retain some kinetic energy. A relativistic calculation indicates that, in the collision of a high-energy proton and a stationary proton, the parti-

FIGURE 41.2 View of the underground tunnel housing the Tevatron accelerator at Fermilab. The long row of magnets encases the beam pipe.

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41.1

The Tools of High-Energy Physics

cles retain most of the kinetic energy, and only a small fraction becomes available for reactions. Such collisions are quite inefficient. The efficiency improves drastically if two high-energy protons are made to collide head-on. The available energy is then the sum of the energies of the two protons. At Fermilab, a second beam pipe has been installed in the tunnel housing the original beam pipe. A beam of antiprotons travels in this beam pipe, in a direction opposite to that of the protons in the original beam pipe. These two opposite beams cross at two intersection points (Fig. 41.3), where the violent head-on collisions create a wide variety of new particles. In order to observe the particles that emerge from the scene of these collisions, physicists have used several kinds of particle detectors. Some of these—scintillation counters and Cerenkov counters—signal the passage of each electrically charged particle by giving off brief (and weak) flashes of light; these flashes of light are detected by sensitive photomultiplier tubes (see Chapter 37, Physics in Practice: Photomultiplier, page 1268). Other detectors—bubble chambers and multiwire chambers—render the tracks of electrically charged particles visible, either on a photographic record or on a computer-generated picture. A bubble chamber is a tank filled with superheated liquid, usually liquid hydrogen, whose temperature is slightly above the boiling point. Such a liquid is unstable— it is about to start boiling but it will usually not start until some disturbance triggers the formation of the first few bubbles. A charged particle zipping through the chamber provides just the kind of disturbance the liquid is waiting for—a fine trail of small bubbles forms in the wake of the particle’s passage. High-speed cameras can take a picture of these bubble tracks before they disperse and disappear in the turmoil of subsequent widespread bubbling and boiling of the liquid. Figure 41.4 shows the BEBC bubble chamber that was used at CERN for many years in a wide variety of experiments. The bubble chamber is surrounded by a large electromagnet, which aids in the identification of the particles passing through the bubble chamber. The magnetic field generated by this magnet pushes the particles into curved orbits as they pass through the chamber. The direction of the curvature depends on the sign of the electric charge, and the radius of the curved orbit is proportional to the momentum

1399

FIGURE 41.3 The Collider Detector at Fermilab (CDF).

bubble chamber Concepts in Context

(a)

FIGURE 41.4 (a) The Big European Bubble Chamber (BEBC) at CERN, surrounded by the magnet, which almost completely hides it from view. (b) The BEBC before its installation into the magnet.

(b)

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CHAPTER 41

Elementary Particles and Cosmology

This collision of a high-energy proton with a proton at rest…

FIGURE 41.5 Tracks of particles in a bubble chamber (the tracks have been colored to distinguish different kinds of particles; in the original bubble-chamber photograph the tracks are all white). A proton enters the bubble chamber (yellow track from top of photograph) and collides with one of the protons at rest in the nucleus of one of the hydrogen atoms in the liquid filling the chamber. The tracks of positively charged particles (red) and negatively charged particles (blue) created in this collision can be seen emerging from the scene of the accident. Besides the charged particles, a neutral lambda was created in the collision. Such a neutral particle leaves no track in the bubble chamber, but it reveals itself when it decays into a proton (yellow track at bottom) and a negative pion (purple track)

Concepts in Context

…creates positive particles: one proton, one kaon, and seven positive pions; …

…and negative particles: seven negative pions; …

…and a neutral lambda that here decays into a proton and a negative pion.

of the particles (Section 30.1). Thus, measurement of the radius of a bubble track tells us the sign of the electric charge of the particle and the momentum of the particle. The photograph of Fig. 41.5 was taken with a bubble chamber. It shows the track of a high-energy proton that entered the bubble chamber and collided with a proton at rest in the core of one of the hydrogen atoms in the liquid filling the chamber. The collision destroyed one of the two protons, but produced a handful of new particles: seven positive pions, seven negative pions, a positive kaon, and a neutral lambda particle. Using the respective symbols for these particles (see Tables 41.2 and 41.3), the reaction can be summarized as follows: p  p S p  7p  7p  K  ¶

(41.1)

The lambda particle subsequently decayed into a proton and a negative pion: ¶ S p  p

(41.2)

The sum of the rest masses of the particles after the reaction (41.1) is much larger than the sum of the rest masses of two protons. The excess mass comes from the conversion of energy into mass—some of the kinetic energy of the incident proton has been converted into mass. This conversion of kinetic energy into mass plays a crucial role in the discovery of new particles. Almost all the new particles discovered during the past 30 years are considerably heavier than protons and neutrons. Physicists need powerful accelerators to produce the large kinetic energies that must be supplied for the manufacture of these new heavy particles.

Elementary-particle physicists prefer to measure the masses of particles in units of MeV/c 2, using the conversion 1 u  931.5 2 MeV/c . Thus a proton has a mass of 938 MeV/c 2, a pion has 139.6 MeV/c2, a kaon has 494 MeV/c 2, and lambda has 1115 MeV/c 2 (see Tables 41.2 and 41.3).

EXAMPLE 1

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41.1

The Tools of High-Energy Physics

1401

Given these masses, calculate the minimum kinetic energy required for each proton if we want to initiate the reaction (41.1) in a head-on collison between two protons of equal kinetic energies. SOLUTION: In a head-on collison between two protons of equal energies, all of the kinetic energy can be converted into rest-mass energy (this would not be true for the collision of a high-energy proton with a stationary proton; that is, it would not be true for the collision as photographed in Fig. 41.5). If the kinetic energy of each proton is K, the kinetic energy of both is 2K, and this must equal the difference of rest-mass energy before and after the reaction:

2K  mpc 2  14[mass of pion]c 2  [mass of kaon]c 2 [mass of lambda]c 2  2mpc 2  14  139.6 MeV  494 MeV  1115 MeV  938 MeV  2625 MeV

(41.3)

Thus, the kinetic energy of each proton must be 2625 MeV 2  1313 MeV.

In a detector region with a uniform magnetic field of 4.38 T, a pion leaves a track with a radius of curvature of 1.46 m. What is the momentum of the pion? The magnetic field is applied in the downward direction, and the motion of the pion is horizontal, clockwise when looking from above. What is the sign of the elementary charge on the pion?

EXAMPLE 2

Concepts in Context

SOLUTION: From Eq. (30.6), the magnitude of the momentum of a pion in cir-

cular motion in a uniform magnetic field is p  qBr  1.60  1019 C  4.38 T  1.46 m  1.02  1018 kgm/s As mentioned in Section 30.1, this expression for the momentum is relativistically correct. In relativistic calculations, the momentum is often expressed in MeV/c, or GeV/c : p  1.02  1018 kgm/s 

1.00 MeV 1.60  1013 J



3.00  108 m/s c

 1.92  103 MeV/c  1.92 GeV/c From this form it is apparent that p 2c 2 is much greater than the square of the pion rest-mass energy, 140 MeV (Example 1). Recalling that E 2  p2c 2  (mc 2)2, we see that E 2 W (mc 2)2, which means that this pion is extremely relativistic. With the magnetic field downward, the magnetic force F  qv  B implies counterclockwise motion for a positive particle. The clockwise motion implies that the motion is that of a negative pion.

Although bubble chambers yield excellent pictures of particle tracks, they are very complex, very large, and very expensive machines, and they cannot take picures fast enough to satisfy the needs of experimenters. Bubble chambers have now been replaced by various types of electronic tracking chambers, for example, multiwire chambers. These chambers are strung with thousands of fine wires, in an array that resembles a

tracking chamber multiwire chamber

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CHAPTER 41

Each wire in array is connected to a voltage source and a current detector…

Elementary Particles and Cosmology

coordinate grid (Fig. 41.6). The wires are connected to voltage supplies and to current sensors. When a charged particle passes through the gas in the chamber, it ionizes the gas along its track, and the electrons released in this ionization are attracted to the wires near the track, where they form miniature electric discharges, which are detected as current pulses by the sensors individually connected to the wires. These current pulses reveal which wires are near the track, and a computer can then immediately reconstruct the location of the track and draw a picture of the track. An important advantage of multiwire chambers over bubble chambers is that the raw data are in the form of current pulses, which can be directly processed and stored in digital form by a computer. In contrast, the raw data from bubble chambers were in the form of photographs, and the tracks recorded on these photographs had to be measured and converted into numbers before they could be processed and analyzed by a computer; the conversion of photographic data into digital data is very time-consuming. The UA1 detector at CERN incorporates several multiwire chambers. Figure 41.7 shows a computer-generated picture of tracks of particles in the multiwire chamber of this detector.

…so that when a passing particle ionizes gas atoms, a sequence of current pulses is detected.

FIGURE 41.6 Wires strung in a cylindrical array for a multiwire chamber. The array is shown during construction, before it was placed inside the vacuum chamber.

FIGURE 41.7 Computer-generated picture of particle tracks in the multiwire chamber of the UA1 detector at CERN. These are the tracks of particles produced in the collision of a proton and an antiproton that entered the chamber horizontally from opposite directions.

✔

Checkup 41.1

Do neutrons make tracks in bubble chambers? In multiwire chambers? QUESTION 2: For what purpose are particle detectors placed in magnetic fields? QUESTION 3: Figure 41.5 shows the tracks of positive and of negative particles in a bubble chamber immersed in a magnetic field. The magnetic field is perpendicular to the plane of the page. Is it directed into the page or out of the page? QUESTION 4: Is the Q value for the reaction (41.1) positive or negative? For the reaction (41.2)? QUESTION 5: What is the advantage of head-on collisions between protons of opposite motion over collisions with a stationary target? (A) The head-on collision is easier to achieve than hitting the stationary target. (B) In the head-on collision, the full kinetic energy is available for reaction. (C) In the head-on collision, the backward recoil makes detection easier. (D) In the head-on collision, both protons and antiprotons can be used. (E) In the head-on collision, exactly twice as much energy is available. QUESTION 1:

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41.2

The Multitude of Particles

1403

4 1 . 2 T H E M U LT I T U D E O F PA RT I C L E S As physicists built accelerators of higher and higher energy, they discovered more and more new particles. By now, several hundred particles are known. Whenever a new, more powerful accelerator makes available more energy for collisions, more particles and heavier particles are produced—there seems to be no end in sight. The known particles fall into three groups: leptons, baryons, and mesons. The leptons are distinguished from baryons and mesons by the interactions in which they participate. Leptons interact via the electromagnetic force and the weak force (we already became acquainted with the weak force in Chapter 40, where we found that the weak force produces some radioactive decay reactions involving neutrinos). In contrast, baryons and mesons interact via the electromagnetic force, the weak force, and the strong force. Because the strong force is stronger than the other forces, it predominates in reactions among baryons and mesons; accordingly, baryons and mesons are called strongly interacting particles, or hadrons (from the Greek hadros, thick). The distinction between baryons and mesons is based on their spin: baryons have half-integer spin ( 12 U, 32 U, 52 U, . . .) and mesons have integer spin (0, U 2U, . . .). Particles with half-integer spin are called fermions, and particles with integer spin are called bosons. Thus, baryons are fermions, and mesons are bosons. There are six different leptons (see Table 41.1). Among these, the electron (e) is the most familiar. The muon (m) is very similar to the electron; it has the same electric charge as an electron but its mass is about 200 times as large. The tau (t) is also similar to the electron, but its mass is even larger than that of the muon. The neutrinos (e ,  , and  ) are particles of zero electric charge and very small mass. Until recently, neutrinos were believed to have zero mass, like photons. But now there is clear evidence that neutrino masses are not zero. In Table 41.1, the masses of the particles are expressed in units of MeV/c 2 and the electric charge and the spin are expressed in multiples of the elementary charge e and in multiples of Planck’s constant U. The choice of MeV/c 2 for the unit of mass is very convenient in calculations involving the rest-mass energies; for instance, from the electron mass of 0.511 MeV/c 2 listed in the table, we immediately see that the electron rest-mass energy is 0.511 MeV. All the leptons have spin 12 . Besides the six leptons of Table 41.1, there are six antileptons: the antielectron (or positron), the antimuon, the antitau, and the three antineutrinos. These antiparticles have the opposite electric charge and some other opposite properties, but they have

TA B L E 4 1 . 1

hadrons

bosons and fermions leptons

THE LEPTONS MASS a

PARTICLE

SYMBOL

electron

e

0.511 MeVc 2

1 2

1

constituent of atoms



105.6 MeVc

1 2

1

produced in decay of pion; abundant in cosmic rays



1784 MeVc

1 2

1

1 2

0

1 2

0

1 2

0

muon tau electron neutrino muon neutrino tau neutrino a

SPIN

2

2

e

 3 eVc



 0.19 MeVc

2



 18.2 MeVc

2

2

CHARGE

COMMENTS

produced in beta decay; abundantly emitted by Sun

Neutrinos are now known to undergo quantum-mechanical oscillations between different mass states. Many experiments are underway to clarify such processes.

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TA B L E 4 1 . 2

PARTICLE

proton neutron lambda sigma-plus sigma-minus sigma-zero xi-minus xi-zero omega-minus charmed lambda

CHAPTER 41

Elementary Particles and Cosmology

SOME BARYONS

SYMBOL

p n ¶ ©



©

 0

©





0







¶c

baryons

mesons

resonances

MASS

SPIN

CHARGE

QUARK CONTENT

COMMENTS

938.3 MeVc2

1 2

1

uud

constituent of nucleus

939.6

1 2

0

ddu

constituent of nucleus

1115

1 2

0

uds

has strangeness 1

1189

1 2

1

uus

has strangeness 1

1197

1 2

1

dds

has strangeness 1

1192

1 2

0

uds

has strangeness 1

1321

1 2

1

dss

has strangeness 2

1315

1 2

0

uss

has strangeness 2

1672

3 2

1

sss

has strangeness 3

2280

1 2

1

udc

contains charmed quark

exactly the same mass and spin as the corresponding particles. (The notation for an antiparticle is a bar over the letter or, alternatively, a superscript indicating the electric charge; thus, the notation for the antielectron is e or e.) The baryons are the most numerous group of particles. The most familiar baryons are the proton and the neutron; these are the baryons of the least mass. Table 41.2 lists some of the other baryons. For every baryon in Table 41.2, there exists an antibaryon. As in the case of leptons, these antiparticles have an opposite charge but the same mass and spin as the corresponding particles. Finally, the mesons are another numerous group of particles. Some of them are listed in Table 41.3. The most familiar mesons are the three pions (with electric charges 0, e, e); these were the first mesons to be discovered. For every meson there exists an antimeson; however, these antiparticles have already been included in Table 41.3. For example, the antiparticle to the  is the p, and vice versa.The antiparticle to the 0 is the 0; this means that when two 0 mesons meet, they can annihilate each other [see Section 40.3 and the discussion before Eq. (36.47)]. Also, the 0 usually decays into two photons, in essence annihilating itself. Most of the particles are unstable; they decay, spontaneously falling apart into several other particles. The only absolutely stable particles are the electron, the proton, and the neutrinos. The lifetimes of the other particles range from a fairly long 15 minutes for the neutron to about 1023 s for many of the exotic particles. A particle that lives only 1023 s is incapable of making a measurable track in a tracking chamber or other detector. Thus, such a short-lived particle cannot be detected directly, but its existence can be inferred from a careful study of the rates of reactions of longer-lived particles engaged in collisions. The short-lived particle participates in these reactions as an intermediate, ephemeral state, which causes a characteristic increase of the reaction rate whenever the energy of the colliding particles coincides with the energy required for the production of the short-lived particle. Because of their enhancing effects on reaction rates, the short-lived particles are often called resonances.

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41.3

TA B L E 4 1 . 3

PARTICLE

Interactions and Conservation Laws

SOME MESONS

SYMBOL

MASS

SPIN

CHARGE

pion-zero

p0

135.0 MeVc 2

0

0

pion-plus

p

139.6

0

pion-minus

p

139.6

kaon-zero

K0

kaon-plus

K

 

kaon-minus

K

J/psi

Jc D

D-zero

0



D-plus

D

upsilon



✔

1405

QUARK CONTENT

COMMENTS

uu and dd

abundant in cosmic rays; carrier of nuclear force

1

ud

abundant in cosmic rays; carrier of nuclear force

0

1

du

abundant in cosmic rays; carrier of nuclear force

498

0

0

ds

has strangeness 1

494

0

1

us

has strangeness 1

494

0

1

su

has strangeness 1

3097

1

0

cc

contains charmed quarks

1865

0

0

cu

contains charmed quark

1869

0

1

cd

contains charmed quark

9460

1

0

bb

contains bottom quarks

Checkup 41.2

Do all leptons have spin 12? All baryons? QUESTION 2: What are the mass, spin, and electric charge of an antitau? An electron antineutrino? QUESTION 3: Which is the heaviest particle listed in Tables 41.1–41.3? QUESTION 4: Particles not listed in Tables 41.1–41.3 include the neutral eta particle (h0), which has spin 12; and the neutral B particle (B0 ), a massive particle with spin QUESTION 1:

0. To what group of particles does the h0 belong? The B0? (A) Lepton, baryon (B) Lepton, meson (C) Baryon, lepton (D) Baryon, meson (E) Meson, baryon

41.3 INTERACTIONS AND C O N S E R VAT I O N L AW S The reactions that occur among the particles in a high-energy collision are governed by the four fundamental forces: the “strong” force, the electromagnetic force, the “weak” force, and the gravitational force. However, at the microscopic level, particles are subject to quantum uncertainties in position and momentum, and the force acting on them is not well defined. Hence, physicists usually prefer to speak of four fundamental types of interactions, instead of forces. The strong force acts on baryons and mesons, but not on leptons. As we will see in Section 41.5, the strong force is a special case of an even stronger force, called the “color” force.

strong force

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CHAPTER 41

electromagnetic force

weak force

gravitational force

Concepts in Context

baryon number

Elementary Particles and Cosmology

The electromagnetic force acts primarily on charged particles, but it also acts on neutral particles—such as the neutron—that contain an internal distribution of electric charge and are endowed with magnetic moments. The weak force acts on leptons, baryons, and mesons. However, its effect on baryons and mesons is often hidden behind the much larger effects produced by the strong or the electromagnetic force. To see the purest manifestation of the weak force, we have to examine reactions involving leptons. The weak force is deeply involved in many reactions that bring about the decay of unstable particles. For example, the weak force is responsible for the beta decay of the neutron. The weak force is also responsible for other beta decays. The gravitational force is of no direct interest in particle physics. Although all particles and all forms of energy interact gravitationally, the gravitational effects produced by individual particles are too feeble to be of any significance at the energies available in laboratories on Earth. However the gravitational force is of great interest to theorists who study the ultrahigh energies of the early Universe. Table 41.4 lists the strength of each of the fundamental forces and also the range, or the maximum distance over which this force can reach from one particle to another. In the table, the strengths of the forces are expressed relative to that of the strong force, to which a strength of 1 has been assigned arbitrarily.1 All the forces, and all the reactions that they produce, obey the usual conservation laws for energy, momentum, angular momentum, and electric charge. Besides these conservation laws for familiar quantities, experiments with high-energy particles have led to the discovery of new conservation laws involving several esoteric quantities, such as baryon number, lepton number, isospin, strangeness, and parity. All reactions obey a conservation law for baryon number, which is a generalization of the conservation law for mass number from nuclear physics. Each baryon has a baryon number of 1, each antibaryon 1, and all other particles have baryon number 0. The conservation law for baryon number then simply states that the net baryon number remains unchanged in any reaction. This is a mathematical expression of the requirement that any baryons that disappear in a reaction must be replaced by an equal number of other baryons, with antibaryons counting as negative baryons. For example, consider the reaction (41.1). The baryon numbers for p, p , K, and are 1, 0, 0, and 1, respectively; hence, in the reaction (41.1), the net baryon number is 1  1 before and 1  0  0  0  1 after; that is, the net baryon number remains unchanged.

TA B L E 4 1 . 4

1

THE FOUR FUNDAMENTAL FORCES

RELATIVE STRENGTH

FORCE

ACTS ON

RANGE

strong/color

baryons and mesons (hadrons)

1

 1015 m

electromagnetic

particles with charge or magnetic moment

102

infinite

weak

leptons, baryons, and mesons

106

 1018 m

gravitational

all forms of matter

1038

infinite

The strengths of the forces depend on the energies of the particles. The values in the table are appropriate for low energies.

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41.3

Interactions and Conservation Laws

Lepton number is for leptons what baryon number is for baryons. Each lepton has lepton number 1, each antilepton 1, and all other particles have lepton number 0. The net lepton number remains unchanged in any reaction. See also Problem 18. Isospin is somewhat more complicated because it is a quantity with several components; that is, isospin is like a vector quantity. The conservation law for isospin states that the net sum of the isospin “vectors” of all the particles involved in a reaction remains unchanged. Strangeness is similar to the baryon and lepton numbers. Each hadron has a strangeness number: the proton has strangeness 0, the kaon-plus has 1, the lambda has 1, the pion has 0, etc. (values of the strangeness are given in the “Comments” column in Tables 41.2 and 41.3). The strangeness of an antiparticle is the negative of the strangeness of the corresponding particle. The conservation law for strangeness states that the net strangeness number remains unchanged in a reaction. For example, in the reaction (41.1), the net strangeness is 0  0 before and 0  0  0  1  1 after; that is, there is no change. Parity characterizes the behavior of a quantum-mechanical wave under a reversal of the x, y, and z coordinates. Such a reversal is physically equivalent to forming a mirror image of the wave. It can be shown that the mirror image of the quantummechanical wave for a stationary state is either equal to the original wave (parity  1) or else equal to the negative of the original wave (parity  1). Conservation of parity means that the net parity (the product of all the individual parities) of all the particles participating in a reaction is unchanged. The conservation laws for energy, momentum, angular momentum, electric charge, baryon number, and lepton number are absolute—no violation of any of them has ever been discovered. In contrast, the conservation laws for isospin, strangeness, and parity are approximate—they are valid for some reactions, but fail in some others. This, of course, raises the question of what possible meaning can be attached to a “law” that works sometimes and fails sometimes. The answer is that whether a conservation law is obeyed or not depends on the kind of interaction, or the kind of force, that drives the reaction. The reactions caused by the strong force obey all the conservation laws, but reactions caused by the electromagnetic or the weak force do not. It is usually easy to tell what force is involved in a reaction: reactions involving the strong force tend to be fast; reactions involving the other forces tend to be (relatively) slow. For example, reaction (41.1) is a fast reaction brought about by the strong force, whereas reaction (41.2) is a slow reaction brought about by the weak force. The difference is evident in Fig. 41.5, where reaction (41.1) appears to occur at a single point whereas reaction (41.2) occurs only after the lambda particle has traveled a measurable distance. Table 41.5 lists the conservation laws obeyed by the strong (color), electromagnetic, and weak forces. TA B L E 4 1 . 5

1407

lepton number

isospin

strangeness

parity

FORCES AND CONSERVED QUANTITIES

ENERGY, MOMENTUM, AND ANGULAR MOMENTUM

CHARGE

BARYON NUMBER

LEPTON NUMBER

STRANGENESS

PARITY

ISOSPIN

strong/color

✓

✓

✓

✓

✓

✓

✓

electromagnetic

✓

✓

✓

✓

✓

✓

weak

✓

✓

✓

✓

FORCE

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EXAMPLE 3

Do the following reactions conserve electric charge? Do they conserve baryon number? Do they conserve strangeness? Are the reactions possible? ¶ S n  p0

(41.4)

 S K  p0

(41.5)

SOLUTION: In the first reaction, all particles are electrically neutral, and in the

second, there is one negative elementary charge before and after, so both reactions conserve electric charge. Reaction (41.4) conserves baryon number, since the baryon number is 1 for the ¶ and 1  0 for n  p0. But Eq. (41.5) does not conserve baryon number, since the baryon number is 1 for the and 0  0 for K  p0. This violation of an absolute conservation law implies that the reaction (41.5) is not a possible reaction. Reaction (41.4) does not conserve strangeness, since the strangeness is 1 for the ¶ and 0  0 for n   0. Reaction (41.4) is possible, but since strangeness is not conserved, it is not mediated by the strong force. Reaction (41.5) does not conserve strangeness, since the strangeness is 2 for the  and 1  0 for K  p0.

✔

Checkup 41.3

What is a hadron? Is the proton a hadron? The neutron? The electron? The photon? The tau neutrino? QUESTION 2: What are the mass, spin, electric charge, and strangeness of an antixi-zero ( 0)? An anti-xi-minus, or xi-plus ( , or )? QUESTION 3: What are the quantities conserved in all interactions? Conserved in electromagnetic interactions, but not in weak interactions? Conserved in strong interactions, but not in electromagnetic? QUESTION 4: Consider the wavefunction c  sin kx. What is the parity of this wavefunction, that is, what happens to it when you change x to x? 0 QUESTION 5: The reaction ¶ S n  p does not conserve strangeness. Which of the strong, electromagnetic, or weak interactions can cause this reaction? (A) Strong only (B) Weak only (C) Electromagnetic only (D) Electromagnetic or weak only (E) Strong, electromagnetic, or weak QUESTION 6: In which of the four interactions does the electron participate? In which does the neutron participate? (A) All, all (B) All, all except electromagnetic (C) All except weak, all except strong (D) All except strong, all (E) All except weak, all except electromagnetic QUESTION 1:

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41.4

Fields and Quanta

4 1 . 4 F I E L D S A N D Q U A N TA

A gamma ray strikes an electron, creating an electron and an antielectron.

We know from our study of electricity and magnetism that forces are mediated by fields. Distant particles do not act on one another directly; rather, each particle generates a field of force, and this field acts on the other particles. As we saw in Section 23.1, the existence of fields is required by conservation of energy and momentum. Fields play the role of storehouses of energy and momentum; the energy and momentum stored in the fields balance any excess or deficit in the energy and momentum of the interacting particles engaged in (nonuniform) motion. The most spectacular example of the conversion of particle energy into field energy occurs in the annihilation of matter with antimatter: if an electron collides with an antielectron, the two particles annihilate one another, giving off a burst of very energetic light, or gamma rays. According to classical physics, such a burst of light consists of electric and magnetic fields; thus, the annihilation converts all of the energy of the particles—including their rest-mass energy—into field energy. The reverse reaction is also possible: if a gamma ray collides with a charged particle, it can create an electron–antielectron pair. In such pair creation, the field energy of the gamma ray is converted into the energy of the pair of particles (Fig. 41.8) Each of the four fundamental forces is mediated by fields of its own. Hence there are gravitational fields, electromagnetic fields, strong fields, and weak fields. According to quantum physics, the energy stored in fields is not smoothly distributed; rather the energy is found in quanta, that is, small packets or lumps of energy that can be regarded as particles. In Chapter 37, we became familiar with the quanta of the electromagnetic field; these quanta are the photons. Each of the other fundamental fields has quanta of its own. Table 41.6 lists the quanta of all four fundamental fields. Like photons, the quanta of the gravitational field, or gravitons, and the quanta of the strong field, or gluons, are both massless. In contrast, the quanta of the weak field, or W, W, and Z0 particles, are endowed with mass. At the quantum level, we can picture the field of force generated by a particle as a swarm of quanta buzzing around the particle. For example, we can picture the electric field surrounding an electron, or any other charged particle, as a swarm of photons.The swarm is in a state of everlasting activity—the charged particle continually emits and reabsorbs the photons of the swarm. Emission is creation of a photon; absorption is annihilation of a photon. Hence we can say that the electric field arises from the continual interplay of three fundamental processes: a photon is emitted by one particle, propagates through

TA B L E 4 1 . 6

1409

FIELDS AND THEIR QUANTA

FIELD

QUANTA

MASS

gravitational

gravitons

weak

W particles

80.4

Z particles

91.2

0 MeVc 2

electromagnetic

photons

0

strong /color

gluons

0

This track was made by recoiling electron.

FIGURE 41.8 The two spiraling tracks in this bubble-chamber photograph were made by an electron (green) and an antielectron (red). These particles were created by a high-energy gamma ray in a collision with the electron of a hydrogen atom in the bubble chamber. gravitons gluons W particles W particles Z0 particles

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Elementary Particles and Cosmology

t This electron emits photon… Time increases vertically.

FIGURE 41.9 Feynman diagram representing the exchange of a virtual photon between two electrons. The solid blue lines indicate the motion of the two electrons, with the t axis plotted vertically and the x axis plotted horizontally. The wavy colored line indicates the motion of the photon.

Feynman diagram virtual photon

carrier

RICHARD PHILLIPS FEYNMAN (1918– 1988) American physicist. His invention of the Feynman diagram revolutionized the computations of relativistic quantum processes. He shared the 1965 Nobel Prize with Julian Schwinger (1918–1994), American physicist, and Sin-Itiro Tomonaga (1906–1979), Japanese physicist, for work on the quantum theory of electrons.

e

…and later, this electron absorbs photon.

e ␥

e

e

x

the intervening distance, and is absorbed by the other particle. This exchange process can be represented graphically by a Feynman diagram (Fig. 41.9), invented by Richard Feynman. The photon exchanged between the two electrons is called a virtual photon, because it lasts only a very short time and, being reabsorbed by an electron, is undetectable by any direct experiment. The steady attractive or repulsive force between two charged particles is generated by continual repetition of the photon exchange process. The photon is the carrier of the electromagnetic forces. This is action-by-contact with a vengeance—at a fundamental level, all the electromagnetic forces reduce to local acts of creation and destruction involving particles and photons in direct contact. In terms of a simple analogy, we can easily understand how the exchange of particles brings about forces. Imagine two boys tossing a ball back and forth between them (Fig. 41.10); it is obvious from momentum conservation that this produces a net repulsive force between the boys due to the recoil they suffer when throwing or catching the ball. Our intuition suggests that no such exchange process can ever produce attraction. However, imagine two Australian boys tossing a boomerang back and forth between them in the manner shown in Fig. 41.11; it is then obvious that this produces an attractive force between the boys. Whether a photon exchanged between two charges behaves like a ball or like a boomerang depends on the signs of the charges. Quantum calculations, which take into account the wave nature of all the particles involved, show that the net force is attractive for unlike charges and repulsive for like charges, as it should be. For this boomerang throw and catch… For this ball throw and catch…

…recoils push boys apart.

FIGURE 41.10 Two boys throwing a ball back and forth.

…recoils push boys together.

FIGURE 41.11 Two boys throwing a boomerang back and forth.

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41.4

Fields and Quanta

The gravitational, weak, and strong forces are also generated by the exchange of virtual particles: gravitons, W and Z particles, and gluons. It is a general rule of quantum theory that the range of the force is inversely related to the mass of the particle that serves as the carrier of the force. Thus, the photons and the gravitons that are the carriers of the electromagnetic and the gravitational force have zero mass—the ranges of these forces are infinite. The W and Z particles that are the carriers of the weak force have a very large mass—the range of this force is very short. The gluons that are carriers of the strong force have zero mass, and accordingly the range of the force ought to be infinite. But because this force is very strong, its action over any appreciable distance triggers the spontaneous creation of particle-antiparticle pairs, and this effectively obstructs the force and limits it to short distances. Although the four fundamental forces seem drastically different, theoretical physicists have sought to formulate a unified theory that treats several or all of these forces as aspects of a single, more fundamental force. Electromagnetism is the most familiar example of a unified field theory; that is, it is a theory that treats electric and magnetic forces as two aspects of a single, underlying force. To appreciate fully the unification of electricity and magnetism we would have to examine what relativity says about electric and magnetic fields; we could then see that electric and magnetic forces are merely two aspects of a single force called the electromagnetic force. But even without adopting a relativistic point of view, we can see from Maxwell’s equations that electricity and magnetism are intimately connected. If we seek to unify the electromagnetic and weak forces, we must regard the carriers of these forces—the quanta whose exchange generates the force—as closely related. This would seem to contradict the large mass difference between these particles: the photon is massless, but the W and Z particles are the heaviest particles now known. The unified theory of the electroweak force formulated by S. Weinberg, A. Salam, and S. Glashow attributes this mass difference to an imperfect symmetry (a “broken

1411

unified field theory

electroweak force

A grand unified force…

…separated into gravity and the combined Big Bang strong/electroweak force; …

10–40 s

temperature 1027 K

10–30 s

…“later,” the strong and electroweak forces separated…

10–20 s age of Universe 10–10 s

1018 K

109 K

tr

1010 s …as finally did the electromagnetic and weak forces.

y vit gra ng c stro agneti om

elec

ak

we

100 s

the present 1020 s

100 K

FIGURE 41.12 The evolution of the Universe after the Big Bang.

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grand unified theory

Antiproton from left and proton from right collide head-on, creating many particles, including one W.

Red arrow points to track of an electron, a decay product of W particle.

FIGURE 41.13 Tracks of particles produced in a very energetic head-on collision between a proton of energy 270 GeV that entered from the right and an antiproton of energy 270 GeV that entered from the left. In this collision a W particle was created. It immediately decayed into an electron and a neutrino. The track of the neutrino is not visible.

Elementary Particles and Cosmology

symmetry”) between the photon and the W and Z particles. The theory asserts that perfect symmetry between these particles can be restored by giving them very high energies, in excess of 100 GeV; at such high energies, the photon and the W and Z particles should become essentially identical. Such high energies are difficult to achieve in our laboratories, but they were readily available during the early stages of the Big Bang (see Section 41.6), when the Universe was younger than 1010 s and had a temperature in excess of 1015 K. It is believed that at these early times, there was no difference between the photon and the W and Z particles, and there was no difference between the electromagnetic and weak forces. Furthermore, many theoretical physicists are now striving to formulate a grand unified theory (GUT) which presumes that the gravitational, electroweak, and strong forces were all originally a single, unified force. It is believed that at about 1034 s after the beginning of the Big Bang, gravity separated from the stillunified strong and electroweak forces, and the strong force separated from the electroweak force slightly later (see Fig. 41.12). The most impressive success of the unified electroweak theory was its prediction of the masses of the W and Z particles. The W and Z particles were detected in 1982 in experiments at the proton–antiproton collider at CERN (and at the electron–position collider at Stanford). The experiments at CERN involved the observation of about a billion head-on collisions between protons and antiprotons of the same energy, 270 GeV. A few dozen W and Z particles were produced in these collisions (Fig. 41.13). The measured masses of the W and Z particles are, respectively, 80 GeV/c 2 and 91 GeV/c 2. These measured values are within a fraction of a GeV/c 2 of the theoretically predicted values. This excellent agreement constitutes a brilliant confirmation of the unified theory of weak and electromagnetic interactions.

✔

Checkup 41.4

Compare the masses of the W and Z particles with the masses listed in Tables 41.1–41.3. Are the W and Z the heaviest of all known particles? QUESTION 2: Order the electromagnetic, strong, and weak forces in order of increasing range. (A) Strong, electromagnetic, weak. (B) Weak, strong, electromagnetic. (C) Electromagnetic, strong, weak. (D) Strong, weak, electromagnetic. (E) Weak, electromagnetic, strong. QUESTION 1:

41.5 QUARKS Let us now return to our initial question. What are the ultimate, indivisible building blocks of matter? We know of more than 300 particles. It is implausable that all these hundreds of particles are truly elementary particles. It is likely that most of them, or maybe all of them, are composite particles made of just a few elementary building blocks. To discover these building blocks, physicists have tried to break protons into pieces by bombarding them with projectiles of very high energy. Unfortunately, if the energy of the projectile is large enough to make a dent in a proton, then it is also large enough to create new particles during the collision.This abundant creation of particles confuses the issue—

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41.5

Quarks

we can never be quite sure which of the pieces that come flying out of the scene of the collision are newly created particles and which are fragments of the original proton. In fact, none of the pieces ever found in such collision experiments seems a likely candidate for an elementary building block.Typically, the particles that emerge from a collision between a high-energy projectile and a proton are pions, kaons, lambdas, deltas, and so on, all of which seem to be less elementary than the proton. Although brute-force collision experiments have failed to fragment protons into elementary building blocks, somewhat more subtle experiments have provided us with some evidence that distinct building blocks do indeed exist inside protons. At the Stanford Linear Accelerator Center (SLAC), very high-energy electrons were shot at protons; these electrons served as probes to “feel” the interior of the protons. The experiments showed that occasionally the bombarding electrons were deflected through large angles, bouncing off sharply from the interior of a proton. These deflections indicate the presence of some lumps or hard kernels in the interior of the proton, just as, in Rutherford’s experiments, the large deflections of alpha particles by atoms indicated the presence of a hard kernel (nucleus) in the interior of the atom. Protons and all the other baryons and mesons seem to be composite bodies made of several distinct pieces. In contrast, electrons and the other leptons seem to be indivisible bodies with no internal structure. In some experiments the electron was probed with beams of extremely energetic particles to within 1018 m of its center. Even at these extremely short distances, no substructures of any kind were found. Thus, the electron seems to be a pointlike particle with no size at all, a truly elementary particle. Even before the experimental evidence for lumps inside protons became available, theoretical physicists had noticed that particles could be classified into groups or families of similar particles on the basis of their quantum numbers and their behavior in reactions. To explain these similarities, they had proposed theories in which all the known baryons and mesons are regarded as constructed out of a few fundamental building blocks. According to these theories, the similarity between particles within a given family reflects the similarity of their internal construction, just as the similarities between atoms in a group of the periodic table reflect the similarity of their internal construction. The most successful of these theories was the quark model proposed by M. GellMann and by G. Zweig. In this model, all particles were constructed of three kinds of fundamental building blocks called quarks. (Gell-Mann took the word quark from Finnegan’s Wake, a book by James Joyce.) The three quarks are labeled up, down, and strange, or simply u, d, and s. They all have spin 12 and they have electric charges of 1 1 2 3 , 3 , and 3 elementary charge, respectively (see Table 41.7). Each quark—like any other particle—has an antiparticle, of opposite electric charge.

THE FIRST THREE QUARKS

TA B L E 4 1 . 7

MASS a

QUARK

u d s a

5 MeVc 2

SPIN

ELECTRIC CHARGE

STRANGENESS

1 2

2 3

0

10

1 2

13

0

200

1 2

13

1

Based on theoretical estimates.

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quarks up quark (u) down quark (d) strange quark (s)

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FIGURE 41.14 Structure of the proton. The sizes of the quarks are not drawn to scale. Their sizes are probably much smaller than the size of the proton.

To make the ordinary particles out of the quarks, the latter must be glued together in diverse ways. For example, a proton is made of two u quarks and one d quark (Fig. 41.14). A neutron is made of two d quarks and one u quark (Fig. 41.15). A positive pion is made of one u quark and one d antiquark (Fig. 41.16), and so on (Tables 41.2 and 41.3 list the quark composition of baryons and mesons). By gluing quarks together, we can build up all the known baryons and mesons and explain their quantum numbers and their similarities. Besides, we can predict some properties, such as magnetic moments and reaction rates, of the composite particles from the assumed properties of the quarks. There is only one snag: all experimental searches for free quarks have been unsuccessful. Physicists now believe that quarks are permanently confined inside the ordinary particles so there is no way to break the quark out of, say, a proton. It seems that the quarks are held in place by an exceptionally strong force, which prevents their escape. This new force is the “color” force. The concept of “color” was first introduced to remedy an unacceptable violation of the Exclusion Principle (see Chapter 39). Quarks, like leptons and other particles of half-integer spin, ought to obey the Exclusion Principle. But investigations of the quantum states of quarks within protons, neutrons, sigmas, etc., disclosed that several apparently identical quarks were often found in the same quantum state. To avoid this apparent violation of the Exclusion Principle, physicists postulated that each of the quarks exists in three varieties, and that whenever two apparently identical quarks are found in the same quantum state, they actually are of different varieties. The varieties of quarks are characterized by a new property called color. Of course this “color” has nothing to do with real color; it is merely a (somewhat unimaginative) name for a new property of matter. The different quark colors are red, green, and blue. Thus there is a red u quark, a green u quark, and a blue u quark, and so on. The antiquarks have anticolors; the different antiquark colors are antired, antigreen, and antiblue. Color is a very subtle property of matter; it usually remains hidden inside the ordinary particles. All the normal particles are “colorless”—they consist of several quarks with an equal mixture of all three colors. For instance, one of the three quarks inside the proton is red, one is green, and one is blue. Nevertheless, color plays a crucial role in the theory of the forces that confine the quarks inside the ordinary particles. The quarks are confined by extremely strong mutually attractive forces. These forces between quarks are called color forces because the source of these forces is the color just as the source of the electric force is the electric charge. Each of the three varieties of color is analogous to a kind of electric charge. A body is color-neutral, or “colorless,” if it contains equal amounts of all three colors or if it contains equal amounts of color and anticolor, just as a body is electrically neutral if it contains equal amounts of positive and negative electric charge. The color force is a fundamental force that is included in our table of fundamental forces together with the strong force (Table 41.4). The color force is closely related to the strong force—the latter is actually a special instance of the former. The relationship between the color force and the strong force is analogous to the relationship between the electric force and the intermolecular force. As we saw in Section 22.1 the force between two electrically neutral atoms or molecules is a residual electric force resulting from an imperfect cancellation among the attractions and repulsions of the charges in the two atoms or molecules. Likewise, the strong force between, say, two “colorless” protons is a residual color force resulting from an imperfect cancellation among the attractions and repulsions of the quarks in the two protons. Thus the “strong” force between two protons is no more than a pale reflection of the much stronger color forces acting within each proton. At a fundamental level, the color force between two quarks is due to an exchange of virtual particles between the quarks. The particle that acts as the carrier of the color

color force

Proton is made of one d quark and two u quarks. d

u

u

proton

Neutron is made of one u quark and two d quarks. u

d

d

neutron

FIGURE 41.15 Structure of the neutron.

Positive pion is made of one d antiquark and one u quark. d

u

positive pion

FIGURE 41.16 Structure of a positive pion (p).

Elementary Particles and Cosmology

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41.5

Quarks

force is the gluon, the same as the carrier of the strong force. Figure 41.17 shows a Feynman diagram representing the exchange of a gluon between two quarks. Such an exchange of a gluon between two quarks is analogous to the exchange of a photon between two charged particles (see Fig. 41.9). Another modification of the simple quark model with three quarks arose from the theory of the unification of the electromagnetic and weak forces. In order to make this theory fit some experimental data, physicists had to postulate the existence of a fourth quark, different from the u, d, and s quarks. This new hypothetical quark was labeled charmed, or simply c. The hypothesis of the charmed quark soon received firm experimental support. In 1974, teams of experimenters at the Brookhaven accelerator and at the Stanford accelerator discovered the Jc meson and some other related mesons (see Table 41.3), all of which have exceptionally long lifetimes. These particles contain charmed quarks with an electric charge of 23 and a mass of 60 MeV/c 2. The proliferation of quarks did not stop with four quarks. In 1977, a team of experimenters at Fermilab discovered the  (Greek Upsilon) mesons. These are by far the most massive mesons known (see Table 41.3). Each of these contains a new quark; this fifth kind of quark has been labeled bottom, or b. Theoretical considerations then suggested that there should exist a sixth quark, labeled top, or t. In 1995, another team of experimenters at Fermilab confirmed the existence of the top quark. Figure 41.18 summarizes all the constituents of particles and the carriers of forces; these form the foundation of the Standard Model of particle physics. This model describes all the known particles in terms of the elementary particles listed in Fig. 41.18, and it describes all interactions in terms of four kinds of carriers (however, the graviton is not fully incorporated into the model, because there is as yet no complete and coherent theory of quantum gravity). The four kinds of carriers are the gluons (which come in various color combinations), the photon, the W , W , Z, and the graviton. The six leptons fall into three families: the electron family, the muon family, and the tau family. Correspondingly, the six quarks also fall into three families: the up–down family, the strange–charm family, and the bottom–top family. Each of the six leptons has an antiparticle. Each of the six quarks (up, down, strange, charmed, top, bottom) comes in three colors (red, green, blue); furthermore, each quark has an antiquark, which comes in three varieties of anticolor (antired, antigreen, antiblue). Altogether, this amounts to 36 quarks, 12 leptons, and 24 carriers (counting all the color combinations of gluons and their

1415

charmed quark (c) bottom quark (b) top quark (t)

Standard Model

This quark emits a gluon, changes color from green to blue; … t Forces

color force

electromagnetic force

weak force

gravitational force

Carriers

gluon

photon

W and Z

graviton

green quark blue quark blue quark

gluon Leptons

green quark x …and this quark absorbs a gluon, changes color from blue to green.

FIGURE 41.17 Exchange of a gluon between two quarks.

electron family muon family tau family

Quarks u

d

s

c

b

t

up-down family strange-charm family bottom-top family

FIGURE 41.18 The forces, carriers, and elementary particles of the Standard Model.

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string theories

Elementary Particles and Cosmology

antiparticles). This proliferation of constituents raises the question whether matter really has such a large number of elementary building blocks. Pushing forward our search for the ultimate building blocks, we have uncovered layers of structures within layers of structures—electrons and nuclei within atoms, protons and neutrons within nuclei, quarks within protons and neutrons. Is there another, more elementary layer within quarks? Seeking to answer this question, some theorists have been exploring string theories, according to which all particles consist of tiny, vibrating strings, only about 1033 cm in length. The different masses, spins, electric charges, and other properties of the particles are supposed to arise from different vibrational patterns of the strings. For this scheme to work, the strings may have to exist in a 9-dimensional space, that is, a space with 6 more dimensions than the 3-dimensional space of our everyday experience. The extra 6 dimensions are thought to be imperceptible because they are tightly curled up, so they extend over distances of only about 1033 cm. It is the ambition of string theorists to deduce the masses and all the other properties of particles from the characteristics of the strings. But in spite of prodigious effort, they have not yet succeeded in this, and they have not demonstrated any clear experimental connection between string theories and our physical world.

✔

Checkup 41.5

QUESTION 1: Do the charges of two u quarks and one d quark add up to the charge of the proton? Do the charges of two d quarks and one u quark add up to the charge of a neutron? 1 QUESTION 2: Each quark has spin 2 . How can the net spin of three quarks add to 1 form the spin 2 of the proton? QUESTION 3: What is a gluon? QUESTION 4: What is the electric charge of the u antiquark? The d antiquark? The s antiquark?

(A)  23e, 13e, 13e (B) 23e, 13e, 13e (C)  23e, 13e, 13e (D) 23e, 23e, 13e (E) 23e, 13e, 13e

41.6 COSMOLOGY Cosmology is the study of the Universe at large, its size, its shape, and its evolution. Seeking to grasp the Universe, our minds must wander over distances as great as 10 billion light-years and over times as long as 10 billion years or longer. Until the early part of the last century, astronomers thought the Universe to be much smaller. They thought that the farthest stars at the edge of our Galaxy were about 30 000 light-years away and that there was nothing but dark, empty space beyond that distance.2 But in the 1920s, Edwin Hubble used the 100-inch telescope on Mount 2

A light-year is the distance traveled by light in 1 year, 9.5  1015 m.

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41.6

Cosmology

FIGURE 41.19 These pictures taken with the Hubble Space Telescope give an idea what our Milky Way Galaxy looks like when viewed (a) face on and (b) edge on. The pictures are actually photographs of two distant galaxies (NGC 5457 and NGC 4631) similar to our Galaxy.

Wilson to establish that the faint, wispy “nebulae” found in all parts of the sky were actually gigantic conglomerations of stars, similar to our own Galaxy but at very great distance from us. Our Milky Way Galaxy contains about 1011 stars arranged in an irregular, disklike region some 105 light-years in diameter. The disk has a central bulge, and it has spiral arms along which stars are concentrated (Fig. 41.19). There are many external galaxies beyond our Galaxy. With our large telescopes we can see altogether about 1011 galaxies! There are supergiant galaxies with 1013 stars each, and there are dwarf galaxies with “only” 106 stars. There are spherical galaxies and elliptical galaxies, like luminous globes and eggs; there are spiral galaxies and barrel-shaped galaxies, like whirling pinwheels; and there are irregular galaxies with the weirdest shapes (see Figs. 41.20a–c).

FIGURE 41.20 (a) Spiral galaxy (NGC 3031) in Ursa Major. The plane of this galaxy is inclined to our line of sight; face on, this galaxy would look circular. (b) Unusual galaxy (NGC 5128) in Centaurus. Note the thick lane of dust surrounding this galaxy. (c) Spiral galaxy (NGC 4565) in Coma Berenices. Note how thin this galaxy is.

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All the galaxies are in motion. Many of them congregate in clusters, orbiting about each other, sometimes accidentally colliding. But let us ignore the fine details of the motion of galaxies and concentrate on the large-scale features of the motion. We then find that on a large scale, the galaxies have a motion of recession—all the distant galaxies are moving away from us. The recessional velocities of galaxies are determined by the red-shift method (see Fig. 41.21). The light from the receding stars is Dopplershifted to lower frequencies, or longer (redder) wavelengths [see Eq. (36.13) in Section 36.3]. To find the recessional velocity, astronomers need only measure how much the wavelength of the light received from the atoms in some distant galaxy is shifted relative to the wavelength emitted by similar atoms in our laboratories on Earth. By measuring the velocities and distances of galaxies, Hubble discovered that the motion of recession obeys a very simple rule: the velocity of each galaxy is directly proportional to its distance. This means that from the point of view of our Galaxy, nearby galaxies move slowly and distant galaxies move fast (see Fig. 41.22). This proportionality between velocity v and distance r is called Hubble’s Law and can be expressed as

red-shift method

v  H0 r

Hubble’s Law

(41.6)

where H0 is the Hubble constant. If r is expressed in billions of light-years, the numerical value of the Hubble constant is H0  2.1  107 (m/s)(billion light-years)

Hubble constant

(41.7)

Astronomers determine the enormous cosmic distances from us to other galaxies by the brightness or “headlight” method. In essence, this method relies on the following. Stars in faraway galaxies look faint to us, and stars in nearby galaxies look bright—just as, on a dark road, the headlights of a faraway automobile look faint and the headlights of a nearby automobile look bright. If all stars generated precisely the same quantity of light, then differences in their apparent brightness as observed by our telescopes would be entirely due to differences in their distances—there would then be a simple mathematical relationship between apparent brightness and distance. (a)

(b)

Absorption lines in a nearby star appear at their rest wavelengths…

A star in the Milky Way

rest

750

H

H

H

H …but absorption lines in the spectrum of a distant galaxy are redshifted by the galaxy’s motion away from us.

A galaxy with v = 7,600 km/s

400

450

500

550 600 Wavelength (nm)

FIGURE 41.21 (a) A distant galaxy seen together with a star in our galaxy. (b) The spectrum of each, shown on the same scale. Note that lines in the galaxy spectrum are red-shifted to longer wavelengths. The H, H, H and H absorption lines are the first four lines of Bolmer Series of hydrogen (see Fig. 38.7).

Recessional velocity v is calculated from the measured redshift.

650

λobserved

700

750

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Cosmology

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Relative to our Galaxy, velocities of other galaxies are radial…

our Galaxy

…and directly proportional to distance.

FIGURE 41.22 The motion of recession of distant galaxies.

However, in practice, there can be some complications; two stars can be at the same distance yet differ in apparent brightness because one generates more light than the other. Such intrinsic differences between stars must be taken into account when using the brightness method. Astronomers have developed clever techniques for selecting stars of standard brightness, but some observational errors remain in the distance determinations. Numerous analyses support the direct proportionality expressed by Eq. (41.6), but there are substantial uncertainties in the numerical value of Eq. (41.7)—the Hubble constant is uncertain by at least — 10%. Although Fig. 41.22 gives the misleading impression that our Galaxy is at the center of the Universe and that all other galaxies are fleeing away from us, our Galaxy does not occupy any special spot in the Universe. The other galaxies are not just fleeing away from us; they are fleeing away from each other. The Universe is expanding. An extraterrestrial astronomer sitting on a distant galaxy would see our Galaxy and all other galaxies fleeing away from her. Hence our spot in the Universe is pretty much the same as every other spot. Cosmologists believe that this overall uniformity holds not only in regard to the expansion, but also in regard to all other general features of the Universe. For instance, the numbers and types of galaxies that the extraterrestrial astronomer finds in her neighborhood will, on the average, be the same as we find in our neighborhood—the Universe is pretty much the same everywhere. This assertion of the large-scale uniformity of the Universe is called the Cosmological Principle. The motion of recession of galaxies can be described by a very crude analogy. When a grenade explodes in midair, the fragments of shrapnel spurt out in all directions. Different fragments may have different velocities, and in a given amount of time, they will reach different distances. After a time t, the position of a fragment having velocity v will be r  vt

(41.8)

v  rt

(41.9)

If we rewrite this as

we see that at any given time the fragments that are at the greatest distances are those with the highest velocities. This proportionality of velocity and distance has the same

Cosmological Principle

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Big Bang

Elementary Particles and Cosmology

form as Hubble’s Law [see Eq. (41.6)]. Thus, Hubble’s Law suggests that the galaxies were set in motion by a primordial cosmic explosion billions of years ago and have been more or less coasting along ever since. Incidentally: In the expansion of the Universe, only the distances between the galaxies increase; the galaxies themselves do not expand. This is also in agreement with the grenade analogy, where, of course, only the distances between the shrapnel fragments increase while the fragments themselves remain of constant size. (But the grenade analogy has serious defects: It posits a center for the explosion, whereas the Universe has no special center. Also, in the modern view, the expansion of the Universe involves the expansion of space itself, not just the motion of matter in ordinary space.) The explosion that started the expansion of the Universe is called the Big Bang. We can calculate how long ago this happened by comparing Eqs. (41.6) and (41.9). We thus see that the inverse of the Hubble constant must coincide with the expansion time: t  1H0

(41.10)

or t

1 2.1  107



billion light-years m/s



1 2.1  107



9.5  1024 m m/s

 4.5  1017 s  1.4  1010 years

(41.11)

However, in this calculation of the age of the Universe, we have ignored the possibility that the velocity of galaxies may change with time. For instance, since gravity pulls the galaxies toward each other, we might expect that gravity tends to inhibit the motion of recession and tends to slow down the expansion of the Universe. This would imply that the velocities of all galaxies were somewhat larger in the past and, consequently, the true age of the Universe ought to be somewhat smaller than the 14 billion years indicated by our naive calculation. But in any case, this number gives us a rough estimate of the age. Since the Universe started some finite time ago, only light from those parts of it that are sufficiently near can have reached us. The speed of light is c  3.00  108 m/s  1 light-year/year, and in the time t light travels a distance ct  (1 light-year/year)  (1.4  1010 year)  1.4  1010 light-years observable universe

This distance is the radius of the observable universe3. Everything within this radius we can see (given sufficiently powerful telescopes); anything beyond we cannot see because the light has not yet had time to reach us. Note that as time increases, the radius ct increases, that is, the observable universe includes more and more of the total Universe. The idea of the Big Bang can be put to several direct tests. First, nothing we observe can be older than about 14 billion years. Our Sun has an age of only 5 billion years. The oldest objects that we can reliably date are the globular clusters of stars found near our Galaxy (see Fig. 41.23). Old stars enter a red-giant stage, turning a reddish color and swelling to several hundred times normal size. A young cluster contains few red giants, and an old cluster contains many. Careful calculations of stellar evolution indicate that the oldest globular clusters have ages of about 10 billion to 13 billion years, in good agreement with the expansion time of Eq. (41.11). 3 This value of the radius of the observable universe is only an approximation. For an exact calculation, we need to use the theory of General Relativity, taking into account that space and time are curved.

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Cosmology

Another method by which we can date the birth of stars is by the age of chemical elements. All of the elements, with the exception of hydrogen and helium, were synthesized in very massive stars soon after our Galaxy came into being. These stars survived only a short time and then exploded as supernovas, scattering these elements into the cloud of gas and dust that was destined to become our Solar System. Radioactive dating tells us that these elements are somewhere between 7 and 15 billion years old, in agreement with the expansion time to within experimental uncertainties. More evidence for the Big Bang emerges from studies of the abundances of hydrogen and helium. These elements were formed in the hot, primordial fireball of the Big Bang. When the Universe was about 3 minutes old, the temperature was 109 K, subjecting hydrogen to nuclear fusion. Theoretical calculations show such fusion led to abundances of about 75% hydrogen and 25% helium, in remarkable agreement with the observed abundances of 74% and 24% (plus traces of other elements).4 This confirms our picture of a hot Big Bang. But the most decisive item of evidence for the Big Bang is that some of the radiant heat given off by the primordial explosion can still be found in the sky today. Originally, the radiant heat emitted by the dense, primordial fireball was in the form of very penetrating gamma rays and X rays. But as the Universe expanded, the electromagnetic radiation expanded with it, and at present the wavelengths of the fireball radiation are much longer. Theoretical calculations indicate that the radiant heat must have a blackbody spectrum (see Chapter 37) with a peak around a wavelength of 1 mm, the wavelength of microwaves. This kind of radiation was discovered in 1964 by Arno A. Penzias and Robert W. Wilson, two scientists working at Bell Laboratories with very sensitive microwave communication equipment (Fig. 41.24). They found that the entire sky is noisy; there is radiation coming at the Earth from all directions. Physicists at Princeton immediately recognized the cosmological significance of this discovery, and identified the radiation as cosmic background radiation, a relic of the Big Bang. This radiation is that of a blackbody with temperature of about 3 K (more precise data give 2.73 K). What has happened here is that the extremely hot radiant heat from the primordial fireball has gradually cooled down as the Universe expanded, and by now its temperature has come pretty close to absolute zero. The cosmic background radiation is residual radiant heat left in the sky after the Big Bang. It is direct material evidence for the Big Bang.

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FIGURE 41.23 The globular cluster (NGC 5272) in Canes Venatici.

cosmic background radiation

FIGURE 41.24 Horn antenna at Bell Laboratories, Holmdel, New Jersey. This antenna was designed for microwave communication experiments with the Echo and Telstar satellites. 4

Fusion within stars continues to produce helium, but the total amount of helium produced by all the stars since the Big Bang is only a few percent.

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A galaxy near the edge of a spherical region…

r

Elementary Particles and Cosmology

The big question in cosmology is this: Will the Universe expand forever? Or will the gravitational attraction between galaxies eventually stop the expansion and cause the Universe to contract, ultimately leading to to a terminal cosmic implosion? To help find the answer, we must calculate the deceleration of the motion of recession caused by gravity. Consider a spherical region of our Universe, and assume that galaxies are uniformly distributed throughout this spherical region and the entire Universe (see Fig. 41.25). Gauss’ Law applies to the inverse-square gravitational force as it does to the inversesquare electric force. Accordingly, we know that the motion of the galaxies in the spherical region is unaffected by the rest of the Universe. Consider now one of the galaxies at the surface of this region, presently at a radial distance r0 from us. That galaxy will continue to recede from us forever provided its present velocity v0 is greater than the escape velocity, Eq. (9.27), v0  12GMr0

(41.12)

where M is the mass inside the spherical region. This mass can be expressed in terms of the average density r0 of mass in the Universe,

…experiences a net gravitational force only from galaxies within the spherical region.

FIGURE 41.25 A spherical region of our Universe.

M  r0

4p 3 r 3 0

(41.13)

and the velocity is given by Eq. (41.6), v0  H0 r0. Thus the condition for a permanently expanding Universe becomes H0 r0 

8p Gr02 r0 A 3

(41.14)

or r0 

3 H2 8pG 0

(41.15)

If we insert the numerical value H0  1(4.5  1017 s) [see Eq. (41.11)] and the numerical value for the gravitational constant, we obtain the conditions r0  9  1027 kg/m3

(permanent expansion)

r0  9  1027 kg/m3

(ultimate contraction)

(41.16)

In principle, this makes it very simple to predict the future evolution of the Universe; but in practice, we are severely handicapped by the uncertainty in the mass density. When reckoning the mass of the Universe, we must take into account the mass belonging to galaxies and also whatever mass is to be found in the intergalactic space between galaxies. Table 41.8 lists all the known contributions to the mas density of the Universe, as a percentage of the critical mass density 9  1027 kg/m3. The visible, luminous matter contained in stars makes only a small contribution, less than 12 % of the critical mass. Thus, the bulk of the mass of the Universe is in nonluminous, invisible forms, which can be detected only by indirected means. The mass of baryonic matter (protons and neutrons) listed in the table is inferred from studies of the abundance of deuterium formed in the fireball of the Big Bang. Apparently, most of this baryonic matter did not get captured into luminous stars during the later evolution of the Universe, and it must still be lurking in and around galaxies, in the form of small, nonluminous stars (“brown dwarfs” and cold stellar

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Cosmology

remnants), perhaps interstellar planets, and large clouds of low-density gas surrounding the galaxies. The mass of the nonbaryonic dark matter is inferred from studies of galactic dynamics. Astronomers can detect the presence of dark matter by its gravitational effects on the rotational motion of galaxies and on the orbital motion of galaxies around each other. From such studies, we know that galaxies are surrounded by large, extragalactic clouds, or halos, of some kind of dark matter. This is sometimes called the “missing mass” because we cannot see it, and we do not know what it consists of. In the absence of any information about this dark matter, astronomers and physicists have felt free to speculate about various possibilities. One such speculation is that this dark matter might consist of clouds of neutrinos. We know that neutrinos have a small, nonzero mass, but the magnitude of this mass has so far proved too small to be measured. If it is about 4 eVc 2, then the clouds of neutrinos left over from the Big Bang in the form of thermal neutrino blackbody radiation might be sufficient to account for the dark matter. Another speculation is that the dark mass might consist of some new, exotic particles invented by theorists working on unified theories of interactions. But no such particles have so far been confirmed experimentally. The final contribution to the mass of the Universe listed in Table 41.8 is the most mysterious. Beginning in 1998, astronomers combined luminosity and red-shift data for a class of very bright supernovas to determine their recessional velocities and distances. These data indicate that instead of the presumed state of slowing expansion, the Universe is actually experiencing an accelerating expansion. To account for an accelerating Universe, cosmologists now assume there exists some sort of dark energy that permeates the space between galaxies. This dark energy apparently makes up about 70% of the critical density and provides an effective repulsive gravitational force that accelerates the expansion. We do not yet have any firm grasp on exactly what this dark energy might be. Will the Universe continue to expand or will it ultimately contract? A foreverexpanding and accelerating Universe gives the best fit to all the facts as we currently know them. But there are enough uncertainties in our measurements and enough loopholes in our arguments that the possibility of an ultimately contracting Universe cannot be dismissed. It will be a while before we know the ultimate fate of the Universe.

✔

Checkup 41.6

QUESTION 1:

Why do astronomers not rely on triangulation to measure the distances

of galaxies? QUESTION 2: If the expansion of the Universe is accelerating or is slowing down, can the Hubble constant be truly a constant? QUESTION 3: Imagine a galaxy very close to the edge of the observable Universe. This galaxy is moving away from us at a speed very close to the speed of light. Does this mean that this galaxy will reach the edge of the observable universe and disappear from sight? QUESTION 4: Where were the atoms of most of the elements inside your body made? (A) In the core of the Earth soon after the formation of the Sun (B) In supernovas soon after our Galaxy was born (C) In the Sun, later transmitted to Earth by cosmic radiation (D) In nuclear fission of heavier elements such as uranium (E) In the first 1034 s of the Big Bang

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dark matter

CONTRIBUTIONS TO THE MASS DENSITY OF THE UNIVERSE

TA B L E 4 1 . 8

FORM OF MATTER

luminous matter (stars)

PERCENTAGE OF CRITICAL DENSITY

0.4%

baryonic matter

4%

dark matter (nonbaryonic)

25%

dark energy

71%

dark energy

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Elementary Particles and Cosmology

S U M M A RY PARTICLES

leptons hadrons

e

baryons mesons carriers

f fermions f bosons

FORCES

strong/color electromagnetic weak gravitational CONSERVED QUANTITIES

Strength

Carrier

1 102 106 1038

gluon photon W and Z graviton

Absolute

Approximate

energy momentum angular momentum electric charge baryon number lepton number

isospin strangeness parity

electron (e), muon (m), tau (t), and neutrinos (e ,  , and  ).

LEPTONS

up (u), down (d), strange (s), charmed (c), bottom (b), and top (t).

QUARKS

d

Forces

color force

electromagnetic force

weak force

gravitational force

Carriers

gluon

photon

W and Z

graviton

u Leptons

u

u

proton

d

electron family muon family tau family

d

neutron

HUBBLE’S LAW (v is the recessional velocity of a galaxy and r is the distance to the galaxy.) COSMOLOGICAL PRINCIPLE

Quarks

e

␯e

u

d

␮

␯␮

s

c

␶

␯␶

b

t

up-down family strange-charm family bottom-top family

v  H0 r H0  2.1  107 (m/s)(billion light-years)

(41.6) (41.7)

On a large scale, the Universe is uniform.

QUESTIONS FOR DISCUSSION 1. Why are high-energy accelerators necessary for the production and discovery of new, massive particles? 2. Physicists were planning to construct the Superconducting Super Collider (SSC), a 20-TeV accelerator, which would have cost some $4 billion. Can such an expenditure be justified?

3. The names baryon, meson, and lepton come from the Greek barys (heavy), mesos (middle), and leptos (thin, slender). These names were originally intended to indicate the masses of the particles. According to the lists of particles and masses given in this chapter, is it true that the baryons have the largest masses and the leptons the smallest?

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Problems

4. How does the antineutron differ from the neutron? 5. In experiments at CERN, an antielectron and an antiproton have been put together to form an antihydrogen atom. In principle, two of these atoms can form an antihydrogen molecule. Could we confine a sample of antihydrogen gas in an ordinary steel bottle? Can you think of any way of confining it? 6. How would you refute the proposition that the Sun is made of antimatter? 7. Why does a particle that lives only 1023 s not make a track in a bubble chamber? (Hint: Suppose the particle moves at the maximum conceivable speed; how far will it travel in 1023 s?) 8. The strengths of the fundamental forces depend on the energies of the particles. In the case of the gravitational force, the strength increases with the energy. Why would you expect this to be true?

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9. The boomerang analogy described in Fig. 41.11 is defective in that the boomerang requires the presence of air. What would be the motion of a boomerang in vacuum? 10. In Fig. 41.7, a large number of particles emerge in the longitudinal direction (toward the right and the left). Why is this expected, whereas the emergence of particles in the transverse direction (upward and downward) is surprising? (Hint: Consider a head-on collision between two rockets; which way do you expect most fragments to spurt out?) 11. Neglecting dark energy, describe the difference between the final states of the Universe for r0  3H02(8pG ) and r0  3H02(8pG ). 12. How will life ultimately end if the Universe continues to expand forever? If it contracts?

PROBLEMS 4 1 . 1 T h e To o l s o f H i g h - E n e r g y P h y s i c s 41.2 The Multitude of Par ticles 1. According to Eq. (30.6), the radius of the orbit of a charged particle of momentum p in a magnetic field is p r qB Expressed in this way, in terms of the momentum, this formula remains valid even if the particle is relativistic (although p  mv for a relativistic particle). (a) Show that for an ultra-relativistic particle the formula becomes E r qcB (b) At a CERN accelerator, protons of energy 450 GeV travel in a circular orbit of radius 1.1 km. Calculate the strength of the magnetic field required to achieve this orbital radius. 2. The Large Hadron Collider (LHC) accelerator at CERN will have a radius of 4.2 km and produce protons of momentum 7.0 TeV/c, or 3.7  1013 kgm/s. (a) What magnetic field is required to hold the protons in a circular orbit of this radius? (Hint: Use the formula given in Problem 1.) (b) What is the period of the orbital motion? (Hint: The speed of the proton is nearly equal to the speed of light.) 3. The relativistic expression 22mc (2mc  K) gives the energy available for inelastic reactions when a particle of mass m and kinetic energy K is incident on a stationary particle of the same mass. Use this formula to calculate the available energy (in GeV) for an antiproton incident on a stationary proton in the following cases: 2

(a) The kinetic energy of the incident antiproton is 10 MeV. (b) The kinetic energy of the incident antiproton is 1.0 TeV (as in the Tevatron). (c) The kinetic energy of the incident antiproton is 7.0 TeV (as in the Large Hadron Collider). 4. Which is the most massive particle listed in the tables of Section 41.2? Express the mass of this particle in atomic mass units and compare the mass with that of the helium atom and that of the lithium atom. 5. Count the number of particles (including antiparticles) that are leptons. *6. A particle detector is in a uniform magnetic field of 1.20 T. For motion perpendicular to the field, what are the radii of curvature for the following particles and kinetic energies: (a) a 5.0-GeV proton; (b) a 1.0-MeV electron; and (c) a 300-MeV positive pion. *7. Show that for a particle of (total) energy E, the time-dilation factor can be expressed as Emc 2. What is the time-dilation factor for a muon of energy 950 MeV? The lifetime of this muon is 2.2  106 s in its own reference frame. What is its lifetime in the laboratory reference frame? *8. Suppose we want to produce a ©  particle by the head-on collision of two protons of equal energies according to the reaction p  p S p  K0  © 

2

What is the minimum kinetic energy required for each proton to initiate this reaction? The masses of the particles are given in Tables 41.2 and 41.3.

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*9. In 1908, a meteoroid struck near Tunguska, in Siberia. The destructive effects of this impact have been estimated as equivalent to the explosion of 12.5 megatons of TNT, or 5.3  1016 J. One possible explanation for this explosion is that the meteoroid was made of antimatter, which annihilated with an equal amount of matter when it came in contact with the Earth’s surface. If so, how much antimatter would account for the explosion? *10. The  particle decays in two alternative ways:  S ¶  K and  S 0  p Which of these reactions releases the most energy and gives the decay products the largest kinetic energy? *11. A ©  particle at rest decays into a pion and a neutron: ©  S p  n What is the net kinetic energy of the decay products? *12. Consider the annihilation of an electron and an antielectron, resulting in two gamma rays, e  e S g  g Suppose that the electron and antielectron are initially at rest. What are the energies of the resulting gamma rays? What are their wavelengths? *13. The p0 meson decays into two gamma rays: 0

p Sg  g If the pion is initially at rest, what are the energies of the two gamma rays? What are their wavelengths? *14. A K0 particle at rest decays into two pions: K0 S p0  p0 What is the kinetic energy of each of these pions? What is the momentum of each? (Hint: Use the relativistic relation between energy and momentum.)

17. Is strangeness conserved in the following reactions? p  n S K  ¶ ¶ S p  p K0  n S p  p 18. Conservation of lepton number is actually three separate conservation laws; experiments indicate that electron lepton number, muon lepton number, and tau lepton number are each individually conserved. For example, the electron lepton number is 1 for e and e, it is 1 for e and e, and it is zero for all other particles. For the following reactions, determine what missing particles are needed to conserve all lepton numbers: m S e  ____  ____ p S n  e  ____ t S m  ____  ____ 19. Show that the emission of a photon by a free electron is impossible because it conflicts with energy conservation. (Hint: Consider the emission process in the reference frame in which the electron is initially at rest.) 20. Show that the annihilation of an electron and an antielectron into a single photon (e  e S g) is impossible, because it conflicts with conservation of momentum. (Hint: Consider the reaction from the reference frame in which the two electrons have opposite velocities of equal magnitudes.) 21. Consider the following reactions produced by a beam of K particles in a bubble chamber filled with liquid hydrogen: K  p S ©   p K  p S © 0  p0 K  p S ¶  p0 K  p S ¶  p  p Verify that all of these reactions conserve baryon number and strangeness. 22. The  particle decays according to the reaction

41.3 Interactions and Conser vation Laws 15. Verify that the reaction (41.2) conserves baryon number. Does the reaction conserve strangeness? 16. Which of the following reactions are forbidden by an absolute conservation law? p  p S ¶  K0 K  p S K  p  p0 p  n S p  p0  p0 K  n S ©   p0 e  ne S p  p0

 S ¶  p Does this reaction conserve baryon number? Strangeness?

41.4 Fields and Quanta 23. The W particle can have either a positive charge (W) or a negative charge (W). Figure 41.26 shows the Feynman diagram for the decay of the neutron (n) via exchange of a W; the end products are a proton (p), an electron (e), and an electron antineutrino ( e). Can you guess the Feynman diagram for the decay of the antineutron?

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Problems

*26. According to a speculative theory of the strong interactions, the proton should be unstable and decay with a lifetime of about 1033 years. Consider the protons in a mass of 1.0  106 kg of water. How many of these protons would decay in one year?

t ␯–e

e

1427

*27. In the hot, early Universe, at a temperature of 1015 K, what was the average kinetic energy of the random thermal motion of gas particles? Compare this energy with the energy of about 100 GeV required to achieve symmetry between W particles and photons.

p W–

41.5 Quarks

n

x

FIGURE 41.26 Feynman diagram for the decay of a neutron.

24. The strong force can also be crudely interpreted as pion exchange. Figure 41.27 shows the Feynman diagram for the exchange of a p between a proton and a neutron; note that the proton changes into a neutron, and vice versa, so electric charge is conserved at each vertex. Draw corresponding diagrams for the exchange of a p and of a p0.

28. Table 41.2 lists the quark composition of baryons. For all the © and particles, verify that the listed quark composition gives the correct values of electric charge, baryon number, and strangeness. 29. The antiproton p is made of three quarks. What kind of quarks are these? 30. How many quarks are there in a hydrogen atom? In a water molecule? (The oxygen nucleus contains eight protons and eight neutrons.) 31. How many quarks are created in the reaction (41.1)? 32. A particle is made of one d quark and one u antiquark. What is the electric charge of this particle? What is this particle?

t

33. Antiparticles contain antiquarks; for example, the proton quark content is uud and the antiproton quark content is u u d. Are any of the baryons in Table 41.2 their own antiparticle? Are any of the mesons in Table 41.3 their own antiparticle? Which ones? What can you say about the quark content of a particle that is its own antiparticle?

p

n ␲

p

34. According to a theoretical prediction based on the quark model, the masses of the nucleon (N  proton or neutron), the , the , and the © should be related by

n

(mN  m )  12(3m¶  m©) x

FIGURE 41.27 Feynman diagram for the exchange of a p between a proton and a neutron. 25. A virtual particle may exist for a time ¢t provided that its energy is less than the uncertainty in energy ¢E. The two are related by an uncertainty principle, ¢E ¢t  U. To mediate the strong nuclear force, how long may a virtual pion of restmass energy 140 MeV exist? To mediate the weak nuclear force, how long may a virtual Z 0 of rest-mass energy 91 GeV exist? How far does each particle travel in this time? (For simplicity, use v  c to estimate the range of the force.)

Since this is intended as an approximate relation, the mass differences between the positive, negative, and neutral kinds of nucleon, , or © are to be neglected. Check this relation against the experimental values of the masses. *35. As described in Section 41.5, high-energy electrons produced by the Stanford Linear Accelerator were used to probe the internal structure of protons. To detect small lumps in the proton, the wavelength of the electrons must be smaller than the lumps. The electrons had an energy of 20 GeV (which is an ultra relativistic energy). What is the de Broglie wavelength, l  hp, of these electrons?

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CHAPTER 41

Elementary Particles and Cosmology

41.6 Cosmology 36. Figure 41.28 shows the positions of four galaxies, with our own Galaxy at the center. Draw a figure showing the positions of these five galaxies at a later time, when the Universe is twice as old.

P

FIGURE 41.29 Motion of a galaxy P and other galaxies relative to our Galaxy ( ). Galaxy) in the reference frame of the galaxy P; do this graphically, by subtracting the velocity vectors.

FIGURE 41.28 Four galaxies ( ) at various distances from our Galaxy ( ).

38. If the velocity of recession of the galaxies were not proportional to distance (v  H0r) but, rather, proportional to the distance squared (v  H0r2) or proportional to some other power of the distance, then our Galaxy would occupy a preferred, central spot in the Universe. Explain.

37. Consider the galaxies shown in Fig. 41.29; the arrows are the velocity vectors of those galaxies in the reference frame of the Earth. According to the Galilean addition law for velocities, find the velocity vectors of these galaxies (including our

39. The value of the Hubble constant that Hubble had deduced from the available data in 1936 was 1.0  108 (m/s)(billion light-years). The corresponding expansion time [see Eq. (41.11)] is 1.9  109 years. How does this compare with the age of the Earth and the age of globular clusters? With what problem was Hubble faced?

REVIEW PROBLEMS *40. A particle at rest decays into two pions, one positive and one negative. In a magnetic field of 1.25 T, each pion leaves a curved track perpendicular to the field with a radius of 54.9 cm. What is the mass of the particle that decayed? What is this particle?

44. Two of the baryons in Table 41.2 have identical quark content. Which two are these? Consider the various conserved quantities discussed in Section 41.3. Which of these quantities do you think might be different for the two baryons?

*41. The neutron decays to a proton, an electron, and an electron antineutrino. Assume that for one particular decay, the neutron is at rest and the neutrino carries away negligible energy and momentum. What is the net kinetic energy of the proton and electron?

45. A particle is made of two d quarks and one s quark. What is the electric charge of this particle? What is this particle?

*42. In Dan Brown’s novel Angels and Demons, 0.25 gram of antimatter is released about 3.0 km above ground level. Assuming that the subsequent annihilation results in isotropic radiation, what is the total energy incident on a square meter of ground directly below the annihilation? If the annihilation occurs over 10 seconds, compare the average energy flux at ground level during the annihilation with that of direct sunlight, approximately 1.0 kW/m2. 43. In each of the following forbidden reactions, what conservation law is violated? p  p S p  p p  nSp  n g  m  n S p0  p

46. The ¢  particle has quark content uuu and decays into two particles in a reaction where only one quark–antiquark pair is created. If one particle is a proton, what is the other particle in the decay? 47. Sketch a Feynman diagram in which two masses are initially moving apart along the x axis, then come to rest after exchanging a graviton, and then approach after exchanging another graviton. 48. The recessional velocity of a galaxy in the constellation Ursa Major is 1.5  107 m/s, and the distance to this galaxy is approximately 1.0  109 light-years. Deduce a value of the Hubble constant from these data alone. How does your value compare with the average accepted value in Eq. (41.7)? Obtain an estimate of the age of the Universe in years from your Hubble constant.

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Answers to Checkups

1429

Answers to Checkups Checkup 41.1 1. No to both. Charged particles are required to boil the liquid in

a bubble chamber to create a bubble track. Similarly, charged particles are required to ionize the gas in a multiwire chamber and produce the electrons that create current pulses. 2. The magnetic field in a particle detector diverts a charged par-

ticle into a circular orbit that reveals both the sign of the charge and the momentum of the particle. 3. The positively charged particles (red tracks) are diverted to the

left as they travel downward. With the initial velocity downward, the magnetic field B must be directed out of the page in order to produce a magnetic force F  q v  B to the left. 4. In reaction (41.1), the masses of the products greatly exceed

the masses of the two initial protons, so the Q value is negative. We saw that reaction (41.2) proceeds spontaneously, so its Q value must be positive. 5. (B) In the head-on collision between protons of opposite

motion, the full kinetic energy is available for reaction. For a stationary target, the nonzero momentum of the projectile must be conserved, requiring much of the final energy to remain kinetic.

Table 41.4, the anti-xi-zero ( 0) then has mass 1315 MeV/c 2, spin 12, zero charge, and strangeness 2; similarly, the ant-ximinus, or xi-plus, has mass 1321 MeV/c 2, spin 12, charge 1, and strangeness 2. 3. All interactions conserve energy, momentum, angular

momentum, charge, baryon number, and lepton number. Strangeness and parity are conserved in electromagnetic (and strong) interactions, but not in weak interactions. Isospin is conserved in strong interactions, but not in electromagnetic (nor in weak) interactions. 4. The parity of sin kx is 1, since the function becomes the

negative of itself when you reverse the x coordinate, sin (ky)  sin kx. 5. (B) Weak only. Only the weak interaction does not conserve

strangeness. 6. (D) All except strong; all. As a lepton, the electron does not

participate in strong interactions. The neutron participates in all four—even though the neutron is net charge-neutral, it has an internal charge distribution and a magnetic moment, and participates in the electromagnetic interaction.

Checkup 41.4

Checkup 41.2

1. Yes. Although heavier particles continue to be sought, the W 1

1. All leptons have spin 2 , but some baryons have other half-

integer spins; for example, the  has spin 32.

2. An antiparticle has the same mass and spin but the opposite

charge as the corresponding particle; thus, from Table 41.1, the antitau has mass 1784 MeVc2, spin 12, and electric charge e, and the electron antineutrino has a mass less than 3 eVc 2, spin 12, and zero electric charge.

and Z are the heaviest particles now known. 2. (B) Weak, strong, electromagnetic. The range of a force is

inversely related to the mass of the particle that mediates the force. The W and Z particles (weak force) have the largest masses, photons (electromagnetic force) and gluons (strong force) are massless, but the range of the strong force is limited by pair production.

3. The heaviest particle listed is the upsilon () particle, one of

the mesons of Table 41.3. 4. (D) Baryon; meson. All known leptons are listed in Table

41.1; there are no others. Other particles with half-integer spin, such as the h0, are baryons; particles with mass and with integer spin, such as the B0, are mesons.

Checkup 41.5 1. Yes to both. Since the charge of the u quark is  3 e and the 2

13e,

charge of the d quark is the charge of 2u  d is e and the charge of 2d  u is zero.

2. Two of the spins are up, one is down, resulting in a net spin

Checkup 41.3 1. A hadron is a particle that participates in the strong interac-

tion, that is, a baryon or a meson. Thus the proton and neutron are hadrons, but the electron and the tau neutrino, both leptons, are not. Similarly, the photon, the quantum of electromagnetism, is not a hadron. 2. An antiparticle has the same mass and spin as the particle, but

the opposite charge and strangeness. From the xi-zero data of

of 12. 3. The gluon is the particle that acts as carrier of the color force

and the strong force. 4. (E) 3 e,  3 e,  3 e. The charge of an antiquark is the negative 2

1

1

of the charge of the corresponding quark.

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Checkup 41.6 1. Other galaxies are so distant that the angles required for trian-

gulation cannot be measured accurately. Even with the diameter of the Earth’s orbit as a baseline, no difference in the angle to a distant galaxy from either end of such a baseline can be detected. 2. No. The Hubble “constant” reflects the present rate of expan-

sion, and its value would increase if the expansion accelerates or decrease if the expansion slows; it would even change sign if contraction ultimately occurs.

3. No. The edge of the observable universe is expanding at the

speed of light, and so a galaxy near the edge, moving slower than the speed of light, becomes further from the edge, and will not disappear. 4. (B) In supernovas soon after our Galaxy was born. These early

supernovas spewed out dust and gas with the heavy elements that later coalesced to become our Solar System.

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Appendix 1: Greek Alphabet A B  ¢

  g d 

alpha beta gamma delta epsilon



™   



zeta eta theta iota kappa lambda mu

h u i k l m

M

o p r

nu xi omicron pi rho

s t y f x c 

sigma tau upsilon phi chi psi omega

n

N 

ß



P © T £  ° 

Appendix 2: Mathematics Review A 2.1 S y m b o l s a  b means a equals b a  b means a is not equal to b a  b means a is greater than b a  b means a is less than b a  b means a is not less than b a  b means a is not greater than b a r b means a is proportional to b a  b means a is approximately equal to b a W b means a is much greater than b a V b means a is much less than b p  3.141 59 . . . e  2.718 28 . . .

A 2.2 P o w e r s a n d R o o t s For any number a, the nth power of the number is the number multiplied by itself n times. This is written as an, and n is called the exponent. Thus, a1  a

a2  a # a

a3  a # a # a

a4  a # a # a # a

etc.

For instance, 32  3  3  9 33  3  3  3  27 34  3  3  3  3  81 etc. A negative exponent indicates that the number is to be divided n times into 1; thus a1 

1 a

a2 

1 2

a

a3 

1 a3

etc. A-1

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APPENDIX 2

Mathematics Review

A zero exponent yields 1, regardless of the value of a: a0  1 The rules for the combination of exponents in products, in ratios, and in powers of powers are an # am  amn an  anm am (an)m  anm For instance, it is easy to verify that 32  33  35 32 3

3

 31 

1 3

2 3

(3 )  323  36 Note that for any two numbers a and b (a # b)n  an # bn For instance, (2  3)3  23  33 The nth root of a is a number such that its nth power equals a. The nth root is written a1>n. The second root a1>2 is usually called the square root, and designated by 1a: a1>2  1a As suggested by the notation a1>n, roots are fractional powers, and they obey the usual rules for the combination of exponents: (a1>n)n  an>n  a (a1>n)m  am>n

A 2.3 A r i t h m e t i c i n S c i e n t i f i c N o t a t i o n The scientific notation for numbers (see the first page of the Prelude) is quite handy for the multiplication and the division of very large or very small numbers, because we can deal with the decimal parts and the power-of-10 parts in the numbers separately. For example, to multiply 4  1010 by 5  1012, we multiply 4 by 5 and 1010 by 1012, as follows: (4  1010)  (5  1012)  (4  5)  (1010  1012)  20  101012  20  1022  2  1023 To divide these numbers, we proceed likewise: 4  1010 12

5  10



4 1010  12  0.8  101012  0.8  102  8  103 5 10

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APPENDIX 2

Mathematics Review

When performing additions or subtractions of numbers in scientific notation, we must be careful to begin by expressing the numbers with the same power of 10. For example, the sum of 1.5  109 and 3  108 is 1.5  109  3  108  1.5  109  0.3  109  1.8  109

A 2.4 A l g e b r a An equation is a mathematical statement that tells us that one quantity or a combination of quantities is equal to another quantity or combination. We often have to solve for one of the quantities in the equation in terms of the other quantities. For instance, we may have to solve the equation xab for x in terms of a and b. Here a and b are numerical constants or mathematical expressions which are regarded as known, and x is regarded as unknown. The rules of algebra instruct us how to manipulate equations and accomplish their solution. The three most important rules are: 1. Any equation remains valid if equal terms are added or subtracted from its left side and its right side. This rule is useful for solving the equation x  a  b. We simply subtract a from both sides of this equation and find xaaba that is, xba To see how this works in a concrete numerical example, consider the equation x75 Subtracting 7 from both sides, we obtain x57 or x  2 Note that given an equation of the form x  a  b, we may want to solve for a in terms of x and b, if x is already known from some other information but a is a mathematical quantity that is not yet known. If so, we must subtract x from both sides of the equation, and we obtain abx Most equations in physics contain several mathematical quantities which sometimes play the role of known quantities, sometimes the role of unknown quantities, depending on circumstances. Correspondingly, we will sometimes want to solve the equation for one quantity (such as x), sometimes for another (such as a). 2. Any equation remains valid if the left and the right sides are multiplied or divided by the same factor. This rule is useful for solving ax  b

A-3

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APPENDIX 2

Mathematics Review

We simply divide both sides by a, which yields ax b  a a or x

b a

Often it will be necessary to combine both of the above rules. For instance, to solve the equation 2x  10  16 we begin by subtracting 10 from both sides, obtaining 2x  16  10 or 2x  6 and then we divide both sides by 2, with the result x

6 2

or x3 3. Any equation remains valid if both sides are raised to the same power. This rule permits us to solve the equation x3  b Raising both sides to the power 13, we find (x3)1>3  b1>3 or x  b1>3 As a final example, let us consider the equation x  12 gt 2  x0 (as established in Chapter 2, this equation describes the vertical position of a particle that starts at a height x0 and falls for a time t; but the meaning of the equation need not concern us here). Suppose that we want to solve for t in terms of the other quantities in the equation. This will require the use of all our rules of algebra. First, subtract x from both sides and then add 12 gt 2 to both sides. This leads to 0  12 gt 2  x0  x and then to 1 2 2 gt

 x0  x

Next, multiply both sides by 2 and divide both sides by g; this yields 2 t 2  (x0  x) g

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APPENDIX 2

Mathematics Review

Finally, raise both sides to the power 12, or, equivalently, extract the square root of both sides. This gives us the final result t

2 (x  x) Bg 0

A 2.5 E q u a t i o n s w i t h T w o U n k n o w n s If we seek to solve for two unknowns simultaneously, then we need two independent equations containing these two unknowns. The solution of such simultaneous equations can be carried out by the method of elimination: begin by using one equation to solve for the first unknown in terms of the second, then use this result to eliminate the first unknown from the other equation. An example will help to make this clear. Consider the following two simultaneous equations with two unknowns x and y: 4x  2y  8 2x  y  2 To solve the first equation for x in terms of y, subtract 2y from both sides and then divide both sides by 4: 8  2y x 4 Next, substitute this expression for x into the second equation: 8  2y 2  y  2 4 To simplify this equation, multiply both sides by 4: 2  (8  2y)  4y  8 and combine the two terms containing y: 16  8y  8 This is an ordinary equation for the single unknown y, and it can be solved by the methods we discussed in the preceding section, with the result y3 It then follows from the above expression for x that 8  2y 823 2 1    x 4 4 4 2

A 2.6 T h e Q u a d r a t i c F o r m u l a The quadratic equation ax2  bx  c  0 has two solutions: x

b ;2b2  4ac 2a

A 2.7 L o g a r i t h m s a n d t h e E x p o n e n t i a l F u n c t i o n The base-10 logarithm of a (positive) number is the power to which 10 must be raised to obtain this number. Thus, from 10  101 and 100  102 and 1000  103 and 10 000  104 we immediately deduce that

A-5

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A-6

APPENDIX 2

Mathematics Review

log 10  1 log 100  2 log 1000  3 log 10 000  4, etc. Likewise log 1  0 log 0.1  1 log 0.01  2 log 0.001  3, etc. Thus, the logarithm of a number between 1 and 10 is somewhere between 0 and 1, but to find the logarithm of such a number, we need the help of a computer program (many calculators have built-in computer programs that yield the value of the logarithm at the touch of a button). For some calculations, it is convenient to remember that log2  0.301 ≈ 0.3 and log5  0.699 ≈ 0.7. The logarithm of the product of two numbers is the sum of the individual logarithms, and the logarithm of the ratio of two numbers is the difference of the individual logarithms. This rule makes it easy to find the logarithm of a number expressed in scientific notation. For example, the logarithm of 2  106 is log (2  106)  log 2  log 106  0.301  6  6.301 Note that the logarithm of any (positive) number smaller than 1 is negative. For example, log (5  103)  log 5  log 103  0.699  3  2.301 The exponential function exp(x) is defined by the following infinite series: exp(x)  1  x 

x2 x3 x4   p 2 32 432

This function is equivalent to raising the constant e  2.718 28 … to the power x: exp(x)  ex The natural logarithm ln x is the inverse of the exponential function, so x  eln x and x  ln(ex) Natural logarithms obey the usual rules for logarithms, ln(x # y)  ln x  ln y x ln a b  ln x  ln y y ln(xa)  a ln x

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APPENDIX 3

A-7

Geometry and Trigonometry Review

Note that ln e  1 and ln 10  2.3026 If we designate the base-10 logarithm, or common logarithm, by log x, then the relationship between the two kinds of logarithm is as follows: ln x  ln(10log x)  (log x)(ln 10)  2.3026 log x

Appendix 3: Geometr y and Trigonometr y Review A 3 . 1 P e r i m e t e r s , A r e a s , a n d Vo l u m e s [perimeter of a circle of radius r]  2r [area of a circle of radius r]  r 2 [area of a triangle of base b, altitude h]  hb2 [surface area of a sphere of radius r]  4r 2 [volume of a sphere of radius r]  4r 33 [area of curved surface of a cylinder of radius r, height h]  2rh [volume of a cylinder of radius r, height h]  r 2h

A3.2 Angles The angle between two intersecting straight lines is defined as the fraction of a complete circle included between these lines (Fig. A3.1). To express the angle in degrees, we assign an angular magnitude of 360 to the complete circle; any arbitrary angle is then an appropriate fraction of 360. To express the angle in radians, we assign an angular magnitude of 2p radians to the complete circle; any arbitrary angle is then an appropriate fraction of 2p. For example, the angle shown in Fig. A3.1 is 121 of a complete circle, that is, 30, or p>6 radian. In view of the definition of angle, the length of arc included between the two intersecting straight lines is proportional to the angle  between these lines; if the angle is expressed in radians, then the constant of proportionality is simply the radius: s  ru

(1)

Since 2p radians  360, it follows that 1 radian 

360 360   57.2958 2p 2  3.141 59

(2)

Each degree is divided into 60 minutes of arc (arcminutes), and each of these into 60 seconds of arc (arcseconds). In degrees, minutes of arc, and seconds of arc, the radian is 1 radian  57 17 44.8–

(3)

r

s

θ

O

FIGURE A3.1 The angle  in this diagram is   30, or /6 radian.

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A-8

APPENDIX 3

P

r

y

θ

x

O

Q

FIGURE A3.2 A right triangle.

Geometry and Trigonometry Review

A 3 . 3 T h e Tr i g o n o m e t r i c F u n c t i o n s The trigonometric functions of an angle are defined as ratios of the lengths of the sides of a right triangle erected on this angle. Figure A3.2 shows an acute angle u and a right triangle, one of whose angles coincides with u. The adjacent side OQ has a length x, the opposite side QP a length y, and the hypotenuse OP a length r. The sine, cosine, tangent, cotangent, secant, and cosecant of the angle u are then defined as follows:

EXAMPLE 1

sine

sin u  y>r

(4)

cosine

cos u  x>r

(5)

tangent

tan u  y>x

(6)

cotangent

cot u  x>y

(7)

secant

sec u  r>x

(8)

cosecant

csc u  r>y

(9)

Find the sine, cosine, and tangent for angles of 0, 90, and 45.

SOLUTION: For an angle of 0, the opposite side is zero (y  0), and the adja-

cent side coincides with the hypotenuse (x  r). Hence sin 0  0 r

sin 90  1

45° x

sin 45 

y

P θ

O

cos 90  0

tan 90  q

(11)

x

1 12

cos 45 

1 12

tan 45  1

(12)

The definitions (4)–(9) are also valid for angles greater than 90, such as the angle shown in Fig. A3.4. In the general case, the quantities x and y must be interpreted as the rectangular coordinates of the point P. For any angle larger than 90, one or both of the coordinates x and y are negative. Hence some of the trigonometric functions will also be negative. For instance, sin 135 

FIGURE A3.4 The angle  in this diagram is larger than 90.

(10)

Finally, for an angle of 45 (Fig. A3.3), the adjacent and the opposite sides have the same length (x  y) and the hypotenuse has a length of 12 times the length of either side (r  12x  12y). Hence

FIGURE A3.3 A right triangle with an angle of 45.

x

tan 0  0

For an angle of 90, the adjacent side is zero (x  0), and the opposite side coincides with the hypotenuse (y  r). Hence

y

y

cos 0  1

1 12

cos 135  

1 12

tan 135  1

Figure A3.5 shows plots of the sine, cosine, and tangent vs. u.

(13)

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APPENDIX 3

A-9

Geometry and Trigonometry Review

(a) sin θ 1 90°

π /2

450°

θ, degrees

2 π 5π /2

θ, radians

180° 270° 360°

π

3π /2

–1

(b) cos θ 1 90° 180° 270°

π /2

0

π

360°

θ, degrees

3π /2 2 π

θ, radians

180°

270°

360°

θ, degrees

π

3π /2

2π

θ, radians

–1

(c) tan θ 2 1 90°

0 0°

π /2

–1 –2

FIGURE A3.5 Plots of the sine, cosine, and tangent functions.

A 3 . 4 Tr i g o n o m e t r i c I d e n t i t i e s From the definitions (4)–(9) we immediately find the following identities: tan u  sin u>cos u

(14)

cot u  1>tan u

(15)

sec u  1>cos u

(16)

csc u  1>sin u

(17)

Figure A3.6 shows a right triangle with angles u and 90  u. Since the adjacent side for the angle u is the opposite side for the angle 90  u and vice versa, we see that the trigonometric functions also obey the following identities: sin (90  u)  cos u

(18)

90° – θ θ

FIGURE A3.6 A right triangle with angles  and 90  .

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A-10

APPENDIX 4

Calculus Review

cos (90  u)  sin u

(19)

tan (90  u)  cot u  1>tan u

(20)

According to the Pythagorean theorem, x2  y2  r 2. With x  r cos u and y  r sin u, this becomes r 2 cos 2 u  r 2 sin 2 u  r 2, or cos2 u  sin2 u  1

(21)

The following are a few other trigonometric identities, which we state without proof: sec2 u  1  tan2 u

(22)

csc2 u  1  cot2 u

(23)

sin 2u  2 sin u cos u

(24)

cos 2u  2 cos2 u  1

(25)

sin(  )  sin  cos   cos  sin 

(26)

cos(  )  cos  cos   sin  sin 

(27)

A3.5 The Laws of Cosines and Sines α

C

In an arbitrary triangle the lengths of the sides and the angles obey the laws of cosines and of sines. The law of cosines states that if the lengths of two sides are A and B and the angle between them is  (Figure A3.7), then the length of the third side is given by C 2  A 2  B 2  2AB cos g

B

The law of sines states that the sines of the angles of the triangle are in the same ratio as the lengths of the opposite sides (Figure A3.7):

β

A

(28)

sin  sin g sin    A B C

γ

FIGURE A3.7 An arbitrary triangle.

(29)

Both of these laws are very useful in the calculation of unknown lengths or angles of a triangle.

Appendix 4: Calculus Review A 4.1 D e r i v a t i v e s We saw in Section 2.3 that if the position of a particle is some function of time, say, x  x(t), then the instantaneous velocity of the particle is the derivative of x with respect to t: v

dx dt

(1)

This derivative is defined by first looking at a small increment ¢x that results from a small increment ¢t, and then evaluating the ratio ¢x>¢t, in the limit when both ¢x and ¢t tend toward zero. Thus dx lim ¢x  ¢tS0 dt ¢t

(2)

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APPENDIX 4

Graphically, in a plot of position vs. time, the derivative dxdt is the slope of the straight line tangent to the curved line at the time t (see Figure A4.1). In general, if f  f (u) is some given function of a variable u, the derivative of f with respect to u is defined by df du

lim  ¢uS0

A-11

Calculus Review

x (t)

¢f (3)

¢u

In a plot of f vs. u, this derivative is the slope of the straight line tangent to the curve representing f (u). Starting with the definition (3) we can find the derivative of any function (provided the function is sufficiently smooth so the derivative exists!). For example, consider the function f (u)  u2. If we increase u to u  ¢u, the function f (u) increases to 2

f  ¢f  (u  ¢u)

(4)

t

FIGURE A4.1 The derivative of x (t) at t is the slope of the straight line tangent to the curve at t.

and therefore ¢f  (u  ¢u)2  f  (u  ¢u)2  u2  2u ¢u  (¢u)2

(5)

The derivative df>du is then TA B L E A 4 . 1 df du

lim  ¢uS0

¢f ¢u

2

lim  ¢uS0

2u ¢u  (¢u) ¢u

lim (2u)  lim (¢u)  ¢uS0 ¢uS0

(6)

SOME DERIVATIVES

(7)

d n u  nun1 du d 1 ln u  u du d u e  eu du

The second term on the right side vanishes in the limit ¢u S 0; the first term is simply 2u. Hence df du

 2u

(8)

or d 2 (u )  2u du

(9)

This is one instance of the general rule for the differentiation of un: d n (u )  nun1 du

d sin u  cos u du d cos u  sin u du d tan u  sec2 u du

(10)

This general rule is valid for any positive or negative number n, including zero. The proof of this rule can be constructed by an argument similar to that above. Table A4.1 lists the derivatives of the most common functions.

A4.2 Impor tant Rules for Differentiation 1. Derivative of a constant times a function: df d (cf )  c du du

(In all the following formulas, u is in radian:)

(11)

d cot u  csc 2 u du d sec u  tan u sec u du d csc u  cot u csc u du d sin1 u  1>21  u2 du d cos1 u  1> 21  u2 du d 1 tan1 u  du 1  u2

t

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APPENDIX 4

Calculus Review

For instance, d d 2 (6u2)  6 (u )  6  2u  12u du du 2. Derivative of the sum of two functions: df dg d ( f  g)   du du du

(12)

For instance, d d d (6u2  u)  (6u2)  (u)  12u  1 du du du 3. Derivative of the product of two functions: df dg d ( f  g)  g f du du du

(13)

For instance, d 2 d 2 d (u sin u)  sin u u  u2 sin u du du du  sin u  2u  u2  cos u 4. Chain rule for derivatives: If f is a function of g and g is a function of u, then df dg d f (g)  du dg du

(14)

For instance, if g  2u and f ( g)  sin g, then d sin(2u) d (2u) d sin(2u)  du d(2u) du  cos(2u)  2 5. Partial derivatives: If f is a function of more than one variable, then the partial derivative of f with respect to one of the variables, say x , is denoted 0f 0x, and is obtained by treating all the other variables as constants when differentiating. For instance, if f  x2y  y2z, then 0f 0x

 2xy,

0f 0y

 x2  2yz,

0f and

0z

 y2

A4.3 Integrals We have learned that if the position of a particle is known as a function of time, then we can find the instantaneous velocity by differentiation. What about the converse problem: if the instantaneous velocity is known as a function of time, how can we find the position? In Section 2.5 we learned how to deal with this problem in the special case of motion with constant acceleration. The velocity is then a fairly simple function of time [see Eq. (2.17)] v  v0  at

(15)

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APPENDIX 4

A-13

Calculus Review

and the position deduced from this velocity is [see Eq. (2.22)] x  x0  v0t  12 at2

(16)

where x0 and v0 are the initial position and velocity at the initial time t0  0. Now we want to deal with the general case of a velocity that is an arbitrary function of time, v  v(t)

(17)

Figure A4.2 shows what a plot of v vs. t might look like. At the initial time t0, the particle has an initial position x0 (for the sake of generality we now assume that t0  0). We want to find the position at some later time t. For this purpose, let us divide the time interval t  t0 into a large number of small time intervals, each of the duration ¢t. The total number of intervals is N, so t  t0  N ¢t. The first of these intervals lasts from t0 to t0  ¢t; the second from t0  ¢t to t0  2¢t ; etc. In Figure A4.3 the beginnings and the ends of these intervals have been marked t0, t1, t2, etc., with t1  t0  ¢t, t2  t0  2¢t, etc. If ¢t is sufficiently small, then during the first time interval the velocity is approximately v(t0); during the second, v(t1); etc.This amounts to replacing the smooth function v(t) by a series of steps (see Fig. A4.3). Thus, during the first time interval, the displacement of the particle is approximately v(t0) ¢t ; during the second interval, v(t1) ¢t ; etc. The net displacement of the particle during the entire interval t  t0 is the sum of all these small displacements: x(t)  x0  v(t0) ¢t  v(t1) ¢t  v(t2) ¢t    

(18)

Using the standard mathematical notation for summation, we can write this as N1

x(t)  x0  a v(ti) ¢t

(19)

i0

We can give this sum the following graphical interpretation: since v(ti) ¢t is the area of the rectangle of height v(ti) and width ¢t, the sum is the net area of all the rectangles shown in Figure A4.3, i.e., it is approximately the area under the velocity curve. Note that if the velocity is negative, the area must be reckoned as negative! Of course, Eq. (19) is only an approximation. To find the exact displacement of the particle we must let the step size ¢t tend to zero (while the number of steps N tends to infinity). In this limit, the steplike horizontal and vertical line segments in Fig. A4.3 approach the smooth curve. Thus, x(t)  x0 

lim

N1

¢tS0 a v(ti) NSq i0

(20)

¢t

v (t)

v (t 2 ) v (t 1) v (t 0 )

FIGURE A4.3 The interval t  t0 has been divided into N equal intervals of duration ¢t, so t1  t0  ¢t, etc.

t0 t1 t2 t3

tN – 2

t tN – 1

v (t)

t0

t

t

FIGURE A4.2 Plot of a function v(t).

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APPENDIX 4

Calculus Review

In the notation of calculus, the right side of Eq. (20) is usually written in the following fashion: v (t)

x(t)  x0 

t

 v(t) dt

(21)

t0

t0

t

t

FIGURE A4.4 The area under the velocity curve.

The right side is called the integral of the function v(t). The subscript and the superscript on the integration symbol  are called, respectively, the lower and the upper limit of integration; and t is called the variable of integration (the prime on the variable of integration t merely serves to distinguish that variable from the limit of integration t). Graphically, the integral is the exact area under the velocity curve between the limits t0 and t in a plot of v vs. t (see Fig. A4.4). Areas below the t axis must be reckoned as negative. In general, if f (u) is a function of u, then the integral of this function is defined by a limiting procedure similar to that described above for the special case of the function v(t). The integral over an interval from u  a to u  b is



b

f (u) du 

a

lim

N1

¢uS0 a NSq i0

f (ui) ¢u

(22)

where ui  a  i ¢u. As in the case of the integral of v(t), this integral can again be interpreted as an area: it is the area under the curve between the limits a and b in a plot of f vs. u. For the explicit evaluation of integrals we can take advantage of the connection between integrals and antiderivatives. An antiderivative of a function f (u) is simply a function F(u) such that dF>du  f. For example, if f (u)  un and n  1, then an antiderivative of f (u) is F(u)  un1>(n  1). The fundamental theorem of calculus states that the integral of any function f (u) can be expressed in terms of antiderivatives: b

 f (u) du  F (b)  F (a)

(23)

a

In essence, this means that integration is the inverse of differentiation. We will not prove this theorem here, but we remark that such an inverse relationship between integration and differentiation should not come as a surprise. We have already run across an obvious instance of such a relationship: we know that velocity is the derivative of the position, and we have seen above that the position is the integral of the velocity. We will sometimes write Eq. (23) as



a

b

f (u) du  F (u) `

b

(24) a

where the notation F (u)ba means that the function F (u) is to be evaluated at a and at b, and these values are to be subtracted. For example, if n  1, b

un1

 u du  n  1 ` n

a

b a



bn1 an1  n1 n1

(25)

Table A4.2 lists some frequently used integrals. In this table, the limits of integration belonging with Eq. (24) have been omitted for the sake of brevity.

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APPENDIX 4

SOME INTEGRALS

TA B L E A 4 . 2 un1

 u du  n  1 1  u du  ln u n

Calculus Review

for n  1 for u  0

eku

 e du  k  ln u du  u ln u  u 1  sin(ku) du   k cos(ku) 1  cos(ku) du  k sin(ku) 1 du  1  ku  k ln(1  ku) ku

 2k

du

2

 2u

 u2

(where ku is in radians)

u  sin1 a b k  ln a u  2u2 ; k2 b

du

2

(where ku is in radians)

; k2

 2k  u du  2 c u 2k  u  k sin du 1 u  k  u  k tan a k b du 1 k  2k ; u b  u 2k ; u   k ln a u du u  (u  k )  k 2u  k 2

1

2

2

2

1

2

u a bd k

1

2

2

2

2

2

2

2

2 3/2

2

2

2

A4.4 Impor tant Rules for Integration 1. Integral of a constant times a function:



b

b

 f (u) du

cf (u) du  c

a

(26)

a

For instance, b

b

b3

a3

 5u du  5  u du  5 a 3  3 b 2

a

2

(27)

a

2. Integral of a sum of two functions:



a

b

[ f (u)  g(u)] du 



a

b

f (u) du 

b

 g(u) du a

(28)

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A-16

APPENDIX 4

Calculus Review

For instance,



b



(5u2  u) du 

a

b



5u2 du 

a

b

b3 a3 b2 a2  b  a  b (29) 3 3 2 2

u du  5 a

a

3. Change of limits of integration:



b



b



f (u) du 

a

c

b

 f (u) du

f (u) du 

a

(30)

c

a

f (u) du  

a

 f (u) du

(31)

b

4. Change of variable of integration: If u is a function of v, then b

 f (u) du   a

v(b)

du dv dv

(32)

v6(2v) dv

(33)

f (u)

v(a)

For instance, with u  v2, b

b

 u du   v du   3

2b

6

a

2a

a

Finally, let us apply these general results to some specific examples of integration of the velocity.

EXAMPLE 1

A particle with constant acceleration has the following velocity as a function of time [compare Eq. (15)]: v(t)  v0  at

where v0 is the velocity at t  0. By integration, find the position as a function of time. SOLUTION: According to Eq. (21), with t0  0,

x(t)  x0 



t

t

v(t) dt 

0

 (v

0

 at) dt

0

Using rule 2 and rule 1, we find that this equals x(t)  x0 



0

t

v0 dt 



t

at dt  v0

0



t

t

dt  a

0

 t dt

(34)

0

The first entry listed in Table A4.2 gives dt  t (for n  0) and tdt  t2>2 (for n  1). Thus, t

1 2 x(t)  x0  v0 t `  2 a t `

0

 v0t  at 1 2

This, of course, agrees with Eq. (16).

2

t 0

(35)

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APPENDIX 4

Calculus Review

The instantaneous velocity of a projectile traveling through air is the following function of time:

EXAMPLE 2

v(t)  655.9  61.14t  3.26t 2 where v(t) is measured in meters per second and t is measured in seconds. Assuming that x  0 at t  0, what is the position as a function of time? What is the position at t  3.0 s? SOLUTION: With x0  0 and t0  0, Eq. (21) becomes t

x(t) 

 (655.9  61.14t  3.26t ) dt 2

0



 655.9

t

dt  61.14

0



t

t

tdt  3.26

0

t

 t dt 2

0

t

 655.9(t) `  61.14(t2>2) `  3.26(t3>3) ` 0

0

 655.9t  61.14t >2  3.26t >3 2

t 0

3

When evaluated at t  3.0 s, this yields x(3.0)  655.9  3.0  61.14  (3.0)2>2  3.26  (3.0)3>3  1722 m

The acceleration of a mass pushed back and forth by an elastic spring is

EXAMPLE 3

a(t)  B cos vt

(36)

where B and  are constants. Find the position as a function of time. Assume v  0 and x  0 at t  0. SOLUTION: The calculation involves two steps: first we must integrate the acceleration to find the velocity, then we must integrate the velocity to find the position. For the first step we use an equation analogous to Eq. (21), t

v(t)  v0 

 a(t) dt

(37)

t0

This equation becomes obvious if we remember that the relationship between acceleration and velocity is analogous to that between velocity and position. With v0  0 and t0  0, we obtain from Eq. (29) v(t) 



t

B cos vt dt  B

0

t 1 sin vt ` v 0

B  sin vt v

(38)

Next, x(t) 



t

v(t) dt 

0



B v2

t

t 1 B B sin vtdt  a  cos vt b ` v v v 0 0



cos vt 

B v2

(39)

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APPENDIX 4

Calculus Review

A 4 . 5 T h e Ta y l o r S e r i e s Suppose that f (u) is a smooth function of u in some neighborhood of a given point u  a, so the function has continuous derivatives of all orders. Then the value of the function at an arbitrary point near a can be expressed in terms of the following infinite series, where all the derivatives are evaluated at the point a: f (u)  f (a) 

df

(u  a) 

du

2 3 1 d f 1 d f 2 (u  a)  (u  a)3     (40) 2 du2 32 du3

This is called the Taylor series for the function f (u) about the point a. The series converges, and is valid, provided u is sufficiently close to a. How close is “sufficiently close” depends on the function f and on the point a. Some functions, such as sin u, cos u, and eu, are extremely well behaved, and their Taylor series converge for any choice of u and of a. The Taylor series gives us a convenient method for the approximate evaluation of a function.

EXAMPLE 4

Find the Taylor series for sin u about the point u  0.

SOLUTION: The derivatives of sin u evaluated at u  0 are

d sin u  cos u  1 du d2

sin u 

d cos u  sin u  0 du

sin u 

d (sin u)  cos u  1 du

sin u 

d (cos u)  sin u  0, etc. du

2

du

d3 du3 d4 du4

Hence Eq. (32) gives sin u  0  1  (u  0)  

1 1  0  (u  0)2   (1)  (u  0)3 2 32

1  0  (u  0)4     432

u

1 3 u   6

Note that for very small values of u, we can neglect all higher powers of u, so sin u  u, which is an approximation often used in this book.

A4.6 Some Approximations By constructing Taylor series, we can obtain the following useful approximations, all of which are valid for small values of u. It is often sufficient to keep just the first one or two terms on the right side. 1 1 1 3 21  u  1  u  u2  u   2 8 16

(41)

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APPENDIX 5

Propagating Uncertainties

1  1  u  u2  u3   1u 1 1 3 5  1  u  u2  u3   2 8 16 21  u n(n  1) 2 n(n  1)(n  2) 3 1  1  nu  u  u   (1  u)n 2 23 1 1 3 eu  1  u  u2  u   2 23 1 1 ln(1  u)  u  u2  u3   2 3

(42) (43) (44) (45) (46)

In all the following formulas, u is in radians: 1 1 5 sin u  u  u3  u   6 120 1 1 1 6 cos u  1  u2  u4  u   2 24 720 1 2 tan u  u  u3  u5   3 15 1 3 3 1 sin u  u  u  u5   6 40 1 1 tan1 u  u  u3  u5   3 5

(47) (48) (49) (50) (51)

Appendix 5: Propagating Uncertainties Experimentalists carefully work to measure physical quantities and to determine the uncertainty in each quantity. We must often calculate a new result from a measured quantity or from several quantities; we must therefore understand the propagation of uncertainties through functions and formulas. To keep things simple, we will make the assumption that the uncertainties in each quantity are symmetrically distributed about its measured value and that the various measured quantities are independent of each other. This is not always true. But by ignoring correlations and assuming symmetry, we can reduce all the necessary propagation of uncertainties to some simple formulas. Suppose we have a measured quantity and its uncertainty, x ± x, where x is a positive quantity and has the same units as x, and is also known as the absolute uncertainty in x. What, then, is the uncertainty of some function, f (x), of this data? Under the assumption that the uncertainty is small, we can obtain the uncertainty from the first terms of the Taylor series expansion of f : f (x  x)  f (x)  (df (x)dx)x … From this we find the uncertainty f  | f (x  x)  f (x)| in the function value f (x) is ¢f  `

df dx

¢x `

(1)

with the derivative evaluated at the point x. We can generalize this result to functions of several variables as follows: given the data x ± x, y ± y, . . ., the function f (x, y, . . .) has the associated uncertainty

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APPENDIX 5

Propagating Uncertainties

¢f  `

0f 0x

¢x `  `

0f 0y

¢y `  

(2)

where all the partial derivatives (see App. 4.2) are evaluated at the point x, y, . . .. If we recall that we defined absolute uncertainties to be positive, we can write this as ¢f  `

0f 0x

` ¢x  `

0f 0y

` ¢y  

(3)

From this relationship, we can derive several simple results for uncertainty propagation.

EXAMPLE 1

Addition and Subtraction.

Given f (x, y)  3x  y  z  5, find f: ¢f  `

0f 0x

` ¢x  `

0f 0y

` ¢y  `

0f 0z

` ¢z

 3 ¢x  1 ¢y  1 ¢z  3¢x  ¢y  ¢z Thus in addition or subtraction, the uncertainties add, and in multiplication by a constant, the uncertainty is multiplied by the same constant.

EXAMPLE 2

Multiplication, Division, and Exponentiation.

Given f (x, y)  x2y(5z), find  f: ¢f  `

0f 0x

` ¢x  `

0f 0y

` ¢y  `

0f 0z

` ¢z

  2xy >(5z) ¢x  x2>(5z) ¢y  x2y>(5z2) ¢z Equivalently, for multiplication and division, we add relative uncertainties (e.g., x/x), and for exponentiation, we multiply the relative uncertainty by the magnitude of the exponent, to get the relative uncertainty of the product, quotient, or power.

EXAMPLE 3

Numerical Application to Ohm's Law, V  I R.

Given V  1.5  0.1 Volt and I  0.50  0.02 A, find R and R: Rearranging we find R  VI  (1.5 Volt)(0.50 A)  3.0 , and

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APPENDIX 6

The International System of Units (SI)

¢R  `

0R 0R ` ¢V  ` ` ¢I 0V 0I

 `

1 V ` ¢V  ` 2 ` ¢I I I

 `

1 1.5 Volt ` (0.1 Volt)  ` ` (0.02A) 0.50 A (0.50 A)2

 0.2   0.12   0.4  Note in the last step that unlike an ordinary calculation, we have rounded this final result up; uncertainties should always be rounded up, never down.

Appendix 6: The International System of Units (SI) A6.1 Base Units The SI system of units is the modern version of the metric system. The SI system recognizes seven fundamental, or base, units for length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity.b The following definitions of the base units were adopted by the Conférence Générale des Poids et Mesures in the years indicated: meter (m) “The metre is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 of a second.” (Adopted in 1983.) kilogram (kg)

“The kilogram is . . . the mass of the international prototype of the kilogram.” (Adopted in 1889 and in 1901.)

second (s)

“The second is the duration of 9 192 631 770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium-133 atom.” (Adopted in 1967.) ampere (A)

“The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed one meter apart in vacuum, would produce between these conductors a force equal to 2  107 newton per meter of length.” (Adopted in 1948.)

kelvin (K)

“The kelvin . . . is the fraction 1/273.16 of the thermodynamic temperature of the triple point of water.” (Adopted in 1967.)

b At least two of the seven base units of the SI system are redundant. The mole is merely a certain number of atoms or molecules, in the same sense that a dozen is a number; there is no need to designate this number 1 as a unit. The candela is equivalent to 683 watt per steradian; it serves no purpose that is not served equally well by watt per steradian. Two other base units could be made redundant by adopting new definitions of the unit of temperature and of the unit of electric charge. Temperature could be measured in energy units because, according to the equipartition theorem, temperature is proportional to the energy per degree of freedom. Hence the kelvin could be defined as a derived unit, with 1 K  12  1.38  1023 joule per degree of freedom. Electric charge could also be defined as a derived unit, to be measured with a suitable combination of the units of force and distance, as is done in the cgs system. Furthermore, the definitions of the supplementary units—radian and steradian—are gratuitous. These definitions properly belong in the province of mathematics and there is no need to include them in a system of physical units.

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TA B L E A 6 . 1

APPENDIX 6

The International System of Units (SI)

NAMES OF DERIVED UNITS

TA B L E A 6 . 2

PREFIXES FOR UNITS

QUANTITY

DERIVED UNIT

NAME

SYMBOL

FACTOR

PREFIX

SYMBOL

frequency

1/s

hertz

Hz

10 24

yotta

Y

force

kgms2

newton

N

10 21

zetta

Z

pressure

Nm2

pascal

Pa

1018

exa

E

energy

Nm

joule

J

1015

peta

P

10

12

tera

T

109

giga

G

106

mega

M

103

kilo

k

10

hecto

h

10

deka

da

101

deci

d

102

centi

c

103

milli

m

106

micro

m

power

J/s

watt

W

electric charge

As

coulomb

C

electric potential

J/C

volt

V

electric capacitance

C/V

farad

F

electric resistance

V/A

ohm



conductance

A/V

siemen

S

magnetic flux

Vs

weber

Wb

2

2

magnetic field

Vs>m

tesla

T

inductance

Vs>A

henry

H

temperature

K

degree Celsius

C

10

nano

n

luminous flux

cdsr

lumen

lm

1012

pico

p

lux

lx

1015

femto

f

2

9

illuminance

cdsr>m

radioactivity

1/s

becquerel

Bq

1018

atto

a

absorbed dose

J/kg

gray

Gy

1021

zepto

z

dose equivalent

J/kg

sievert

Sv

yocto

y

10

24

mole

“The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12.” (Adopted in 1967.)

candela (cd)

“The candela is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540  1012 Hz and that has a 1 radiant intensity in that direction of 683 watt per steradian.” (Adopted in 1979.) Besides these seven base units, the SI system also recognizes two supplementary units of angle and solid angle:

radian (rad)

“The radian is the plane angle between two radii of a circle which cut off on the circumference an arc equal in length to the radius.”

steradian (sr) “The steradian is the solid angle which, having its vertex in the center of a sphere, cuts off an area equal to that of a [flat] square with sides of length equal to the radius of the sphere.”

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APPENDIX 7

A-23

Best Values of Fundamental Constants

A6.2 Derived Units The derived units are formed out of products and ratios of the base units. Table A6.1 lists those derived units that have been glorified with special names. (Other derived units are listed in the tables of conversion factors in Appendix 8.)

A6.3 Prefixes Multiples and submultiples of SI units are indicated by prefixes, such as the familiar kilo, centi, and milli used in kilometer, centimeter, and millimeter, etc. Table A6.2 lists all the accepted prefixes. Some enjoy more popularity than others; it is best to avoid the use of uncommon prefixes, such as atto and exa, since hardly anybody will recognize those.

Appendix 7: Best Values of Fundamental Constants The values in the following table are the “2002 CODATA Recommended Values” by P. J. Mohr and B. N. Taylor Listed at the website physics.nist.gov/constants of the National Institute of Standards and Technology. The digits in parentheses are the one–standard deviation uncertainty in the last digits of the given value.

TA B L E A 7 . 1

BEST VALUES OF FUNDAMENTAL CONSTANTS

RELATIVE UNCERTAINTY (PARTS PER MILLION)

QUANTITY

SYMBOL

VALUE

UNITS

UNIVERSAL CONSTANTS speed of light in vacuum magnetic constant

c 0

electric constant 1 0c2

0

299 792 458 4  107  12.566 370 614 ...  107 8.854 187 817 ...  1012

m.s1 N·A2 N·A2 F·m1

gravitational constant Planck constant in eV.s h>2 in eV.s

G h U

6.6742(10)  1011 6.626 0693(11)  1034 4.135 667 43(35)  1015 1.054 571 68(18)  1034 6.582 119 15(56)  1016

m3·kg1·s2 J·s eV·s J·s eV·s

1.5  104 1.7  107 8.5  108 1.7  107 8.5  108

ELECTROMAGNETIC CONSTANTS elementary charge

e

1.602 176 53(14)  1019

C

8.5  108

magnetic flux quantum h2e quantum Josephson constant

0 2e 2h 2eh

2.067 833 72(18)  1015 7.748 091 733(26)  105 483 597.879(41)  109

Wb S Hz·V1

8.5  108 3.3  109 8.5  108

Bohr magneton e U 2me in eV.T1

B

927.400 949(80)  1026 5.788 381 804(39)  105

J·T1 eV·T1

8.6  108 6.7  109

nuclear magneton e U 2mp in eV.T1

N

5.050 783 43(43)  1027 3.152 451 259(21)  108

J·T1 eV·T1

8.6  108 6.7  109

(exact) (exact) (exact)

(continued )

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A-24

APPENDIX 7

QUANTITY

Best Values of Fundamental Constants

UNITS

RELATIVE UNCERTAINTY (PARTS PER MILLION)

SYMBOL

VALUE



7.297 352 568(24)  103

3.3  109

1

137.035 999 11(46)

3.3  109

Rq

10 973 731.568 525(73)

m1

6.6  1012

a0

0.529 177 2108(18)  1010

m

3.3  109

me me c 2

9.109 3826(16)  1031 5.485 799 0945(24)  104 0.510 998 918(44)

kg u MeV

memp

5.446 170 2173(25)  104

ATOMIC AND NUCLEAR CONSTANTS General fine-structure constant e24p0Uc inverse fine-structure constant Rydberg constant 2mec/2h Bohr radius 4p0 U2>mee2

Electron electron mass in u energy equivalent in Me V electron-proton mass ratio electron charge to mass quotient

eme

11

1.758 820 12(15)  10

12

1.7  107 4.4  1010 8.6  108 4.6  1010

C·kg

1

8.6  108 6.7  109 1.0  108

Compton wavelength h>mec classical electron radius 2a0

lC re

2.426 310 238(16)  10 2.817 940 325(28)  1015

m m

Thomson cross section (8p3)re2

e

0.665 245 837(13)  1028

m2

electron magnetic moment to Bohr magneton ratio to nuclear magneton ratio electron magnetic moment anomaly | e | B  1 electron g-factor 2(1  ae)

e e B e N

928.476 412(80)  10 1.001 159 652 1859(38) 1838.281 971 07(85)

ae ge

1.159 652 1859(38)  103 2.002 319 304 3718(75)

Muon muon mass in u energy equivalent in MeV

m

1.883 531 40(33)  1028

kg

1.7  107

m c 2

0.113 428 9264(30) 105.658 3692(94)

u MeV

2.6  108 8.9  108

muon-electron mass ratio

m me

206.768 2838(54)

26

J·T

2.0  108 1

8.6  108 3.8  1012 4.6  1010 3.2  109 3.8  1012

2.6  108

muon Compton wavelength h/m c

C,

11.734 441 05(30)  10

m

2.5  108

muon magnetic moment to Bohr magneton ratio

 B

4.490 447 99(40)  1026 4.841 970 45(13)  103

J·T1

8.9  108 2.6  108

muon magnetic moment anomaly | |(e U /2m )  1 muon g-factor 2(1  a )

a g

1.165 919 81(62)  103 2.002 331 8396(12)

mp mpc 2

1.672 621 71(29)  1027 1.007 276 466 88(13) 938.272 029(80)

proton-electron mass ratio proton-neutron mass ratio proton charge to mass quotient

mp me mp mn emp

proton Compton wavelength h/mpc proton magnetic moment to Bohr magneton ratio to nuclear magneton ratio

Proton proton mass in u energy equivalent in MeV

Neutron neutron mass in u energy equivalent in MeV neutron-electron mass ratio neutron-proton mass ratio

15

5.3  107 6.2  1010 kg u MeV

1.7  107 1.3  1010 8.6  108

1836.152 672 61(85) 0.998 623 478 72(58) 9.578 833 76(82)  107

C·kg1

4.6  1010 5.8  1010 8.6  108

C, p

1.321 409 8555(88)  1015

m

6.7  109

p p  B p N

1.410 606 71(12)  1026 1.521 032 206(15)  103 2.792 847 351(28)

J·T1

8.7  108 1.0  108 1.0  108

mn mn c2

1.674 927 28(29)  1027 1.008 664 915 60(55) 939.565 360(81)

kg u MeV

1.7  107 5.5  1010 8.6  108

mn me mn mp

1838.683 6598(13) 1.001 378 418 70(58)

7.0  1010 5.8  1010 (continued)

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APPENDIX 7

A-25

Best Values of Fundamental Constants

RELATIVE UNCERTAINTY (PARTS PER MILLION)

QUANTITY

SYMBOL

VALUE

UNITS

neutron Compton wavelength h>mnc neutron magnetic moment to Bohr magneton ratio to nuclear magneton ratio

C, n n n B n N

1.319 590 9067(88)  1015 0.966 236 45(24)  1026 1.041 875 63(25)  103 1.913 042 73(45)

m J·T1

6.7  109 2.5  107 2.4  107 2.4  107

md mdc 2

3.343 583 35(57)  1027 2.013 553 212 70(35) 1875.612 82(16)

kg u MeV

1.7  107 1.7  1010 8.6  108

deuteron-electron mass ratio deuteron-proton mass ratio

mdme mdmp

3670.482 9652(18) 1.999 007 500 82(41)

deuteron magnetic moment to Bohr magneton ratio to nuclear magneton ratio

d d B d N

0.433 073 482(38)  1026 0.466 975 4567(50)  103 0.857 438 2329(92)

J·T1

8.7  108 1.1  108 1.1  108

Alpha Particle alpha particle mass in u energy equivalent in MeV

m mc 2

6.644 6565(11)  1027 4.001 506 179 149(56) 3727.379 17(32)

kg u MeV

1.7  107 1.4  1011 8.6  108

alpha particle to electron mass ratio alpha particle to proton mass ratio

mme mmp

7294.299 5363(32) 3.972 599 689 07(52)

PHYSICO-CHEMICAL CONSTANTS Avogadro constant atomic mass constant

NA

6.022 1415(10)  1023

mole1

1.7  107

mu

1.660 538 86(28)  1027

kg

1.7  107

muc 2

931.494 043(80)

MeV

Deuteron deuteron mass in u energy equivalent in MeV

mu  121 m(12C)  1 u energy equivalent in MeV

4.8  1010 2.0  1010

4.4  1010 1.3  1010

8.6  108 1

Faraday constant NAe

F

96 485.3383(83)

C·mole

8.6  108

molar gas constant

R

8.314 472 (15)

J·mole1·K1

1.7  106

Boltzmann constant RNA in eV.K–1

k

1.380 6505(24)  1023 8.617 343(15)  105

J·K1 eV·K1

1.8  106 1.8  106

molar volume of ideal gas RT>p T  273.15 K, p  101.325 kPa Loschmidt constant NA>Vm

Vm n0

22.413 996(39)  103 2.686 7773(47)  1025

m3.mole–1 m3

1.7  106 1.8  106

Stefan-Boltzmann constant (260)k4 U3c 2



5.670 400(40)  108

W·m2·K4

7.0  106

Wien displacement law constant b  maxT

b

2.897 7685(51)  103

m·K

1.7  106

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APPENDIX 8

Conversion Factors

Appendix 8: Conversion Factors The units for each quantity are listed alphabetically, except that the SI unit is always listed first. The numbers are based on “American National Standard; Metric Practice” published by the Institute of Electrical and Electronics Engineers, 1982.

Angle 1 radian  57.30  3.438  103   (1/2p) rev  2.063  105  1 1 degree ()  1.745  102 radian  60  3600–  360 rev 4 1 1 minute of arc ()  2.909  10 radian  60  4.630  105 rev  60– 1 revolution (rev)  2p radians  360  2.160  104   1.296  106  1 1 second of arc (–)  4.848  106 radian  3600   601   7.716  107 rev

Length  1 meter (m)  1  10 9 nm  1  1010 A  6.685  1012 AU  100 cm  15 1  10 fm  3.281 ft  39.37 in.  1  103 km  1.057  1016 light-year  1  106 mm  5.400  104 nmi  6.214  104 mi  3.241  1017 pc  1.094 yd  )  1  1010 m  1  108 cm  1  10 nm  1 angstrom (A 1  105 fm  3.281  1010 ft  1  104 mm 1 astronomical unit (AU)  1.496  1011 m  1.496  1013 cm  1.496  108 km  1.581  105 light-year  4.848  106 pc 1 centimeter (cm)  0.01 m  1  108 A  1  1013 fm  3.281  102 ft  0.3937 in.  1  105 km  1.057  1018 light-year  1  104 mm 1 fermi, or femtometer (fm)  1  1015 m  1  1013 cm  1  105 A 1 foot (ft)  0.3048 m  30.48 cm  12 in.  3.048  105 mm  1.894  104 mi  13 yd 1 inch (in.)  2.540  102 m  2.54 cm  121 ft  2.54  104 mm  361 yd 1 kilometer (km)  1  103 m  1  105 cm  3.281  103 ft  0.5400 nmi  0.6214 mi  1.094  103 yd 1 light-year  9.461  1015 m  6.324  104 AU  9.461  1017 cm  9.461  1012 km  5.879  1012 mi  0.3066 pc   1  104 cm 1 micron, or micrometer (mm)  1  106 m  1  104 A 6 5  3.281  10 ft  3.937  10 in. 1 nautical mile (nmi)  1.852  103 m  1.852  105 cm  6.076  103 ft  1.852 km  1.151 mi 1 statute mile (mi)  1.609  103 m  1.609  105 cm  5280 ft  1.609 km  0.8690 nmi  1760 yd 1 parsec (pc)  3.086  1016 m  2.063  105 AU  3.086  1018 cm  3.086  1013 km  3.262 light-years 1 1 yard (yd)  0.9144 m  91.44 cm  3 ft  36 in.  1760 mi

Time 1 1 second (s)  1.157  105 day  3600 h  601 min  5 1.161  10 sidereal day  3.169  108 yr 1 day  8.640  104 s  24 h  1440 min  1.003 sidereal days  2.738  103 yr 1 hour (h)  3600 s  241 day  60 min  1.141  104 yr

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APPENDIX 8

Conversion Factors

1 minute (min)  60 s  6.944  104 day  601 h  1.901  106 yr 1 sidereal day  8.616  104 s  0.9973 day  23.93 h  1.436  103 min  2.730  103 yr 1 year (yr)  3.156  107 s  365.24 days  8.766  103 h  5.259  105 min  366.24 sidereal days

Mass 1 kilogram (kg)  6.024  1026 u  5000 carats  1.543  104 grains  1000 g  1  103 t  35.27 oz  2.205 lb  1.102  103 short ton  6.852  102 slug 1 atomic mass unit (u)  1.6605  1027 kg  1.6605  1024 g 1 carat  2  104 kg  0.2 g  7.055  103 oz  4.409  104 lb 1 1 grain  6.480  105 kg  6.480  102 g  2.286  103 oz  7000 lb 3 23 1 gram (g)  1  10 kg  6.024  10 u  5 carats  15.43 grains  1  106 t  3.527  102 oz  2.205  103 lb  1.102  106 short ton  6.852  105 slug 1 metric ton, or tonne (t)  1  103 kg  1  106 g  2.205  103 lb  1.102 short tons  68.52 slugs 1 ounce (oz)  2.835  102 kg  141.7 carats  437.5 grains  28.35 g  161 lb 1 pound (lb)c  0.4536 kg  453.6 g  4.536  104 t  16 oz  1 2 slug 2000 short ton  3.108  10 1 short ton  907.2 kg  9.072  105 g  0.9072 t  2000 lb 1 slug  14.59 kg  1.459  104 g  32.17 lb

Area 1 square meter (m2)  1  104 cm2  10.76 ft2  1.550  103 in.2  1  106 km2  3.861  107 mi2  1.196 yd2 1 barn  1  1028 m2  1  1024 cm2 1 square centimeter (cm2)  1  104 m2  1.076  103 ft2  0.1550 in.2  1  1010 km2  3.861  1011 mi2 1 square foot (ft2)  9.290  102 m2  929.0 cm2  144 in.2  3.587  108 mi2  19 yd2 1 1 square inch (in.2)  6.452  104 m2  6.452 cm2  144 ft2 2 6 2 10 2 1 square kilometer (km )  1  10 m  1  10 cm  1.076  107 ft2  0.3861 mi2 1 square statute mile (mi2)  2.590  106 m2  2.590  1010 cm2  2.788  107 ft2  2.590 km2 1 square yard (yd2)  0.8361 m2  8.361  103 cm2  9 ft2  1296 in.2

Vo l u m e 1 cubic meter (m3)  1  106 cm3  35.31 ft3  264.2 gal  6.102  104 in.3  1  103 liters  1.308 yd3 1 cubic centimeter (cm3)  1  106 m3  3.531  105 ft3  2.642  104 gal  6.102  102 in.3  1  103 liter 1 cubic foot (ft3)  2.832  102 m3  2.832  104 cm3  7.481 gal  1728 in.3  28.32 liters  271 yd3 c

This is the “avoirdupois” pound. The “troy” or “apothecary” pound is 0.3732 kg, or 0.8229 lb avoirdupois.

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A-28

APPENDIX 8

Conversion Factors

1 gallon (gal)d  3.785  103 m3  0.1337 ft3 1 cubic inch (in.3)  1.639  105 m3  16.39 cm3  5.787  104 ft3 1 liter (l)  1  103 m3  1000 cm3  3.531  102 ft3  0.2642 gal 1 cubic yard (yd3)  0.7646 m3  7.646  105 cm3  27 ft3  202.0 gal

Density 1 kilogram per cubic meter (kg>m3)  1  103 g>cm3  6.243  102 lb>ft3  8.345  103 lb>gal  3.613  105 lb>in.3  8.428  104 short ton>yd3  1.940  103 slug>ft3 1 gram per cubic centimeter (g>cm3)  1  103 kg>m3  62.43 lb>ft3  8.345 lb>gal  3.613  102 lb>in.3  0.8428 short ton>yd3  1.940 slugs>ft3 1 lb per cubic foot (lb>ft3)  16.02 kg>m3  1.602  102 g>cm3  0.1337 lb>gal  1.350  102 short ton>yd3  3.108  102 slug>ft3 1 pound-per gallon (1 lb>gal)  119.8 kg>m3  7.481 lb>ft3  0.2325 slug>ft3 1 short ton per cubic yard (short ton>yd3)  1.187  103 kg>m3  74.07 lb>ft3 1 slug per cubic foot (slug>ft3)  515.4 kg>m3  0.5154 g>cm3  32.17 lb>ft3  4.301 lb>gal

Speed 1 meter per second (m>s)  100 cm>s  3.281 ft>s  3.600 km>h  1.944 knots  2.237 mi>h 1 centimeter per second (cm>s)  0.01 m>s  3.281  102 ft>s  3.600  102 km>h  1.944  102 knot  2.237  102 mi>h 1 foot per second (ft>s)  0.3048 m>s  30.48 cm>s  1.097 km>h  0.5925 knot  0.6818 mi>h 1 kilometer per hour (km>h)  0.2778 m>s  27.78 cm>s  0.9113 ft>s  0.5400 knot  0.6214 mi>h 1 knot, or nautical mile per hour  0.5144 ms  51.44 cms  1.688 ft>s  1.852 km>h  1.151 mi>h 1 mile per hour (mi>h)  0.4470 ms  44.70 cms  1.467 fts  1.609 km h  0.8690 knot

Acceleration 1 meter per second squared (m>s2)  100 cm>s2  3.281 ft>s2  0.1020 g 1 centimeter per second squared (cm>s2)  0.01 m>s2  3.281  102 ft>s2  1.020  103 g 1 foot per second squared (ft>s2)  0.3048 m>s2  30.48 cm>s2  3.108  102 g 1 g  9.807 m>s2  980.7 cm>s2  32.17 ft>s2

Force 1 newton (N)  1  105 dynes  0.2248 lb-f  1.124  104 short ton-force 1 dyne  1  105 N  2.248  106 lb-f  1.124  109 short ton-force 1 1 pound-force (lb-f )  4.448 N  4.448  105 dynes  2000  short ton-force 3 8 1 short ton-force  8.896  10 N  8.896  10 dynes  2000 lb-f

d

This is the U.S. gallon; the U.K. and the Canadian gallon are 4.546  103 m3, or 1.201 U.S. gallons.

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APPENDIX 8

Conversion Factors

Energy 1 joule (J)  9.478  104 Btu  0.2388 cal  1  107 ergs  6.242  1018 eV  0.7376 ftlb-f  2.778  107 kWh 1 British thermal unit (Btu)e  1.055  103 J  252.0 cal  1.055  1010 ergs  778.2 ftlb-f  2.931  104 kWh 1 calorie (cal) f  4.187 J  3.968  103 Btu  4.187  107 ergs  3.088 ftlb-f  1  103 kcal  1.163  106 kWh 1 erg  1  107 J  9.478  107 Btu  2.388  108 cal  6.242  1011 eV  7.376  108 ft lb-f  2.778  1014 kWh 1 electron-volt (eV)  1.602  1019 J  1.602  1012 erg  1.182  1019 ftlb-f 1 foot-pound-force (ftlb-f )  1.356 J  1.285  103 Btu  0.3239 cal  1.356  107 ergs  8.464  1018 eV  3.766  107 kWh 1 kilocalorie (kcal), or large calorie (Cal)  4.187  103 J  1  103 cal 1 kilowatt-hour (kWh)  3.600  106 J  3412 Btu  8.598  105 cal  3.6  1013 ergs  2.655  106 ftlb-f

Power 1 watt (W)  3.412 Btu>h  0.2388 cal>s  1  107 ergs>s  0.7376 ftlb-f>s  1.341  103 hp 1 British thermal unit per hour (Btu>h)  0.2931 W  7.000  102 cal>s  0.2162 ft lb-f>s  3.930  104 hp 1 calorie per second (cal>s)  4.187 W  14.29 Btu>h  4.187  107 ergs>s  3.088 ftlb-f>s  5.615  103 hp 1 erg per second (erg>s)  1  107 W  2.388  108 cal>s  7.376  108 ftlb-f>s  1.341  1010 hp 1 foot-pound-force per second (ftlb-f>s)  1.356 W  0.3238 cal>s  4.626 Btu>h  1.356  107 ergs>s  1.818  103 hp 1 horsepower (hp) g  745.7 W  2.544  103 Btu>h  178.1 cal>s  550 ftlb-f>s 1 kilowatt (kW)  1  103 W  3.412  103 Btu>h  238.8 cal>s  737.6 ftlb-f>s  1.341 hp

Pressure 1 newton per square meter (N>m2), or pascal (Pa)  9.869  106 atm  1  105 bar  7.501  103 mm-Hg  10 dynes>cm2  2.953  104 in.-Hg  2.089  102 lb-f>ft2  1.450  104 lb-f>in.2  7.501  103 torr 1 atmosphere (atm)  1.013  105 N>m2  760.0 mm-Hg  1.013  106 dynes>cm2  29.92 in.-Hg  2.116  103 lb-f>ft2  14.70 lb-f>in.2 1 bar  1  105 N>m2  0.9869 atm  750.1 mm-Hg 1 dyne per square centimeter (dyne>cm2)  0.1 N>m2  9.869  107 atm  7.501  104 mm-Hg  2.089  103 lb-f>ft2  1.450  105 lb-f>in.2

e

This is the “International Table” Btu; there are several other Btus. This is the “International Table” calorie, which equals exactly 4.1868 J. There are several other calories; for instance, the thermochemical calorie, which equals 4.184 J. g There are several other horsepowers; for instance, the metric horsepower, which equals 735.5 W. f

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APPENDIX 8

Conversion Factors

1 inch of mercury (in.-Hg)  3.386  103 N>m2  3.342  102 atm  25.40 mm-Hg  0.4912 lb-f>in.2 1 pound-force per square inch (lb-f>in.2, or psi)  6.895  103 N>m2  6.805  102 atm  6.895  104 dynes>cm2  2.036 in.-Hg  7.031  102 kp>cm2 1 torr, or millimeter of mercury (mm-Hg)  1.333  102 N>m2  1>760 atm  1.333  103 bar  1.333  103 dynes/cm2  0.03937 in.-Hg  0.01934 lb-f/in.2

Electric Chargeh 1 coulomb (C) 3 2.998  109 statcoulombs, or esu of charge 3 0.1 abcoulomb, or emu of charge

Electric Current 1 ampere (A) 3 2.998  109 statamperes, or esu of current 3 0.1 abampere, or emu of current

Electric Potential 1 volt (V) 3 3.336  103 statvolt, or esu of potential 3 1  108 abvolts, or emu of potential

Electric Field 1 volt per meter (V>m) 3 3.336  105 statvolt/cm 3 1  106 abvolts/cm

Magnetic Field 1 tesla (T), or weber per square meter (Wb>m2)  1  104 gauss

Electric Resistance 1 ohm (Ω) 3 1.113  1012 statohm, or esu of resistance 3 1  109 abohms, or emu of resistance

Electric Resistivity 1 ohm-meter (m) 3 1.113  1010 statohm-cm 3 1  1011 abohm-cm

Capacitance 1 farad (F) 3 8.988  1011 statfarads, or esu of capacitance 3 1  109 abfarad, or emu of capacitance

Inductance 1 henry (H) 3 1.113  1012 stathenry, or esu of inductance 3 1  109 abhenrys, or emu of inductance

h

The dimensions of the electric quantities in SI units, electrostatic units (esu), and electromagnetic units (emu) are usually different; hence the relationships among most of these units are correspondences ( 3 ) rather than equalities ().

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APPENDIX 9

The Periodic Table and the Chemical Elements

A-31

Appendix 9: The Periodic Table and Chemical Elements TA B L E A 9 . 1

THE PERIODIC TABLE

IA (1) 1

2

Periods

3

1 H

VIIIA (18)

1.00794

IIA (2)

3 Li

4 Be

6.941

9.012182

11 Na

12 Mg

22.98977 24.3050

4

19 K 39.0983

5

6

20 Ca

Group designation Atomic number Symbol for element Atomic mass VIIIB IIIB (3)

IVB (4)

VB (5)

VIB (6)

VIIB (7)

(8)

(9)

21 Sc

22 Ti

23 V

24 Cr

25 Mn

26 Fe

27 Co

40.078 44.955910 47.867

51.9961 54.938049 55.845

VIA (16)

VIIA (17)

4.002602

5 B

6 C

7 N

8 O

9 F

10 Ne

10.811

12.0107

14.0067

13 Al

14 Si

15 P

72.64

49 In

50 Sn

114.818

118.710

121.760

81 Tl

82 Pb

83 Bi

204.3833

207.2

91.224

92.90638

95.94

98.9072

72 Hf

73 Ta

74 W

75 Re

76 Os

77 Ir

178.49

180.9479

183.84

186.207

190.23

192.217

104 Rf

105 Db

106 Sg

107 Bh

108 Hs

109 Mt

110 Ds

111 Uuu

112 Uub

114 Uuq

262.12

265.1306

(268)

(271)

(272)

(285)

(289)

60 Nd

61 Pm

62 Sm

63 Eu

64 Gd

65 Tb

144.9127

150.36

151.964

93 Np

94 Pu

95 Am

†Actinides

54 Xe

69.723

48 Cd

57 *La

92 U

83.798

53 I

65.409

47 Ag

88.90585

91 Pa

79.904

52 Te

63.546

56 Ba

90 Th

78.96

51 Sb

33 As

87.62

140.116 140.90765 144.24

74.92160

32 Ge

55 Cs

59 Pr

36 Kr

31 Ga

46 Pd

26.98154 28.0855 30.97376

85.4678

58 Ce

39.948

35 Br

30 Zn

43 Tc

*Lanthanides

35.453

34 Se

29 Cu

42 Mo

223.0197 226.0277 227.0277 261.1089 262.1144 263.118

32.065

28 Ni

41 Nb

89 †Ac

18 Ar

(10)

40 Zr

88 Ra

17 Cl

IIB (12)

39 Y

101.07 102.90550 106.42

78 Pt

107.8682 112.411

79 Au

80 Hg

195.078 196.96654 200.59

66 Dy

67 Ho

15.9994 18.99840 20.1797

16 S

IB (11)

58.93320 58.6934

45 Rh

VA (15)

38 Sr

87 Fr

44 Ru

IVA (14)

37 Rb

132.90545 137.327 138.9055

7

50.9415

127.60 126.90447 131.293

84 Po

85 At

97 Bk

98 Cf

99 Es

86 Rn

208.98037 208.9824 209.9871 222.0176

68 Er

69 Tm

70 Yb

157.25 158.92534 162.50 164.93032 167.26 168.93421 173.04

96 Cm

2 He

IIIA (13)

100 Fm

101 Md

102 No

71 Lu 174.967

103 Lr

232.0381 231.0359 238.0289 237.0482 244.0642 243.0614 247.07003 247.0703 251.0796 252.083 257.0951 258.0984 259.1011 262.110

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A-32

TA B L E A 9 . 2

APPENDIX 9

The Periodic Table and the Chemical Elements

ATOMIC MASSES AND ATOMIC NUMBERS OF CHEMICAL ELEMENTS

Data were obtained from the National Institute for Standards and Technology; values are for the elements as they exist naturally on Earth or for the most stable isotope, with carbon-12 (the reference standard) having a mass of exactly 12 u. The estimated uncertainties in values between ± and ± 9 units in the last digit of an atomic mass are in parentheses after the atomic mass. (Source: http://physics.nist.gov/PhysRefData/Compositions/index.html)

ELEMENT Actinium Aluminum Americium Antimony Argon Arsenic Astatine Barium Berkelium Beryllium Bismuth Bohrium Boron Bromine Cadmium Calcium Californium Carbon Cerium Cesium Chlorine Chromium Cobalt Copper Curium Darmstadtium Dubnium Dysprosium Einsteinium Erbium Europium Fermium Fluorine Francium Gadolinium Gallium Germanium Gold Hafnium Hassium Helium Holmium Hydrogen Indium Iodine Iridium Iron Krypton Lanthanum Lawrencium Lead Lithium Lutetium Magnesium Manganese Meitnerium Mendelevium

SYMBOL

ATOMIC NUMBER

ATOMIC MASS (u)

ELEMENT

Ac Al Am Sb Ar As At Ba Bk Be Bi Bh B Br Cd Ca Cf C Ce Cs Cl Cr Co Cu Cm Ds Db Dy Es Er Eu Fm F Fr Gd Ga Ge Au Hf Hs He Ho H In I Ir Fe Kr La Lr Pb Li Lu Mg Mn Mt Md

89 13 95 51 18 33 85 56 97 4 83 107 5 35 48 20 98 6 58 55 17 24 27 29 96 110 105 66 99 68 63 100 9 87 64 31 32 79 72 108 2 67 1 49 53 77 26 36 57 103 82 3 71 12 25 109 101

227.027 7 26.981 538 (2) 243.061 4 121.760 (1) 39.948 (1) 74.921 60 (2) 209.987 1 137.327 (7) 247.070 3 9.012 182 (3) 208.980 38 (2) 264.12 10.811 (7) 79.904 (1) 112.411 (8) 40.078 (4) 251.079 6 12.010 7 (8) 140.116 (1) 132.905 45 (2) 35.453 (9) 51.996 1 (6) 58.933 200 (9) 63.546 (3) 247.070 3 271 262.114 4 162.500 (1) 252.083 167.259 (3) 151.964 (1) 257.095 1 18.998 403 2 (5) 223.019 7 157.25 (3) 69.723 (1) 72.64 (1) 196.966 55 (2) 178.49 (2) 265.130 6 4.002 602 (2) 164.930 32 (2) 1.007 94 (7) 114.818 (3) 126.904 47 (3) 192.217 (3) 55.845 (2) 83.798 (2) 138.905 5 (2) 262.110 207.2 (1) 6.941 (2) 174.967 (1) 24.305 0 (6) 54.938 049 (9) 268 258.098 4

Mercury Molybdenum Neodymium Neon Neptunium Nickel Niobium Nitrogen Nobelium Osmium Oxygen Palladium Phosphorus Platinum Plutonium Polonium Potassium Praseodymium Promethium Protactinium Radium Radon Rhenium Rhodium Rubidium Ruthenium Rutherfordium Samarium Scandium Seaborgium Selenium Silicon Silver Sodium Strontium Sulfur Tantalum Technetium Tellurium Terbium Thallium Thorium Thulium Tin Titanium Tungsten Ununbium Unununium Ununquadium Uranium Vanadium Xenon Ytterbium Yttrium Zinc Zirconium

SYMBOL

ATOMIC NUMBER

ATOMIC MASS (u)

Hg Mo Nd Ne Np Ni Nb N No Os O Pd P Pt Pu Po K Pr Pm Pa Ra Rn Re Rh Rb Ru Rf Sm Sc Sg Se Si Ag Na Sr S Ta Tc Te Tb Tl Th Tm Sn Ti W Uub Uuu Uuq U V Xe Yb Y Zn Zr

80 42 60 10 93 28 41 7 102 76 8 46 15 78 94 84 19 59 61 91 88 86 75 45 37 44 104 62 21 106 34 14 47 11 38 16 73 43 52 65 81 90 69 50 22 74 112 111 114 92 23 54 70 39 30 40

200.59 (2) 95.94 (1) 144.24 (3) 20.179 7 (6) 237.048 2 58.693 4 (2) 92.906 38 (2) 14.006 7 (2) 259.101 1 190.23 (3) 15.999 4 (3) 106.42 (1) 30.973 761 (2) 195.078 (2) 244.064 2 208.982 4 39.098 3 (1) 140.907 65 (2) 144.912 7 231.035 88 (2) 226.025 4 222.017 6 186.207 (1) 102.905 50 (2) 85.467 8 (3) 101.07 (2) 261.108 9 150.36 (3) 44.955 910 (8) 263.118 6 78.96 (3) 28.085 5 (3) 107.868 2 (2) 22.989 770 (2) 87.62 (1) 32.065 (6) 180.947 9 (1) 98.907 2 127.60 (3) 158.925 34 (2) 204.383 3 (2) 232.038 1 (1) 168.934 21 (2) 118.710 (7) 47.867 (1) 183.84 (1) 285 272 289 238.028 9 (1) 50.941 5 (1) 131.293 (2) 173.04 (3) 88.905 85 (2) 65.409 (4) 91.224 (2)

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APPENDIX 10

Formula Sheets

A-33

Appendix 10: Formula Sheets Chapters 1–21 dL  rF dt

v  dx>dt

E  mc 2

a  dv>dt  d 2x>dt 2

P  dW>dt

x  x0  v0t  12at 2

PF

a(x  x0) 

1 2 2 (v



v20)

Ax  A cos u A  2Ax2  Ay2  Az2 A

 B  AB cos f  Ax Bx  Ay By  Az Bz

A  B  AB sin f a  v2>r v  v  VO



x  A cos(vt  d)

v

T  2p>v; f  1>T  v>2p 2

F  GMm>r

m d 2x>dt 2  kx

g  GME >R2E

  2k>m

v2  GMS >r

v  2g>l ; T  2p2l>g

U  GMm>r

v  2mgd>I

p  mv rCM  I





y  A cos k(x ; vt)  A cos(kx ; vt)

1 r dm M

l  2p>k; f  v>l; v  2pf

¢t

F dt

v  2F>(M>L)

0

m a  Fnet

m1  m2

f beat  f 1  f 2

w  mg fk  mk N

  df>dt

fs  ms N

  dv>dt  d 2f>dt 2

sin u  v>VE

F  kx

v  R

p  p0  rgy

W  Fx ¢x

s W  F  ds WF

m1  m2

K

1 2 2 I

I

R

2

v1; v2 

2m1

v1 

1 2 2 rv

dm

ICM  MR2 (hoop); 12MR2 (disk); 2 2 5 MR

U  mgy

I  ICM  Md 2

E  K  U  constant

t  FR sin u

x

 F (x) dx x0

Fx  

x

dU dx

f   f (1 ; VR>v)

f   f >(1 < VE >v)

K  12mv2

U(x)  

m1  m2

v1

(sphere); 121 ML2 (rod) .

I  t P  tv L  Iv

 rg y  p  constant

pV  NkT TC  T  273.15 vrms  23kT>m TV g1  [constant];

pV g  [constant]; g  Cp >CV

¢E  Q  W e  1  T2 >T1 ¢S 



B

dQ>T

U  12kx 2

L  rp

g  9.81 m>s2

me  9.11  1031 kg

NA  6.02  1023>mole

mp  1.67  1027 kg

k  1.38  1023 J>K

c  3.00  108 m>s

1 cal  4.19 J

G  6.67  1011 N.m2>kg2 ME  5.98  1024 kg RE  6.37  106 m

A

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Chapters 22–41 dF  I d l  B

1 qq 4p0 r 2

F

Interference maxima: d sin u  0, l, 2l, . . .

  B U  

E  s20

B

Diffraction minima: a sin u  l, 2l, 3l, . . .

E  vBl

p  lQ

E

t  pE

E  E  dA

U  p !E

d sin u  12l, 32l, 52l, 

m  I  area of loop

1 q 4p0 r 2

E

Interference minima:

£B 



Qinside E d A  E d A  H H 0

H

d£ B

a sin u  1.22l

dt

 



E ds 

H

1c f B 1 ; v>c

f 

B dA E ‘ ds  

d£ B

x  Vt

x 

dt

21  V 2>c 2

y  y

£ B  LI

1 q V 4p0 r

E  L

0V 0V 0V Ex   , Ey   , Ez   0x 0y 0z

U  12 LI 2

¢t 

1 2 u B 2m0

L  21  V 2>c 2 L

0  1>2LC

vx 

U

1 2 Q1V1



1 2 Q2V2



1 2 Q3V3

u  120E 2 C  Q>¢V C  0A>d

Z

E  Efree>k H

 

kE d A 

Qfree, inside

H

1 b vC

p

N2



H

I  ¢V>R

P  IE ; P  I ¢V m0 qvI F 2p r m0 I ds  r dB  4p r3 H

H

B ‘ ds  m0I

B  m0nI r

1  vxV>c 2 mv

21  v2>c

H H

B ds  m0I  m00

[pressure]  S c

p qB

0x2

 m00

; E 2

0 2E

f  ; 12R

d£ E dt

¢y ¢py  h4p L  nU En  

mee4

m0  1.26  106 Hm

0  8.85  1012 Fm

c  3.00  108 m/s h  2p U  6.63  1034 Js

1

2(4p0)2 U2 n2



13.6 eV

l  h>p mspin  E

eU 2me

J ( J  1) U2 2I

R  (1.2  1015 m)  A13 n  n0ett; t  t12 0.693

1 1 1   s s f

e  1.60  1019 C

21  v2>c2

p  hf >c

0t 2

c  1 2m00 v  c/n n1 sin u1  n2 sin u2

mc2

E  hf

B1dA  0

1 S EB m0 0 2E

F  qv  B



vx  V

B  E>c

R  rl>A

B ds 



B ds 

21  V 2>c 2

E 2  p2c 2  m2c4

N1

B dA 

21  V 2>c 2 ¢t

2

0

u  12k0E2

t  Vx>c 2

t 

R2  a vL 

E2  E1

0

¢Y   PP F  ds

B

dI dt

me  9.11  10 31 kg mp  1.67  10 27 kg

n2

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ANSWERS

A-35

Appendix 11: Answers to Odd-Numbered Problems and Review Problems

1. 5.87 ft; 1.78 m (Assuming a height of 5 ft 10 in) 3. 48.7 m 5. 66 picas long and 51 picas wide 7. 12.7 mm; 6.35 mm; 3.18 mm; 1.59 mm; 0.794 mm; 0.397 mm 9. a) 1 mm; 3  106 m (Assuming grapefruit diameter  0.1 m); b) 7 mm; 0.5 km (Assuming head diameter  0.2 m) 11. 1 mm 13. 4.41 m; 6.94 m 15. 6.3  106 m 17. 1.4  1017 s 19. 7761 s 21. 23 h 56 min 23. 12 days 25. 3.7  107 beats/year 27. 0.25 min of arc; 0.463 km 29. 0.134 % in planets; 99.9% in sun 31. 0.021 % electrons; 99.98 % nucleus 33. 373.24 g 35. a) 8.4  10 24 molecules; b) 4.3  10 46 molecules; c) 1680 molecules 37. 28.95 g/mol 39. 6.9  108 m 41. 2.1  1022 m 43. a) 1 pc  2.06  105 AU; b) 1 pc  3.08  1016 m; c) 1 pc  3.25 ly 45. 35.31 ft3 47. 2.72 m 49. 8.9  103 kg/m3; 5.6  102 lb/ft3; 0.32 lb/in3 51. 8.0 m3/day 53. 108; 1013 55. a) 7.4  102; b) 1.855  102; c) 8.47  103 57. 6.0  107 metric tons/cm3 59. 5.00  103 m3/s; 5.00 kg/s 61. 7.1  1015 m; 3.0  1015 m 63. 354 m2 65. 11; 5.7; 570 atoms 67. 359.76; 1440.0 69. 8.9 m; 9295 tons 71. 3.902  1025 kg; 235.0 u 73. 2.8  1019 molecules 75. 0.125 mm 77. 88.5 km/h; 80.7 ft/s; 24.6 m/s 79. 3.81  109 s

81. yes, because the distance traveled while gliding  18.7 km 83. a) 3840 km; b) 296 km; c) 0.315 or 1:3.2

Chapter 2 1. 0.3 s 3. 6.3  107 m/s; 5.4 cm/day 5. 32.5 km/h 7. 600 km/h 9. 14 km/h 11. 2.5  104 yr; 2.5  107 yr 13. 12.8 m/s; 46 km/h 15. 5.87 h; 150 h 17. 0.06 m 19. a) 14 s; 380 m; b) 72 m 21. 4.83 m/s 23. 2.0 m/s 25. a) PLANET

Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto

ORBIT CIRC (km)

PERIOD (s)

SPEED

LOG SPEED

3.64  108

7.61  106

6.79  10

8

9.42  10

8

1.43  10

9

4.89  10

9

8.98  10

9

1.80  10

10

2.83  10

10

3.71  10

10

47.8

1.68

8.56

1.94  10

35.0

1.54

8.83

3.16  10

7

29.8

1.47

8.97

5.93  10

7

24.1

1.38

9.16

3.76  10

8

13.0

1.11

9.69

9.31  10

8

9.65

0.985

9.95

2.65  10

9

6.79

0.832

10.26

5.21  10

9

5.43

0.735

10.45

7.83  10

9

4.74

0.676

10.57

2 1.5 1 0.5 0 8.5

9

9.5 10 log (circumference)

slope  2.01 27. 20 m/s; 16.3 m/s

LOG CIRC

7

b)

log (speed)

Chapter 1

10.5

11

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A-36

ANSWERS

29. 1.2 m/s; 0.5 m/s 31. 12 m/s; 0 m/s 33. 0.67 m/s; 0.53 m/s 35. 32.4 m/s 37. 3.4  103 m/s2 39. a) t(s) a(m/s2) a(in g) 0 10 20 30 40 b) t(s) 0 10 20 30 40

6.1 1.4 0.83 0.56 0.49 a(m/s2) 0.74 0.44 0.44 0.31 0.22

49. a) 2.0 1.0 x (m)

0.62 0.14 0.085 0.057 0.050

Method: i) draw tangent to curve ii) get slope of line by counting squares iii) to find Δv and Δt convert from km/h to m/s

a(in g) 0.075 0.045 0.045 0.032 0.022

41. 0 s; 1 s; x(0)  0 m; x(1)  1.2 m 43. 1 m/s2; 0.9 m/s2; 1.3 m/s2 45. at t  0, a  0; at t  2 s, a  2.5 m/s2; at t S ", a S 0 47. a)

0 –1.0 – 2.0

1

2

3

4

5

6

7

t (s)

b) 1.6 s; 4.7 s; c) 0 s; 3.1 s; 6.3 s; v(0)  v(3.1)  v(6.3)  0 m/s; a(0)  a(6.3)  2 m/s2; a(3.1)  2 m/s2 51. 2.4 m/s2 53. 6.36  107 s; 6.2  108 m/s 55. 350 m/s2; will probably survive 57. 7.1 m/s2; 3.8 s 59. 30 m/s; 300 m 61. 16 s 63.

v (m/s)

18 660 640 Speed, v (m/s)

620 600

1

2

3

4

5 6 t (s)

7

8

9

10

1

2

3

4

5 6 t (s)

7

8

9

10

580 560 540

60 54

500

0

0.6

1.2

1.8

2.4

x (m)

520 3.0

Time, t (s)

b) TIME INTERVAL (s)

AVG SPEED (m/s)

DISTANCE TRAVELED (m)

0–0.3

647.5

194

0.3–0.6

628.5

189

0.6–0.9

611.5

183

0.9–1.2

596.0

179

1.2–1.5

579.5

174

1.5–1.8

564.0

169

1.8–2.1

549.5

2.1–2.4

65. 32.9 m/s; 40.4 m/s 67. 0.875 m/s2; 4.4 m/s 69. 

v20

(m)

TOTAL STOPPING DISTANCE (m)

v 0(m/s)

v 0 t (m)

15

4.17

8.3

1.1

9.4

165

30

8.33

16.7

4.3

21.0

535.0

161

45

12.5

25.0

10

35.0

2.4–2.7

521.0

156

60

26.7

33.3

18

51.3

2.7–3.0

508.0

152

75

20.8

41.7

27

68.7

90

25.0

50.0

39

89.0

c) 1722  2 m

v 0(km/h)

2a

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ANSWERS

71. 15.5 m/s 73. a   get/ 75. 66 m 77. 6.1 m/s 79. 44 m 81. 7.96 m or 3 floors; 22.1 m or 8 floors; 43.5 m or 15 floors 83. 2.8 s; 14 m/s up 85. 3.7 m above launching point; 8.6 m/s 87. 1.1 m/s; 5.5 m/s 89. 0.22% 91. 1.6  104 m/s2 93. 1.9  103 m/s; 2.6  102 s 95. 802 m/s; 1.9 s 97. 14.9 m/s; 5.1 m/s 99. a) n12 h/g ; b) (3/4)h above the ground; c) (2/3)h 101. 18.3 m/s; 26.7 m/s; 33.3 m 105. 13.7 m 107. average speed  1.3 m/s; average velocity  0 m/s 109. 0.95 s; 28.8 m 111. 2.9 s 113. a) 4.3 m; b) 3.0 m/s; c) 6.0 m/s2 115. 21.1 m/s 117. 33.1 m/s; 2.21  103 m/s2 119. a) 8.10 m above ground; 11.1 m above ground; b) 9.8 m/s down; c) 0 m/s2

Chapter 3 1. 11.8 km, 30 N of E 3. 11.2 km, 27.7 S of E 5. 612 m, 11.3 W of S 7. 436 km, 7.4 W of N 9. B  (1.26 m)i  (3.2 m)j 11. 13.6 nmi, 88 E of N 13. 1.88  104 km, 1.98  104 km 15. 6.07 mi, 78.3 W of S 17. 9.19 km N, 7.71 km W 19. 1.7 m 21. (2i  5j) cm 23. Az   4.2 units 25. a) 3i  2j 2k b)  7i 4j  4k c)  16i  9j  11k 27. x   9.9 m, y  9.9 m 29. (1/3) i  (2/3) j  (2/3) k 31. c1  8/7, c2  9/7 33. 4940 km 35. (6/7) i  (12/7) j  (4/7) k 37. 9 39. 8, 112 41. 56.1 43. 45

A-37

45. (3.9  106 m)k 47. Because the vectors are nonzero, a zero result for the dot product means they must be perpendicular. 49. Bx  6.83, Bz  4.5, Cz  1.34 51. 0.44i  0.22j  0.87k 53. 0.49i  0.81j  0.32k 55. 24 59. 12i  14j  9k 61. 0.45i  0.59j  0.67k 65. Coordinate system rotated at 26.6 67. 415 m, 29.8 W of N 69. x  1.0, y  1.7 71. A  B  5.4i  12.7j; A  B  5.4i  6.5j 73. 4.58 75. 304 m2 77. 4.0, 5.0

Chapter 4 1. a) 7 km, 5 degrees E of N; b) 5.6 km/h, 5 degrees E of N; c) 8.24 km/h 3. 3.93 m 5. a) 2 i  (5  8t) j  (2  6t) k; b) 8 j  6 k, magnitude  10 m/s2, direction  37 below the y-axis in the y-z plane 7. 19.6 m at 90 below the direction of travel of the airplane at 2 s; 24.7 m at 83 below the direction of travel of the plane at 3 s 9. 13.3 km/h i  123 km/h j 11. velocity  (90 i  15 j) m/s, speed  91 m/s; direction  9.5 below the x-axis 13. a) v  (3t i  2t j) m/s; b) r  [(3t2/2)] i  t2 j m 15. 2.4 m/s 17. 38 m/s 19. 65.8 m/s, 93.4 m/s 21. a) 7.25 b) 13 m 23. 1.74 sec, 14.9 m, 59.5 m 25. 3.13  103 m/s, 2.5  105 m/s, 452 sec 27. 64.8 m, 3.04 sec 29. 76 31. 12 m/s, r  21 m i  55 m j 33. 21 m/s 35. The lake surface is 34.3 m below the release point and the horizontal distance from release point is 68.8 m 37. Yes, puck passes 2.2 m above the goal, 0.391 sec 41. 63.4 43. 9.29 and 80.5 45. 5.19, when angle off 0.03 in vertical direction arrow still hits bull’s-eye (arrow hits 4.6 cm off center, which is still within 12 cm diameter), when angle off 0.03 in horizontal

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A-38

ANSWERS

direction arrow still hits bull’s-eye (arrow hits 4.7 cm off center, which is still within 12 cm diameter) 47. 7.49 49. No, projectiles will never collide 51. 45.8 km 1 53. u  (d  p/2) 2 55. h  R(1  sin u) 

u2cos2u ; the maximum possible 2g

height is 4.4 m 57.   9.94, 17 km 59. 2.1 rev/min 61. 9.4 m/s2 63. 3.95 m/s2, 4  105 g 65. 8.99  1013 m/s2, 9.16  1012 g 67. 5.9  1013 m/s2 69. aeq  3.39  102 m/s2, a45  2.4  102 m/s2 71. aM  0.0395 m/s2, aV  0.0113 m/s2, aE  0.00595 m/s2 73. 5.5 m/s, 2.5 m/s 75. 633 m/s i  226 m/s j 77. V  12 m/s,   83 79. 4.60 m/s 81. 60 cm/s at 34 above the horizontal 83. speed  27 km/h,   33 85. a) 50  103 m; b) 33  103 m, 67  103 m 87. 15.1 km/h at 15 E of N 89. vrel 

2v0 2

(h



4v20t 2

 1)1>2

91. 528 km/h at 8.5 N of W 93. a) vertical component  62.1 km/h down, horizontal component  232 km/h in direction of plane’s travel; (b) 1.9 min 95. vE  3.9 km/h, vN  1.4 km/h (1.4 km/h south) 97. a) 13 m/s; b) 56.3 99. a) 25 km; b) 50 km; c) No 101. 26 m/s 103. 8.9 m/s2 105. 10.8 sec

Chapter 5 1. 442 kg 3. 2.69  1026 kg; 3.7  1025 atoms 5. 3.8 m/s2; 6.2  103 N 7. 6.6  103 N; 12 times the weight 9. 1.2  103 N

11. 1.8  103 m/s2; 1.3  105 N 13. 35 N 15. 4.2 m/s2; 2.4 m/s2; 1.0  103 N; 5.7  102 N 17. 0.063 m/s2 19. 2.90  103 N; 1.82  103 N 21. v  bx 0 sin(bt); a  b 2 x 0 cos (bt); F  mb 2 x 0 cos (bt); F  mb2 (x  x0) 23. No, since the tension in the rope  150 000 N  breaking tension 25. 36o south of east; 260 N 27. 3.7 m/s2; 23.4o east of north 29. 4.7  1020 N; 25o clockwise from the Moon-Sun direction 31. 770 N in the positive x-direction 33. 2.6 kg; 34 N 35. 285 N on Mars; 1900 N on Jupiter 37. a) 9.9904  104; b) no 39. 128 N in the upper cord; 29.4 N in the lower cord 41. F  T1  (m1  m2  m3)g ; T2  (m2  m3)g ; T3  m3g 43. Mg; Mg/2 pd 2lrg

pd 2lrg

at the upper end; T  at the midpoint 4 8 47. 1.2 m/s2; 36 N to the right; 36 N to the left 49. 600 N 51. 5.2  103 N 53. 165 N; 19.5o clockwise from positive x 57. 1.8  103 N upward m2  m3 59. F in the first cable; F a b in the second m1  m2  m3 m3 cable; F a b in the third cable; m1  m2  m3 F  (m1  m2  m3)g a m1  m2  m3 61. 1.14  103 N or 265 lb; 820 N or 184 lb 63. 0.51 N 65. 1.9 m/s2; 14 m/s 67. 64 m; 5.1 s mgR 69. 2l(l  2R) 45. T 

y T θ

R

y

θ θ

N

x

R

Fnet Direction of motion

mg

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ANSWERS

71. N  680 N, F  340 N

87. a)

A-39

N

N P F cos 30° 30°

mg cos θ

F

θ

F sin 30°

588 sin 30°

30°

mg

mg sin θ 588 cos 30°

60 × 9.8 = 588 N

73. a) Incline forward; b) 22.3 m/s2 ¢x B g 2g 77. a  x ; x(t)  x0e22g/l t l 79. 7.9  105 m/s2; 0.14 m 81. a) Fnet  2i  3j  4k N; b) a  0.33i  0.50j  0.67k 75. π/4; 2

2

m/s , 0.9 m/s 83. a)

2

B T cos θ T sin θ

7000 N

Chapter 6 1. 5.7  103 people 3. 0.83 5. 1.6  102 m 7. 53 m 9. 3.4 m, so he will reach the plate; 1.9 m/s 11. 0.48 13. 2.1  103 N 15. 0.27 17. 2.0  102 N 19. 2.8 m/s 21. 1.9 23. 0.78 25. 1.4  103 m/s (1.4 mm/s) 27. 39.5 m 29. a)

y

θ

b) 1.8  104 N; c) same as b) 89. a) 590 N; b) 700 N; c) 590 N; d) 0 N 91. a) 0.98 m/s2; b) 99 m; c) 50 km/h, same as speed when first decoupled m1m2 m2 93. a) a  g; b) T  g m1  m2 m1  m2

x

T mg

b) 7.1  103 N; c) 2.6  104 N 85. a)

F cos θ Ff = μ s N

boy

Frope on girl

Frope on boy

girl

θ

F sin θ

N

Fground on boy

Ff = μs cos θ F

Fground on girl

N

F cos θ

θ

mg F boy on rope

T

T

Fgirl on rope

F sin θ

b) 250 N; c) 250 N; d) 250 N

mg

F

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A-40

ANSWERS

mg

; c) u  tan1 ms; cos u  ms sin u mg F ; d) u  tan1 ms, but now  is the 2 2(1  ms ) angle between the force F and the positive x direction 31. The quantity cos u  mk sin  must be positive in order to find a solution for P 1 This gives the condition tan u  mk 33. 3.6 m/s2 mkm1m2 g cos u 35. T  (m1  m2) cos f  (m1  2m2)mk sin f 37. 23 N; 11 N 39. No, since k is not constant 41. 50 N/m 43. 8.8  102 N/m 45. 1.1  103 m 49. 4.4  102 N 51. 7.8  102 N at top; 8.1  102 N at bottom 53. 6.3 55. 0.13 m 57. 0.224 N 59. 6.9 m/s 61. 22 m/s 63. 68 65. 2gl sin u tan u 67. 1.40  103 m 69. The equilibrium conditions when the balls are at maximum angular displacement is cos u2 cos u1  m1 m 2, and the condition when they are both vertical is (m 1  m 2 )g m2v22  m1v21 . These conditions cannot both be satisfied,  l so the motion described is impossible b) F 

71. T 

b) 3.9  102 j N; c) 3.1  102 i N.; d) 3.9  102 j N; 3.1  102 i N; e) a  7.85i m/s2; 31 m 81. 0.15 m m2  mkm1 g 83. a1  m1  m2 85. 4.1 cm; 2.5 cm, 1.6 cm 87. a) 1.2  102 m; b) 1.7 m/s2 up the incline; c) 1.0  102 m 89. Yes, since the centripetal force exceeds the maximum frictional force

Chapter 7 1. 1.5  103 J 3. 252 J 5. 3.7  103 J 7. 2.35  105 J; 357 J/s 9. 2.2  107 J by first tugboat; 1.0  107 J by second tugboat; 3.2  107 J total 11. 2.6  103 J 13. 7  104 J by gravity; 7  104 J by friction 15. 54 17. a) 1.3  104 J; b) 290 N; 1.3  104 J 19. a) 1.4  104 N; 8.3  103 N; b) 4.3  103 J; c) 2.2  104 J 21. a) 7.1  103 N; b) 2.2  105 J; 8.1  103 N 23. 6 J 25. 26 J 27. 3W0; (2N  1)W0 l2 l 29. a) k c y2   2(l/2)2  y2 d ; 2 2 (l/2) b) P  2ky c 1  d 2(l/2)2  y2 31. 17 J

mv2 2pr

A B2 ; b)  BB 12A 35. 2.7  1033 J 37. 1.3  105 J; 5.8  103 J; 22 39. a) 4.0  105 J; b) 2.5  104 J; c) 1.2  106 J 41. 4.1  106 J 43. Kball  46 J; Kperson  38 J; they are of the same order of magnitude 45. 1.9 J; 0.44 m 47. 6.2  109 J 49. 196 m/s 51. 3.4  1018 J 53. a) 80 J; b) 1.295  103 J; c) 1.375  103 J 55. 7.4  103 J; 1.6 % of the energy acquired by eating an apple 57. 8.2  106 m3 33. a) xeq 

73. f  tan1 a

tan u 1  4p2 Rt 2g

75. 0.89 m/s 77. 40 m; 3.2 s 79. a) N

fk

mg

b u; at   45,   0.099

6

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ANSWERS

59. 5.1 m 61. 99 m/s; 9.8  1010 J; 23 tons 63. 0.16 65. 79% 67. 7.7 m/s 69. 1.1  104 J 71. 53 N 73. 100 m/s 77. a) 24 m/s; b) 7.1 m; c) 26 m 79. 48.2 81. 2.1  103 J 83. 1.69  105 J; 2.06  105 J 85. a) 8.8 m/s2; 35.4 m; b) 1.06  104 N; 3.75  105 J 87. a) U  2.35  107 J; K  2.89  106 J; b) U  17.7  106 J; K  8.7  106 J; 120 m/s (or 430 km/h) 89. a) 150 J; b) 150 J; c) 122 m/s; d) 1520 m; e) 122 m/s 91. 4.1  104 J 93. a) 25.8 m/s; b) 10.3 m/s

Chapter 8 1. 0.076 m Ax4 , assuming that x0  0; 5.6 m/s 4 x4 7. U (x)  x2  , assuming that x0  0; 1.3 J; 8 J; 29.3 J 4 9. F  4x  4x3 11. 64.5 m/s 13. a) bxy; b) bxy 15. 2.61  106 J A B 17. U(x)   6 12 12x 6x 19. a) 13 kN; b) 13 kN. The force is independent of the rope length xi  yj 21. F  a 2 (x  y2)3/2 5. U (x) 

23. 1.89  105 N 25. a) 4.58 m; 14.9 m/s; b) 14.7 m; 21.7 m/s2 27. a)

U (x) 0.5

1.0

x

–1.0

b) x1,2  (2 ; 22) (b/c); c) x  (26  2)(b/c)

A-41

mv2 2A 31. x   1.0 m; unbound for E  0 J 33. a) 0.382 nm; b) 1.67  1019 J; c) 0.34 nm, 0.89 nm 35. 6.3 eV/molecule 37. 29. x  ;

VEHICLE

ENERGY PER MILE (J/mi)

Motorcycle

1/60 gal/mi  1.3  108 J/gal  2.2  106 J/mi

Snowmobile 1/12  1.3  108  1.1  107

ENERGY PER PASSENGER PER mi (J/ passenger-mi)

2.2  106 1.1  107

Automobile

1/12  1.3  108  1.1  107

1.1  107/4  2.7  106

Bus

1/5  1.3  108  2.6  107

2.6  107/ 45  5.8  105

Jetliner

1/0.12  1.3  108  1.1  109 1.1  109/ 110  9.8  106

Concorde

1/0.1  1.3  108  1.3  109

1.3  109/ 360  3.6  106

Most efficient is the bus, least efficient is the snowmobile.

39. 183 m, assuming a mass of 70 kg for the climber 41. 1.05  104 kJ 43. Walking 1.7 kcal/kg; Slow running plus standing 2.8 kcal/kg; Fast running plus standing 2.8 kcal/kg 45. 1.88  109 eV 47. Thermal energy is 0.0001 % of mass energy 49. 511 keV; 939 MeV 51. 1.4  109 kg; .00000005% of the mass of the gasoline 53. 9.40  108 eV 55. 542 kcal 57. 18 kWh 59. 2.88  108 J 61. 1.5  103 hp; 23 kcal 63. a) 769 gal; b) 5.8 kW 65. 526 kWh/year; $79 67. 1100 W; 0 W 69. 0.61 hp 71. 746 J 73. 50 75. 2.0 W 77. 4.24  105 W 79. 2500 km2 81. a) 1.7  1010 J; b) 17 min; c) 3.73 km 83. 1.2  104 W 85. 195 m diameter 87. a) 487 hp; b) 2593 hp 89. 52 hp 91. 37% 93. a) 3.2  104 W; b) 784 W; c) 3.1  104 W 95. 2.3 W 97. 2.1  107 kW 99. 3.4 kW

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A-42

ANSWERS

101. a) 4.3  1012 J; b) 6.39  1014 J; c) 6.1  1011 kg/s; d) 7.8  1010 years 103. a) 5 J; b) 4 J; no 105. a) 7.2  106 N; b) 0.414 m/s 107. 4.2  109 W 109. a) P  1.82  106  3.63  105 t  2.71  104 t2  964 t3; b) 9.757  106 J; 5.714  106 J; c) 1.35  106 W 111. 14 min 113. a) 1.6  107 kWh; b) 3.8  102 m3/s

Chapter 9 1. 8.2 N 3. 3.46  108 m 5. FSun  0.41 N, FMoon  2.3  103 N 7. FAlpha  1.5  1017 N, FEarth  3.5  1022 N 9. aJ  24.9 m/s2, aS  10.5 m/s2, aU  8.99 m/s2 11. 1  109 N 13. 2.54  1010 N at 52 15. 2.76  104 g 17. aEarth-Moon  2.21  106 m/s2, aEarth-Moon /g  2.25  107, aJupiter -Io  0.0123 m/s2, aJupiter - Io /g  0.00687 19. 101 m/s 21. 3.08  103 m/s 23. 5.8  1015 sec  1.8  108 years, 3.1  105 m/s 25. TIo  1.77 days, TEuropa  3.55 days, TGanymede  7.15 days 27. 0.927 days 29. About 10 times 31. Same latitude 22.6 West, around Lincoln, Nebraska 33. m1/m2  1.6 35. 3.0  1010 m 37. a) 7.50  10 3 m/s, 8.32  10 3 m/s; b) 3.94  10 8 J, 4.85  108 J 39. 8.2  103 m/s 41. TS  96.5 min, TE  115 min 43. 5.33  10 10 km, about 10 times Pluto’s mean orbital distance from sun 45. 7.8  103 m/s, 1.4  1011 J 47. 22 49. U  1.04  106 J, K  5.2  105 J, E  5.2  105 J 51. 8.86 mm 53. 0.253 55. a) 1.11  104 m/s; b) 1.23  1011 J  29 tons of TNT; c) 1.23  105 m/s2 57. 2270 m/s, 1.11  104 m/s 59. elliptical 61. a) speed  1680 m/s, time  6510 sec  109 min; b) this will give an elliptical orbit; c) this orbit will not be closed because the launch speed is greater than the escape speed

63. a) No, speed is less than that needed for circular orbit; b) 1.22  104 m/s 65. ES  4.4  109 J, EE  6.06  108 J 67. vperigee  6.96  104 m/s, vapogee  5.75  103 m/s 71. h  4.26  106 m, v  817 m/s 73. a) 2.6  10 3 m/s; b) 2.8  10 3 m/s; c) 0.401 years; d) Venus moves 234.7, Earth moves 144 75. 1 rev/min 77. 1.6  109 N 79. 4.9 years 81. 1.90  1027 kg 83. a) vrel  1.53  104 m/s; b) 3.91 g 85. height above earth  9.89  106 m 87. a) –3.30  109 J; b) 3.30  109 J

Chapter 10 1. 9.0 kg·m/s; 3.2 kg·m/s. 3. 1.8  1029 kg·m/s; 4.3  107 kg·m/s; 3.8  104 kg·m/s; 95 kg·m/s; 2.0  1024 kg·m/s 5. (1.6  1025 i  7.7  1026 j) kg·m/s 7. a) (9.7i  5.6j) kg·m/s; b) (9.7i  0j) kg·m/s; c) (9.7i  5.6j) kg·m/s 9. 9.81 kg·m/s down; 98.1 kg·m/s down 11. 9.0 i m/s 13. 2.2  105 j 15. 2.02  105 i m/s 17. 8.26  103 m/s; 1.29  1017 J 19. (1.3 m/s)i  0j 21. 66 N; 1.3  106 J 23. 150 Ns 25. (4.10  103i  929j) m/s 27. 5.2 N 29. 5  1011 kg/s; 5  105 N m n 1 31. u a M k1 1  km/M 33. 1.9 m from woman 35. 7.42  105 m; 0.107% of the sun’s radius 37. h/3 along the height, away from the unequal side 39. 0.027 cm directly away from the 40 g piece 41. 0.23 nm from the hydrogen atom 43. (950  103, 180  103, 820  103) light-years 45. (L/3, L/3, L/3) 47. (0.061L, 0, 0) 49. 950 m from the base 51. 31.5 J; 63 J 53. 9.0  107 J 55. CM is on the axis of symmetry, a distance R/2 away from either the base or the top of the hemisphere 57. 6.9  106 m/s in the direction of motion of the proton 59. 953 kg

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ANSWERS

61. 1.05 m 63. 1.6 m/s in the direction of motion of the bullet 65. 4D from the launch point 67. (17.4 km/s, 17.4 km/s) 69. KCM  3.8  104 J; KTOT  6.38  105 J 71. 4.76  104 J; 3.6  105 J; 4.76  104 J 73. 3.9  103 J; 4.0  103 J 75. 0.955  103 77. 1.1 m/s 79. aboy  5.0 m/s2 toward girl; agirl  6.7 m/s2 toward boy; 1.7 m from boy 81. 8 km/h 83. 0.0927 nm 85. 59 cm 87. If the stack of two books is at the top of the triangle, the CM is at a point halfway between the other two books and 0.43 m above the line connecting them 89. Halfway along the line joining the centers of the plates

Chapter 11 3

4

1. 1.13  10 kgm/s, 2.3  10 N 3. a) 12 m/s, 7m/s, 3 m/s, 1 m/s, 1 m/s, 250 m/s2, 200 m/s2, 100 m/s2, 100 m/s2; b) 4.2  105 N, 3.4  105 N, 1.7  105 N, 1.7  105 N; c) 1.1  105 Ns 5. 12.6 kgm/s, 4200 N 7. –1400 N 9. –1.8 kgm/s, 1.35  104 N 11. 8.1 kgm/s, 0.045 s 13. 18 kgm/s 15. 7.5  102 s, 2.8  104 N 17. a) vproj  0.27 m/s, vtarg  0.53 m/s; b) Kproj  1.9 102 J, Ktarg  0 J Kproj  2  103 J, Ktarg  1.7  102 J 19. 39 m/s 21. 0.57 J 23. 0.22v M 25. 7 27. v1  15 m/s, v2  17 m/s 29. Last ball has velocity  v and other two balls have velocity  0 4h h 31. a) mass m rises to , mass 2m rises to ; b) mass m rises 9 9 to h, mass 2m stops and does not rise 33. 13.5 m/s 35. a) The 1400 kg mass has a velocity  1.3 m/s and the 800 kg mass has a velocity  6.1 m/s; b) t  0.98 s x  1.7 m 37. Yes 39. a) 7.5 m/s; b) 15 m/s; c) 15 m/s 41. 0.17 m/s, 0.18 m/s, 0.41 m/s, 0.34 m/s

A-43

43. 9.3 m/s 45. a) 9.8 m/s; b) 4.8  105 J; c) 130 m/s2, 850 m/s2 47. 4.0  1013 J 49. a) 3.9  105 J, 3.9  105 J; b) 7.8  105 J, 3.9  105 J 51. 210 m/s 53. a) 440 m/s; b) 1200 J; c) 9.6 J; d) missing kinetic energy is energy that shows up as heat in bullet and block, compression/deformation, and noise 55. 620 m/s 57. 860 m/s 59. (a) 3 m/s ˆi (b) 79 m/s2 61. 21 m/s b b2 b ; b) 2mv sin1 a 1  b 2R 4R2 v v v 65. v1  (1  cos) i  sin j; v2  (1  cos) i 2 9 2 v v2  sin  j v1 v2  (1  (sin 2   cos 2 ))  0 2 4 so v1  v2 63. a) p  2 sin1 a

67. 3.2 kgm/s, 3200 N 69. v1  2.6 km/h, v2  13 km/h 71. 1.2  1012 J 73. a) v1  0 and v2  20 m/s; b) The ball that had an initial velocity lands on ground next to fence and the ball with no initial velocity lands 11 meters away from the fence 75. 0.964, 0.036 77. a) 3.3  1012 m/s; b) 2.4  107 tons of TNT; c) 8.1  1014 N 79. a) Ball height h  (1 – cos) where  is the angle with the vertical; b) height h  L4 (1 – cos ) where   cos1Q3414 cos uR 80. a) 21 m/s and 11 west of north; b) 1.4  105 J

Chapter 12 1. 1.7  103 rad/s; 3.5  104 m/s 3. 81.5 rad/s; 13 rev/s 5. a) 0.52 m/s; 0.17 m/s; b) 1.8 m/s2; 0.61 m/s2 7. 22.0 rad/s; 1.28 m/s; 55.4 rad/s; 8.83 rev/s 9. 9.4 rad/s; 0.94 m/s; 1.9 rad/s2 11. 88 rev/s for aluminum; 2.1 rev/s for steel 13. At t  0,   0;   0;   40 rad/s2; at t  1.0 s,   15 rad;   25 rad/s;   10 rad/s2; at t  2.0 s,   40 rad;   20 rad/s;   20 rad/s2 15. 5.5  103 cm/s; 51 cm/s; 22 cm/s 17. 611 rad/s2; 70 revolutions 19. 1.5  102 rad/s2 21. 4.36  102 rad/s2; 0.89 revolutions

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A-44

ANSWERS

23. 23 rad/s2; 1.7 s 25. 9.5  1022 rad/s2 27. 10 rad/s; 8 rad 31. 3.8  104 J 33. 1.21  1010 m 35. 6.50  1046 kg·m2 37. 0.44 kg·m2 39. 0.46 kg·m2 41. 2.13  1029 J 43. 0.96 kg·m2; 1.5  107 J 45. 1.10  1022 kg·m2 47. 0.379MER2E 49.

ML2 sin2 u 12

Ml 2 3 53. 0.426 MR2 51. I  55. a 57.

1 1  b MR2 2 10  30h4R

5 5 2 M(R2  R1) 5 R32  R31

3 2 M 5 I 61. a) 225 kg·m2; b) 4.4  103 J 63. 3.49  105 J; 3.9% 65. 2  1025 J/s; 1.05  1015 s or 3.3  107 yr 59. I 

67.

MR2 4

Ml 2 6 MR2 71. 4 3 73. mR2 10 69.

75. (160 km/h) i at top; 0 km/h at bottom; (80 km/h) i  (80 km/h) j at front 77. 12 rad/s2; 74.1 rad/s; 35 turns 79. 0.012 J; 0.37 m/s 2 4 81. I1  2  MR2  MR2; 5 5 2 14 MR2 I2  2 a MR2  MR2 b  5 5 5 1 83. MR2 a  b 8 27p 85. 4.4 m/s

Chapter 13 1. (4610 N·m)R, 613 kg, 940 kg 3. 310 N 5. 59 N·m 7. 130 hp, 176 N·m 9. 2900 N·m 11. 5.4 m/s, 7.7 m/s 13. 230 W 15. 1.9  106 J 17. 19 J, 75 J 19. 5.6  104 W, 1.1  104 N·m 21. 4.6 rad/s 23. a) 1140 N·m; b) 2.1 m/s2 25. 820 N 27. 2.7  104 N·m 29. 9.7 m/s2 31. Proof required. 33. 9.6 m/s2 35. rolls to the right, f  2F/3 37. 17 rev 39. 0.024 N·m, 0.16 N 41. Proof required. 43. 2.83 m/s2, 2.94 m/s2, 3.89 % 45. 1.6 m 47. 2.8  1034 kg·m2/s 49. 1.6 kg·m2/s 51. 1.05  10 34 kg·m 2/s, 2.11  10 34 kg·m 2/s, 3.15  1034 kg·m2/s 53. a) 1.8  109 kg·m2/s, upward; b) zero kg·m2/s 55. 7.9  1022 rad/s, 7.9  107 m/s, 2.1  1012 J, 1.5  1010 J 57. 0.051 kg·m2/s2 59. 5.6  1041 kg·m2/s, 3.14  1043 kg·m2/s, 1.8 % 61. 0.57 rad/s, 5.5 rad/s 63. 14 rad/s 65. 1.5  1019 rev/day GME GME and v2  . The satellite closer to r B 1 B r2 earth has the greater speed. L1  m1r1GME and L2  m1r2GME . The satellite closer to earth has the smaller angular momentum. 69. 4.3  1022 rad/s2, 3.5  1016 N·m, 2.6  1012 W 71. 7 73. a) 5.0 m/s; b) 0.009 rad/s; c) 6860 or 19 rev m2v2 75. 1 2g (m  3M )(m  12M )

67. v1 

77.

mv30

,

mv30

g132 g12 79. 0.37 rad/s

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ANSWERS

81. The instantaneous change in angular momentum opposes the original direction of the angular momentum and makes the tilt worse 83. a) 110 kg·m2/s; b) 34 N·m; c) 34 N·m 85. 1.6  108 kg·m2/s, east or west, 1.2  104 N·m, 1.0  104 N 87. 3100 N·m 89. 3100 J 91. a) 76.2; b) 290 N, 250 N 93. a) 4.10  105 kg·m2; b) 0.12 rad/s2; c) 4.8  106 N·m; d) 0.0023 J 95. 420 N 97. 8.50 s 99. a) 2.0  1010; b) 4.9  109; c) 4.9  109 rev/month or 1860 rev/s 101. 1.3  1014 kg·m2/s, north

Chapter 14 1. 590 N 3. 5.5  106 N; 5.1  106 N 5. 5200 N 7. 8 cm; 18 cm 9. 5.88 kg 11. 3500 N; 6800 N 13. 51 N; 29 N 15. 1420 N; 2500 N 17. 30 19. 7.65 m 21. Proof required. 23. F  Mg

2R2  (R  h)2 R ; F  Mg 2 (R  h) 2R  (R  h)2

25. a) 9  103 N; b) 2.6  103 N 27. 0.408 Mg 29.T Mg a

LR 2(L  R)2  R2

b ; N Mg

31. 0.62 mg 33. a); b) 26.6; c) 56 J

R 2(L  R)2  R2

A-45

39. 7.2 m/s2 41. a) 8.8 m/s2; b) 5.4 m/s2; c) 3.8 m/s2 mg mg ; 43. 2mg; 13 13 45. tan1 (2 s) 47. T1 

t 1 t 1 a b and T2  a b R emsp  1 R 1  emsp

49. 1580 N; 1340 N 51. 400 N 53. 240 N 55. 0.010 micrograms resolution; 0.20000 milligrams max load 57. 736 N (R1  R2) ; 59. a) 12 mg (R1  R2); b) F  12 mg l 2l F  c) F R1  R2 61. 8.9 63. 393 65. 36 67. 2  103 m 69. 3.5  104 m 71. a) 4  103 m; b) 3.5  103 N 73. 0.057 m 75. 0.033% 77. 4.3  106 m 79. 0.52 cm 81. 1.5 83. 3.96  108 N/m2 85. 624 rad/s 87. 425 rad/s 89. 360 N 91. 1.23  104 N; 2.13  104 N 93. 2400 Nm 95. 0.577 Mg; 0.289 Mg 97. 490 N 99. 1200 N 101. 0.71 mm 103. 5.0  103 m 105. 4.85  105 N

U 53 J

Chapter 15

1/2 × 53 J

10° 20° 30° 40° 50° 60°

35. 1.04 L 37. 17.5 m/s

θ

1. a) 3.0 m; 0.318 Hz; 2.0 rad/s; 3.14 s; b) tmidpoint  0.785 s; tturn.point  1.57 s 3. a) 0.83 Hz; 5.2 rad/s; b) 0.20 m; c) 0.30 s; 0.40 s; d) 1.05 m/s; 0.91 m/s 5. a) 251 m/s; b) 251 m/s 7. 211 N; 4.2 m/s

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A-46

ANSWERS

9. A 

B

x20 

11. 3.98  10

6

v02

; d  tan1 a 2

v

v0 b vx0

75. a) mg B1 

m b)

3p 13. d  2 2

15. a) x  0.292 cos(6 t  0.815); b) 0.043 s; 103 m/s 17. 2.8  106 N/m; 3.16 Hz 19. 1.9  104 N/m 21. 1.13  1014 Hz 23. 0.20 m; 7.3 rad/s; 0 rad; 1.46 m/s; 10.7 m/s2 25. 5.51  103 Hz; 1.0  104 g; 6.7  106 cm 27. x  0.27 cos 6t, with the axis chosen so the initial position of the unstretched spring is at x  0.27 m. 29.

mg sin u k

31. f 

;

1 k 2p B m

1 k 2p B (m  6M)

33. 3.6 J 35. 2.12 Hz; E same; A same; 2.0 m/s; 26.7 m/s2 37. 0.34 m 39. Fd 

kd 2 2

43. 10.4 s 45. 34.8 s 47. 0.188 Hz 49. a) 0.73 min/day; b) 1.0 mm 51. 24.8 m 53. 3.0 s 55. a)

2k B MR

2

57. 2.09

; b) 0

2k B MR2

; c) 0  1 radian

L Bg

59. a) 1.26  103 J; b) 0.145 m/s 61. 1.64 s 63. 9.8  103 m/s2 65. 1.6 s 67. 2p

L m1  m2 a b B g m1  m2

69. 3.6 rad/s; 3.3 m/s k 71. m  g 73. 2p

4L B 5g

g A2 3A 2 2  sin a tbR ; 2 2 Bl

p l ; mg (1  A 2) 2 Bg

77. 1.0  103 79. 92; 0.32 W 81. 30 83. 395 85. 21 87. 1.5 cm; 66.7 Hz 89. a) midpoint at 2 s; 6 s, 10 . . .; turning point at 0 s, 4 s, 8 s, . . .; b) midpoint at 0 s, 4 s, 8 s, . . .; turning point at 2 s, 6 s, 10 s, . . . 91. 26.7 m/s; 1.68  104 m/s2; 2.0  104 N 93. 2.12 Hz 95. 1.25 Hz 97. 18.75 J; 3.54 m/s 99. 0.375 m 103. 0.35 Hz

Chapter 16 1. 4.3  1014 Hz to 7.5  1014 Hz (violet) 3. 2.08 cycles/hour, 356 km 5. a) 4.4 m/s; b) 205 m/s 7. a) 10.8 hour; b) 2.5 /hour 9. a) vmax  0.27 m/s as it passes through equilibrium between crests; b)  amax   6.2 m/s2 at the wave crests. 11. a) 0.02 m; b) 1.4 Hz; c) 10 m 13. a) 0.2 sec, 5.0 Hz, 31 rad/sec, 5.2 m1; b) y  0.020 cos (5.2 x 31.4 t) 15. wavelength decreases by 20 cm 17.  amax   2.41 m/s2; wavelength  156 m 19. 0.45 m/s 21. 0.97 sec 23. 2.0 kg/m 25. 250 m/s 27. 1280 N 29. 184 km 31. v  v

m Bm

33. 0.017 sec 35. 22l /g 37.

mV 2 ; Proof required. 2pR

p ; b) 0.20 m 2 43. 6.0 m, .6 m, 0.80 Hz; 3.2 m 45. 1.14 m, 6.28 m 41. a)

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ANSWERS

47. a) 0.0194 m; b) 0.0179 m 49. a) y  0.0060 cos(400t)  0.0040 cos(1200t); b) y

4L 4L 4L , , , . . ., where 4L is the longest possible 3 5 7 wavelength

93. l  4L,

Chapter 17

0.010 m

0.005 0.003

0

0.001

0.002

0.004 0.005 s

t

–0.005

–0.010

51. 0.028 is the fractional increase or decrease. We cannot tell which from the given information. 53. 392 Hz, 588 Hz, 784 Hz, 980 Hz 55. 1.58 Hz, 3.16 Hz 57. 28 Hz 59. 7.07 m, 66.6 m/s, 628 m/s2 61. 9.3 ms, down, 54 Hz 63. 8.16 Hz, 16.3 Hz, 24.5 Hz, 32.7 Hz, 40.8 Hz 65. 71 N 67. 1.3  107 m, 2.5  107 m 69. 3.7 m/s, 6.8  103 m/s2 71. a) 4.82  103 N; b) 8.2 Hz 73. y  (0.20 mm) sin(2x  880t)  (0.20 mm) sin(2x  880t), A  0.20 mm, v  440 m/s 75. a) v(x)  c) f 

A-47

F 1 F ; b) l(x)  ; B A  Bx f B A  Bx

n F 2L B A  BL

77. Large amplitude is at x 

np . Smallest amplitude is at k

(2n  1)p where n is an integer. k 79. a) 0.030 m; b) 5.2 m; c) crests: 0 m, 5.2 m, 10.4 m, . . .; troughs: 2.6 m, 7.9 m, 13.5 m, . . . 81. 2.1 m/s2, 4.9 m/s 83. 13 m/s, 7.9  103 m/s2 85. 0.0731 kg/m, 261 m/s x

87. a) T1 

22 1  23

T, T2 

2 1  23

T;

b) v1  7.2 m/s, v2  8.6 m/s 89. 12 Hz 91. a) v(x)  2g(L  x), 14 m/s, 9.9 m/s, 0 m/s; b) 2.9 sec

1. 17 m (20 Hz) to 1.7 cm (20 kHz) 3. 765 m; 166 m 5. about 9 cm 7. 1.9 mm and 0.10 mm 9. 6.8 m/s 11. D–D# 1.9 cm, D#–E 1.8 cm, E–F 1.7 cm, F–F# 1.6 cm, F#–G 1.5 cm, G–G# 1.5 cm, G#–A 1.3 cm, A–A# 1.3 cm, A#–B 1.2 cm, B–C 1.1 cm, C–C# 1.1 cm, C#–D 1.0 cm 13. 1.0  103 W/m2 15. D#, 6 octaves above the one listed in Table 17.1 17. 3.0 dB 19. 83 dB 21. 0.11 W 23. 130 times (intensity measured in W/m2), 21 dB 25. 9.1 sec 27. 249 m 29. a) f  3.0  105 Hz; T  3.3  104 s (about 9 h); b) It’s possible because the period of the first overtone is close to 1/4 of the tidal period. 31. a) 3.0  105 s; b) glass 33. 272 Hz, 3.9% 35. about 4000 Hz 37. 2.0 km 39. 3.3 km in sea water 41. 2.8 km 43. 92.4 m 45. 337 m/s 49. a) 0.632 m; b) C–C# 3.5 cm, C#–D 3.4 cm, D–D# 3.2 cm, D#–E 3.0 cm, E–F 2.8 cm, F–F# 2.7 cm, F#–G 2.5 cm, G–G# 2.4 cm, G#–A 2.2 cm, A–A# 2.1 cm, A#–B 2.0 cm, B–C 1.9 cm 2L 51. ln  n nv , n  1, 2, 3, . . . fn  2L 53. 619 Hz 55. 21.5 m/s, 0.215 Hz 57. 381 m/s 59. 2.63 m/s 61. 30 63. 405 Hz 65. 594 Hz, 595 Hz 67. 476 Hz 69. 481 Hz 71. 29.4

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A-48

ANSWERS

73. a) Proof required; b) 165 m/s 75. 0.15 mm 77. 1.5 mm, 0.33 mm 79. 5.0 m/s, 3.2  107g 81. 3 women 83. 3.0 dB 85. a) 3.0* 103 sec; b) The bat will think distances are 0.77 times the real distances. 87. a) 33.5; b) 30.2 sec 89. a) 660 Hz; b) 691 Hz; c) 723 Hz

Chapter 18 1. In the hose: 1.39 m/s; 2.8 m/s; 4.2 m/s; In the nozzle: 22.3 m/s; 25.1 m/s; 23.9 m/s 3. 7.23  105 W 5. 12 m/s 7. 8.84 cm/s; 8.84 m/s 9. a) 11.5 m; b) 8.1 cm; 11.7 cm 11. 84 m 13. 1370 lbf  6090 N 15. 132 cm2 17. 2.34  105 Pa 19. 5.08  104 N 21. 48.6 cm2 23. 2.0  104 Pa; 7.5  106 Pa 25. a) 360 N; b) 330 N 27. 3.56  105 Pa; 3.60  105 Pa; 4  103 N 29. 0.85 m 31. 10.3 m 33. 3.3  104 Pa 35. 2.94  105 Pa 37. 2.1% 39. 3.1  108 N 41. a) Proof required; b) 5.0  109 Pa 43. a) Fb  rghp R22; F  Ftotal  

prgh(R2  R1)(R1  2R2) 3

prghQR21  R1R2  R22R 3

b) Fb  rghp R21; F   Ftotal   7

;

prgh(R2  R1)(2R1  R2)

prghQR21  R1R2  R22R 3 3

;

45. a) 4.7  10 m ; b) 4.3  1010 kg 47. Yes 49. 0.32 m 51. 61 m/s2 53. 113 kg 55. 31 m

3

;

57. 12.6 cm 59. a) Proof required; b) 2g; c) " 61. 4.4 cm 63. a) Proof required; b) 0.094 Hz 65. 1.13  105 Pa 67.

B

2gh  2

patm r

69. 1.9  104 N 71. v 

2ptank B r

73. 0.013 m/s 75. a) 332 N; b) Average rate  4.3 kW; Peak rate  8.6 kW 77. 12.4 hp 79. 2.7 m/s; 190 liters/s 81. 7.3  103 N 83. 1.12  1012 Pa 87. 8.06 mm-Hg 89. 1.21  105 Pa; 1.01  105 Pa; 2  103 N 91. 0.73 cm 93. 2.5  105 kg 95. 5.3  103 m3 97. 1.07  103 kg/m3 99. 2.0  103 m3/s 101. The pressure inside increases by 209 Pa; Smaller

Chapter 19 1. 32F,  380F,  423.4F,  452.2F,  454F,  459.67F 3. 5.3  1021 atoms 5. 78% N2, 22% O2 7. The frequency decreases by 14 Hz 9. 1.9 atm 11. a) pO2  7.5  104 Pa, pN2  8.6  104 Pa; b) 1.6  105 Pa 13. 3.4 atm 15. i/0  1.05 17. 1.4  109 Pa 19. 12 kg/m3 21. 4.3 atm 23. 500 kg/m3 25. 100 kg/m3 27. 1.3 cm 29. 4.5  107 Pa 31. 96.3 g 33. 3150 kg, 2.8  103 m3 35. a) Water rises 1.2 m; b) 6.8 kg at 2.6  105 Pa 37. 29 g/mol 39. Differentiating p  p0  gy yields dp  gy. Using the Ideal-Gas Law and n 

pM m m . ,r  M V RT

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ANSWERS

Substituting  into the differential equation: dp   or,

dp p



Mg RT

pMg RT

dy

81. 615 m/s 83. a) 22; b) 1; c) 1; d) 0.5 85. 375 K

dy.

41. 615 m/s 43. 1200 K 45. 4100 m/s, 1.4  1020 J 47. 0.12 m/s 49. 5.65  1021 J 51. For O2, vrms  428 m/s, For O3, vrms 349 m/s, For O2, (translational) K  4.87  1021 J, for O3, (translational) K  4.85  1021 J 53. 0.43% 55. 9.7  106 K 57. 0.47 m/s 59. Using the hint, the volume swept out per molecule with an effective radius 2R0 going a distance l is cylindrically shaped with volume V/N   (2R0)2 l. Solving for l yields the desired result. 61. a) for 1 atm:  0.091%; b) for 1000 atm:  91% 63. 1.9  105 J, translational 0.6, rotational 0.4 65. A 7% increase in kinetic energy by changing temperature, no change in the kinetic energy by changing the pressure. 67. 1.0  105 J 69. 291 J 71. 1.88  1025 molecules 73. 1.3  103 N 75. 1.1  1023 nitrogen molecules, 2.9  1022 oxygen molecules, 1.4  1023 total 77. a) 1.8  1032 particles/m3 3.7  1016 Pa; b) 9.0  1031 particles/m3, 1.9  1016 Pa; c) 4.5  1031 particles/m3, 9.3  1015 Pa 79. From the Ideal-Gas Law at constant temperature pV  pV. So, p p  p ¢p ¢Z V  V V .  1 1   V V V p p p V  ¢V a

p ¢p

b,

This can be rearranged to where V is the decrease in volume and p is the corresponding increase in pressure. p  ¢p p p Furthermore, and  1 ¢p ¢p ¢p V  ¢V a 1 

A-49

p ¢p

b  ¢V a

p ¢p

b for p  p.

Using the specifics of the problem, Vc  Vs  V or Vs  Vc  V  Vc  ¢V a

p ¢p

b

Chapter 20 1. 540 s 3. 0.28C 5. 1.6 km 7. 750 s 9. 8500 steps 11. 0.17C 13. 1.7C 15. 1.7  104 C/km 17. The heat produced from electric power 2.6% of the incident solar heat. This is enough to slightly increase the local temperature. 19. 1.1  103 m3/s 21. a) 1.7  103 Nm; b) 2.3  103 W; c) 4.0  104 C 23. 27C 25. 136C 27. 0.67C 29. 38C 31. 0.18 liter 33. 0.028 J of work done by iron, 2.7  107 J of heat absorbed by iron, amount of work is 1.0  106 times the heat absorbed 35. 4.9  104 m, 17 N 37. a) 3.8  104, 1.9  104; b) 16 s 39. Proof required. 41. 100.28C 43. 23000 W, the rate through window is 13 times greater than the rate through the wall 45. a) 2.4 m2sC/J; b) 13.6 ft2hF/BTU 47. 4.2  103 W 49. The solution is a proof. 51. 11 W, 79C 53. The solution is a proof. 55. 0.51 cm/h 57. a) 4.26  1014 J; b) 5.3 bombs 59. 270 g 61. 1.1  109 J, 1.2  104 W 63. 3.9 kg 65. a) 2.0  1011 kg; b) 1.1  1017 cal; c) 2.9  1015 J  7.0  1014 cal; d) 1.0  1013 J  2.4  1012 cal. The kinetic energy is smaller than the potential energy due to frictional losses with the air. 67. 0.092 kg 69. 4.3 km3/h 71. 41C

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A-50

73.

ANSWERS

GAS

C V (J/(kG . K)

He

3.12  103

Ar

3.13  102

N2

7.42  102

O2

6.50  102

CO

7.39  102

NH3

1.60  103

CH4

1.69  103

The gas with the highest specific heat per kilogram is helium; and that with the lowest is argon. 75. 971 m/s 77. V  3.7  102 m3, W  3.7  103 J 79. Cp  26.3 J/(molK), CV  18.0 J/(molK) 81. 110 kcal/h 83. 1 K 85. a) 0.072 m3; b) 0.42 m3, 145 K 87. 214 K 89. 3.6C 91. 160 liters/h 93. 0.33 m, 0.050 m, 46 m2 95. 2.3  105 kg/s 97. 0.52 kg 99. 880 J, 1500 J 101. a) 1700 J; b) 1200 J; c) 0.029 m 3 , 1.7  10 4 N/m 2 ; d) 5.0  102 J, 2.4  103 J

Chapter 21 1. Q  1.9  103 J, E  1.1  103 J 3. a) W  0, Q  E  610 J; b) W  810 J, Q  2.0  103 J, E  1.2  103 J 5. 470 J 7. a) 4.29 moles; b) W  1010 J, E  2490 J; c) 5/2, diatomic 9. a) W  9.19 J, E  3.34  105 J; b)  18.4 J, the heat of vaporization remains unchanged 11. W  37.9 J, E  37.1 J 13. 4.87  104 J, 2.56 J 15. 4.3  104 J 17. 0.014 m3, 7.2  105 Pa 19. 2.43  106 J/kg 21. 35% 23. 14% 25. 60%, 1.2  107 J 27. 44.5% 29. 5.5%, W  myou (9.81 m/s2) (3.0 m),

Q

myou(9.81 m/s2) (3.0 m) 0.055 6

a

1 kcal b 4187 J

31. 8.2  10 J/s, 0.19 kg/s 33. 0.999 999 997 35. 1.4  105 J 37. 44%, 1100 W 39. 75 W 41. 19.5 43. 8.5  103 J, 3.4 45. a) 1.3  103 J; b) Heat is absorbed during step 2 and rejected in step 3; c) heat is rejected by the system in step 1, d) 0.39 47. a) 0.067; b) 1.39  107 W; c) 180 kg/s 49. a) 48 W; b) 20 times 51. a) eturbine  0.34, eengine  0.42; b) 0.62, the two efficiencies are the same 53. 9.5  103 J/K 55. 3.0 W/K 57. 12 400 W/K 59. 3  103 J/K 61. SAl  430 J/K, SFe  150 J/K, SAg  80 J/K, SHg  47 J/K. The change in entropy seems to decrease with increasing atomic number. Largest is aluminum; smallest mercury. 63. 41 W/K 65. 9.5  106 W/K 67. 0.94 W/K 69. 5.8 J/K 71. a) Proof required; b S  780 J/K 73. 37 J/K 75. Proof required. 77. 120 K 79. a) 4.16  105 Pa, 2.27  105 Pa; b) W  Q  3.0  102 J 81. 24%, 4.8  104 W 83. 4.0  102 W 85. a) Beginning with the point at the upper left, the gas undergoes an isobaric expansion in step 1 as the volume increases, followed by a isovolumetric reduction of pressure in step 2 as the temperature is reduced. The gas is then compressed isobarically in step 3 by reducing the volume, before an isovolumetric increase in pressure in step 4 by increasing the temperature. b) W 1  2100 J, W 2  0 J, W 3  700 J, W 4  0 J; c) Q1  5260 J, Q2  3160 J, Q3  1750 J, Q4  1050 J; d) 44% 87. 48%, 1  107 W 89. SN  2600 J/K, SO  2300 J/K, SH  22 500 J/K. Hydrogen is largest and oxygen smallest. 91. a) S  1.1 J/K; b) Q2  340 J, S  1.1 J/K; c) S  0 J/K

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ANSWERS

A-51

Chapter 22

Chapter 23

1. 5.8  105 N 3. 58 N, 3.5  1028 m/s2 5. 51 N 7. 1.6  1020 electrons 9. 2.39  107 N 11. Fg  6.7  1013 N, Fe  2.3  108 N 13. 9.63  104 C 15. 4.7  1013 electrons 17. 3.2  1019 N 19. 1.3 21. 99.9 % 23. 2.9  109 N/m 25. (2.3  105 N) i  (3.5  105 N) j, (2.3  105 N) i  (3.5  105 N) j 27. 6.81  1032 electrons on Earth and 1.9  1032 electrons on Moon 29. 1.0  109 at 1 m, 1.0  105 at 1  104 m 31. 3.8  1039 C, ratio  2.8  106 (attractive)

1. F  5.4  1014 N, a  6.0  1016 m/s2 3.   20 5. E  2.1  105 N/C j 7. Fe  0.6 Fg 9. 6.3  107 m 11. E  5.1  1011 N/C 13. E  5.1  1012 N/C 15. 28 N/C N m2 Q 17. EA  a1.15  1010 b i, C2 L2

33.

2kqxQ 2

d4  x23/2

, x direction

Q2 L2

i1.35k

Q2 L2

49. k

Q2 a2

2

x



4 (d  2x)2

C

b

Q L2

EC   a1.15  1010

N m2

ED   a1.15  1010

N m2

j 1.35k

Q2 L2

£ (ij)

2 45. 212 Q 13 2  1R  9.85

1

2

b

2

C

2

C

21. E(x)  2  a8.99  109

41. (3.1  1015 N) i  (6.9  1016 N) j 43. 1.2  104 kg or 0.12 g

47. kQq c

N m2

j, Q L2

b

i,

Q L2

j

19. EP  9.5  103 N/C i  2.8  104 N/C j

35. (1.9  107 N) i(1.7  107 N) j 37. 1.0  107 C 39. 1.35k

EB  a1.15  1010

d

(1.116 i1.75 j0.5k)

51. 0.35d 53. p  p → n  n  π+, p  p → n  p  π0, p  p → n  p  π 0 + π 55. 1 57. 5.6  1021 electrons 59. 1.9  109 kg 61. any negative charge 63. 3.6  108 N, 6.9  102 m/s2 65. 1.0  106 C, 6.5  1012 electrons 67. 0 C, e, e, 0 C

(40 C)(10000 m) 2

2 3/2

x  (10000 m) 

23. E ≈

N m2



C2

b

(30 C)(4000 m) x2  (4000 m)23/2

§

Q 2p0x2

25. a) Emax is at y  R 122 b) The field distribution for the ring is the same as that for two positive charges rotated around the y axis. 27. E0.5  7.2  104 N/m, E1.0  3.6  104 N/m, E1.5  2.4  104 N/m 29. E  a2.6  1010 31. E  33. Ex 

N m2 C2

b

(i  j) L

1 l (cos 30) j p0 2d 1 l l x c d , Ey  c1  d 2 4p0 2x2  y2 4p0y 2x  y2

35. EA  1.13  105 N/C j, EB  1.13  105 N/C j, EC  3.39  105 N/C j, ED  1.13  105 N/C j 37. E  2.4  107 N/C directed at an angle of 45 with respect to each sheet zs 39.Ez   20 2z2  R2

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A-52

ANSWERS

41. a) EP 

55. The solution is a sketch of the electric field.

1 Q 1 1 ¢  ≤ i, 4p0 l x x  l

b) EP 

1 1 Q j 2pe0 y 2l 2  4y2

+

+

2Q 43. a) l0  , l b) Ep 

1 Q 1 xl 2 x ≤  B ln ¢ 1R i, x 2p0 l 2 2 xl

c) When x  l, the electric field resembles that of a point charge and goes to zero as x goes to infinity. l 22 l22 45. E  i  j p0l p0l 47. E  

y l 1 1 b B ¢1 2 2 1/2 ≤  2 Ri x 4p0 (x y ) (x  y2)1/2

B

x 1 1 ¢1  2 ≤ 2 R jr 2 1/2 y (x  y ) (x  y2 )1/2

49. E  0.0609

l 0R

– 57. E  3.1  105 N/C 59. v0  1.4  105 m/s 61. v0  110 m/s 63. 1085 electrons 65. p  2.0  105 C m 67. a) p  1.6  1029 C m, b) The dipole moment is reduced due to the motion of electrons. 69. E  520 N/C 71. E  1.1  107 N/C,   3 with respect to the y-axis 73. E  1.21  104 N/C,   74 with respect to the y-axis 75. (a) E 

51. The solution is a sketch of the electric field.

q 1 l l ¢ ≤ , (b) F  ¢ ≤, 4p0 d 4p0 d

(c) The magnitude of the force is the same as in (b), but its direction is opposite that of the rod on the charge.

+2q

–3q

77. E 

Q 1 1 £  3≥ j 2 2p0 y2 (y  d 2)2

79. E 

1 Q 4p0 y2

Chapter 24 1. 1.1  1012 N m2/C 3. 0.16 N m2/C 23 s 2 23 s 2 a, a 8 0 24 0 7. E,1  E,6  0, E,2  E,4  2.0 N m2/C, E,3  E,5  3.5 N m2/C, total  0 q 9. f  0 5.  53. The solution is a sketch of the electric field.

11. 1.4  103 N m2/C 13. 0.038 N m2/C sq q sq s 15. (a) (b) (c) F  ,E 20 20 20 20

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ANSWERS

17. (a)

q 40

(b)

s 40

 8

2

19. 45 N m /C, 5.5  10 C/m 21. 2.3  102 N m2/C 23. G  4Gminside 25. Imagine a small cube of volume dV . If the cube itself contains no charge and it’s located in a uniform electric field, the net flux through the six faces of the cube must be zero, no matter how the cube is oriented in the field. If the cube contains charge, then the flux through it cannot be zero and the field cannot be uniform. If the field is uniform, Gauss’ law tells us that the charge density inside the cube, which is the charge inside divided by the volume dV, must be zero. 27. 160 N qenc 1 qenc  0 ‹ E A  0 29. E A  0 since A  0, E  0 lr 31. in direction perpendicular to axis 2pR2 0 r (r3  a3) , 33. r a E  0, a r b E  30 r2 3

rbE 

r (b  a ) 30 r2 (r3  a3)

Q

4p0r2 (b3  a3)

,

4p0r2 Q 4p0r2

r3

e2r

S

the magnitude of that force is F 

2pR2

(b) E 

e2r 4p0R3

Q 4p0R2

15

. (b) 7.2  10 Hz

(c) E 

4pk n3 kr n1 41. (a) Q  (b) E  r n3 0(n  3) E  k>2 (d) if n  3, Q  q

Q



q

q

47. E 



q

q

Eydx 

q q v 2p0R

q q v 2p0R

Fydt 

Q 1 ° 2 4p0 r

1 R 2 8ar  b 2

¢r

47rd ˆ rd ˆ S 1 i , r  2d iˆ E  i, 49. r  diˆ E  4 60 960 S rd 1 1 1 1 ˆ S r  diˆ  djˆ E  ca  b iˆ  jd 0 2 2 3025 1525

51. Ebigsphere 

qr

and Esmallsphere 

3

4p0R

qr 4p0R3

where r  d r Etotal  Ebigsphere  Esmallsphere ‹ Etotal 

qd

‹ E

3

4p0R

qd 4p0R3



q

q 45. f  Ey 2pRdx  2pR Eydx   0 q q

61. (a) Qsmall,inner  Q1, Qsmall,outer  Q1 (b) E 

(c) n  1,

Q1 4p0r2

(c) Qlarge,inner  Q1,

Qlarge,outer  Q1  Q2 (d) E  63. fbase 

4p0r2

d Cx3 d Cd 3 43. For x , E 

For x  E 

2 30 2 240 q

2pR0

q v

59. 8.85  1010 C/m2

r where

b . Substituting yields F   rˆ R3 4p0R3 where the minus sign indicates this is a restoring force and

39. (a) C 



‹

l , E0 2p0x q , a  r  b E  0, r  b 57. r  a E  4p0r2 q q q E , sa   , sb  2 2 4p0r 4pa 4pb2

Q

Q

‹

q

55. E 

37. (a) F  qE where q e. E  Q  e a

q

Eydx 

53. 1.1  105 C, 0 C, 1.1  105 C

3

35. r a E  0, a r b E  rb E

q

A-53

65. (a) f 

q 20

, fcurved 

q 40

(b) f 

Q1  Q2 4p0r2

q 20 3q 40

67. (a) 1.1  1012 N m2/C (b) 1.1  1012 N m2/C (c) 1.7  1012 N m2/C (d) 0 69. r R E  

rl 3pR20

where the minus sign indicates

the field points towards the center of the sphere. r  R

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A-54

ANSWERS

E

R 3l

Take any point (x,y,z) on the surface of the sphere. Then x2  y2  z2  R2 Q Q(R/h) 1 V c  d r 2 4p0 1

where again, the minus sign indicates

3p0R2r2

the field points towards the center of the sphere. 71. m  73. E 

4pR3rs 6g0

r1 and r2 as shown r1  2x2  y2  (h  z)2

S l 2l S , E total  i 2p0R 5poR

75. r a E  

x2  y2  a

r2 

B Therefore,

l b2  r2 l , a r bE , 2p0r 2p0r b2  a2

rb E0 13

7

77. 1.2  10 m/s, 2.3  10

34

J, 1.1  10

V(x,y,z) 

2

kg m /s,

2 R2  zb h

Q 1 ≥ 2 4p0 2x  y2(h  z)2

20

6.6  10 Hz

Chapter 25



B

1. 2.4  106 J 3. 60,000 V



6

5. 6.9  10 m/s 7. a) 2.1  106 m/s; b) 1.45  106 m/s

(x  y  z  2R2zh  R4 h2)12

11. 9.2  102 V



13. 2.05  105 m/s

2

2

hr(R  2R zh  R4 h2)12 2

17. 3.86  1012 J 19. 5.8  1012 J Q 2 1 21. a1   b ap0 3 25



2

d

d

Q 1  c 2 4p0 (R  2z h  h2)12 1

(h  2z h  R2)12 2

lu 23. 4p0

¥

Q 1 c  4p0 (R2  2zh  h2)12 1

15. 7.36  108 C

d 0

Therefore, potential constant is at 0 on entire surface. 39. (a) r  R2, Va  0 (b) R1  r  R2, Vb  0 Q Q (c) R  r  R1, Vc   4p0R1 4p0r

25. a) 5.5  1012 J; b) 4.0  107 m/s 27. 23 V Q 1 2  23 ln a b l 4p0 2  23 2

(d) R0  r  R, Vd 

2

Q 1 l  2l  (x  l) lx b  ln a c ln a b x l 4p0 lx

x  l  2l  (x  l ) l  2x  l b  ln a bd x x  2x2  l 2 1 l a2 ln (ba) 33. c  d 2p0 b2  a2 2 35. 1.16  104 V/m for the wire and the same for the cylinder. 4k 37. V(r)  Volt 0 1r

47. Ex  

2

2

2

2

Q Q  4p0R1 4p0r

3Q 3Q 3Q Q Q ; ;   ; 4p0r 4p0c 4p0c 4p0r 4p0b

41.

 ln a

2 R2  zb h

1

9. 2.7  10 m/s

31.

xy a

Q 1 R c  4p0 (x2  y2  z2  2zh  h2)12 h 2

7

29.

1

R h

Q 1 1 3 r2 1 a    3 b a 4p0 c b 2a a 2py 2p 2px 2pz sin cos ; cos a a c b 2py 2p 2px 2pz Ey   cos sin ; cos a c b b

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ANSWERS

2py 2p 2px 2pz cos cos sin c a c b p 3xz 49. Ex  ; 4p0 (x2  y2  z2)52 Ez  

Ey 

3yz

p

4p0 (x2  y2  z2)52

89. V 

;

3z2

c

b) 

d; 4p0 (x2  y2  z2)52 (x2  y2  z2)32 p On the z axis: Ex  Ey  0, Ez  ; On the x axis: 2p0z3 p Ex  Ey  0, Ez   2p0x3 p

Ez 



1

Q 1 x  l2  2R2  (x  l2)2 c ln a b d; l 4p0 x  l2  2(x  l2)2  R2 Q b)  4p0

51. a)

4l 2l 2  4lx  4(R2  x2)Ql  2x  2l 2  4lx  4(R2  x2) R 18

53. 8.1  10

4

55. 6.3  10

2

57. 3  10

J

J Jm ; b) 2.3  10 J 3

59. a) 4.4  10

11

3

61. 1.8  10 J/m

65. 5.8  106 eV 67. 50 eV 8p0

a

7 b 10R

73. 8.6  105 eV 75. b) Ugrav  Ugrav Urest mass

3 GM 2  1.24  1029 J 5 R



(35)(GM2 R) 2

Mc



3 GM  1.9  1011 5 Rc2

77. 4.1  107 J. 79. a)

4p0l 2

91. Utotal 

2

2R  x

c xln



(x  2l )x (x  l )2

x 2

R  x2 B 4

 2l ln

≤i

(x  2l ) d (x  l )

Q2

6 9 a b6  3a3b3  a5b b 5 8p0b(b3  a3)2 5

Chapter 26 1. Q2  3Q1, V1  3V2 3. C  1.1  1011 F, Q  1.1  106 C 5. C  1.1  109 F, Q  1.3  108 C 7. n  4.5  1014 electrons 9. Amin  5.6  104 m2, Amax  6.8  103 m2 11. Q  125 C 13. C  1.8  1017 F, Q  1.8  1018 C

19. Q  1.7  103 C 21. Cnet 

2 C 3

25. Q  1.1  105 C, ¢V2.5mF  4.5 V, ¢V5.0mF  2.3 V 27. The only arrangement of capacitors to give the same net capacitance is two pairs of two capacitors in series, connected in parallel. 29. Q1  3.0  104 C, Q2  4.1  104 C, Q3  4.8  104 C, ¢VPP  68 V 31. C  4p0kR 33. k  5100 35. k  1.7 e0 A 37. C  (k1  k2) a b 2d 39. a) Efree 

l2

l2 1 b b) ; a  ln b 2 2 a 4p0 4 8p 0r

81. 9.96  102 m; 2.5  105 V 83. 0 V 85. 3.1  107 m/s R2 87. a) a 2R  x   x2 b ; B 4 3p0R2 2Q

Q2

x 2

23. Q  12.0 mC, ¢V  3.5 V

Q2 63. (4  22) 4p0d

Q2

3p0R

±

17. C  9.9 mF

8

69.

2

15. C  8.0  1012 F/m

J

9

2Q

A-55

2

2

2¢V k 1 2¢V a b , b) E  a b, d 1k d 1k

c) sbound 

20 ¢V d

a

1 b 1k

41. a) C0  2.0  1011 F, b) C  2.9  1011 F 43. Q  9.0  106 C, ¢Vfilled  1.8V, ¢Vempty  4.5 V 45. a) sbound,inside  2.6  106 C/m2, sbound,outside  1.7  106 C/m2,

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A-56

ANSWERS

b) Einner  1.6  105 V/m, Eouter  1.1  105 V/m, c) Efree  3.0  105 V/m 47. F  0.050 N 1 1 1 49. C  2p0 a  b (ktop  kbottom) R1 R2 51. a) C  1.6  1010 F, b) ¢V  380 V, c) E  7.5  104 V/m, d) u  0.025 J/m3, e) U  1.1  105 J 53. Q  0.2 C, U  2000 J 55. ¢V  2000 V 57. Qsuper  17 C, Qsupply  0.33 C (52 times more charge in the supercapacitor), Usuper  21 J, Usupply  66 J (3 times more energy in the supply capacitor) 59. a) C  7.1  1010 F, Q  8.5  109 C, ¢V  12V, U  5.1  108 J, b) C  2.1  109 F, Q  2.5  108 C, ¢V  12 V, ¢U  1.0  107 J, c) C  7.1  1010 F, Q  2.5  108 C, ¢V  36 V, d) C  2.1  109 F, Q  8.5  109 C, ¢V  4.0 V 61. The solution is a proof. 63. Q1  3.5  104 C, Q2  1.5  104 C, Q3  2.0  104 C, U1  3.1  102 J, U2  1.9  103 J, U3  2.5  103 J 65. F  11 N 67. Q1  Q2  9.6  104 C, Q3  1.2  103 C, U1  0.115 J, U2  0.077 J, U3  0.24 J 69. a) Cnet  8.0  106 F, b) Cnet  8.0  107 F 71. k  2.0 73. a) U  13 J, b) u  4.8  104 J/m3, Volume  2.7  104 m3 75. C2  8.0 mF, U1  7.2  105 J, U2  3.6  105 J ¢ c b k2ln a b  k1ln a b a c 2 79. ¢V  4.7  10 V Q2 Q2 Q2 81 . a ) F  ¢d, , b) W   ¢d, c ) ¢U  2A0 2A0 20A 77.

C  2p0 ° l

k1k2

d) Compare the answers from parts a) and c).

Chapter 27 1. Q  1800 C/h, n  1.1  1022 electrons/h 3. t  1.2  104 s 5. I  4.0 A, E  6.0 V/m 7. I  7.5 A, Q  15 C 9. I(0.0s)  1.0 A, I(1.0s)  0.25 A, Q(2.0s)  0.67 C 11. t  3.8  1014 s, vd  0.054 m/s

13. R  3.0  15. I  1.3 A, n  8.2  1018 electrons/s 17. Rnet  R>32 19. R  0.40  21. E  0.069 V/m 23. R  0.87  25. r  5.7  107  m 27. ¢R  0.92  29. j  5.7  106 A/m2, E  0.097 V/m 31. s  5.9  107( m)1 33. T  22C 35. ¢V  9.9 V 37. mCu  380 kg, mAl  190 kg 39. d  0.16 cm 41. I  8.0  105 A 43. R  4.4  1010 , I  6.8  109 A 45. Req  2.2 , I  5.4 A, I1  2.4 A, I2  1.7 A, I3  1.3 A 47. Iiron  1.7 A, Ibrass  4.3 A 49. In series: Req  9 , in parallel: Req  0.92 , in combinations of series and parallel: Req  3.7 , 4.3 , 5.2 , 1.6 , 2.0 , 2.2  51. I  27 A 53. a) ICu  210 A, b) Irubber  2.7  1019 A 55. d  1.5 km from point A 57. R  3.2  103  59. R  5.4  102 /m 61. a) Rnet  4.4 , b) I  1.8 A, c) I1  1.8 A, I2  1.1 A, I3  0.7 A, ¢V1  3.6 V, ¢V2  ¢V3  4.4 V 63. ¢V  36 V 65. R  2.73  67. Q  1.4  105 C, n  9.0  1023 electrons 69. E  2300 V/m ¢R  0.089 or 8.9% 71. R0 73. R  5.2  10 4 , ¢V  0.31 V 75. Connecting two sets of two resistors in parallel gives Rnet  1.0  77. I  0.30 A 79. j  3.3  106 A/m2, vd  2.4  104 m/s 81. a) Req  1.3  106 , b) I1  9.2 mA, harmless, I2  88 mA, harmless, I3  0.18 A, fatal

Chapter 28 4 1. smallest battery: 3.3  103 kW h kg  1.2  10 kg ; J

J

4 largest battery: 1.9  102 kW h kg  6.7  10 kg

3. 6.9  106 J

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ANSWERS

5. 1.3  105 J 7. 0.38 A, 4.0  103 J 9. 6.0 A, I2  I1  3.0 A 11. current through R1 does not change, but it increases through the other two resistors 13. (a) 2.4  104 A; (b) 4.8  106 V 15. 62 , 115 V, 4.8 V 17. (a) 1.7 A, (b) 0.86 A through resistors 2, 3, and 4 19. 0.49 V 21. 0.11 A 23. 31.0 A, out of the junction 25. 8.0  27. I1  15 A, I2  25 A, I3  5 A, I4  15 A, I5  20 A, R5  4.0  29. I1  1.25 A, I2  1.75 A, I3  0.50 A 31. 16 V 33. The current through R1 is 2.58 A, through R2 is 1.71 A, through R3 is 1.28 A, through R4 is 2.15 A, through R5 is 0.85 A. V  2.6 V 35. 6.0  104 W 37. $0.36 39. 3.7  103 A 41. 1100 W 43. 4.9 A 45. 6.3  1012 protons/s, 7.0  102 W 47. 6.7 h 49. 58 h 51. 1.2  106 W 53. 10% 55. R115  13.23 , R230  52.90  57. 0.50 W 59. The solution is a proof. 61. (a) 3.3%, (b) 33%, lower current is more efficient 63. (a) 180 V, (b) 1.8  106 W 65. 19 liters/min, 67. 7.0  102  69. 3.958 A 71. RCu  5.0 , Rconstantan  5.6  73. 3.2  104 s 75. 4.0  104 s 77. 0.011 s 79. Q  Ce a 1  et>(R1 R2)C b , ER1

(et>(R1 R2)C) (R1  R2) 81. (a) A  E2C, B  C(E1  E2), t  R2C, Q(t)  E2C  (E1  E2)Cet/R2C ¢V 

(b) A  E1C, B  C (E2  E1), t  R1CI, Q(t)  E1C  (E2  E1)Cet/R1C 83. 0.020 s

A-57

85. 5.3 s, 5.0  106 A 87. (a) 1380 , (b) 430 A, (c) 2670 A 89. 3.0  101 W, 1.7  101 W ER i E ,I  91. I1  , Ri R  RR  RRi 2 R R i  R a  1b Ri 93. I1  0.45 A, I2  1.31 A, I3  0.85 A, P1  5.4 W, P2  14 W 95. (a) 0.024 m, (b) 3.4  106 W 97. (a) 47 A, 7.1  103 m 99. 2.2% 101. (a) 7.6  105 C, (b) 1.2  104 C, (c) 6.0  102 A

Chapter 29 1. F  1.07  1016 N, opposite the direction of the current 3. F   1.04  1016 N, a  1.14  1014 m/s2 5. F   1.38  1024 N 7. u  19.5 9. B  4.18 T, pointing downward 11. B  1.44  105 T, Bearth  4.2 B 13. F  8.2  1016 N 15. F  (2.98  1018 i  6.8  1019 j  3.33  10 18 k) N 17. F  9.98  1018 N and points in direction 16 above the horizontal in North direction. 19. a) I  lv, m0lv m0v m0I b) B    2p0Er  0m0vE 2pr 2pr 2pr 21. B  0.19 T 23. B  4  108 T, F  1.92  1018 N 25. B  5.03  103 T m0I 27. B ds  R



29. a  1.5  1013 m/s2 31. a) B  5  106 T, b) u  16 m0 I 3 33. B  22 2p d 2 m0I m0I 35. B   , pointing into the page. 2R 2pR m0 Ir m0 I 37. r  r1 : B  , r  r  r3 : , r  r  r2 : B  2 1 2pr 2 2pr1 B

m0I r23  r2

, r  r3 : B  0. 2pr r23  r22 39. For y  0, z  0, q  x  q : B  0, for y  0, z  0, q  x  q : B  m0s i, for y  0, z  0, q  x  q :B  0, for y  0, z  0, q  x  q : B  m0s i

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A-58

ANSWERS

41. B 

m0 I 1 b tan a b p b 2z

43. For z  2R, B1  B2 B3 B4 B5 B6

m0IR

, for R  z  2R, p(z2  R2) m0I z  R 1  a  b , for 0  z  R, 2 2p z  R R m0I R  z 1  a  b , for R  z  0, 2 2p z  R R m0I 1 Rz  a  b , for 2R  z  R, 2p R  z R2 m0I 1 zR  a  b , for z  2R, 2p R  z R2 m 0 IR m0I  , Bmax  B3(z  0)  2 2 pR p(z  R )

45. I  26.5 A 47. B  1.26  102 T 49. Bmax  3.34 T, Bmin  1.43 T 51. B  m0nn (r2  r1)I for r  r1; B  m0nn (r2  r)I for r1  r  r2 4m0I 53. B  22pL 55. B  7.9  105 T, directed into the paper m0I m0I 57. B   , directed into the paper 2pR 8R m0I 59. B  0.11 , directed into the paper L 8m0I 61. a) BP  , 2125R b) At any point z, the fields produced by the coils are m0I R2 Bbottom  2 (z2  R2)3>2 Btop 

R2

m0I

2 [(R  z)2  R2]3>2 Their first derivatives are 3m0I dBbottom zR2  dz 2 (z2  R2)5>2 dBtop



3m0I

(R  z)R2

dz 2 [(R  z)2  R2]5>2 Both of these derivatives cancel each other at z  R>2. The second derivatives are 3m0IR2(R2  4z2) d 2Bbottom  dz2 2(z2  R2)7>2 d 2Btop dz2



3m0IR2 SR2 4(R  z)2T 2 S(R  z)2  R2T 72

Both of these derivatives cancel each other at z  R>2.

63. F  4.8  1017 N, a  2.87  1010 m/s2, both directed opposite the current. 65. For currents in the same direction, B  7.33  105 T, for currents in the oposite direction, B  1.05  105 T. m0I 67. B  1.05 , the angle between B and the straight wire R is u  17.7 69. For r  R1, B  2m0nI; for R1  r  R2, B  m0NI ; for r  R2, B  0. m0nI , directed in a plane parallel to the plane of the 71. B  2p wires and perpendicular to the current, with the exception along the edges. 73. B 

m0I25

, directed into the paper. 2pL m0I 1 3 a  b , directed out of the paper. 75. B  p 2R 8

Chapter 30 1. p  1.1  1017 kg m/s 3. p  3.4  1017 kg m/s 5. B  3.3 T 7. p  3.8  1019 kg m/s 9. B  0.036 T r12 11. 1 r14 13. I  0.39 A 15. (a) The electron will follow a circular path that spirals in the direction of the magnetic field; (b) f  1.4  107 Hz, T  7.2  108 s; (c) x  0.29 m. 17. B  3.2 T 19. (a) vp  1.6  103 m/s; (b) up  3.3  103 rad 21. F  6.7  105 N 23. Consider an element dl of the long, straight wire. As the current I flows through the loop, the magnetic field B produced by the loop exerts a magnetic force F on the element d l given by d F  I d l ? B. This force is perpendicular to element d l. Applying Newton’s third law of motion, the element d l exerts a force that is directed oppositely to d F, which is also perpendicular to element d l. 25. F  13 N j 27. I  1.0  106 A 29. t  2.4  107 N m 31. (a) t  (1.1  1026 sin u) N m, sinusoidal behavior; (b) W  2.25  1026 J

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ANSWERS

33. m  0.023 A m2 35. m  14QR2 The magnetic and electric fields surrounding the spinning paper disk are illustrated as follows:

37. (a) t  (1.71  107 sin u) N m; (b) f  0.47 Hz m 39. (a) M  1.7  106 A/m; (b)  2.0  1023 A m2 V 41. (a) x  1; (b) m  0 43. ¢VH  3.1 mV 45. B  2.5  102 T 47. (a) vd  2.7  103 m/s; (b) ¢VH  1.6  105 V 49. RH  4.7  1010 m3/C 51. (a) F  0 N; (b) F  0.18 N due north; (c) F  0.18 N due east; (b) F  0.16 N due east. 53. B  0.010 T, f  1.6  105 Hz L 2 F eEL2 ba b  , m v0x 2mv2 2yE e eEL2 , therefore  2 2 (b) y  m 2mv BL 4mv 57. B  qd 59. a  3.1  103 rad/s2 55. (a) y  12ayt2  12 a

Chapter 31 1. 3.0  103 V; left side is positive and right side is negative 3. 0.5 m/s 5. 2.2  103 V 7. v  0.48 m/s, current flows counterclockwise, F  10 N to the left 9. 9.8  103 V 11. 5.5  105 T m2 13. 22 rev/sec 15. 9.7  103 V; patient does not need to be pushed more slowly 17. E  6.88 V, I  0.606 A, W  146 J 19. 0.047 V 21. 3.5  107 T m2 23. (a) 6.6  106 V/m; (b) 7.4  106 V/m 25. I  0.565 A, t  5.59  103 N m 27. 3.0  102 A

A-59

29. 0.63 A 31. At r  0.80 m: E0  140 V/m e0  704 V; at r  1.5 m: E0  117 V/m e0  1100 V 33. 0.010 C 35. 0.40 H 37. (a) 0.5 H; (b) 10 V 39. 1.1  104 A/s 41. 1.9 V 43. M  200m0npR2; the shape of the coil wire does not matter 45. 1.1  107 H 49. 1.2  105 J 51. 1.0 J/m3 53. B-field (T) u (J/m3) 108 4.0  1021 103 4.0  1011 45 8.1  108 8 2.5  107 2 1.6  106 1.5 9.0  105 55. U  5.29  104 J; V  1.64  103 m3 57. 211 J/km 59. 1.0  103 J dI E 61. IR  L  0; I  et>t satisfies the differential dt R equation if  LR 63. (a) 3.75 ; (b) 1.58  104 W; (c) 9.51  104 J 65. (a) 2.0 A; (b) 3.0  103 V L L 67. (a) ; (b) R2 R1  R2 dI2 dI1 69. (a)  0.75 A/s,  1.5 A/s; (b) IResistor  0.50 A, dt dt IL1  0.17 A, IL2  0.33 A 71. R  0.125 , L  0.0396 H 73. 2.4  104 V 75. 0.014 J/m3 77. 7.5  105 V 79. (a) 2.6  103 V; (b) 0; (c) 9.2  1015 C/m2 81. (a) 0.0188 T/s; (b) 6.4  103 V; (c) 6.4  103 V 83. 102 V 85. (a) Imax  48 A, t  14 s; (b) Umax  576 J, percentage of energy that remains  25%

Chapter 32 1. (a) Irms  10.4 A, Imax  14.8 A; (b) Pmax  2400 W, Pmin  0 W

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A-60

ANSWERS

3. (a) Emax  3.25  105 V, Imax  1.05  103 A; (b) Pmax  3.4  108 W, Pave  1.7  108 W 5. Irms  10.4 A, Imax  14.8 A, R  11.1  7. Irms  31.1 A, Pave  48.5 kW 9. 0.032 A, 0.064 A 11. 2.0  106 , 1.5  106  13. (a) 208 ; (b) 9.6  104 A; (c) I(t  0)  0 A, I(t  4p>)  6.8  104 A 15. 1.0  109 F 17. For an amplitude of 1.00 V: t (s) 0.0012cos2(600t) (W) 0.001 8.17E-04 0.002 1.58E-04 0.003 6.19E-05 0.004 6.52E-04 5 19. 5.6  10 F 21. 1.0  105 A/s 23. f  1.6  103 Hz 25. 0.033 A, 0.017 A 27. 0.049 A 29. 0.343 A/s, 0.00218 Hz 31. 0.018  at 60 Hz, 3.0  104  at 100 MHz 33. 3.6  103 Hz 35. (a) Imax  5.0  106 A, (b) C  1.66  106 F, L  6.6  102 H 37. 9.4  103 rad/s 39. 379 Hz 41. 663 Hz 43. 2.11 Hz, 259 H 45. 2.76  108 F, 7.5  108 H 47. 8.23  106 F, 6.25 W 49. (a) P 

1 R Imax Emax a b 2 Z

(b) From the impedance triangle, cos f  (c) cos f  0.642, f  ;50. No.

R Z

51. (a) 1.9  103 V; (b) 12 V; (c) 24 W; (d) 1.6  102 R 53. 0.995Emax; Emax>10;   L 55. IC Imax  2IR2  (IC  IL)2 IR

IL

Emax Emax 2  a b a  b B R XC XL Emax

57. 4348 turns 59. Transformer 1: Transformer 3:

N1 N2 N1 N2

 0.044, Transformer 2:  16.5, Transformer 4:

N1 N2

N1 N2

 7.6,

 34.8

61. Igenerator  9.09  104 A, Iline  5.0  103 A 63. 36 W 65. 0.139 A 67. Imax  1.63 A. One-quarter cycle after the maximum I  0. One-half cycle after maximum I  1.63 A, three-quarters cycle after maximum I  0. 69. (a) I  (2.4  0.43cos 360t)A; (b) 29.3 W 71. (a) Q  6.5  108 sin (120 pt) F; (b) First max. at t  1>240 sec, first min. at t  0; (c) 5.9  108 J, 2.95  108 J 73. Imax  0.89 A. One quarter cycle later, I  0. One half cycle later, I  0.89 A. Three-quarters of a cycle later, I  0. 75. (a) IR(t)  1.5  103 A  (7.5  104 A) cos (6000pt), IL(t)  60t  (1.6  103) sin (6000pt) A. (b) PR  4.5  103  1.1  103cos 2(6000pt) W, PL  180t  90t cos (6000pt)  (4.8  103) sin (6000pt)  (2.4  103 sin (6000pt cos (6000pt) W 77. 1.4  103 J; 7.9  104 s for fully magnetic energy; 1.6  103 s for fully electric energy. 79. (a) 1.36  103 Hz, (b) 1.2  103 V, (c) 78 81. 0.026 A at the power plant, 0.0012 A in the transmission line 83. E2  15 sin (3000pt) V

Chapter 33 1. (c) Idisp  e0

d£ E dt



dQ dt

3. (a) 4.0 A; (b) 4.5  1011 V m/s 5. (a) 0.10 A; (b) 6.3  103 A 9. (a) 2.0  103 A; (b) 1.0 sec; (c) 8.9  104 A, 8.9  109 T, 1.8  108 T, 2.7  108 T 11. (a) E 

2

13.

rCt 2

pR

; (b) B 

m00rCr 2

2pR

; (c) B 

m00rC 2pr

 E # dA  k ,  B # dA  0,   E # ds   dt , d£ B

Q

0

 B # ds  m a I  ke 0

0

d£ E dt

b

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ANSWERS

15. E is  to acceleration. B is  to acceleration and propagation. E

B a

E

B

17.

1

2

3

19. west

55. 2.8  1028 W 57. 4.0  1026 W 59. 6.42  105 W/m2, 2.02  104 W 61. 0.065 V/m, 0.043 V/m 63. 6.37  103 W/m2, 2.19  103 V/m, 1.67  1010 J 65. 1.49 W/m2, 7.81  1011 N 67. 8.0  103 m2 69. 51 71. 6.0  108 N, 3.5  1022 N 73. (a) 1.77  1017 J/m3 each; (b) 4.0  103 V/m, and 0 T, or 0 V/m and 1.33  1011 T; (c) one is 7.08  1017 J/m3, the other is zero. 75. 80.0 m, 0.0785 m1, 2.36  107 s1, minus z direction, Ey  cB0 cos (kx  t) V0 sin t A A  e0 V0 cos t, 79. V0 sin t/R, e0 V0 cos t, d R d e0pr m0 V0 sin t a  V cos t b 2p rR d 0 81. 0.067 sec 83. 0.31 W/m2 85. 5.00  107 T north 87. 9.60  103 V/m, 3.2  105 T 89. 5.5  102 V/m, 1.8  1010 T 91. 2.9  1020 W, 3.6  1027 W 93. (a) 3.4  1015 W; (b) 5.9  1017 m  63 ly, 3600 stars

21. 1.1  105 A/m 23. 3.3  1011 s 25. 5.00  103 s (8.33 min) 27. 3.3  109 s, 0.30 GHz, no 29. North, 2.0  109 T 31. 1/4 33. 1.5% 35. 39 37. (b) u  0, u  90, u  180, u  270; The field rotates clockwise with a period of 2p/ 39.

I0 cos 2 f 2(sin 2f cos 2   cos 2 f)

41. 3.0  1011 Hz 43. AM: 566 m to 187 m; FM: 3.41 m to 2.78 m 45. (a) 1.5  1010 m (X-ray); (b) 1.0 cm (microwave radio); (c) 5.0  106 m (“electric” wave) 47. 0.027 m 49. 2.8  108 J/m3 51. 9.5  106 W/m2 53. 1.7  109 T down; 6.6  104 W/m2 north

A-61

Chapter 34 1. 40 3. 90  u 5. 20 7. 7 9. H  0.90 m; W  0.31 m 11. 4.74  1014 Hz, 416 nm, 1.97  108 m/s 13. 1.09 15. 50 17. 39.014 19. Fused Quartz 21. 41.7 23. 42 25. 24.4 27. 55.6, 56.1, 57.7 29. 64.5 31. 41.1, 61.0 33. 1.000 21 35. 36.9 37. 0.77 mm

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A-62

ANSWERS

39. 53.1

111.

41. 1 n 22 43. 48.8, no dependence on n 45. 1.4002, 40.25%. 47. 13729 , 13914 49. 10.9 cm

lens

51. s  12R, s  12R 53. 2.5 cm 55. 16.7 cm 57. 120.0 cm, concave 59. 60.0 cm, 30.0 cm 61. 8.3 mm 63. 21 cm 65. 12 cm 67. 8.6 cm, virtual, upright, smaller, 0.57 69. 14 cm, 14 71. 45 cm, inverted, enlarged 73. The solution is a proof. 75. 60 cm, 21 cm 77. The solution is a proof. 79. 475 cm behind the lens (also behind the mirror) 81. The solution is a proof. 83. 3.9 cm to the right of the center of the ball 85. 25.8 cm to infinity

convex

object distance

image characteristics

concave

sf

virtual, upright, and magnified real, inverted, and magnified real, inverted, and reduced

convex

109. 1.7 m, 0.24 m

all

sf

virtual, upright, and reduced virtual, upright, and reduced

sf

mirror

s  2f

image characteristics

concave

87. 8.9  102 s 89. 22 cm, diverging lens, 0.72 cm 91. The solution is a proof. 93. 1/64 95. 1340 97. The solution is a proof. 99. 31, 50 101. 56 103. 7.5 cm 105.   38.7 107.

f  s  2f

object distance

virtual, upright, and reduced

sf f  s  2f s  2f

virtual, upright, and magnified real, inverted, magnified real, inverted, reduced

113. 30 cm 115. 1.9 mm, 108

Chapter 35 1. 3.2  108 m/s 3. 127 nm for both 5. 244 m 7. 78 nm l 2l 3l 9. 2d  , , , ; yes n2 n2 n2 1 3 5 11. 2d  l, l, l, . . . 2 2 2 13. (a) Only one reflected ray suffers a phase reversal; (c) 1.22 mm 15. 0.257 mm 17. 5.76 mm/s 19. 1.000277 21. 7.4 mm 23. 1.9 mm 25. 0.994 27. measured angles: 5.5, 20.5, 35.5, 51.5; predicted angles: 6.1, 18.6, 32.2, 48.2 29. 0, ; 0.0074, ; 0.0149, ; 0.0223, . . . 31.  (degrees)  (radians) IImax 1.0 0.0175 0.626 2.0 0.0349 0.064 3.0 0.0524 0.154 7 33. 5.4  10 m 39. 19.0, 40.5, 77.2 41. First order: u400  13.7, u700  24.4; Second order: u400  28.2, u700  55.7 Third order: u400  45.1 ; no third order max for 700 nm Second and third orders overlap 43. 1.38  105 rad 45. 0.34 mm

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ANSWERS

75. (a) 9.6  105 radian; (b) tremor is 9.7  105 radian to 1.5  104 radian, eliminating tremor would improve resolution somewhat 77. (a) 2.24  106 radian; (b) 0.34 m; (c) 0.0054 mm 79. cat barely distinguishes the mice 81. film is at least 120 nm thick 83. 454 nm 85. ;14.5, ;48.6 87. (b) 95.9 MHz 89. 13 km 91. 4.2 cm, 2.1 cm 93. 1.63  103 radian; 5.7 km 95. (a) 1.95 km; (b) 5.02  104 W/m2

47.

I/Imax

1.0

0.5

0.0 0.0

δ=0

1.0

2.0 Delta (radians)

3.0

2E0

E0

E0

E = 4E 0

Chapter 36

E0

1. (a) 0, 60 km/s; (b) 30 km/s, 90 km/s

E = 3.41E0 2E 0 δ=

3. ¢t1 

L2 2 V 2 c (L1  L2)  a L1  b a b d; c c 2

¢t2 

L1 2 V 2 c (L2  L1)  a L2  b a b d; c c 2

π 4

E0 E0

¢t1  ¢t2  E = 0.59E 2E0

E = 2E0

δ=

π 2

E0

δ=

E0

3π 4

2E 0 E E0

E0

E=0

49. (b) f  0 ; (c) u  f 51. 2.5 cm 53. 0.042 mm 55. 47 57. (a) not resolved; (b) resolved; (c) colors smeared, not resolved 59. 0.045Imax 61. (b) m interference maxima between adjacent diffraction minima 63. 0.952(3p) 65. about 0.22 mm in diameter 67. 94 69. about 0.06 arcsec, about eight times better than from Earth 71. (a) 6.7  105 rad; (b) about 1 cm in diameter 73. minimum separation  6  107 m; it is possible to resolve the moons but they can only be seen as points of light

2(L1  L2) 3 V 2 c2  a b d c 2 c

5. 0.995c 7. 0.866c (2.60  108 m/s) 9. 0.16 s. 11. 0.447c 13. 4.8  103c (1.4  106 m/s)

2E0 E0 δ=π

A-63

15.

f f

 0.58; 0.42; 0.23

17. 366 m/s; time dilation factor  1  7.4  1013 19. time dilation factor  1  1.17  1012; the clock at the North Pole will be ahead by 3.7  105 s in one year. 21. 0.866 m 23. 0.19 m 25. 0.866c 27. 0.44c 29. 3.3  106 C/m3 31. Relative to the upper part of the belt, the lower belt segment is moving with a speed greater than V and appears shortened even more than the base. Therefore the belt is tightened in both reference frames. 35. 0.65c 37. (a) x  7.67  1016 m, y  1.2  1017 m, t  1.56  108 s; (b) The light from the nova will arrive at earth before the radio message.

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A-64

ANSWERS

45. Krel  5.04  103mc2, KN  5.00  103mc2, deviation  0.8% 47. 0.85c 49. 0.94c 53. 2.7 m/s; 0.016 m/s; 1.3  104 m/s. 55. 1.06  1017 kg m/s 61. 1.02  106 eV mthermal 63.  1.1  104 % msun

65. 1.26  1013 J, 1.5 times the rest mass energy of the electron. 67. 0.828c 69. 9.33 light years from earth, message arrives 21 years after departure 71. (a) To the ship observer, the 600 nm pulse is emitted 2.60 ms before the 400 nm pulse; (b) 133 nm instead of 400 nm, 1800 nm instead of 600 nm; (c) 1.2 km 73. Dimensions will be 1.0 m  1.0 m  0.8 m. The area of the two faces perpendicular to the direction of motion will be 1.0 m2. The four faces whose planes are parallel to the direction of motion will have area  0.80 m2. The volume of the cube will be 0.80 m3. 75. The deviation is 2%. 77. 0.98c if forward, 0.54c if backward 79. 6.96  1020 J; 7.74 metric tons 81. 1.2  1017 J; 2.9  107 tons of TNT 83. 0.33 m/s 85. (a) 0.906c; (b) vmuon  0.98c, vantimuon  0.67c

Chapter 37 1. 3.3  1034 J 3. 5.3  1022 J 5. 0.0259 nm; 0.0366 nm 9. 1.60  106 m 11. 5.8  1013 m 13. 43 W; 0.2 W; liquid nitrogen 15. 6.1  109 K 17. 1.38  103 W>m2 21. 44 K 23. a) 7.16  1015 Hz; b) 0.07 nm 25. 7  1024 photons 27. 1.9  1031 photons/s 29. 3.2  1018 photons/s 31. 1.77 eV; 3.10 eV 33. 0.29 eV; 0.60 V 35. 1.86 eV 37. Red light (700 nm): None; Blue light (400 nm): K; UV light (280 nm): K, Cr, Zn

39. 6.8  1034 J s 41. 2.76  1015 J; 9.21  1024 kg m/s 43. 0.00243 nm 45. 0.0312 nm; 0.0336 nm 47. 0.37%; 3.3%; 49% 49. 0.016 nm 51. 62.5 eV 53. 4.0  1015 J 55. 1.24 kV 57. 8.042 kV 59. 1.3  1028 kg m/s 61. 0.031 nm 63. 0.872 Hz; 1.96  1016 m 65. a) 1.07 mm, microwave; b) 1.54  109 W 67. 2  1020 photons/s 69. 0.0486 nm; 2.12  107 m/s 71. 1.7  108 V/m 73. 7.8 eV 75. 0.062 nm and 0.070 nm; only 0.070 nm will be present

Chapter 38 1. 121.568 nm, 102.573 nm, 97.254 nm, 94.975 nm 3. 4.34 m/s 5. a) 410.3 nm, 434.2 nm; 1.0034; 0.0023; b) 1.0  106 m/s 7. a) d  0.0013; p  0.00407; b) “Principal” is analogous to Lyman series, “Diffuse” is analogous to Balmer series. 9. 5.16  1012 J 11. 3.14  1026 m2; 103 13. a) 1.2  1018 atoms ; b) 1.2  104; c) 1.2  106 15. 4.3  1032 § 17. 2.2  106 m/s 19. 0.0283 mm; 2.5  105 eV 21. 10.2 eV 23. From the n  3 to the n  1 state. 25. 7.2  103 27. n  3, the first two lines of the Lyman series and the first line of the Balmer series. 29. E  f

nUe 24pe0me R3 (n2  n1) 2p

a



e2 3 a b; 4pe0 2R e

24pe0meR3

b ; R  4a0

33. 4.98  106 eV 35. 2pa0; 4pa0 37. 0.00614 nm; 1.1  105 39. If l2>l1  4, then E2>E1  1>16

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ANSWERS

41. 5.8 mm 43. 2.7  1020 m/s 2 2 45. , 0, L L 47. n  1 locations where the probability per unit length  0, mL at x  , m  0,1,2, p n; n locations where the proban (2m  1)L bility per unit length  1, at x  , m  1,2, p n 2n 49. 0.091 51. 0.0014 nm n2h2 2L nh 55. a) , n  1,2,3, p ; b) ; c) ; d) 37.7 eV, 151 eV, n 2L 8mL2 339 eV, 603 eV 57. 91.176 nm; 820.58 nm; 1458.8 nm 59. 122 nm 61. 0.212 nm; 1.10  106 m/s; 2.12  1034 J s; 5.71  1021 m/s2 63. 0.0454 65. 1.67  1014 m; 4.87  106 m/s 67. 12.8 eV; 97.3 nm; 0.661 eV; 1.88  103 nm (GM)2m3 n2 U 2 69. rn  ; ; n  323; 5.9 E   n Gm2M 2n2 U 2  1017eV 71. 5.3  1023 kg m/s; 1  1021 m/s

Chapter 39 1. n  1, l  0 only, for n  2, l  0 or 1, and for n  3, l  0, 1, or 2 3. 4U,3U,2UU, 0, U, 2 U, 3 U, or 4U 5 3eU 5. U , 2 2me 7. (a) 18 states, (b) argon 9. U and ms  1, 0,  1 11. 30 13. ;msgmB, where g is approximately equal to 2 for the spin moment and equal to 1 for the orbital moment. 15. mp  2.79mN, mC  0.703 mN

21. n

l

m

ms

Number of States

1

0

0

12

2

0

0

12

2

2

1

1

12

2

1

0

12

2

1

1

12

2

23. K shell: n  1, l  0, m  0, ms  3>2, 3/2, 1>2, and 1/2 (4 states) L shell: n  2, l  0, m  0, ms  3>2, 3/2,1>2, and 1/2 (4 states) and l  1, m  0 and 1, ms  3>2, 3/2, 1>2, and 1/2 (12 states) In comparing the hypothetical periodic table with the known periodic table, we see that the two additional spin states cause significant changes. For example, the completed shells would no longer be the present noble gases, but would be Be, Mg, Ca, Sr, … Instead of two elements with n  1 ground states, there would be four: H, He, Li, Be. Instead of Li through Ne (8 elements) having n  2, there are now 16 elements, C through Ca. 25. 9.48  103 eV, 1.31  1010 m 27. 30 (zinc), 60 (neodymium) 29. 0.128 nm 31. 74 (tungsten) 33. 9.27  1013 Hz 35. (a) 5.43  1046 kg # m2, (b) 1.3  104 eV, 3.9  104 eV, 7.8  104 eV 37. 1.60  1014 Hz 39. (a) 37.9 eV, (b) 0.0150 eV 41. 9.26  1047 kg # m2, 0.089 nm 43. 1.1  106 m 45. Because of the steep slope of the curve, a small change in voltage across the emitter and base produces a large increase in voltage crossing the emitter-base junction. This increased current appears in the current IC leaving the collector. 47. +

V1

V2

17. L2x  L2y  l (l  1)U 2  m2 U 2and l(l  1)  m2 Lx  Ly  U B 2 19. In its ground state, two electrons occupy the n  1 level with l  0, m  0, and ms  ;1>2, two electrons occupy the n  2 level with l  0, m  0, and ms  ;1>2, and the remaining electron has n  2, l  1, m  0, and ms  1>2.

A-65

+

+

E

R1

E p

n B

B

p

n p

n C

C R2

V3

R3

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ANSWERS

49. 0.060 W, 12% 51. (a) for n  1, two electrons and for n  2, eight electrons, (b) two, six, and zero 53. m  1, 0, or 1, 45, 90 and 45 55. two electrons occupy the n  1 level with l  0, m  0, and ms  ;1>2, two electrons occupy the n  2 level with l  0, m  0, and ms  ;1>2, and six electrons have n  2, l  1, m  0 and ;1, and ms  ;1>2. The remaining two electrons have n  3, l  0, m  0, and ms  ;1>2. 57. 5 (boron), 111 (transuranic elements) 59. 28 (nickel) 61. 4.14  1022 J, 4.80  104 m, 8.27  1022 J, 2.40  104 m, 1.24  1021 J, 1.60  104 m 63. (a) 1.32 eV/bond, (b) 5.12 eV

Chapter 40 1. 16O: Z  8, A  16, N  8; 56Fe: Z  26, A  56, N  30; 238 U: Z  92, A  236, N  146 3. Z

A

NA-Z

24

11

24

13

27

13

27

14

52

24

52

28

52

25

52

27

63

29

63

34

ISOTOPE

Na Al Cr Mn Cu

63

30

63

33

124

Zn

54

124

70

138

57

138

81

Xe La

5. Z  8 13O 14O 15O 16O 17O 18O 19O 20O 21O 22O 23O N  A - 8 5 6 7 8 9 10 11 12 13 14 15 7. First reaction is not possible (Z is OK, A is wrong); Second reaction is not possible (Z is wrong, A is OK); Third reaction is not possible (Z is wrong, A is OK) 9. 2.5  1015 m; 3.2  1015 m 44 118 207 11. 31 15P is correct; 22Sc is not correct; 50Sn is correct; 83Pb is 89 197 not correct; 39Y; 79 Au 13 13. C; 16O 15. A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 No. of isobars 1 1 2 1 2 3 2 4 4 3 5 4 4 5 4 4 6 1 H, 2H, 4He have no isobars 17. 4.3  1015 19. 1.46  108 u 21. 28.3 MeV 23. 128 MeV for 16O, 112 MeV for 16F 25. 6.3 MeV/neutron vs 8.38 MeV/nucleon in 140Ce

27. ZTOT  9 on both sides, NTOT  9 on both sides 29. 1.64 MeV 31. (a)180 MeV; (b) 1.51  107 m/s for Kr; 9.69  106 m/s for Br; 110 MeV for Kr; 70.1 MeV for Br 33. 212Pb; 235U 35. 22Ne; 64Ni 37. 0.820 MeV 39. 4.0 km/s 41. 0.782 MeV 43. 14N; 0.156 MeV 45. 140 yr 47. 7.7  1015 Bq 49. 2.2  108 Bq 51. 271 days; 1.44  106 g 53. 11 J; 0.15 Gy; 0.26 Sv (26 rem) 55. 20 Gy; 2.3 Gy 57. 1.0 kg 59. 6.94  103 kg 61. 123 MeV 63. 4.78 MeV 65. (a) 3.24  1015 m; (b) 7.10  1014 J; (c) 1.7  108 K 67. (a) 6.6  1011kg/s; (b) 7.2  1010 yr 69. 3.27 MeV; 4.03 MeV; 17.6 MeV; 18.4 MeV; 7.2 MeV/nucleon 71. 6.4 MeV 73. 0.157 MeV 75. 3.00 MeV 77. 9.91 MeV 79. 1.5 Bq 81. 8.1  107 g 83. 1.0087 u 85. (a) 16,000 reactors; (b) 42 yr.

Chapter 41 1. (b) 1.36 T 3. (a) 1.88 GeV, (b) 43.4 GeV, (c) 115 GeV 5. 12 leptons altogether 7. Time dilation factor  9.0; t  1.98  105 s 9. 0.294 kg 11. 118 MeV 13. 67.5 MeV, 1.84  1014 m for each of the g-rays 15. Baryon number is conserved; Strangeness is not conserved. 17. Strangeness is conserved in the first reaction; strangeness is not conserved in the second and third reaction. 19. In the rest frame of the electron, the energy before the emission equals the rest mass energy of the electron. After the emission, the total energy equals the rest mass energy of the electron plus the energy of the photon. This violates the conservation of energy law.

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ANSWERS

21. left side right side baryon # baryon #

left side right side strangeness strangeness

011

101

101 101

K pS 

011

101

101 101

K   p S   0

011

101

101 101

011

1001 101 1001

Reaction K   p S    

0



0





K  p S    

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35. 9.87  109 nm 37. P

+

23. t e–

p–

v

w+ n– x

25. ¢t0  4.70  1024 s, d0  1.41  1015 m; ¢tZ  7.23  1027 s, dZ  2.17  1018 m 27. 129 GeV, greater than the energy required to achieve symmetry between W and photons. 29. uud 31. 30 quarks are created. 33. In table 41.2, no particle is its own antiparticle. In table 41.3, p0 is its own antiparticle, J>c is its own antiparticle,  is its own antiparticle. The latter three consist of a quark and the corresponding antiquark.

39. The Earth and the oldest globular clusters were older than the estimated total age of the universe according to Hubble’s Law. 41. 0.789 MeV 43. Baryon number is not conserved in the first reaction, electric charge is not conserved in the second reaction, energy is not conserved in the third reaction. 45. Electric charge is 1, the particle is © . 47.

t

x

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H. David Seawell/Corbis; p. 710: North Carolina Museum of Art/Corbis; fig. 22.15: Lightingmaster, Streamer Delaying Air Terminal; fig. 22.17: Larry Stepanowicz/Visuals Unlimited. Chapter Opener 23: Kent Wood/Photo Researchers, Inc.; p. 735: (Pip fig. 2): PPC Industries; fig. 23.22: Courtesy of Ionoptika Ltd., England. Chapter Opener 24: Trustees of Princeton University; p. 763: American Institute of Physics; fig. 24.22: Trustees of Princeton University; fig. 24.46: Tom Pantages. Chapter Opener 25: Hank Morgan/Photo Researchers, Inc.; p. 793: Bettmann/Corbis; p. 805: (Pip fig. 3) IT Stock Int’l/indexphoto.com; p. 805: (Pip fig. 4) Euclid Garmet/Georgia Power; fig. 25.32: Stanford University. Chapter Opener 26: (left and right) National Ignition Facility; fig. 26.18: Edward Kinsman/Photo Researchers, Inc.; fig. 26.23: Edward Kinsman/Photo Researchers, Inc. Chapter Opener 27: William Taufic/Corbis; fig. 27.2: American Journal of Physics 30, 19, 1963; fig. 27.3: American Journal of Physics 30, 19, 1963; Table 27.1a: National Center for Atmospheric Research; Table 27.1b: Patrick Bennett/Corbis; Table 27.1c: Brownie Harris/Corbis; Table 27d: Courtesy of Bosch; Table 27e: Corbis; p. 866: Bettmann/Corbis; fig. 27.9: © Crown copyright 1999. Reproduced by permission of the Controller of HMSO and the Queen’s Printer for Scotland; fig. 27.21: John Wilkes Studio/Corbis; fig. 27.26: Principals of Physics by Hans Ohanian, 1994; fig. 27.28: Courtesy of Superpower, Inc. Chapter Opener 28: Martyn F. Chillmaid/Photo Researchers, Inc.; fig 28.6b: Loren Winters/Visuals Unlimited; fig. 28.7b: NASA/Johnson Space Center; fig. 28.7c: AP Photos; fig. 28.22: Loren Winters/Visuals Unlimited; fig. 28.34: Mark C. Burnett/Photo Researchers, Inc.; fig. 28.56: Kim Kulish/Corbis. Chapter Opener 29: © 1990 Richard Megna, Fundamental Photographs, NYC; fig. 29.5: Tom Pantages; p. 930: American Institute of Physics, Emilio Segrè Visual Archives.; p. 934: Science Museum/Science and Society Picture Library; Table 29.1a: Maximilian Stock Ltd./Photo Researchers, Inc.; Table 29.1b: John Chumack/Photo Researchers, Inc.; Table 29.1c: Adrianna Williams/zefa/Corbis; Table 29.1d: NASA/HST/ASU/J. Hester et al; fig. 29.15: Loren Winters/Visuals Unlimited; fig. 29.17: Loren Winters/Visuals Unlimited; p. 940: Science Museum/Science and Society Picture Library; fig. 29.24: © 1990 Richard Megna, Fundamental Photographs, NYC; p. 943: © 1990 Richard Megna, Fundamental Photographs, NYC; p. 950: Science Museum/Science and Society Picture Library. Chapter Opener 30: Lawrence Berkley Laboratory; fig. 30.1: Science Museum/Science and Society Picture Library; fig. 30.4: © 2002 Richard Megna, Fundamental Photographs, NYC; fig. 30.21: Neil Borden/Photo Researchers, Inc.; fig. 30.22: © 1986 Richard Megna, Fundamental Photographs, NYC; p. 978 (Pip fig. 1): Up The Resolution (uptheres)/Alamy; p. 979: (Pip fig. 2) Tek Image/Photo Researchers, Inc.; p. 979: (Pip fig. 3) Courtesy of John Markert; p. 981: AIP Emilio Segrè Visual Archives; fig. 30.24a: Appl. Phys. Lett., Vol. 61, No. 16, 19 October 1992; fig. 30.24b: Appl. Phys. Lett., Vol. 61, No. 16, 19 October 1992; fig. 30.27: LBNL/Photo Researchers, Inc.

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Photo Credits

Chapter Opener 31: Porterfield-Chickering/Photo Researchers, Inc.; p. 998: Hulton-Deutsch Collection/Corbis; Table 31.1: Phototake Inc./Alamy; Table 31.2: Image courtesy of Indigo (R) Instruments, www. indigo.com; Table 31.4: AllOver Photography/Alamy; p. 1011: Bettmann/Corbis; fig. 31.31: Maximilian Stock Ltd./Photo Researchers, Inc.; fig. 31.45: Marty Snyderman/Visuals Unlimited. Chapter Opener 32: (left) Lester Lefkowitz/Getty Images; (right) Chris Knapton/Photo Researchers, Inc.; fig. 32.3: David Michael Zimmerman/Corbis; fig. 32.18: Loren Winters/Visuals Unlimited; fig. 32.35: Steve Callahan/Visuals Unlimited; fig. 32.42: Tom Pantages. Chapter Opener 33: The Imagebank/Corbis; p. 1080: American Institute of Physics p. 1086: Library of the Academy of Science, Paris. Courtesy of AIP, Niels Bohr Library; fig. 33 18: © 1992 Diane Hirsch, Fundamental Photographs, NYC; fig. 33 20: Philip Bailey/Corbis; fig. 33.21a: Corbis; fig. 33.21b: Food pix/Getty Images; fig. 33.29: Jim Sugar/Corbis; fig. 33.31: Courtesy of Hans Ohanian; fig. 33.32: Maximilian Stock Ltd./Photo Researchers, Inc.; fig. 33.33: Steve Percival/Photo Researchers, Inc.; fig. 33.34: Courtesy of Hans Ohanian. Chapter Opener 34: NASA/Science and Society Picture Library; fig. 34.1: Marli Miller/Visuals Unlimited; fig. 34.2: Digital Vision/Getty Images; fig. 34.6: © 1997 Richard Megna, Fundamental Photographs, NYC; fig. 34.11: Courtesy of Hans Ohanian; fig. 34.20: © 1990 Richard Megna, Fundamental Photographs, NYC; fig. 34.22 : Hugh Turvey/Photo Researchers, Inc.; p. 1124: (Pip fig. 1) Hank Morgan/Photo Researchers, Inc.; p. 1124: (Pip fig. 2): PHT/Photo Researchers, Inc.; fig. 34.24: Junenoire Photography; fig. 34.28: Will/Demi McIntyre/Photo Researchers, Inc.; fig. 34.31: © 1987 Ken Kay, Fundamental Photographs, NYC; fig. 34.41b: Courtesy of John Markert; fig. 34.43: David Parker/Photo Researchers, Inc.; fig. 34.44: David Parker/Photo Researchers, Inc.; fig. 34.61: Reuters/Corbis; fig. 34.66a: Roger Ressmeyer/Corbis; fig. 34.66b: McDonald Observatory; fig. 34.67: The Lady and the Unicorn: ‘Sight’ (tapestry), French School, (15th century)/Musée National du Moyen Age et des Thermes de Cluny, Paris/The Bridgeman Art Library; fig. 34.68: Courtesy of Hans Ohanian; fig. 34.69: © 1990 Paul Silverman, Fundamental Photographs, NYC; fig. 34.70: Point Reyes National Seashore; fig. 34.88: DK Limited/Corbis; fig. 34.90: Roger Ressmeyer/Corbis; fig. 34.95: Bo Zaunders/Corbis. Chapter Opener 35 (left): Steve Percival/Photo Researchers, Inc.; (right): Peter Steiner/Alamy; fig. 35.1: Peter Aprahamian/Photo Researchers, Inc.; p. 1170: The Granger Collection, NY; fig. 35.7: Sciencephotos/Alamy; fig. 35.10: Bureau of International des Poids et Mesures, Sevres, France; p. 1175: The Granger Collection, NY; fig. 35.16: Erich Schrempp/Photo Researchers, Inc.; fig. 35.17: Courtesy of Chris C. Jones; fig. 35.21: Courtesy of Chris C. Jones; fig. 35.23: Roger Ressmeyer/Corbis; fig. 35.28: Principals of Physics by Hans Ohanian, 1994; fig. 35.29: © 1987 Ken Kay, Fundamental Photographs, NYC; p. 1191: Stefano Bianchetti/Corbis; fig. 35.33: Courtesy of Chris C. Jones; fig. 35.38: Courtesy of Chris C. Jones; fig. 35.39: Courtesy of Chris C. Jones; fig. 35.40: Courtesy of Chris C. Jones; fig. 35.41: Denis Scott/Corbis; fig. 35.42: Stephanie Maze/Corbis; p. 1198: Mary Evans Picture Library/Alamy; fig. 35.43: Courtesy of Michael Lockwood, University of Illinois Urbana-Champaign.; fig. 35.45: Courtesy of Chris C. Jones; fig. 35.46: Northwestern University/Photo Researchers, Inc.; fig. 35.47: Eye of Science/Photo Researchers, Inc.; fig. 35.57: Principals of Physics by Hans Ohanian, 1994. Part Opener 6: NASA/Marshall Space Flight Center.

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Chapter Opener 36: ESA/CE/Eurocontrol/Photo Researchers, Inc.; p. 1217: Hulton Archive/Getty Images; p. 1235: Science Photo Library/Photo Researchers, Inc. Chapter Opener 37: Larry Landolfi/Photo Researchers, Inc.; fig. 37.1: Simon Lewis/Photo Researchers, Inc.; fig. 37.2: Edward Kinsman/Photo Researchers, Inc.; p. 1260: AIP Emilio Segrè Visual Archives; fig. 37.8: Photo Researchers, Inc.; p. 1268: (Pip fig. 2): Courtesy of Burle Industries; p. 1269: AIP Emilio Segrè Visual Archives; fig. 37.15: Science Museum/Science and Society Picture Library; fig. 37.16: © Mediscan/Visuals Unlimited; fig. 37.17: © Science VU/Visuals Unlimited; fig. 37.21a-c: Courtesy of E.R. Huggins; fig. 37.21d: Courtesy of Chris C. Jones; p. 1277: AIP Emilio Segrè Visual Archives; p. 1278: AIP Niels Bohr Library. Chapter Opener 38: (top) Courtesy of Eastman Kodak, Inc. (bottom) Courtesy of Hans C. Ohanian; fig. 38.1: Prof. C.K. Shih, Department of Physics, The University of Texas at Austin; fig. 38.3: Adam Jones/Visuals Unlimited; p. 1289: Courtesy of Eastman Kodak; fig. 38.5: Courtesy of Hans C. Ohanian; fig. 38.6: Courtesy of Hans C. Ohanian; p. 1294: Cavendish Laboratory, Cambridge; p. 1295: Princeton University/American Institute of Physics/Science Photo Library; p. 1301: AIP Emilio Segrè Visual Archives; p. 1304: AIP Emilio Segrè Visual Archives; p. 1310: (Pip fig. 1b): Richard J. Green/Photo Researchers, Inc.; p. 1310: (Pip fig. 2b): Andrew Syred/SPL/ Photo Researchers, Inc.; p. 1311: (Pip fig. 3b): © IBM Research; p. 1311: (Pip fig. 4): Delft University of Technology/Photo Researchers, Inc. Chapter Opener 39: Professor C.K. Shih, Department of Physics, The University of Texas at Austin; fig. 39.3a-d: A.F. Burr and A. Fisher, New Mexico State University; p. 1325: AIP Emilio Segrè Visual Archives; p. 1326: Granger Collection, NY; p. 1332: Courtesy of The University of Oxford, Museum of the History of Science; fig. 39.10: Courtesy of John Markert; fig. 39.24: Ton Kinsbergen/Photo Researchers, Inc.; fig. 39.25: Astrid & Hanns-Frieder Michler/Photo Researchers, Inc.; fig. 39.29: NASA/Corbis; fig. 39.30: Reuters/Corbis; fig. 39.31: Corbis; fig. 39.32: Roger Ressmeyer/Corbis. Chapter Opener 40: U.S. Department of Energy/Photo Researchers, Inc.; fig. 40.7: John Cockcroft, Cambridge, UK; p. 1366: Jean-Loup Charmet/Science Photo Library; p. 1373: Physics Today Collection/AIP/Science Photo Library; p. 1376: (Pip fig. 1): Reuters/Corbis; p. 1377: Ullstein bild/The Granger Collection, NY; p. 1380: AIP Emilio Segrè Visual Archives; fig. 40.18: Corbis; fig. 40.21: Corbis; p. 1384: Corbis; fig. 40.22: Roger Ressmeyer/Corbis; fig. 40.23: National Ignition Facility (NIF). Chapter Opener 41: CERN, P. Lopez/Photo Researchers, Inc.; fig. 41.1: Fermilab/Photo Researchers, Inc.; fig. 41.2: Fermilab/Photo Researchers, Inc.; fig. 41.3: David Parker/Photo Researchers, Inc.; fig. 41.4a: CERN/SPL/ Photo Researcher, Inc.; fig. 41.4b: CERN/SPL/ Photo Researcher, Inc.; fig. 41.6: Stanford Linear Accelerator Center/SPL/Photo Researchers, Inc.; fig. 41.7: CERN/Photo Researchers, Inc.; fig. 41.8: LBNL/Photo Researchers, Inc.; p. 1410: Harvey of Pasadena/American Institute of Physics/Science Photo Library; fig. 41.19a: JeanCharles Cuillandre/CFHT/Photo Researchers, Inc.; fig. 41.19b: NOAO/Photo Researchers, Inc.; fig. 41.20a: Dr. Jean Lorre/Photo Researchers, Inc.; fig. 41.20b: Celestial Image Co./Photo Researchers, Inc.; fig. 41.20c: C. Butler/Photo Researchers, Inc.; fig. 41.21a: NASA; fig. 41.23: Courtesy of Jeff Hester; fig. 41.22: NASA; fig. 41.23: NOAO/AURA/NSF/Photo Researchers, Inc.; fig. 41.24: Roger Ressmeyer/Corbis.

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Index Page numbers in italics refer to biographies. Page numbers in boldface refer to figures. Page numbers followed by “t” refer to tables.

aberration: chromatic, 1144 spherical, 1128, 1144 absolute acceleration, 132 absolute motion, 1217 absolute temperature scale, 604 absolute thermodynamic temperature scale, 609, 686 absolute zero, entropy at, 680 absorbed dose, 1375 Acapulco, divers, 688 accelerated charge: electric field of, 1075–79, 1077 radiation field of, 1075–76, 1076, 1077 acceleration, 39–54 absolute, 132 angular, see angular acceleration average, 39, 39t, 40, 60 average, in three dimensions, 101–2 average, in two dimensions, 96 of center of mass, 323 centripetal, 113–14, 113, 114, 132, 184–90, 195, 371, 372 components of, 95–98, 101 as derivative of velocity, 41 formulas for, 39, 41 instantaneous, 40–41, 41 instantaneous, components of, 97 instantaneous, in two dimensions, 96–97 motion with constant, 42–49, 43, 63, 102–4, 103, 104, 122 motion with variable, 54–56 negative, 39 positive, 39–40 standard g as unit of, 52 tangential, 371–72 translational, 402 vectors, 100–101 acceleration of free fall, 49–54, 64 universality of, 49, 49 acceleration of gravity, 52–53, 64, 274–75 measurement of, 52–53 variation of, with altitude, 274–75 accelerators: linear, 1413, 1415 for particles, 1363, 1363, 1398–99 acceptor impurities, 1339

accidents, automobile, 339, 343, 355 AC circuits, 1030–67 AC current, 1031 hazards of, 913–14 acoustic micrograph, 539 action and reaction, 144–51, 144, 145, 146 action-at-a-distance, 274, 722 action-by-contact, 722, 723 action-reaction pairs, 144–51, 144, 145, 146, 149 AC voltage, 1004 Adams, J. C., 272 addition law for velocity, Galilean, 1218 addition of vectors, 72–76, 72, 73, 74, 89 commutative law of, 74 by components, 78–79 addition rule for velocities, 115–16, 117 adiabatic equation, for gas, 649 adiabatic expansion, 668, 689 adiabatic process, 647–49 air, composition by element and mass, 620, 623, 657 air bag, 343 air conditioner, 672, 673 airfoil, flow around, 570, 582–83 airplane: motion, pitch, roll, yaw, 366 propeller, 392 air resistance, 49, 51, 61, 181, 181 in projectile motion, 111 Al’Aziziyah, Libya, hottest temperature, 621 Alpha Centarui, A and B, xli, 296 alpha decay, 1365–67 alpha particles, 1363–64 scattering of, 1293–94 alpha rays, 1365 alternating current, 1031 hazards of, 913–14 alternating emf, 1004, 1032–33, 1035, 1046–53 Alvin, DSV, 565, 565, 574, 574, 577, 582 ammeter, 905, 916 Amontons, Guillaume, 174 Ampère, André Marie, 941 amperes, 697 Ampère’s Law, 939–40 displacement current and, 1073–74 electric flux and, 1073

modified by Maxwell, 1071, 1073, 1074, 1080, 1096, 1097, 1097 amplitude: of motion, 470 of wave, 511 Analytical Mechanics (LaGrange), 236 analyzer, 1085 Andromeda Galaxy, xliv Angers, France, bridge collapse at, 491, 491 angle: elevation, 109, 111, 111 of incidence and of reflection, 1115–16, 1115 angle in radians, 368 angular acceleration: average, 370 constant, equations for, 374 instantaneous, 370 rotational motion with constant, 374–76 time-dependent, 376–78 torque and, 400 angular frequency, 470–71, 471 of simple harmonic oscillator, 477 of wave, 512, 513–16 angular magnification: of magnifier, 1147 of microscope, 1149 of telescope, 1150 angular momenta, some typical values, 407t angular momentum, 284, 407t for circular orbit, 409 in elliptical orbit, 291–92 orbital, 409 quantization of, 1322–23 spin, 409 torque and, 410–16 angular momentum, conservation of: in planetary motion, 284 in rotational motion, 406–10 angular-momentum quantum number, 1296, 1322, 1324–26 angular momentum vector, 411, 411 angular motion, 375 angular position, for time-dependent angular velocity, 376–77 angular resolution, of telescope, 1196–99 angular velocity, 369t, 471 average, 369 A-77

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angular velocity (continued) instantaneous, 369 for time-dependent angular acceleration, 376–77 annulus of sheet metal, 381 antibaryons, 1404 antielectron, see positron antimatter-matter annihilation, 706 antimesons, 1404 antineutrino, 1368–70 antinodes and nodes, 520–21, 521, 544–45, 544–45 antiparticles, 706, 1403–4 antiquarks, 1413 aphelion, 282, 284, 409 of planets, xl, 285 apogee: of artificial satellites, 286 of moons, xxxix of planet, 286 Apollo 16, xxxix Apollo astronaut, 295 apparent weight, 187–88, 187 apple, chemical energy of, 632, 652 Archimedes, 581 Archimedes’ Principle, 580–82, 599 area, 13 areas, law of, 283–84 Arecibo radiotelescope, 1198–99, 1198 argon: compression, 658 Lennard-Jones potential, 263 monatomic kinetic energy, 616 thermal window of, 656 artificial satellites, 271–72, 281, 286–87, 1344, 1344, 1421 apogee of, 286 perigee of, 286 astigmatism, 1147 astrology, 295 astronaut: weightlessness training, 589 see also Apollo; International Space Station; Skylab mission astronomical unit (AU), 24 Atlas rocket, guidance system, 414 atmosphere, 573 atmospheric electric field, 806–8 atmospheric pressure, 577–78 gauge, 573 atom, 1397 electron configuration of, 1328–32 electron distribution in, 695 nuclear model of, 1294, 1294 nucleus of, see nucleus quantum structure of, 1320–40 stationary states of, 1299 structure of, 695, 1287–95 atomic clock, Cesium, 9 atomic force microscope (AFM), 475, 475, 1311, 1311 atomic mass, 11–12 atomic mass unit, 11, 20, 1355 atomic number, 1356 atomic standard of mass, 11 atomic standard of time, 9 atomic states, quantum numbers of, 1328–32 atomic structure, 695, 1287–95 atom smashers, 1397

Index

attractors, 492 Atwood’s machine, 403, 421 automobile battery, 707, 890–91, 891, 892 automobiles: collisions, 339, 343, 355 crash tests of, 339, 340, 355 efficiency of, 674, 674 electric fields and, 805 energy conversions, 674 engine cycle, 674 impact speed, 343t automobile stopping distances, 45, 46, 47, 47 average acceleration, 39, 39t, 40, 60 formula for, 39 in three dimensions, 101–2 in two dimensions, 96 average angular acceleration, 370 average angular velocity, 369 average power, 253 average speed, 29–31, 30t average velocity, 32–35, 33, 101–2 in two dimensions, 95 Avogadro’s number, 11, 20, 607 axis of symetry, 380–82, 382t back emf, 1012 balance, 136–37 beam, 136–37, 137 Cavendish torsion, 277 Coulomb’s, 698, 700 spring, 136, 136, 151 watt, 11, 11 ballistic curve, 111 ballistic pendulum, 349–50, 350 balloons: hot air, 126, 581, 594, 602, 605, 612, 618, 622 Raven S-66A, 593 research, 622 Balmer, Johann, 1291 Balmer series, 1291, 1291t banked curve, 186–87, 187 barometer, mercury, 577, 577 baryon, 1403, 1404, 1404t, 1405, 1406, 1422 baryon number, 1406, 1407 conservation law for, 1406 base units, 13 bathyscaphe, 589, 589 battery: automobile, 707, 890–91, 891, 892 dry cell, 891, 891 internal resistance of, 895–96 lead-acid, 707, 890–91, 891, 892, 893 Bay of Fundy, 531, 531, 559–60 beam balance, 136–37, 137 beam dump, 659 beat frequency, 518 beats, of a wave, 518 becquere (Bq), 1375 Becquerel, Antoine Henri, 1365, 1366 Bell Laboratories, 1421, 1421 Bernoulli, Daniel, 585 Bernoulli’s equation, 582–85, 586, 587, 598, 599 beta decay, 1368–70 beta rays, 1365 bicycle: rounding curve, 456 suspended, 431 upright, 433 Big Bang, xliv, 626, 1420–21

Big European Bubble Chamber (BEBC), CERN, 1399 bimetallic strip thermometers, 610, 610, 636, 637 binary star system, 297 resolution of telescope and, 1197 binding energy of nucleus, 1359–65 curve of, 1361, 1377 binoculars, 1156 Biot, Jean Baptiste, 950 Biot-Savart Law, 948–50, 948, 949 blackbody, spectral emittance of, 1259 blackbody radiation, 1255–58, 1259–61 black holes, 299 block-and-tackle, 443–44, 444 blood pressure, 579 blood vessels, xlvii blowhole, 546 blue, 1414–16 body-mass measurement device, 134, 134, 468, 468, 478, 482, 490 Bohr, Niels, 1295, 1296, 1321 Bohr magnetron, 976 Bohr radius, 1297 Bohr’s postulates, 1296 boiling points, common substances, 642t Boltzmann, Ludwig, 608 Boltzmann’s constant, 607 bomb, hydrogen, 1380 bomb calorimeter, 250, 250 bonds, interatomic, 1333 bones as lever, 442, 442 boom, sonic, 552–53, 552 Born, Max, 1277, 1277, 1302 Bose-Einstein condensate, 623 boson, 1403 bottom quark, 1415 boundary conditions, 522, 522 bound charges, 838 bound orbit, 245 Boyle, Robert, 606 Boyle’s Law, 606 Brackett series, 1292 Brahe, Tycho, 285 brake, hydraulic, 575–76 brake, power, 456 breeder reactor, 1383 Bremsstrahlung, 1090, 1273–74 Brewster’s Law, 1124 bridge, 433, 433 thermal expansion and, 637, 637 bridge collapse: at Angers, France, 491, 491 at Tacoma Narrows, 523–24, 524 British system of units, 6–7, 12 British thermal unit (Btu), 630 Brown Mountain hydroelectric storage plant, 242–43, 242, 243, 249, 257–58, 258 bubble chamber, 1396, 1399–1400, 1400, 1402, 1409 bulk modulus, 447–48, 447t bullet: impact on block, 350 measuring speed of, 356 bungee jumping, 246–47, 246, 247 buoyant force, 580–81 cable, superconducting, 883 cable capacitance, 843 Cailletet, liquify oxygen, 658

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Index

calculus (review), A10–21 antiderivative in, A14 approximation of small values in, A18–19 chain rule for derivatives in, A12 derivatives in, 38, A10–12, A11t integral, A12–17, A15t integration rules in, A15–16 partial derivatives in, A12 Taylor series in, A18 uncertainties, propagation of, A19–21, application to Ohm’s Law of, A20–21 Calder, Alexander, mobile, 317 Caledonian Railway wheel set, 454 California Speedway, 343 calorie, 250, 630, 631 calorimeter, bomb, 250, 250 camera, photographic, 1144–45, 1144 camera, ultrasonic range finder in, 560 Canes Venatici, 1421 capacitance, 811–16, 829–38 of earth, 830 of single conductor, 829 capacitative reactance, 1036 capacitor microphone, 833 capacitors, 811–16, 829–38 circuit with, 1035–38 energy in, 844–47 guard rings for, 848, 848 multiplate, 851, 851, 852, 852 in parallel, 834 parallel-plate, 831–32, 851, 851, 852, 852 in series, 834–35 Caph, spectrum of, 1288 carbon: isotopes of, 1355–59, 1356 mass of, 1358 Carnot, Sadi, 667 Carnot cycle, 668–69, 669, 671–73, 675–76 Carnot engine, 667–73 efficiency of, 671–73 Second Law of Thermodynamics and, 676 Carnot’s theorem, 675–76 carrier, 1410 Cartesian diver, 589, 589 Cavendish, Henry, 277, 698 Cavendish torsion balance, 277, 277 cavity radiation, 1257 ceiling fan, 367, 368, 371–72 cell, triple-point, 609, 609 cello, notes available on, 562 cells (of eye), rods, cones, xlvii Celsius temperature scale, 611, 612 Centaurus, xli center of force, 240 center of mass, 313–23, 320 acceleration of, 323 of continuous mass distribution, 316 gravitational force acting on, 430–33 motion of, 323–27 velocity of, 323–24, 348 centrifugal compressor, 99 centrifugal force, 188–89, 189 centrifuge, 114, 114, 365, 365, 373, 383 centripetal acceleration, 113–14, 113, 114, 132, 184–90, 195, 371, 372 Newton’s Second Law and, 185 centripetal force for circular motion, 185 centroid, 316 Cerenkov counters, 1399

CERN (Organisation Européenne pour la Recherche Nucléaire), 1225, 1238, 1398–99, 1402, 1402, 1411 Cesium atomic clock, 9 Cesium standard of time, 9 cgs system of units, 713 Chadwick, James, 1355 chain reaction, 1378–79, 1378 Chamonix waterfall, 652 Champlain Canal, 334 changes of state, 642–43 chaos, 492–93 characteristic spectrum, 1274 characteristic time, 1016–17 of RC circuit, 909 characteristic X rays, 1303 charge, electric, 698–702, 729–30 bound, 838–39 bound, in dielectrics, 838–39 conservation of, 706–7 of electron, 696–97, 698 of electron, measurement of, 747 of elementary particles, 1403t, 1404t, 1414 of particles, 696–97, 698, 706, 1414 point, 699 of proton, 696–97, 698 quantization of, 706 SI unit of, 972 static equilibrium of, 774–75 surface, on dielectric, 840–41 charge distribution, electric field of, 732 Charles’ Law, 606 charm, of quarks, 1415 chemical elements, Periodic Table of, 1328–32, 1329t chemical reactions, conservation of charge in, 707–8 Chernobyl, 1383 Chicago (Sears Tower), 653, 654 chromatic aberration, 1144 chromatic musical scale, 539, 539 chronometer, 21, 21 circuit, electric: AC, 1030–67 with capacitor, 1035–38 DC, 1031, 1032 frequency filter, 1037 with inductor, 1038–41 LC, 1041–46 loop method for, 898 multiloop, 897–900 RC, 907–12 with resistor, 1031–35 RL, 1015, 1015, 1018 single-loop, 893–97 circular aperture: diffraction by, 1196–1999, 1196 minimum in diffraction pattern of, 1196 circular motion: centripetal force for, 185 translational speed in, 374 circular orbits, 278–82, 278, 1321 angular momentum for, 409 energy for, 290–91 in magnetic field, 966 circular polarization, 532, 532 clarinet, sound wave emitted by, 538 classical electron radius, 1315 classical mechanics, quantum mechanics vs., 1287

A-79

Clausius, Rudolph, 678 Clausius statement of Second Law of Thermodynamics, 676 Clausius’ theorem, 678 clock: Cesium atomic, 9 grandfather, 219, 219 pendulum, 487, 487, 495, 495 synchronization of, 4, 5, 133n, 1220–23 Coast and Geodetic Survey, U.S., 500 coaxial cable, 843 Cockroft-Walton accelerator, 1363 coefficient of kinetic friction, 175–78, 175t coefficient of linear thermal expansion, 633, 637 coefficient of performance (heat), 673 coefficient of restitution, 358 coefficient of static friction, 175t, 179–80 coefficient of volume thermal expansion, 634–35 coefficients of friction, 174–81, 175t coherence of light, 1177 “cold resistance,” 869 Collider Detector, Fermilab, 1399, 1399 colliding beams, 1398–99 collisions, 338–64 automobile, 339, 355 impulsive forces and, 339–44 collisions, elastic, 342–47 conservation of energy in, 344–45, 351–52, 353 conservation of momentum in, 344–45, 351–52, 353 in one dimension, 344–47 one-dimensional, speeds after, 345–47 in three dimensions, 351–53 in two dimensions, 351–53 collisions, inelastic, 348 conservation of energy in, 351–52, 353 conservation of momentum in, 351–52, 353 in three dimensions, 351–53 totally, 348 in two dimensions, 351–53 color: of quarks, 1414–15 of visible light, 1092 “color” force, 1405 color-strip thermometer, 610, 610 Coma Berenices, xliii combination principle, Rydberg-Ritz, 1293 comets, 291 Hale-Bopp, 299, 299 Halley’s, 291, 298, 298 perihelion of, 291 period of, 291 Shoemaker-Levy, 299 communication satellites, 271–72, 281, 290–91 commutative law of vector addition, 74 compact disc, 367 compass needle, 927, 927 components, of vectors, 77–86, 78, 95–98, 97, 99, 101 formulas for, 77 compression, 446, 448–49 compressor, centrifugal, 99 Compton, Arthur Holly, 1269, 1269 Compton effect, 1269–72 Compton wavelength, 1315 concave spherical mirror, 1128, 1130 Concorde SST, 553, 557, 563 sonic boom of, 553 concrete, thermal expansion of, 637

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A-80

condenser, 833 conduction band, 1338 conduction of heat, 638–42 conductivity, thermal, 638–41, 639t conductors, 708, 708, 709–10, 1337 in electric field, 774–76 insulators vs., 871 potential energy of, 812–13 conservation laws, 205 for baryon number, 1406 elementary particles and, 1406–7 for mass number, 1406 conservation of angular momentum: in planetary motion, 284 in rotational motion, 406–10 conservation of electric charge, 706–7 in chemical reactions, 707–8 Conservation of Electric Charge, Law of, 710 conservation of energy, 205, 207, 223, 235–70, 290, 797 in analysis of motion, 223 general law of, 248, 249, 252, 662 in inelastic collision, 351–52, 353 law of, 790 in one-dimensional elastic collision, 345 in rotational motion, 397 in simple harmonic motion, 483 in two-dimensional elastic collision, 351–52, 353 in two-dimensional inelastic collision, 351–52, 353 conservation of mass, 205, 252 conservation of mechanical energy, 238 equation for, 239 law of, 221–23, 221, 222, 223, 238 conservation of momentum, 307–12, 310, 345, 348 in elastic collisions, 344–45, 351–52, 353 in fields, 723 in inelastic collisions, 351–52, 353 law of, 309 conservative electric field, 804–5 conservative force, 236–43, 238 gravity as, 288 potential energy of, 236–43 constant, fine-structure, 1315 constant angular acceleration, equations for, 374 constant force, 205, 208 constant-volume gas thermometer, 609–10, 609 Constitution, USS, 331 constructive and destructive interference, 517, 517, 1169 in Michelson interferometer, 1175–76 for wave reflected by thin film, 1169–73 contact force, 142–43, 143 “contact” forces, 697 continuity equation, 570 control rod, in nuclear reactor, 1381 convection, 641 conversion factors, 17–19, 20 conversion of units, 16–17, 18 convex spherical mirror, 1129 cooling, evaporative and laser, 624 Coordinated Universal Time (UTC), 9 coordinate grid, 3–4, 115 coordinates, Galilean transformation of, 1218, 1234, 1241 coordinates, origin of, 3, 3, 4, 44, 45, 46, 47 Copernicus, Nicholas, 279 corner reflector, 1116

Index

Corona Borealis, 59 corona discharge, 710, 710 corona wire, 709 cosecant, A8–10 cosine, 19, 473–74, 486, A8–10 formula for derivatives of, 473 law of cosines, A10 cosmic background radiation, 1421 Cosmological Principle, 1419 cosmology, 1416–23 cotangent, A8–10 Coulomb, Charles Augustin de, 700 coulomb (C), 696–97, 972 Coulomb constant, 699 Coulomb potential, 794–95 Coulomb’s balance, 698, 700 Coulomb’s Law, 698–99, 703, 711, 762, 790, 810, 1074 in vector notation, 699 crane (tower), see K-10000 tower crane critical angle, for total internal reflection, 1123 cross product, 83–86, 84, 85, 410–11, 743 of unit vectors, 85 crystals, atomic arrangement of, xlix Curie, Marie Sklowdowska, 1373 Curie, Pierre, 1373 current, electric: AC, 1031 AC, hazards of, 913–14 in capacitor circuit, 1035–38 DC, 859–63, 862t, 887–925, 1031 DC, hazards of, 913–14 displacement, 1073–74 generating magnetic field, 939–40, 940, 948 in inductor circuit, 1038–41 “let-go,” 913 SI unit of, 972 time-dependent, 907–12 current density, 869 current loop: potential energy of, 975–76 right-hand rule for, 942 torque on, 972–76 current resistor circuit, 1031–35 curve: ballistic, 111 banked, 186–87, 187 of binding energy, 1361, 1377 of potential energy, 244–47, 244 cutoff frequency, 1275 cutoff wavelength, 1274–75 cyclic motion, 469 cyclotron, 964, 967–68, 968, 979–80 cyclotron emission, 1090 cyclotron frequency, 967 Cygnus X-1, 298 cylindrical symmetry, 767–68 da Costa, Ronaldo, 14 damped harmonic motion, 489 damped oscillations, 488–91, 491 damped oscillator: driving force on, 490 harmonic, 1045 resonance of sympathetic oscillation of, 489, 490–91, 491 dark energy, xlv, 1423 dark matter, 1423 daughter material, 1366

da Vinci, Leonardo, 174–75, 174 Davisson, C. J., 1303 day: mean solar, 9 sidereal, 294 solar, 9 DC-3 airplane: efficiency, 686 engines, 267 take off speed, 266 DC-10 airliner, accident (Orly, France), 591 DC current, 887–925, 1031, 1032 hazards of, 913–14 instruments used in measurement of, 905–7 DC voltage, 1004 de Broglie, Louis Victor, Prince, 1302, 1303 de Broglie wavelength, 1302, 1336 decay constant, 1373 decay of particles, 1365–70 decay rates of radioisotopes, 1374–76 deceleration, 40 decibel, 541–42, 542t dees, 967–68 degree absolute, 604 density, 13, 316–17, 566 of field lines, 739 of fluid, 566–67, 567t of nucleus, 1359 depth finder, 557 depth of field, 1145 derivative, of the potential, 806–8 derivatives, rules for, 38 derived unit, 13–14, 20, A20–21, A21t destructive and constructive interference, 517, 1169 in Michelson interferometer, 1175–76 for wave reflected by thin film, 1169–73 determinant, 86 deuterium, 498 dialectic strength, 841 diamagnets, 977 diatomic gas, energy of, 617–18 diatomic molecule, 244 dielectric, 829, 838–47 electric field in, 838–40 energy density in, 844–45 Gauss’ Law in, 842 linear, 838 dielectric constant, 840, 841t diffraction, 553–55, 1190–98 at a breakwater, 553 by circular aperture, 1196–99, 1196 by a single slit, 1190–96 of sound waves, 554 of water waves, 553 diffraction pattern, 554, 554 of single slit, 1194 diffuse series, 1314 dimensional analysis, 16 dimensionless quantities, 17 dimensions, 16 diodes, 1340–42 dioxyribonucleic acid (DNA), xlviii dip angle, 954 dip needle, 954 dipole, 742–44, 756 potential energy of, 743 torque on, 743

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Index

dipole moment, 743–44 magnetic, 973 permanent and induced, 887–925 Dirac, Paul Adrien Maurice, 1302, 1325, 1326 direct current, 887–925, 1031, 1032 hazards of, 913–14 Discoverer II satellite, 297 discus thrower, 109, 109 dispersion, of light, 1125 displacement current, 1073–74 displacement vector, 69, 70–72, 70, 71, 88, 96 dog, hearing, 558 dominoes, toppling, 525 donor impurities, semiconductors with, 1339 door (swinging), 367, 396 Doppler, Christian, 549 Doppler shift, 547–53, 547, 548, 561 of light, 1228–29 dot product: in definition of work, 208–9 of vector components, 82–83, 82, 86 of vectors, 81–83, 81, 83, 86, 208–9 double-well oscillator, 492, 493 “doubling the angle on the bow,” 89, 89 down quark, 1413–14 drag forces, 180–81 drift velocity, of free-electron gas, 864 driving force, on damped oscillator, 490 dry cell battery, 891, 891 dumbell, rotation, 411, 412 dynamics, 29, 130–72 fluid, 582–87 of rigid body, 394–428 dynodes, 1267, 1269 Earth, xxxix–xl, 285t, 286, 286 angular momentum of, 409, 427 capacitance of, 830 coldest and hottest temperature, 621 densities and pressures in upper atmosphere, 621t escape velocity from, 292 mass distribution within, 390t moment of inertia of, 388, 389, 390–91 perihelion of, 295 polar ice cap melting, 425 reference frame of, 132 rotational motion of, 120 rotation of, 9, 132, 476 translational motion of, 120 earthquake, Tangshan, China, 533 echolocation, 536, 536 Echo satellites, 1421 Eddington, Arthur, 1276 eddy currents, 1003 efficiency: of automobiles, 674 of Carnot engine, 671–73 of engines, 666–67 Eiffel Tower, 654 eigenfrequencies, 523, 545, 546 Einstein, Albert, 251, 394, 1217, 1217, 1264–65, 1361 photoelectric equation of, 1266–67 elastic body, 182, 445 elongation of, 445–49 elastic collision, 342–47 conservation of energy in, 344–45, 351–52, 353 conservation of momentum in, 344–45, 351–52, 353

in one dimension, 344–47 speeds after one-dimensional, 345–47 in three dimensions, 351–53 in two dimensions, 351–53 elasticity of materials, 445–49 elastic moduli, some values, 447t elastic potential energy, 236–37, 238 electrical measurements, 905–7 electric charge, 698–702, 729–30 bound, 838 bound vs. free, in dielectrics, 838 conservation of, 706–7 of electron, 696–97, 698 of electron, measurement of, 747 of elementary particles, 1403t, 1404t, 1414 of particles, 696–97, 698, 706, 1414 point, 699 of proton, 696–97, 698 quantization of, 706 SI unit of, 972 static equilibrium of, 774–75 surface, on dielectric, 840–41 Electric Charge, Law of Conservation of, 710 electric circuit, see circuit, electric electric constant, 699 electric dipole, 725, 742–44, 756 electric energy: in capacitors, 844–47 of spherical charge distribution, 813 electric energy density in dielectric, 844–45 electric field, 721–55, 722, 724t, 1078 of accelerated charge, 1075–79, 1077 atmosphere, 806–8 of charge distribution, 732 conductors in, 774–76 conservative, 804–5 definition of, 722–23 in dielectric, 838–40 electric dipole in, 742–44 electric force and, 723, 724 energy density of, 815 as fifth state of matter, 722 of flat sheet, 734–36, 759 of harmonic traveling wave, 1098 induced, 994 motion in, 740–44 of plane wave, 1079–80, 1079 of point charge, 723–24, 728 superposition of, 772–73, 773 of thundercloud, 725–28 in uniform wire, 860 electric field lines, 736–39 made visible, with grass seeds, 859 of point charge, 736–37, 737, 739, 936 sources and sinks of, 739–40 electric flux, 757–61 Ampère’s Law and, 1073 electric force, 695–99 compared to gravitational force, 696, 699 Coulomb’s law for, 698–99 electric field and, 723, 724 in nucleus, 1355, 1359 qualitative summary of, 695–97 superposition of, 703 and xerography, 709 electric fringing field, 848 electric generators, 892 electric ground, 803 electricity:

A-81

frictional, 695, 711 Gauss’ Law for, 1074 electric motor, 974 electric resistance thermometer, 610, 610 electric shock, 913–14 electrolytes, 709 electromagnet, 946 electromagnetic flow meter, 996 electromagnetic force, 191, 695, 1405, 1406, 1406t electromagnetic generator, 1000, 1003–4 electromagnetic induction, 993–1029 electromagnetic interactions, 1405, 1406, 1406t electromagnetic radiation: kinds of, 1088–91 wavelength and frequency bands of, 1090–91, 1091 electromagnetic wave, 1070–1110, 1071, 1075–76, 1076, 1077 energy flux in, 1093–94 generation of, 1088–91 momentum of, 1094–96 right-hand rule for, 1078–79, 1079 speed of, 1080 electromagnetic wave pulse, 1080–83, 1082 electromagnetism, 1411 electromotive force, see emf (electromotive force) electron: charge of, 696, 698 distribution of, in atoms, 695 elliptical orbit of, 1321–22 free, 708, 866–67 free of gas, 711, 863–65 magnetic moment of, 1325, 1355 mass of, 137, 137t, 1356 measurement of charge of, 748 neutron and proton vs., 1356 quantum behavior of, 1302–9, 1320–43 spin of, 1324–25, 1328–29, 1355–56 states of, 1327t electron-attracting wire, 709 electron capture, 1390 electron configuration: of atoms, 1328–32 of solids, 1337–39 electron field, 1409–10 electron-holography microscope, 694 electron microscopes, see microscope, electron electrons, of neon, xlix-l electron scanning microscope, 1310, 1310 electron-volts (eV), 248, 796, 1302, 1359, 1397–98 electrostatic equilibrium, 774 electrostatic force, 695–97 electrostatic induction, 711 electrostatic potential, 790–98 calculation of, 798–803 electrostatic precipitators, 735 electroweak force, 1411 elektron, 695 elementary particles, 1396–1430 collisions between, 342–44 conservation laws and, 1406–7 electric charges of, 1403t, 1404t, 1414 masses of, 1403t, 1404, 1404t spins of, 1403t, 1404t elements, chemical: age of, 1421 during Big Bang, 1421 transmutation of, 1363

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A-82

elephant, rumble (communication), 558 elevation angle, 109, 111, 111 elevator with counter weight, 154–55, 154, 155, 157 ellipse, xxxix, 283 major axis of, 282 semimajor axis of, 282, 285, 291 elliptical orbits, 282–86, 282 angular momentum in, 291–92 of electron, 1321–22 energy in, 291–92 vs. parabolic orbit for projectile, 287 elongation, 445–49, 445 emf (electromotive force), 887–93 alternating, 1004, 1032–33, 1035, 1046–53 induced, 994, 998–1007 motional, 993–97 power delivered by, 901 in primary and secondary circuits of transformer, 1053–54 sources of, 890–92 steady, 1004 emission, stimulated, 1090 Empire State Building, 63, 227 energy, 204–23 alternative units for, 248–49 in capacitor, 844–47 for circular orbit, 290–91 conservation of mechanical, 221–23, 221, 222, 223, 238, 239 dark, xlv, 1423 of diatomic gas, 617–18 electric, see electric energy in elliptical orbit, 291–92 equivalent to atomic mass unit, 1361 gravitational potential, 218–23, 219, 220, 238, 288–93 gravitational potential, of a body, 321 of ideal gas, 616–19 internal, 616 kinetic, see kinetic energy law of conservation of, 790 in LC circuits, 1042 Lorentz transformations for, 1234–37 magnetic, 1012 magnetic, in inductor, 1013–15 and mass, 1242–44 mass and, 251–53 mechanical, see mechanical energy momentum and, relativistic transformation for, 1242–44 in orbital motion, 288–93 of photon, 1264–65 of point charge, 796 potential, see potential energy rate of dissipation of, 264 rest-mass, 1242 rotational, 1334 of rotational motion of gas, 617–18 sample values of some energies, 249t in simple harmonic motion, 480–83 of stationary states of hydrogen, 1299 of system of particles, 327–28 thermal, 248, 616, 629 threshold, 355 total relativistic, 1243 vibrational, 1333 in wave, 1092–96 zero-point, 1307

Index

energy, conservation of, 205, 207, 223, 235–70, 290, 797 in analysis of motion, 223 general law of, 248, 249, 252, 662 in inelastic collision, 351–52, 353 in one-dimensional elastic collision, 345 in rotational motion, 397 in simple harmonic motion, 483 in three-dimensional elastic collision, 351–52, 353 in two-dimensional elastic collision, 351–52, 353 energy bands, 1337 energy banks, in solids, 1336–40 energy density: in dielectric, 844–45 in electric field, 815 in magnetic field, 1014 energy flux, in electromagnetic wave, 1093–94 energy level, 244 of hydrogen, 1299 in molecules, 1333–36 energy-level diagram, 1298–99, 1299, 1334, 1337, 1337 energy quantization, of oscillator, 1259–60 energy quantum, 1258–63 energy-work theorem, 215, 236, 400 engine: automobile, 665, 674 Carnot, 667–73 efficiency of, 666–67 flowchart, 667 heat, 665 steam, 665, 671, 671 enriched uranium, 1381 entropy, 678 at absolute zero, 680 change, in isothermal expansion of gas, 668 disorder and, 680 irreversible process, 679–80 negative, 683 equation of motion, 151–53, 151, 174 integration of, 54–56 of simple harmonic oscillator, 477 of simple pendulum, 485 see also Newton’s Second Law equilibrium: electrostatic, 774 of fluid, 575 of mass, 155 nuetral, 432, 432 static, see static equilibrium unstable, 432 equilibrium point, 245, 245 equilibrium position, 476–77 equipartition theorem, 617–18 equipotential surface, 808–9 escape velocity, 292 from Earth, 292 from Sun, 292 Escher, M. C., waterfall, 662 ether, 1218–19 ether wind, 1219 evaporative cooling, 624 evaporative loss, Mediterranean, 657 excited states, 1299 Exclusion Principle, 1321, 1328–32, 1337 expansion, free, of a gas, 664, 668 expansion, thermal: of concrete, 637

linear, coefficient of, 634, 637 of solids and liquids, 633–37, 633 of water, 635, 635 expansion, volume, 634 expansion joints, bridge, 636 expansion of railroad rails, 636 Explorer I, 286, 298–99 Explorer III, 286 Explorer X, 298 exponential function, 910 external field, 742 external forces, 311 eye: astigmatic, 1147 compound, 1202, 1202 insect, 1202, 1202 myopic and hyperopic, 1146 nearsighted and farsighted, 1146, 1146 as optical instrument, 1145–47 eye, components of: cone cells, xlvii iris, xlvi retina, xlvii rod cells, xlvii Fahrenheit temperature scale, 611, 612 fallout, from nuclear accident, 1383 farad (F), 830 Faraday, Michael, 994, 998 Faraday cage, 804–5 Faraday’s constant, 715 Faraday’s Law, 997–1001, 1005, 1006, 1008, 1009, 1071, 1071, 1080, 1096–97, 1097 farsightedness, 1146, 1146 Fermi, Enrico, 1358, 1380 fermi (fm), 1358 Fermilab, 125, 422 Fermi National Accelerator Laboratory (Fermilab), 1398–99, 1398, 1415 Collider Detector, 1399, 1399 Tevatron, 1398–99, 1398 fermion, 1403 Ferris, George, 388, 421 Ferris wheel, 388, 421, 527 ferromagnetic materials, 947 ferromagnets, 977 Feynman, Richard P., 1410, 1410 Feynman diagram, 1410, 1410 fibrillation, 913 field, 722 depth of, 1145 electric, see electric field electron, 1409–10 as fifth state of matter, 722 magnetic, see magnetic field quanta and, 1409–11, 1409t radiation, of accelerated charge, 1075–76, 1076, 1077 field lines, 736–39 density of, 739 of point charge, 736–37, 737, 739 sources and sinks of, 738–39 fifth state of matter, 722 filter, polarizing, 1084–86, 1084, 1086 fine-structure constant, 1315 fire extinquisher, 595 fire hose: pressure within, 596 rate of water flow, 590t

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Index

First Law of Thermodynamics, 662–64 first overtone, 522, 522 fission, 252, 1355, 1377–79 in alpha-decay reaction, 1366, 1366 as source of nuclear energy, 1242 flat sheet, electric field of, 734–36, 759 flip coil, 1019 flow: around an airfoil, 570, 582–83 of free electrons, 866–67 of heat, 638–39, 638 incompressible, 569 laminar, 569 methods for visualizing, 570 in a nozzle, 586 from source to sink, 570 steady, 569 streamline, 569 turbulent, 571, 571 velocity of, 566, 568 around a wing, 571, 584 flowmeter, Venturi, 585–86, 596, 596 fluid, 566 density of, 566–67, 567 equilibrium of, 575 incompressible, 569 nuclear, 1360 static, 575–79, 575 fluid dynamics, 582–87 fluid mechanics, 565–99 flute, 519, 546, 560, 561 flux: electric, 757–61 magnetic, 997–1001, 999 f number, 1144, 1152 focal length, of spherical mirror, 1128 focal point: of lenses, 1136 of mirror, 1128 foot, 6–7 force, 133, 133, 135, 135t, 136 buoyant, 580–81 calculated from potential energy, 241 center of, 240 centrifugal, 188–89, 189 “color,” 1405, 1414–15 conservative, 236–43, 238, 288 conservative vs. nonconservative, 238–39 contact, 142–43, 143, 697 as derivative of potential energy, 241 electric, see electric force electromagnetic, 191, 695, 1405, 1406, 1406t electromotive, see emf (electromotive force) electrostatic, 695–97 electroweak, 1411 external, 311 fundamental, 191 gravitational, see gravitational force impulsive, 339–44 internal and external, 311 inverse-square, 240 moment arm of, 400 motion with constant, 151–59 net, 138–40, 138, 139 normal, 143, 143, 144, 147 nuclear, 1355 power delivered by, 255 between quarks, 1414–15 restoring, 182–84, 183

resultant, 138 of a spring, 182–84, 183, 184 “strong,” 191, 1355, 1359–65, 1405, 1406, 1406t, 1414 torque and, 395–97 units of, 135–36, 141 “weak,” 191, 1369, 1405, 1406, 1406t work done by constant, 205, 208 work done by variable, 211–13, 211, 212, 213, 214 force, magnetic, 695, 697, 927–31, 927 due to magnetic field, 931 magnitude of, 933 on moving point charge, 928–30 right-hand rule for, 934, 935, 936 vector, 933 on wire, 969–72 forced oscillations, 488, 490–91, 1046 Fornax constellation, xliv Foucault pendulum, 499 Fourier’s theorem, 519, 519, 538 fractal striations, 492, 493 frames of reference, 3, 4, 20, 114, 115 in calculation of work, 208 of Earth, 132 freely falling, 142, 142 inertial, 132, 132, 133, 1219 for rotational motion, 366 Franklin, Benjamin, 710, 710 Fraunhofer diffraction pattern, 1194 Fraunhofer Lines, 1290 “free-body” diagram, 146, 146, 147, 148, 153, 157, 158, 159, 176, 180, 181, 184, 186, 187, 188 for automobile tire, 403 for backbone as lever, 442 for box on truck, 440 for box titled, 439 for bridge, 433 for ladder, 438 for pulley, 402, 443 for string-bob system, 484 for tower crane, 435–36 free charges, 844 free-electron gas, 711 friction in, 863–65 free electrons, 708 flow of, 866–67 free expansion of a gas, 664, 689 entropy change in, 668 free fall, 49–54, 51, 52, 53, 64, 141, 142, 142, 495–96 formulas for, 49, 50 universality of, 49, 49 weightlessness in, 142, 142 French Academy, speed of sound, 560 French horn, 546 freon, 672–73 frequency, 369–70 beat, 518 normal, 523 proper, 523 resonant, 1043 of simple harmonic motion, 470–71 threshold, 1266 of wave, 510–11 frequency bands, of electromagnetic radiation, 1090–91 frequency filter circuit, 1037

A-83

Fresnel, Augustin, 1191 Fresnel lens, 1156 friction, 171–81 air, 49, 51, 61, 181, 181 coefficient of kinetic, 175–78, 175t coefficients of, 174–81, 175t of drag forces, 180–81 equation for kinetic, 175 equation for static, 179 in flow of free–electron gas, 863–65 heat produced by, 248 kinetic (sliding), 174–78, 175, 176, 177, 190 loss of mechanical energy by, 238–39 microscopic and macroscopic area of contact and, 174–75 as nonconservative force, 238–39 static, 178–80, 179, 180, 190 static, coefficient of, 175t, 179–80 of viscous forces, 181 frictional electricity, 695, 711 fringes, 1170 fringing field, 792, 848 fuel cells, 892–93 on Skylab, 892 fuel consumption, 263t fuel rods, of nuclear reactor, 1381 function, oscillating, 1047, 1051 fundamental forces, 191 strength of, 191 fundamental frequency, 523 fundamental mode, 522, 522 fusion: heat of, 642, 642t nuclear, 1421 g, 40, 52–53, 189 measurement of, 52–53, 277–78 NASA centrifuge, 592 standard, 52–53 Gagarin, Yuri, 299 gain factor, 1343 galaxies, 1417–18 Milky Way, 1417, 1417 Galilean addition law for velocity, 1218 Galilean coordinate transformations, 1218, 1234, 1241 Galilean telescope, 1142 Galilean transformation: for momentum, 1239 for velocity, 1218 Galilean velocity transformation, 116 Galilei, Galileo, 51, 131, 495 claim on isochronous pendulum, 495, 501 experiment on free fall, 495 experiments on universality of free fall by, 495 isochronism of pendulum and, 501 pendulum experiments by, 495 tide theory of, 120 Galvani, Luigi, 793 galvanometers, 975 gamma emission, 1369–70 gamma rays, xlix, 1090, 1365, 1421 gas: adiabatic equation for, 649 diatomic, energy of, 617, 617, 620, 625 diatomic, molar heat, 645, 646t distribution of molecular speeds in, 615 energy of ideal, 616–19 entropy change in isothermal expansion of, 668

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A-84

gas (continued) of free electrons, 711, 863–65 free expansion of, 664, 664, 668 ideal, see ideal gas Law of Boyle for, 606 Law of Charles and Gay-Lussac for, 606 monatomic, energy of, 616–18, 620 monatomic, molar heat, 645, 646t polyatomic and non-linear, 618, 620 polyatomic and non-linear, molar heat, 646t root-mean-square speed of, 614–15 specific heat of, 644–47 gas constant, universal, 604 gas thermometer, constant-volume, 609–10, 609 gauge, pressure, 578 gauge blocks, 6, 6 gauss (G), 934 Gaussian surface, 762 Gauss’ Law, 274, 757, 762–72, 790, 1422 in dielectrics, 842 for electricity, 1074 for magnetic field, 937, 940 for magnetism, 1074 Gay-Lussac’s Law, 606 Gedankenexperiment, 287 Geiger, H., 1294 Gell-Mann, Murray, 1413 General Relativity theory, 394 general wavefunction, 511 generator: electric, 892 electromagnetic, 1003–4 homopolar, 1004–5 geometric optics, 1112 geometry (review), A3–27 angles, A3 areas, perimeters, volumes, A3 geostationary orbit, 271–72, 281 geostationary satellite, 280, 281, 290–91 geosynchronous orbit, 271–72 geothermal power plant (Waiakei, NZ), 690 Germer, L., 1303 GeV, 1398–99, 1411 gimbals, 414 Glashow, Sheldon Lee, 1411 Global Positioning System (GPS), 25, 1216, 1220, 1228 gluon, 1409, 1411 golf ball, club impact, 347 Goudsmit, Samuel, 1325 gradient, of the potential, 806–10 grandfather clock, 219, 219 grand unified theory (GUT), 1411 graphite, xlix grating, 1184–85, 1184 principal maximum of, 1184–85 reflection, 1185 resolving power of, 1186 gravitation, 271–303 law of universal, 131, 272–76, 278 gravitational constant, 273 measurement of, 277–78 gravitational force, 191, 272–76, 1405, 1406, 1406t acting on center of mass, 430–33 compared to electric force, 696, 699 gravitational interactions, 1405, 1406, 1406t gravitational potential energy, 218–23, 219, 220, 238, 288–93 of a body, 321

Index

graviton, 1409, 1411, 1415 gravity: action and reaction and, 146, 146 as conservative force, 288 of Earth, see weight galaxies and, 1420 work done by, 207, 207 gravity, acceleration of, 52–53, 64, 274–75 measurement of, 52–53 variation of, with altitude, 274–75 Gravity Probe B satellite, 394, 394, 401, 401, 406, 414 gray, 1375 Great Pyramid (Giza), 318, 334 Greek alphabet, A1 green, 1414–16 Greenwich time, 9 Griffiths-Joyner, Florence, 59 ground, electric, 803 ground state, 1299 guard rings, for capacitors, 848, 848 Guericke, Otto von, 161, 591 guitar, 523, 526, 530, 531 lowest note available, 558 gyrocompass, 414 gyroscope, 394, 394, 401, 406, 414, 414 precession of, 415–16 hadron, 1403 Hahn, Otto, 1377 Hale-Bopp comet, 299, 299 half-life, 1372–73 Hall, Edwin Herbert, 983 Hall coefficient, 990 Hall effect, 980–83 Halley’s comet, 298, 298 Hall sensors, 982 Hall voltage, 981 hammer in free fall, 366 harmonic function, 470 harmonic motion, damped, 489 see also simple harmonic motion harmonic oscillator, 477, 492 harmonic wave, 510–13, 1098 harmonic wavefunction, 513–16 Hawking radiation, 1281 hearing: dog, 558 threshold of, 538 hearing trumpet, 559 heat, 248, 628–60 as energy transfer, 662–63 of fusion, 642, 642t latent, 642 mechanical equivalent of, 631–32 specific, see specific heat temperature changes and, 630–31 transfer, by convection, 641 transfer, by radiation, 641 transfer; see also R value of transformation, 642 of vaporization, 642, 642t heat capacity, specific, 630 heat conduction, 638–42 equation of, 638–39 two-layer, 640, 640 heat current, 638 heat engine, 655, 665

heat flow, 638–39, 638 across electric insulator, 656 heat pump, 672, 672, 673 heat reservoir, 665 height, maximum, of projectiles, 109–11 Heisenberg, Werner, 1278, 1278, 1302 Heisenberg’s uncertainty relation, 1278 heliocentric system, 279 helium: ion collision with O2, 352 and hydrogen, abundance of in universe, 1421 monatomic kinetic energy, 616 helium liquid, 624 helium nuclei, 1366 Helmholtz, Hermann von, 248 henry, 1010 Henry, Joseph, 1011 hermetic chamber, 626 Hertz, Heinrich, 1080, 1080, 1265 hertz (Hz), 471 high-voltage transmission line, power dissipated in, 904 Hobby-Eberly telescope, 1152, 1152 “holes,” 982 in semiconductors, 1338–39, 1340–42 homopolar generator, 1004–5 Hooke’s law, 182–84, 445–46, 476 Hoover Dam, 993 horizontal velocity, 103–8, 103, 106, 107, 108 horse, work efficiency, 689, 689 horsepower (hp), 254, 254, 256 house walls, heat flow in, 639–40, 652, 659 Hubble, Edwin, 1416–17, 1418 Hubble’s Law, 1418–19, 1420 Hubble Space Telescope, 1111, 1130, 1135, 1152, 1198, 1198, 1209 human: aorta, average blood flow, 590 chemical energy conversion, 685–86 circulatory system, 571, 571 ear, audible frequencies, 558 heart pressure, 591 muscle striation, 685 power ed bicycling, 666, 666 powered flight (over English Channel), 264 venous presssure, 597, 597 voice, pitch, 559 human body: lever-like motion of, 442, 451, 458–59 mean tissue density, 598, 626 speed of sound in, 558 temperature, 612 ultrasound frequencies used in, 558, 562 hurricane, barometric pressure, 598 Huygens, Christiaan, 221, 495 Huygens’ Construction, 1113–14, 1113 Huygens-Fresnel Principle, 1190–91 Huygens’ tilted pendulum, 495, 495 Huygens’ wavelets, 1113, 1120 Hyades Cluster, xliii Hyatt Regency hotel, collapse of “skywalks” at, 451, 451 hydraulic brake, 575–76, 576 hydraulic car jack, 576 hydraulic press, 575, 575 hydroelectric pumped storage, see Brown Mountain hydrogen: atom, stationary states of, 1298, 1300

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Index

atomic mass, xlix electron orbit, 423 energy levels of, 1300 and helium, abundance of in universe, 1421 interstellar gas, density and temperature, 621 isotopes of, 1357 quantum numbers for stationary states of, 1328 spectral series of, 1291–93, 1299, 1324 hydrogen bomb, 1380 hydrogen molecule: diatomic gas, 625 oscillations of, 483, 483 vibration frequency, 498 hydrogen spectrum, 1291–92, 1291t produced by grating, 1184 hydrometer, 594 hyperbola, 291 hyperbolic orbit, 291 hyperopic eye, 1146 ice: density of, 581, 581 density v. liquid water, 634 iceberg, 593, 593 ideal gas, 604 energy of, 616–19 kinetic theory and, 602–27 Ideal Gas Law, 603–8, 627, 646, 670 ideal-gas temperature scale, 609–12 ideal particle, 3, 20 ideal solenoid, 943–44, 943 image: real, 1133 virtual, 1116 image charges, 727 image orthicon, 1269 impact parameter, 1294, 1294 impedance, for RLC circuit, 1050 impulse, 339–40 impulsive force, 339–44 impurities, acceptor, 1339 incidence, angle of, 1115–16, 1115 inclined plane, 153, 157 incompressible flow, 569 incompressible fluid, 569, 576 indeterminacy relation, 1278 index of refraction, 1118, 1119t, 1125–26, 1125 induced dipole moments, 744 induced electric field, 994, 996 induced emf, 994, 998–1007 induced magnetic field, 1071 in capacitor, 1072, 1082 inductance, 1008–12 mutual, 1009–10 self-, 1011 induction: electromagnetic, 993–1029 electrostatic, 711 Faraday’s Law of, 997–1001 induction furnace, 1024 induction microphone, 1000–1001 inductive reactance, 1039–40 inductor: circuit with, 1038–41 current in, 1038–41 inelastic collision, 348 conservation of energy in, 351–52, 353 conservation of momentum in, 351–52, 353 in three dimensions, 351–53

totally, 348 in two dimensions, 351–53 inertia: law of, 132, 132 moment of, see moment of inertia inertial reference frames, 132, 132, 133 infrared radiation, 1090 initial speed, 111 instantaneous acceleration, 40–41, 41 components of, 97 in two dimensions, 96–97 instantaneous angular acceleration, 370 instantaneous angular velocity, 369 instantaneous power, 253 instantaneous velocity, 35–39, 36, 61 components of, 96 as derivative, 38 formulas for, 37, 38 graphical method for, 37 numerical method for, 37–39 as slope, 35–37 in two dimensions, 96, 96 instantaneous velocity vector, 98, 98 insulating material, see dielectric insulators, 708 conductors vs., 871 electron configurations of, 1338 resistivities of, 871 integrals, for work, 212–13 integrated circuit, 1345 integration, of equations of motion, 54–56 INTELSAT, 281, 281 intensity: of sound waves, 538, 540–43, 542t, 543 for two-slit interference pattern, 1179 interaction: “color,” 1405 electromagnetic, 1405, 1406, 1406t gravitational, 1405, 1406, 1406t “strong,” 1405, 1406, 1406t “weak,” 1369, 1405, 1406, 1406t interatomic bonds, 1333 interference, 1169–86 constructive and destructive, 517, 1169, 1169–75 maxima and minima, for multiple slits, 1184, 1187–89 maxima and minima for, 1179 from multiple slits, 1183–89 in thin films, 1169–73 two-slit, pattern for, 1180, 1180 from two slits, 1177–83, 1177 interferometer, Michelson, 1174–77, 1175 interferometry, very-long-baseline, 1202 internal energy, 616 internal forces, 208, 311 internal kinetic energy, 348 internal resistance, of batteries, 895–96 International Bureau of Weights and Measures, 1175 International Space Station, 389, 468 international standard meter bar, 5–6, 5 International System of Units (SI) 5, 14, 972 see also system of units (SI) interstellar hydrogen gas, density and temperature, 621 invariance of speed of light, 1220–21 inverse Lorentz transformation equations, 1236 inverse-square force, 240

A-85

inverse-square law, 739 Io, 64, 64 ion gun, 740–41, 740 ionosphere, temperature and density, 620 ions, 697 in electrolytes, 711 iris, of eye, xlvi irreversible process, 678 entropy change in, 679–80 isobars, 1389 isochronism, 477, 501 of simple pendulum, limitations of, 486 isospin, 1407 isothermal compression of gas, 689 isothermal expansion of gas, entropy change in, 668 isotopes, 1355–59 of carbon, 1355–59, 1356 chart of, 1357t of hydrogen, 1357 radioactive, 1372–76 of uranium, 1362, 1383 J/ meson, 1415 Jodrell Bank radiotelescope, 1209 Jordan, P., 1302 Joule, James Prescott, 207, 629, 652 joule ( J), 206, 254 Joule heat, 902–5 Joule’s experiment and apparatus, 631–32, 631 Jupiter, 285t, 286 moons Europa, Ganymede, Io, 296 K-10000 tower crane, 429, 429, 435–37, 435–37, 448, 448 Keck Telescope, 1151 kelvin, 604 Kelvin, William Thomson, Lord, 458, 604 Kelvin-Planck statement of Second Law of Thermodynamics, 675–76 Kelvin temperature scale, 604, 609–10 Kepler, Johannes, 285 Kepler’s Laws, 282–86 of areas, 282–84 First, 282 limitations of, 288 for motion of moons and satellites, 286–88 Second, 282–84 Third, 285–86 kilocalorie, 248–49, 250 kilogram, 5, 11, 13, 14 multiples and submultiples of, 13t, 134 standard, 11, 134 kilometers per hour (km/h), 30 kilowatt-hours, 248, 250 kinematics, 29 kinetic energy, 214–17, 216, 238 equation for, 215 of ideal monatomic gas, 616–18 internal, 348 relative examples of, 216t relativistic, 1240–41 of rotation, 378–84 in simple harmonic motion, 480–83 of a system of particles, 327–28 kinetic friction, 174–78, 175, 176, 177, 190 coefficient of, 175–78, 175t equation for, 175 kinetic pressure, 613–16 kinetic theory, ideal gas and, 602–27

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A-86

Kirchhoff, Gustav Robert, 894 Kirchhoff ’s current rule, 897–900 Kirchhoff ’s rule, 1031, 1032, 1038, 1044, 1046, 1054 Kirchoff ’s voltage rule, 894, 898, 900, 908 Knot meter, 568 krypton, monatomic kinetic energy, 616 Lagrange, Joseph Louis, Comte, 236 lamda particle, 1400, 1400 laminar flow, 569 La Paz, Bolivia, airport, 620 Large Magellanic Cloud (galaxy), xliv laser, 1090 stabilized, 6, 6 laser cooling, 624 Laser Interferometer Gravitational Observatory (LIGO), 1175 laser light, 1178 laser printers, 694, 709 latent heat, 642 launch speed, 111 Lave, Max von, 1273 Lave spots, 1273 law of areas, 283–84 Law of Boyle, 606 Law of Charles and Gay-Lussac, 606 law of conservation of angular momentum, 407–9 Law of Conservation of Electric Charge, 710 Law of Conservation of Energy, 662, 790 Law of Conservation of Energy, general, 248, 249, 252 law of conservation of mechanical energy, 221–23, 221, 222, 223, 238, 584 law of conservation of momentum, 309 law of inertia, 132, 132 law of Malus, 1086 law of radioactive decay, 1372–76 law of reflection, 1114–15 law of refraction, 1121 law of universal gravitation, 131, 272–76, 278 Lawrence, Ernest Orlando, 968 laws of planetary motion, 282–86 Lawson’s criterion, 1391 LC circuit, 1041–46 energy in, 1042 freely oscillating, 1041–46 natural frequency of, 1042 lead-acid battery, 707, 890–91, 891, 893 leaf spring oscillator, 492 length, 5–8 precision of measurement of, 6 standard of, 5–6, 5, 6 length contraction, 1230–32 visual appearance and, 1230–32 Lennard-Jones potential, 262, 263 lens: focal point of, 1136 magnification produced by, 1139 objective, 1149–50 ocular, 1149–50 thin, 1135–39 lens equation, 1138 lens-maker’s formula, 1135–36 Lenz’ Law, 1001–2, 1006 lepton, 1403, 1403t, 1406 lepton number, 1406, 1407 “let-go” current, 913 Leverrier, U. J. J., 272

Index

levers, 441–45, 445, 458–59 human bones acting as, 442, 442, 451, 458–59 mechanical advantage and, 441, 441, 444 Lichtenstein, R., 294 lift, 584 light: coherent, 1177 dispersion of, 1125 Doppler shift of, 1228–29 from laser, 1178 polarized, 1126–27 pressure of, 1094–95 quanta of, 1254–55 reflection of, 1112, 1114–17 refraction of, 1112, 1117–27 spectral lines of, 1127 spectrum of, 1126, 1184, 1291–93 ultraviolet, 1090 unpolarized, 1084, 1084 visible, 1090–91 light, speed of, 6, 1218–20 in air and water, 527 invariance of, 1220–21 in material medium, 1118 measurement of, 1080 universality of, 1220–21, 1238 light-emitting diode (LED), 1343–44 lightning, 710, 721, 805 distance of, 560 lightning rod, 710–11 light waves, 1079–1110 coherent, 1090 incoherent, 1090 light-year, xlv, 24 linear dielectric, 838 linear magnification, 1141–42 linear polarization, 532, 532 lines of electric field, 736–39 made visible, 859, 859 liquid drop model of nucleus, 1360 liquids: bulk moduli for, 447–48 thermal expansion of, 633–37, 633 locomotive steam engine, 661, 666, 671 Loki (volcano on Io), 64 Long Island, xxxviii longitudinal wave, 508, 509, 509 long wave, 1088 looping the loop, 187–88, 187 loop method, for circuits, 898 Lorentz, Hendrik Antoon, 1235 Lorentz force, 933 Lorentz transformations, 1232–38 for momentum and energy, 1234–37 Lyman series, 1292 Mach, Ernst, 552 Mach cone, 552, 552 Mach number, 562 macroscopic and microscopic parameters, 603 Magdeburg hemispheres, 161, 161 magic nuclei, 1389 magnet, permanent, 977 magnetic constant, 929 magnetic dipole moment, 973 magnetic energy, 1012 in inductor, 1013–15 magnetic field, 931–38, 935t circular motion in, 965–69

circular orbit in, 966 energy density in, 1014 Gauss’ Law for, 937 generated by current, 939–40, 940, 948 induced, in capacitor, 1072, 1072 made visible with iron filings, 937, 942 magnetic force and, 931 of plane wave, 1081–83, 1082, 1097 of point charge, 936 of point charge, represented by field lines, 936–37 principle of superposition for, 938 right-hand rule for, 932, 932, 933 of solenoid, 943–44 of straight wire, 940–41, 941 magnetic field lines: of circular loop, 942, 942 made visible with iron filings, 942, 942, 943 magnetic flux, 997–1001, 999 magnetic force, 695, 697, 927–31, 927 magnetic field and, 931n magnitude of, 933 on moving point charge, 928–30 right-hand rule for, 934, 935, 936 vector, 933 on wire, 969–72 magnetic levitation, 1008–9 magnetic moment, 1355 of electron, 1325 spin, 976 magnetic permeability, 977 magnetic quantum number, 1321, 1328 magnetic recording media, 978–79, 978–79 magnetic resonance imaging (MRI), 656, 977, 977 magnetism, Gauss’ Law for, 1074 magnetization, 977, 984 magneto, 1019 magnification, 1139 magnifier, 1147–49 major axis of ellipse, 282 Malus, Étienne, 1086 Malus, law of, 1086 mandolin, 546 frequencies, 530, 531, 558 manometer, 578, 578 marker point, 29, 29 Mars, xl, 285t, 286, 286 Mars Climate Orbiter, 16, 16 Marsden, E., 1294 mass, 11–13 atomic, 11–12 atomic standard of, 11 of carbon, 1358 center of, see center of mass conservation of, 205, 252 definition of, 134–35 of electron, 137, 137t, 1356 of elementary particles, 1403t, 1404t and energy, 1242–44 energy and, 251–53 equilibrium of, 155 molecular, 11 moment of inertia of continuous distribution of, 379–80 of neutron, 137, 137t, 1356 of proton, 137, 137t, 1355 relative examples of, 12t standard of, 134–35

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Index

of universe, 1422–23, 1423t weight vs., 141 mass defect, 1362 mass-energy relation, 1242–44, 1361 mass number, 1356 binding energy per nucleon vs., 1361 conservation law for, 1406 mass spectrometer, 986, 1393 mathematics (review), A1–27 algebra, A3–25 arithmetic in scientific notation, A2–23 equation with two unknowns, A5 exponent function, A5–27 logarithms, common (base–10) and natural, A5–27 powers and roots, A1–22 quadratic formula, A5 symbols, A1 see also calculus; geometry; trigonometry; uncertainties (propagation of ) matter, dark, 1423 matter-antimatter annihilation, 706 maximum height, of projectiles, 109–11 Maxwell, James Clerk, 1071, 1080, 1411 formula for speed of light by, 1080 Maxwell-Ampère’s Law, 1071, 1073, 1074, 1080, 1096, 1097, 1097 Maxwell distribution, 615, 615 Maxwell’s demon, 683 Maxwell’s equations, 1071–72, 1074–75, 1112 Mayer, Robert von, see von Mayer, Robert mean free path, 624 mean lifetime, 1373 mean solar day, 9 mechanical advantage, 441, 441, 443–44 mechanical energy, 238 law of conservation of, 221–23, 221, 222, 223, 238 loss of, by friction, 238–39 mechanical equivalent of heat, 631–32 mechanics, 29 classical vs. quantum, 1287 fluid, 565–99 Mediterranean, evaporative loss, 657 medium, wave propagation, 508 medium wave, 1088 Meissner effect, 989 Meitner, Lise, 1377, 1377 melting points, common substances, 642t Mercedes Benz automobile crash test, 340 Mercury, xl, 284, 285t, 286, 286 mercury barometer, 577 mercury-bulb thermometers, 610, 610, 636, 636 merry-go-round, 367 meson, 1403, 1404, 1405, 1405t, 1406, 1415 metal: resistivities, 870 temperature dependence of resistivity of, 868–69 Meteor Crater (Arizona), 362 meteoroid incidents, 301 meter, 5–6, 13–14 cubic, 13 cubic, multiples and submultiples of, 14t square, 13, 13 square, multiples and submultiples of, 14t meters per second (m/s), 30 meters per second squared (m/s2), 39 metric system, 5

MeV, 1359, 1361–62, 1364, 1397, 1415 Michelson, Albert Abraham, 1175, 1219, 1220 Michelson interferometer, 1174–77, 1175 Michelson-Morley experiment, 1175–76, 1219, 1220 microelectromechanical system (MEMS), 503, 503 microfarad, 830 micrograph, acoustic, 539 microscope, 1149 angular magnification of, 1149 atomic-force (AFM), 475, 475, 1311, 1311 electron-holographic, 694 scanning capitance, 849 microscope, electron: scanning (SEM), xlvii, 1310, 1310 scanning tunneling (STM), xlviii, xlix, 1311, 1311, 1320 transmission scanning (TEM), xlviii microscopic and macroscopic parameters, 603 microwaves, 1084, 1090 Midas II satellite, 296 middle C, 539, 539 Milky Way Galaxy, xliii, xliv, 1417, 1417 Millikan, R. A., 1267 Millikan’s experiment, 747 mirror: focal point of, 1128 image formed by, 1116–17 magnification produced by, 1139 plane, 1116 reflection by, 1116–17 spherical, 1128–35 in telescope, 1151 mirror equation, 1131 mirror nuclei, 1388 mode, fundamental, 522 moderator, in nuclear reactor, 1381 modulation, of wave, 518 modulus, see bulk modulus; elastic moduli; shear modulus; Young’s modulus mole, 11, 20, 604–8 molecular mass, 11 molecular speeds: distribution of, in gas, 615 Maxwell distribution, 615 most probable speed, 615 root mean square, 614 molecule, diatomic, 244 molecules: energy levels in, 1333–36 quantum structure of, 1321, 1333–36 rotational energy of, 1334 vibrational energy of, 1333 water, 389, 743 moment, magnetic dipole, 973 moment arm, 400 moment of inertia: of continuous mass distribution, 379–80 of Earth, 388, 389, 390–91 of nitric acid molecule, 388 of oxygen molecule, 388 of system of particles, 378–84 of water molecule, 389 momentum: angular, see angular momentum of an electromagnetic wave, 1094–96 energy and, relativistic transformation for, 1242–44

A-87

Galilean transformations for, 1239 Lorentz transformations for, 1234–37 of a photon, 1270 rate of change of total, 312 relativistic, 1239–40 of a system of particles, 306–13, 324 momentum, conservation of, 307–12, 310, 345, 348 in elastic collisions, 344–45, 351–52, 353 in fields, 723 in inelastic collisions, 351–52, 353 law of, 309 monatomic gas, kinetic energy of ideal, 616–18 Moon, xxxix moons of Saturn, 296, 297t Morley, E. W., 1175, 1219, 1220 most probable speed, 615, 615 motion: absolute, 1217 along a straight line, 28–68, 32 amplitude of, 470 angular, 375 of center of mass, 323–27 circular, translational speed in, 37 with constant acceleration, 42–49, 43, 63 with constant acceleration, in three dimensions, 102–4, 103, 104, 122 with constant force, 151–59 cyclic, 469 energy conservation in analysis of, 223 equation of, see equation of motion free-fall, 49–54, 51, 52, 53, 64, 141, 142, 142 harmonic, 489 Newton’s Laws of, 130–72 one-dimensional, 28–68, 32 parabolic, 108–9, 109 periodic, 469 planetary, 282–86 position vs. time in, 32–33, 33, 34, 35, 35, 36, 60 of projectiles, 104–12, 122, 124 of projectiles, formulas for, 104 as relative, 31, 115–18, 1217 of rigid bodies, 366–67 rotational, see rotational motion simple harmonic, see simple harmonic motion three-dimensional, 95 with time-dependent angular acceleration, 376–78 translational, 29, 95, 95, 120, 366, 404 two-dimensional, 94–129 uniform circular, 112–15, 112, 113, 125, 184–90 in uniform electric field, 740–44 with variable acceleration, 54–56 wave, 508–9 motional emf, 993–97 Mount Fuji, 334 Mt. Everest, descent, 127 Mt. Palomar telescope, 1151, 1151 Mt. Pelée volcano, 622 multiloop circuits, 897–900 multiplate capacitor, 851, 851, 852, 852 multiwire chambers, 1401–2, 1402 muon, 1403 musical instruments, standing waves, 546, 546 mutual inductance, 1009–10 muzzle velocity, 287, 331 myopic eye, 1146

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A-88

nanoelectromechanical system (NEMS), 503 NASA centrifuge, 592 NASA Spacecraft Center, 114, 114 NASA weightlessness training, 589 National Ignition Facility (NIF), 828, 845 National Institute of Standards and Technology, 9, 500, 609 National Ocean Survey buoys, 527 National Research Council of Canada, speed of sound, correction, 560 navigation, vectors in, 71 near point, 1147 nearsightedness, 1146, 1146 negative acceleration, 39 negative vectors, 75, 75 negative velocity, 39 negentropy, 683 neon, 696 atomic structure of, xlix nucleus of, l–li Neptune, 285t, 286 discovery of, 272 Nernst, Walther Hermann, 681 net force, 138–40, 138, 139 neutral equilibrium, 432, 432 neutrino, 1368–69, 1403, 1423 neutron, 696, 1355, 1397 beta-decay reaction and, 1368 isotopes and, 1356 mass of, li, 137, 137t, 1356 nuclear binding energy and, 1359–62 in nuclear reactor, 1381 quark structure of, 1414 Newton, Isaac, 131, 131, 287, 1358 on action-at-a-distance, 722 bucket experiment of, 588 experiments on universality of free fall by, 496 pendulum experiments by, 496 newton (N), 135, 141 newton-meter, 396 newton per coulomb, 724 Newton’s cradle (pendulum), 355 Newton’s Laws: angular momentum and, 400 First, 131–33, 131 of Motion, 130–72 in rotational motion, 395 Third, 144–51 of universal gravitation, 131, 272–76, 278 Newton’s rings, 1173 Newton’s Second Law, 133–38, 171, 205, 214, 400, 413, 514, 604 centripetal acceleration and, 185 empirical tests of, 135 see also equation of motion Newton’s theorem, 274 New York City, xxxvii–xxxviii, 386 electric power, 653 Public Library, xxxvii NGC 2997 (spiral galaxy), xliv Niagara Falls, 266, 688 nitric acid molecule: distance and angles between component atoms, 333 moment of inertia of, 388 nitrogen molecule: diatomic kinetic energy, 617, 617 nodes and antinodes, 520–21, 521, 544–45 noise, white, 538

Index

noise reduction, 557 normal force, 143, 143, 144, 147 normal frequencies, 523 normalization condition, 1308 n-p-n junction transistor, 1342, 1342 n-type semiconductor, 1338–42, 1341, 1343–45 nuclear bomb, 1355 nuclear density, 1359 nuclear explosion, overpressure of blast wave, 590 nuclear fission, see fission nuclear fluid, 1360 nuclear force, see “strong” force nuclear fusion, 1421 nuclear model of atom, 1294, 1294 nuclear power plant, 1382–83 nuclear radius, 1358 nuclear reactions, 1371 nuclear reactor, 1381–82, 1381 containment shell of, 1382, 1382 control rods in, 1381 fission, 1380–81 fuel rods in, 1381 heavy water, 358 mass and, 1293 moderator in, 1381 neutrons in, 1381 plutonium (Pu) produced in, 1383 rods in, 1381 water-moderated, 1383 nucleons, 1355, 1355t, 1359 see also neutron; proton nucleus, 1355 binding energy of, 1359–65, 1359, 1360 density of, 1359 electric force and, 1355, 1359 of helium, 1366 liquid-drop model of, 1360 magic, 1389 mirror, 1388 stable and unstable, 1360 Oak Ridge National Laboratory, 387 object, 1131 objective lens, 1149–50 observable universe, 1420 ocean, energy extraction, 687 ocean waves: amplitude of, 558 diffraction of, 553 see also seiche; tides; tsunami octave, 539 ocular lens, 1149–50 Oerlikon Electrogyrobus, 391 Oersted, Hans Christian, 930 Ohm, Georg Simon, 866 ohm (), 868 ohmmeter, 916, 916 Ohm’s Law, 866, 1034 oil pipeline, lateral loops, 651 oil tanker: cross section, 591 see also supertanker ommatidia, 1202 opera house acoustics, European, 557 optical fiber, 1124 optical pyrometer, 610, 610 optics: geometric, 1112

geometric vs. wave, 1169 wave, 1169 orbit: bound, 245 circular, see circular orbits of comets, 290–91 elliptical, 282–86, 282, 287, 291–92, 1321–22 geostationary, 271–72, 281 geosynchronous, 271–72 hyperbolic, 291 parabolic, 287, 291 period of, 279 planetary, 279, 286 planetary, data on, 285–86, 285t synchronous, 271–72 unbound, 245 orbital angular momentum, 409 orbital motion, energy in, 288–93 orbital quantum number, 1321, 1328 organ, pipe, 546, 546, 561 Organisation Européenne pour la Recherche Nucléaire (CERN), 1225, 1238, 1398–99, 1402, 1402, 1411 origin of coordinates, 3, 3, 4, 44, 45, 46, 47 orthicon, 1269 oscillating beads, 497 oscillating function, 1047, 1051 oscillating mass on spring, 473 oscillations, 468–506 forced, 488, 490–91, 1046 oscillator, 493 damped, see damped oscillator double-well, 492 energy quantization of, 1259–60 harmonic, 492 leaf spring, 492 simple harmonic, 476–79, 481 torsional, 500 overpressure, 578 nuclear blast wave, 590 overtones, 522 Oxygen, liquify, 658 oxygen molecule: collision with He, 352 diatomic kinetic energy, 617, 617 moment of inertia of, 388 palladium, atoms, xlix parabola, 291 parabolic motion, 108–9, 109 parabolic orbit, 291 vs. elliptical orbit for projectile, 287 parallel-axis theorem, 382–83 parallel-plate capacitor, 831–32, 851, 851, 852, 852 paramagnetism, 977 parbuckle, 459 parent material, 1366 parity, 1407 parsec, 24 particle: accelerators for, 1363, 1363, 1398–99 alpha, 1293–94 center of mass of system of, 313–23 decays of, 1365–70 electric charges of, 696–97, 698, 706, 1414 elementary, see elementary particles ideal, 3 kinetic energy of system of, 327–28

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Index

moment of inertia of system of, 378–84 momentum of system of, 306–13, 324 in quantum mechanics, 1302–9 scattering of, 1294 stable, 1404 system of, see system of particles unstable, 1404 W, 1409–11 w, 1409, 1411, 1415 W, 1409, 1411, 1415 wave vs., 1276–79 Z, 1409–11 Z0, 1409, 1411, 1415 pascal, 573 Pascal, Blaise, 574, 588 Pascal’s Principle, 575, 575, 599 Paschen series, 1292 Pauli, Wolfgang, 1325, 1328–32, 1329t, 1368 pendulum, 476, 495, 496, 500 ballistic, 349–50 Foucault, 499 Huygens’ tilted, 495 physical, 487, 501 see also Foucault pendulum; Huygens’ tilted pendulum; Newton’s cradle; simple pendulum pendulum clocks, 487, 654–55, 655 penetration depth, 989 Pentecost Island “land divers,” 356 Penzias, Arno A., 1421 perigee: of artificial satellites, 286 of comets, 291 of Earth, 295 of Moon, xxxix perihelion, 282 of planets, 286 perihelion, 282, 284, 409 of comets, 291 of Earth, 295 of planets, xl, 285 period, 369 of comets, 291 of orbit, 279 of planets, 285t of simple harmonic motion, 470–71 of wave, 510, 510 periodic motion, 469 Periodic Table of Chemical Elements, 1328–32, 1329t periodic waves, 509–16 permanent dipole moments, 743 permanent magnet, 977 permeability constant, 929 permittivity constant, 699 perpetual motion machine: of the first kind, 662 hypothetical, 662 M. C. Escher, 662 of the second kind, 662 Perseus Cluster, xliii phase, 471 phase constant, 471 phase , RLC current of, 1050 phasor, 1047, 1047, 1049, 1187 Phobos, moon of Mars, 424 photocopiers, 694 photoelectric effect, 1264–69 photoelectric equation, 1266 photographic camera, 1144–45, 1144

photomultiplier, 1268 photomultiplier tube, 1266–67 photons, 1255, 1264–69 in Compton effect, 1270–71 emitted in atomic transition, 1300–1301 energy of, 1264–65 momentum of, 1270 virtual, 1410 physical pendulum, 487, 501 piano: frequencies, 531 notes avaiable on, 558 picofarad, 830 pion, 1400, 1401, 1404, 1414 pitch, 366 Pitot tube, 596, 596 planar symmetry, 770–72 Planck, Max, 675, 1259–61 Planck’s constant, 976, 1259, 1403 Planck’s Law, 1259–60 plane mirror, 1116 planetary motion: conservation of angular momentum in, 284 Kepler’s laws of, 282–86 planetary orbit, 279 aphelion of, 285 data on, 285–86, 285t perihelion of, 285 periodicity of, 492 period of, 285t plane wave, 537 electric and magnetic fields of, 1081–83, 1082, 1097 plane wave pulse, electromagnetic, 1080–83, 1082 plasma, 710 Pleiades Cluster, xliii “plum-pudding” model, 1293 Pluto, xli, 285t, 286 xli, mass of, 294 plutonium (239Pu), produced in nuclear reactor, 1383 in chain reaction, 1379 p-n junctions, 1340–42, 1341, 1343–45, 1343 point charge, 699 electric field lines of, 736–37, 737, 739, 936 electric field of, 723–24, 728 energy of, 796 magnetic field of, 936 moving, magnetic force and, 928–30 potential of, 794–99 Poisson spot, 1202, 1202 polarization, 839–40, 1083, 1126–27 circular and linear, 532, 532 polarized light, 1126–27 polarized plane waves, electric and magnetic fields of, 1081–83, 1082 polarizing filter, 1084–86, 1084, 1086 Polaroid, 1085, 1086 polyatomic and non-linear molecules, energy in, 618, 620 position, time vs., 32–33, 33, 34, 35, 35, 36, 60 position vector, 76–77 positive acceleration, 39–40 positive velocity, 39 positron, 1369 positronium, 1316 potential: derivatives of, 806–8 electrostatic, 790–98, 798–803

A-89

gradient of, 806–10 of point charge, 794–99 of spherical charge distribution, 799–800 potential difference, 793 potential energy, 218 of argon (Lennard-Jones potential), 263 of conductor, 812–13 of conservative force, 236–43 of current loop, 975–76 curve of, 244–47, 244 of dipole, 743 in double-well oscillator, 492 elastic, 236–37, 238 of force, equation for, 239 force calculated from, 241 gravitational, 218–23, 219, 220, 238, 288–93 gravitational, of a body, 321 in simple harmonic motion, 480–83 of a spring, 236–37, 237, 241 turning points and, 244–45 potentiometers, 872, 906 pound (lb), 12, 134 pound-force (lbf ), 136 pound-force per square inch, 574 Powell, Asafa, 31 power, 253–58 average, 253 delivered by force, 255 delivered by source of emf, 901 delivered by torque, 397 dissipated by resistor, 1033 dissipated in high-voltage transmission line, 904 instantaneous, 253 sample values of somepowers, 257t time-average, 1033 transported by a wave, 516 power brake, 456 power cable, 444 copper contraction, 461 Poynting, John Henry, 1093 Poynting vector, 1093–94 precession, 415 of a gyroscope, 415–16 pressure, 448, 566, 567, 573–75, 574t, 590 atmospheric, 577–78 barometric in hurricane, 598 blood, 579 gauge, 578 human vein, 597, 597 in incompressible fluid, 576 kinetic, 613–16 standard temperature and, 604, 607–8 in a static fluid, 575–79 primary, of a transformer, 1053–54 principal maximum, of grating, 1184–85 principal quantum number, 1321, 1328–29 principal rays, 1129 of lens, 1137 Principia Mathematica (Newton), 131, 287 principle of relativity, 1220–23 principle of superposition, 138, 703 for magnetic field, 938 for waves, 516–20, 518 prism, 1123, 1126 probability interpolation of wave, 1277 problem-solving, guidelines for, 50 projectile: maximum height of, 109–11 parabolic vs. elliptical orbit for, 287

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A-90

projectile (continued) range of, 109, 111, 111, 287 time of flight of, 109–10 trajectory of, 105, 106, 107, 111, 111 projectile motion, 104–12, 122, 124 with air resistance, 111 propagation of uncertainties, see uncertainties proper frequencies, 523 proton, 1355 charge of, 696, 698 collision of, 361 isotopes and, 1356 mass of, li, 137, 137t, 1356 nuclear binding energy and, 1359–62 quark structure of, 1414 Proxima Centauri, xli, 52 p-type semiconductor, 1338–42, 1341, 1343–44 pulleys, 443–44, 458, 459 mechanical advantage and, 443–44 pulsar (neutron star), rotation, 391, 427 pupil, of eye, xlvi p-V diagram, 668–69, 678 four step, 687, 690 one step, 689 three step, 686 pyrometer, optical, 610, 610 Q, or quality factor, of oscillation, 489–91, 502–3 quality factor (Q), 1045 quantization: of angular momentum, 1296, 1322–23 of electric charge, 706 of energy, 1260 quantum: of energy, 1260 fields and, 1409–11, 1409t of light, 1254–55 quantum Hall effect, 982 quantum jump, 1296 quantum mechanics, 1287, 1302–9 quantum numbers, 1260 angular-momentum, 1296, 1322, 1324–26 of atomic states, 1328–32 electron configuration and, 1328 magnetic, 1328 orbital, 1321, 1328 principal, 1321, 1328 spin, 1324, 1328 for stationary states of hydrogen, 1328 quantum structure, 1320–53 quark, 1397, 1404t, 1412–16 bottom, 1415 charmed, 1415 charm of, 1415 color of, 1414–15 down, 1413–14 electric charges of, 706, 1414t force between, 1414–15 strange, 1413–14 top, 1415 up, 1413–14 up and down, within proton, li quark structure of proton and neutron, 1414 Quito, Ecuador, 386 rad, 1375 radar “guns” (Doppler), 558 radians, position angle, 368 radiation, 641

Index

blackbody, 1255–58, 1259–61 cavity, 1257 electromagnetic, 1090–91 electromagnetic, wavelength and frequency bands of, 1090–91, 1091 Hawking, 1281 heat transfer by, 641 infrared, 1090 thermal, 1255, 1256–57, 1256 radiation field, of accelerated charge, 1075–76, 1076, 1077 radioactive dating, 1376, 1421 radioactive decay, law of, 1372–76 radioactive series, 1367–68 radioactivity, 1365–72 radioisotopes, decay rates of, 1374–76 radio station WWV, 9 radio telescope, 1186 at Arecibo, 1198–99, 1198 at Jodrell Bank, 1209 Very Large Array, 1186 radio waves, 1079–1110, 1089, 1094 railroad tracks, thermal expansion and, 637, 637 range, of projectiles, 109, 111, 111, 287 rate of change, of momentum, 312 Rayleigh, John William Strutt, Lord, 1198, 1258 Rayleigh’s criterion, 1197, 1209 rays, 1115, 1115 alpha, 1365 beta, 1365 gamma, 1090, 1365 principal, 1129, 1137 rays, X, 1090, 1303 RC circuit, 907–12, 907, 911, 912 reactance: capacitative, 1036 inductive, 1039–40 reaction and action, 144–51, 144, 145, 146, 149 reactor, nuclear, 1381–82, 1381 breeder, 1383 containment shell of, 1382, 1382 control rods in, 1381 fission, 1380–81 fuel rods in, 1381 moderator in, 1381 neutrons in, 1381 rods in, 1381 water-moderated, 1383 real image, 1133 recoil, 309–10 rectangular coordinates, 3, 3 rectifier, 1340–42 red, 1414–16 red-shift, 1418 reference circle, 472 reference frames, 3, 4, 20, 114, 115 in calculation of work, 208, 208 of Earth, 132 freely falling, 142, 142 inertial, 132, 132, 133 for rotational motion, 366 reflection, 1112, 1114–17, 1114 angle of, 1115–16, 1115 critical angle for total internal, 1123 law of, 1114–15 by mirror, 1116–17 polarization by, 1086, 1125 total internal, 1123 reflection grating, 1185

reflector, corner, 1116 refraction, 1112, 1117–27, 1126 index of, 1118, 1119t, 1125–26, 1125 law of, 1121 refrigerators, use of freon and, 672–73, 673 relative biological effectiveness (RBE), 1375 relative measurement, 4 relativistic kinetic energy, 1240–41 relativistic momentum, 1239–40 relativistic total energy, 1243 relativistic transformation, for momentum and energy, 1242–44 relativity: of motion, 31, 115–18, 1217 principle of, 1220–23 of simultaneity, 1220–22 special theory of, 1216–53 of speed, 31, 31 of synchronization of clocks, 1222–23 Relativity, General, theory of, 394 rem, 1375 resistance, 866 air, 49, 51, 61, 111, 181, 181 in combination, 872–76 of human skin, 913 internal, of batteries, 895–96 of wires connecting resistors, 875–76 resistance thermometers, 880 resistivity, 865, 866t of insulators, 871t of materials, 868–72 of materials, temperature dependence of, 869 of metals, 870 of semiconductors, 871t temperature coefficient of, 880 resistors, 872–76, 872, 873t circuit with, 1031–35 in parallel, 874 power dissipated by, 1033 power dissipated in, 902 resistance of wires connecting, 875–76 in series, 874 resolution, angular, of telescope, 1196–99 resolving power, of grating, 1186 resonance, 523, 1404 of bridge (Tacoma, WA), 523, 524 of damped oscillator, 490–91, 491 in musical instruments, 546 tides, 531 resonant frequency, 1043 rest-mass energy, 1242 restoring force, 182–84, 183 resultant, of vectors, 72, 74 resultant force, 138 retina, of eye, xlvii reversible process, 667 Reynolds number, 595 rheostats, 872, 872 right-hand rule, 84, 84, 85, 743 for a current loop, 942 for electromagnetic waves, 1078–79, 1079 for magnetic field, 932, 932, 933 for magnetic force, 934, 935, 936 for magnetic moment, 974 for solenoids, 944 right triangle, 19 rigid body: dynamics of, 394–428 kinetic energy of rotation of, 378–84

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Index

moment of inertia of, 378–84 motion of, 366–67, 366–67 parallel-axis theorem for, 382–83 rotation of, 365–93 some moments of inertia for, 382t statics of, 430–33 translational motion of, 366 RLC circuit, 1044–45, 1044, 1045 impedance Z of, 1050 phase f of, 1050 RL circuit, 1015–18, 1015, 1046 roll, 366 root-mean-square (rms) speed, of gas molecules, 614–15 root-mean-square voltage, 1033 rotation: of the Earth, 9, 132, 476 frequency of, 369–70 kinetic energy of, 378–84 period of, 369 of rigid body, 365–93 rotational energy, of molecule, 1334 rotational motion: analogies to translational motion, 374t, 407t conservation of angular momentum in, 406–10 conservation of energy in, 397 with constant angular acceleration, 374–76 of Earth, 120 equation of, 399–406 about a fixed axis, 367–73 of gas, 617–18 Newton’s laws in, 395 reference frame for, 366 torque and, 405 work, energy, and power in, 395–99 roulette wheel, 367 Rumford, Benjamin Thompson, Count, 629, 629 Rutherford, Sir Ernest, 1287, 1293–95, 1294, 1355, 1363, 1365 Rutherford backscattering, 357 Rutherford scattering, 1293 R value, 640, 641, 655 three layers, 659 Rydberg constant, 1292, 1300–1301 Rydberg-Ritz combination principle, 1293 Sagittarius, constellation, xliii Salam, Abdus, 1411 Sandia National Laboratory centrifuge, 365, 365 satellites: artificial, 271–72, 281, 286–87, 1344, 1344, 1421 communication, 271–72, 281, 281, 290–91 geostationary, 280, 281, 290–91 Kepler’s laws for motion of moons and, 286–88 see also specific satellites Saturn, 285t, 286 moons of, 296, 297t scalar, 72 scalar (dot) product, of vectors, 81–83, 81, 83, 86, 208–9 scale, chromatic musical, 539, 539 scanning capitance microscope, 849 scanning tunneling microscope, 1311, 1311, 1320 scattering, of alpha particles, 1294 scattering cross section, 1315 Schlieren photograph (of bullet), 552 Schrödinger, Erwin, 1302, 1303 Schrödinger equation, 1303–6 Schwinger, Julian, 1410

scientific notation, xxv, 14–15, A2–23 scissorjack, 460 Sears Tower, 653, 654 seat belt, 343 secant, A8–10 second, 5, 9–10, 13, 14 multiples and submultiples of, 10t secondary, of a transformer, 1053–54 second harmonic, 523 Second Law of Thermodynamics, 675–80 second overtone, 522, 522 seiche, 558 seismic waves (P and S), 533, 558 seismometer, 558, 558 selection rule, 1333 self-inductance, 1011 semiconductors, 871, 1340–45 with donor and acceptor impurities, 1339 electron configurations of, 1338–39, 1339 “holes” in, 1338–39, 1340–42 n-type, 1338–42, 1341, 1343–45 p-n junctions of, 1340–42, 1343–44 p-type, 1338–42, 1341, 1343–45 resistivities of, 871t temperature dependence of resistivity of, 1338 semimajor axis of ellipse, 282, 285 related to energy, 291 series limit, 1291 Sèvres, France, 5 shear, 445–46, 447t, 449 shear modulus, 447–48, 447t sheerlegs, 463 shells, 1330, 1330t ship collison, Andrea Doria and Stockholm, 361 shock wave, of bullet, 552 Shoemaker-Levy comet, 299 short wave, 1088 sidereal day, 294 sievert, 1375 significant figures, 14–15, 18 silicon: nanoparticle, 264 wafer, 654 silicon solar cells, 893 silicon structures, micromachined, 195 silk, 711 silo, grain, 588 simple harmonic motion, 469–76 conservation of energy in, 483 frequency of, 470–71 kinetic energy in, 480–83 period of, 470–71 phase of, 471 potential energy in, 480–83 simple harmonic oscillator, 476–79 angular frequency of, 477 equation of motion of, 477 as timekeeping element, 479 simple pendulum, 484–88, 484 equation of motion for, 485 isochronism of, 486, 501 simultaneity, relative, 1220–22 sine, 19, 473–74, 486, A8–10 formula for derivatives of, 473 law of sines, A10 single-loop circuits, 893–97 single slit: diffraction by, 1190–96 diffraction pattern of, 1191

A-91

minima in diffraction pattern of, 1191–92 sinks, of field lines, 738–39 siphon, 595, 596 Sirius, star, xliii SI units, see system of units (SI) skater, figure, 408 skiing, speed record, 267 skin, human, resistance of, 913 sky divers, 53 Skylab mission, 134, 134, 229, 295, 424, 504 body-mass measurement device on, 134, 134 fuel cell on, 892 sliding friction, 174–78, 175, 176, 177, 190 slope, 17, 18, 32–33 instantaneous velocity as, 35–37 slug, 136n small angle approximations, 485 Small Mass Measurement Instrument, 504, 504 Snell’s Law, 1121, 1124 sodium gas, 624 solar cells, 893, 1343–45 silicon, 893 solar day, 9 solar heat collector, 653 Solar System, xli, 282 data on, 285–87 solenoid, 943–46, 946 magnetic field of, 943–44 self-inductance of, 1012 solid: compression of, 446, 448 elasticity of, 445–49 elastic moduli for, 447t electron configuration of, 1337–39 elongation of, 445–49, 445 energy banks in, 1336–40 quantum structure of, 1321, 1336–45 shear of, 445–46, 448 thermal expansion of, 633–37, 633 ultimate tensile strength of, 447t sonic boom, 552–53, 552 sonography, 536, 536, 544 sound: intensity level of, 542t speed of, 543–45, 544t, 559 speed of (French Academy), 560 speed of correction (NRC, Canada), 560 speed of in air, 559–60 speed of in freshwater, 527, 559 speed of in human body, 558 theoretical expression for speed in air, 543 theoretical expression for speed in gas, 657 sound waves: in air, 538–39 diffraction of, 554 intensity of, 538, 540–43, 543 longitudinal, 538, 538 maximum speed in air, 558 sources: of electromotive force, 890–92 of field lines, 738–39 Space Shuttle, 125, 142, 280, 280, 290–91, 426 Space Telescope, 1209 Special Relativity theory, 1216–53 specific heat: of common substances, 630t at constant pressure, 644 at constant volume, 644 of a gas, 644–47, 646t

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A-92

specific heat capacity, 630 spectra, color plate, 1288 spectral band, 1335 spectral emittance, 1257 of blackbody, 1259 spectral lines, 1127, 1286, 1287–88, 1289, 1290, 1290 splitting of, 1324–26 spectral series, of hydrogen, 1291–93, 1299, 1324 spectrum, 1288, 1290 of Caph, 1288 of hydrogen, 1291–92, 1291t produced by grating, 1184, 1184 produced by prism, 1126 speed: average, 29–31, 30t of bullet, 356 in circular motion, 371 of electromagnetic wave, 1080 initial, 111 launch, 111 of light, see speed of light of light in air and water, 527 molecular, in gas, 615 most probable, 615, 615 after one-dimensional elastic collision, 345–47 as relative, 31, 31 of sound, 543–45, 559, 560 standard of, 6 terminal, 53 unit of, 30 velocity vs., 33–34, 96 of wave, 510 of wave in Bay of Fundy, 559–60 of waves, on a string, 513–16 speed of light, 6, 1218–20 invariance of, 1220–21 in material medium, 1118 measurement of, 1080 universality of, 1220–21, 1238 spherical aberration, 1128, 1144 spherical charge distribution: electric energy of, 813 potential of, 799–800 spherical mirrors, 1128–35 concave, 1128, 1130 convex, 1129 focal length of, 1128 sphygmomanometer, 579 spin: of electron, 1324–25, 1328–29, 1355–56 of elementary particles, 1403t, 1404t spin angular momentum, 409 spin magnetic moment, 976 spin quantum number, 1324, 1328 Spirit of America, 136, 136 spokes of wire wheels, 389, 392, 426, 526, 531 spring balance, 136, 136, 151 spring constant, 183 springs, 476–79 force of, 182–84, 183, 184 potential energy of, 236–37, 237, 241 spring tides, 296 Sputnik I, 286, 299, 301 Sputnik II, 286 Sputnik III, 286 stabilized laser, 6, 6 stable equilibrium, 432, 432 stable particle, 1404

Index

standard g, 52–53 standard kilogram, 11, 134 standard meter bar, international, 5–6, 5 standard model, 1415 standard of length, 5–6, 5, 6 standard of mass, 11, 134–35 standard of speed, 6 standard of time, 9 standard temperature and pressure (STP), 604, 607–8 standing wave, 520–24 of bridge (Tacoma WA), 523, 524 mode and overtone, 522–23, 522, 545 in a tube, 545, 545 Stanford Linear Accelerator (SLAC), 1411, 1413, 1415 beam dump, 658, 659 star, 55 Cancri system, 297 star, binary system: Cyguns, 298 Kruger 60, 297 PSR 1913+16, 297 states of aggregation (solid, liquid, gas), xxxvi states of atoms, stationary, 1299 states of matter, fifth, 722 static equilibrium, 430–41 condition for, 430 of electric charge, 774–75 examples of, 432–41, 432 static fluid, 575–79, 575 static friction, 178–80, 179, 180, 190 coefficient of, 175t, 179–80 equation for, 179 statics, 174 of rigid body, 430–33 stationary state, 1296 of hydrogen, 1298, 1300 steady emf, 1004 steady flow, 569 steam engines, 671, 671 steel, maximum tensile stress and density, 528 steel rods, deformation, 449 Stefan-Boltzmann Law, 1261–62 step-down transformer, 1054 step-up transformer, 1054 stimulated emission, 1090 stopping distances, automobile, 45, 46, 47, 47 stopping potential, 1266 STP (standard temperature and pressure), 604, 607–8 straight line, motion along, 28–68, 32 strangeness, 1407 strange quark, 1413–14 Strassmann, Fritz, 1377 streamline flow, 569 streamlines, 569–71, 569, 570, 584, 585 velocity along, 584 stream tube, 569–70 stress, thermal, 461 stringed instruments, 546 see also specific stringed instruments string theory, li, 1416 “strong” force, 191, 1405, 1406, 1406t, 1414 nuclear binding energy and, 1359–65, 1359, 1360 in nucleus, 1355 strong interactions, 1405, 1406, 1406t strontium, radioactive decay of, 1372–76 sublimation, 642

suction pump, 592 Sun, xl age of, 1420 escape velocity from, 292 sunglasses, Polaroid, 1086 superconducting cable, 883 superconductivity, 858, 861, 870 superconductors, high-temperature, 870 superposition: of electric fields, 772–73, 773 of electric forces, 703 superposition principle, 138, 703, 729, 734 for magnetic field, 938 for waves, 516–20, 518 supersonic aircraft, sonic boom and, 552–53, 552 supertanker: empty and loaded, 593 Globtik Tokyo, 266 Seawise Giant, 659 see also oil tanker surface tension, 566 symmetry, axis of, 380–82, 382t symmetry, in physics: cylindrical, 767–68 planar, 770–72 synchronization of clocks, 4, 5, 133n, 1220–23 relativity of, 1222–23 synchronous (geostationary) orbit, 271–72 synchrotron, 1398 Syncom communications satellite, 271–72, 290–91 system of particles, 305–37 center of mass in, 313–23 energy of, 327–28 kinetic energy of, 327–28 moment of inertia of, 378–84 momentum of, 306–13, 324 system of units (SI) 5, 14, 20 base and derived, A21–23 of current, 972 of electric charge, 972 for radioactive decay rate, 1374–75 Tacoma Narrows, 523–24, 524 Tampa Bay, Fla., 99 tangent, 19, A8–10 tangent galvanometer, 954, 954 tangential acceleration, 371–72 telescope, 1149–52 angular magnification of, 1150 angular resolution of, 1196–99 Arecibo radio-, 1198–99, 1198 Galilean, 1142 Hobby-Eberly, 1152, 1152 Jodrell Bank radio-, 1209 Keck, 1151 mirror, 1151 Mt. Palomar, 1151, 1151 radio-, 1186, 1198–99, 1198, 1209 Space, 1209 Very Large Array radio–, 1186 television cameras, 1269 Telstar satellites, 1421 temperature: coefficient of resistivity, 880 on Earth, hottest and coldest, 620 some typical values, 611t standard pressure and, 604, 607–8 at Sun’s surface and center, 622, 625

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Index

temperature scales: absolute, 604 absolute thermodynamic, 609 Celsius, 611, 612 comparison of, 611, 612 Fahrenheit, 611, 612 ideal-gas, 609–12 Kelvin, 604, 609–10 Tennessee River, frictional losses, 267, 653 tensile strength, of solids, 447t tension, 149–50, 149, 151, 155 terminal speed, 53 terminal velocity, 53 Tesla, Nikola, 935 tesla (T), 934 Tethys, moon of Saturn, 297 TeV, 1398 Tevatron, Fermilab, 1398–99, 1398 thermal conductivity, 638–41, 639t thermal energy, 248, 616, 629 thermal engine, efficiency of, 666–67 thermal expansion, 637 of concrete, 637 linear, coefficient of, 634, 634t, 637 of solids and liquids, 633–37, 633 volume, coefficient of, 634t of water, 635, 635 thermal radiation, 1255, 1256–57, 1256 thermal resistance (R value), 640, 641, 655 thermal stress, 461 thermal units, 249 thermocouples, 610, 610 thermodynamics, 661–91 calculation techniques, 672 First Law of, 662–64 Second Law of, 675–80 Third Law of, 680 thermodynamic temperature scale, absolute, 609 thermograph, 628, 652 thermometer: bimetallic strip, 610, 610, 636, 637 color-strip, 610, 610 constant-volume gas, 609–10, 609 electric resistance, 610, 610 mercury-bulb, 610, 610, 636, 636 resistance, 880 thermocouple, 610, 610 thermonuclear bomb, 260 thermos bottle, 1257 thickness monitor, 498 thin films, interference in, 1169–73 thin lenses, 1135–39 third harmonic, 523 Third Law of Thermodynamics, 680 Thomson, Sir Joseph John, 1293 thread of a screw, 22 threshold energy, 355 threshold frequency, 1266 threshold of hearing, 538 thundercloud, electric field of, 725–28 tidal flow, 99 tides, 120, 530 height at Bay of Fundy, 531, 531 height at Pakhoi, 530 resonance, 531 spring, 296 tightrope walker, 418 time: atomic standard of, 9

Cesium standard of, 9 Coordinated Universal, 9 position vs., 32–33, 33, 34, 35, 35, 36, 60 standard of, 9 unit of, 9–10 velocity vs., 61 time constant, 1373 of RC circuit, 909 time-dependent angular acceleration, 376–78 time dilation, 1224–28 time of flight, of projectiles, 109–10 time signals, 9 TNT, 263 Tomonaga, Sin-Itiro, 1410 top, 415 top quark, 1415 tornado, air pressure within, 590 toroid, 945, 945 torque, 395–99 angular acceleration and, 400 angular momentum and, 410–16 on a current loop, 972–76 on dipole, 743 equation of, 400 power delivered by, 397 rotational motion and, 405 static equilibrium and, 430–41 torque vector, 410, 411 torr, 578 Torricelli, Evangelista, 578 Torricelli’s theorem, 585 torsional oscillator, 500 torsion balance, 277, 277 total internal reflection, 1123 totally inelastic collision, 348 tower crane, see K–10000 tower crane tracking chambers, 1401–2 traction apparatus, 149, 149 train à grande vitesse (French TGV), 563, 563 train box car collisions: elastic, 346 inelastic, 349 trajectory of projectiles, 105, 106, 107, 111, 111 transfer of heat, 638–41, 641 transformation, heat of, 642 transformation equation: Galilean, 1218, 1234, 1239, 1241 Lorentz, 1232–38 transformation of coordinates, 1218, 1234, 1241 transformer, 1010, 1053–57, 1055, 1056 step-down, 1054 step-up, 1054 transistors, 1340, 1342–43, 1343 translational acceleration, 402 translational motion, 29, 95, 95, 366, 404 analogies to rotational motion, 374t, 407t of Earth, 120 transmission line, power dissipated in, 904 transmutation of elements, 1363 transverse electric field, 1078 transverse radiation field of accelerated charge, 1076–77 transverse wave, 508, 508 triangle, right, 19 Triangulum Galaxy, xliv Trieste, 589, 589 trigonometry (review), A8–10 functions of (sine, cosine, tangent, secant, cosecant, cotangent), A8–10

A-93

identities, A9–10 laws of cosines and sines, A10 of right triangle, 19, A8 small angle approximation, 486 triode, 1340, 1342–43 triple-point cell, 609, 609 triple point of water, 609, 609 trombone, 546 trumpet, 546 sound wave emitted by, 538 tsunami, height (Mexico), 526, 527 tuning fork, 530 tunneling, 1311 turbulent flow, 571, 571 turning points, 244–45 of motion, 470 TV waves, 1088 twin paradox, 1226 two-slit interference, 1177–83, 1177 pattern for, 1180, 1180 UA1 detector, CERN, 1402, 1402 UFOs, 557 Uhlenbeck, George, 1325 ultimate tensile strength, 447t ultrahigh vacuum, 624, 624 ultrasound, 536, 538, 539 frequencies used in human body, 558, 562 ultraviolet catastrophe, 1259 ultraviolet light, 1090 unbound orbit, 245 uncertainties: application to Ohm’s Law of, A20–21 propagation of, A19–21 uncertainty relation, 1278 unified field theory, 1411 unified theory of weak and electromagnetic forces, 1411 uniform circular motion, 112–15, 112, 113, 125, 184–90 Unionville, Md. rainfall, 332 unit, derived, see derived unit unit of length, 5–8 units: consistency of, 16, 18, 20 conversions of, 16–17, 18 prefixes for, A21t Units, International System of (SI), 5, 14 of current, 972 of electric charge, 972 units of force, 135–36, 141 unit vector, 79–80, 79 cross product of, 85 universal gas constant, 604 Universal Gravitation, Law of, 131, 272–76, 278 universality of acceleration of free fall, 49, 49 universality of free fall, 49, 49 universality of speed of light, 1220–21, 1238 universe: contraction of, 1422 expansion of, xlv, 1419–23 mass of, 1422–23, 1423t observable, 1420 unpolarized light, 1084, 1084 unstable equilibrium, 432, 432 unstable particles, 1404 ununquadium, atomic mass of, xlix up quark, 1413–14

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A-94

uranium, 1355 alpha decay of, 1365–67, 1366 enriched, 1381 fission of, 1377–79, 1377 isotopes of, 1362 as nuclear fuel, 1242 uranium isotopes (235U, 238U, UF6), 623–24 Uranus, 272, 285t, 286 Ursa Major, 1417 valence band, 1338 van der Waals equation of state, 624 Vanguard I, 286 vaporization, 642, 642t variable force, 211–13, 211, 212, 213, 214 Vasa (Swedish ship), 320 vector, magnetic force, 933 vector addition, 72–76, 72, 73, 74, 89 commutative law of, 74 by components, 78–79 vector product, 83–86, 84, 85 vectors, 69–93 acceleration, 100–101 addition of, see addition of vectors components of, 77–86, 78, 95–98, 97, 99, 101 cross product of, 83–86, 84, 85 definition of, 72 displacement, 69, 70–72, 70, 71, 88, 96 dot (scalar) product of, 81–83, 81, 83, 86, 208–9 instantaneous velocity, 98, 98 multiplication of, 75, 81–86 in navigation, 71 negative, 75, 75 notation of, 71 position, 76–77 Poynting, 1093–94 resultant of, 72, 74 subtraction of, 75 three–dimensional, 79–81, 79, 80 unit, 79, 79 velocity, 98–100 vector triangle, 73 velocity: acceleration as derivative of, 41 addition rule for, 115–16, 117 along streamlines, 584 angular, 369t, 376–77, 471 angular, average and instantaneous, 369 average, 32–35, 33 average, in three dimensions, 101–2 average, in two dimensions, 95 of center of mass, 323–24, 348 components of, 95–98, 197, 199 escape, 292 of flow, 566, 568 of galaxies, 1422 Galilean addition law for, 1218 horizontal, 103–8, 103, 106, 107, 108 instantaneous, see instantaneous velocity magnitude of, 96 muzzle, 287, 331 negative, 39 positive, 39 speed vs., 33–34, 96 terminal, 53 time vs., 61 transformation equations for, 1218 vectors, 98–100

Index

Velocity Peak, Colorado, speed skiing, 267 velocity transformation, Galilean, 1218 venous pressure, 597, 597 Venturi flowmeter, 585–86, 596, 596 Venus, 285t, 286, 286, x. Verne, Jules, 300 Very Large Array (VLA) radiotelescope, 1186 very-long-baseline interferometry (VLBI), 1202 vibrational energy, of molecules, 1333 violin, 546, 546 frequencies, 530, 531, 558 sound wave emitted by, 538 vibration of standing wave, 546 wavefront emitted, 538 Virgo Cluster (galaxies), xliv, xlv virtual image, 1116 virtual photon, 1410 viscosity, 566, 595 viscous forces, 181, 196 visible light, 1090–91 colors and wavelengths of, 1090–91, 1091 volcanic bombs, 94, 100, 100, 105 volt (V), 792 Volta, Alessandro, Conte, 793 voltage, DC, 889–90, 1004 AC, 1004 voltmeter, 905, 906, 916, 917 volume, 13 compression of, 446 volume expansion, 634 volume expansion coefficient, 634, 634t Von Mayer, Robert, 629, 653 vortex, 546, 546, 571 Vostok, Antartica, coldest temperature, 621 Voyager spacecraft, 358 W particle, 1409, 1411, 1415 Wairakei, New Zealand (geothermal plant), 690 Warsaw Radio tower, 634 water: thermal expansion of, 635, 635 triple point of, 609, 609 volume as function of temperature, 634 waterfall: Chamonix, 652 hypothetical (M. C. Escher) see also Niagara Falls water-moderated nuclear reactor, 1383 water molecule, 743 distance and algle between component atoms, 332 moment of inertia for, 388, 389 motion, 566 water park, 507 waterwheel, 215–16, 216, 217 overshot, 267 undershot, 267, 653 Watt, James, 254 watt (W), 254, 540 watt balance, 11, 11 wave, 507–64 amplitude of, 511 angular frequency of, 512, 513–16 beats of, 518 constructive and destructive interference for, 517 crests, 509–10 deep water, 526

electromagnetic, see electromagnetic wave energy in, 1092–96 equation, 513 freak (North Atlantic), 533 frequency of, 510–11 harmonic, 510–13, 511 intensity of, 540, 1093–94 light and radio, 1079–1110 long, 1088 longitudinal, 508–9, 509, 538, 538 measurement of momentum and position of, 1277–78 medium, 1088 modulation of, 518 motion, 508–9 number, 511 ocean, 553 ocean wavelength and speed, 528 particle vs., 1276–79 periodic, 509–16, 510 period of, 510, 510, 513 plane, 537, 537, 1081–83, 1082, 1097 pool, 507, 511 power transported by, 516 probability interpolation and, 1277 pulse, 508–9, 514, 522 radio, 1079–1110, 1094 shallow water, 515–16, 527, 528 short, 1088 sound, see sound waves speed, 510 speed of, on a string, 513–16 standing, 520–24, 524 standing, in a tube, 545, 545 superposition of, 516–20, 518 transverse, 508, 508 trough, 509–10, 510 TV, 1088 water, 515–16, 526–28 wave equation, 1096–99 wave fronts, 537, 541–42, 541, 1113, 1113, 1115 circular, 537 plane, 537 spherical, 537 wavefunction, 511 harmonic, 511, 513–16 wavelength, 510–13, 510, 511 of visible light, 1090–91, 1091 wavelength bands, of electromagnetic radiation, 1090–91, 1091 wavelength shift, of photon, in Compton effect, 1270–71 wavelets, 1113, 1120, 1276 wave mechanics, 1309, 1322 wave optics, 1169 wave pulse, electromagnetic, 1080–83, 1082 waves, radio, 1089 wavicle, 1255, 1276 measurement of, 1277–78 “weak” force, 191, 1369, 1405, 1406, 1406t “weak” interactions, 1369, 1405, 1406, 1406t Weber, Franz, speed skiing, 267 weber (Wb), 997 weight, 11, 141–42, 141 apparent, 187–88, 187 mass vs., 141 weightlessness, 142, 142 simulated, 589 Weinberg, Stephen, 1411

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Index

whale: breaching, 263, 264 song, 558 Wheatstone bridge, 906–7 wheel pottery, 409 white noise, 538 Wien’s Law, 1261 Wilson, Robert W., 1421 winches, 441, 441, 458–59, 458 geared, 459 wind instruments, 546 wing, flow around, 571, 584 wire: magnetic force on, 969–72 straight, magnetic field of, 940–41, 941 uniform, electric field in, 860

work, 205–18 calculation of, 218 definition of, 205 done by constant force, 205, 208 done by gravity, 207 done by variable force, 211–13, 211, 212, 213, 214 dot product in definition of, 208–9 frame of reference in calculation of, 208, 208 integrals for, 212–13 internal, in muscles, 208 in one dimension, 205–8, 206 in rotational motion, 395–99 in three dimensions, 208–10, 209, 210 zero, 238–39 work-energy theorem, 215, 236, 400

A-95

work function, 1266 Wparticle, 1409, 1411, 1415 WWV (NIST radio station), 9, 23 xerography, 709 X-ray, scattering, xlviii X rays, 1090, 1269, 1273–75, 1273, 1274, 1421 characteristic, 1303 yaw, 366 Young, Thomas, 1169, 1170, 1177 Young’s modulus, 447–48, 447t, 460, 479 zero-point energy, 1307 zero work, and conservative force, 238–39 Z0 particle, 1409, 1411, 1415 Zweig, G., 1413

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FUNDAMENTAL CONSTANTS

MISCELLANEOUS PHYSICAL CONSTANTS

(See Appendix 7 for More) Speed of light

c  3.00  108 ms

Standard acceleration of gravity

1 g  9.81 ms2  32.2 fts2

Planck’s constant

h  6.63  10 34 Js U  h2p  1.05  1034 Js

Molecular mass of air

28.98 g/mole

Gravitational constant

G  6.67  1011 Nm2 kg2

Density of dry air

1.29 kgm3 (0C, 1 atm)

Permeability constant

m0  1.26  106 Hm

Speed of sound in air

331 m/s (0C, 1 atm)

m04p  1.00  107 Hm

Density of water

1000 kg>m3

Permittivity constant

0  8.85  1012 Fm

2.26106 J/kg  539 kcal/kg

Coulomb constant

1 4p0  8.99  109 m F

Heat of vaporization of water

Electron charge

e  1.60  1019 C

Heat of fusion of ice

3.34105 J/kg  79.7 kcal/kg

Electron mass

me  9.11  1031 kg

1 cal  4.187 J

Proton mass

mp  1.673  1027 kg

Mechanical equivalent of heat Solar constant Index of refraction of water

1.4 kW>m2 1.33

27

Neutron mass

mn  1.675  10

Bohr radius

4p0 U mee  5.29  10 11 m

Compton wavelength

h mec  2.43  1012 m

-2

Bohr magneton

mB 

kg

2

eU  9.27  1024 J T 2me

k  1.38  1023 JK

Boltzmann constant Avogadro’s number

NA  6.02  1023 mole

Universal gas constant

R  NAk  8.31 J/mole·K

PREFIXES FOR UNITS

SPECIAL UNITS AND CONVERSION FACTORS (See Appendix 8 for More) QUANTITY

UNIT

SYMBOL

CONVERSION

length

inch

in.

1 in.  2.540 cm

length

angstrom

Å

1 Å  108 cm  1010 m

length

light year

—

1 light year  9.461  1015 m

FACTOR

PREFIX

SYMBOL

1018

exa

E

1015

peta

P

10

12

tera

T

109

giga

G

6

mega

M

3

kilo

k

10

volume

cubic centimeter

cm3, cc

1 cm3  1 ml  103 l

10

volume

liter

liter, l

1 l  103 cm3  103 m3

102

hecto

h

deka

da

7

time

year

y

1 y  365.25 d  3.56  10 s

10

mass

pound

lb

1 lb  0.453 6 kg

101

deci

d

102

centi

c

milli

m

10

micro

m

109

nano

n

pico

p

femto

f

atto

a

mass

atomic mass unit

u

1 u  1.661  10

27

kg

19

energy

electron-volt

eV

1 eV  1.602  10

force

pound-force

lb-f

1 lb-f  4.448 N

pressure

atmosphere

atm

1 atm  1.013  105 Pa

pressure

torr

torr

1 torr  1 atm>760

temperature

Celsius scale

C

C  K  273.15

temperature angle

Fahrenheit scale radian

F rad

F  95(C)  32 1 rad  57.30

J

3

10

6

12

10

15

10 10

18

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SOLAR SYSTEM DATA

EARTH

MOON

Mass

ME  5.98  1024 kg

Mass

7.35  1022 kg

Equatorial radius

RE  6.378  106 m

Radius

1.74  106 m

Polar radius

R E  6.357  106 m

Mean density

3340 kg m3

5520 kgm3

Surface gravity

1.62 m s2

Period of rotation

27.3 days

Mean distance from Earth

3.84  108 m

Mean density

g  9.81 ms2  32.2 fts2

Surface gravity

Period of rotation 1 sidereal day  23 h 56 min 4 s  8.616  104 s

Period of revolution 1 sidereal month  27.3 days

Moment of inertia: about polar axis about equatorial axis Mean distance from Sun

I  0.331 MER2E

SUN

I  0.329MER2E 1.50  10 m

Period of revolution (period of orbit) 1 year  365 days 6 h  3.16  107s Orbital speed

PLANET Mercury

29.8 km/s

MEAN DISTANCE FROM SUN 57.9  106 km

MS  1.99  1030 kg

Mass

11

PERIOD OF REVOLUTION

Radius

6.96  108 m

Mean density

1410 kgm3

Surface gravity

274 m s2

Period of rotation

 26 days

Luminosity

3.9  1026 W

EQUATORIAL RADIUS

MASS

SURFACE GRAVITY

0.241 year

3.30  1023 kg

2 439 km

0.38g

PERIOD OF ROTATION 58.6 days

Venus

108

0.615

4.87  1024

6 052

0.91

Earth

150

1.00

5.98  1024

6 378

1.00

0.997

Mars

228

1.88

6.42  1023

3 393

0.38

1.026

Jupiter

778

11.9

1.90  1027

71 398

2.53

0.41

26

243

Saturn

1430

29.5

5.67  10

60 000

1.07

0.43

Uranus

2870

84.0

8.70  1025

25 400

0.92

0.65

Neptune

4500

165

1.03  1026

24 300

1.19

0.77

Pluto

5890

248

1.5  1022

1 500

0.045

6.39

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Online Concept Online Concept Tutorials Tutorial www.wwnorton.com/physics

3

(chapter section) 1 Unit Conversion 1.5, 1.6 2 Significant Digits 1.6 3 Acceleration 2.4, 2.5, 2.6 4 Vector Addition and Vector Components 3.1, 3.2, 3.3 5 Projectile Motion 4.4 6 Forces 5.4 7 “Free-Body” Diagrams 5.3, 5.5, 5.6 8 Friction 6.1 9 Work of a Variable Force 7.1, 7.2, 7.4 10 Conservation of Energy 8.1, 8.2, 8.3 11 Circular Orbits 9.1, 9.3 12 Kepler’s Laws 9.4 13 Momentum in Collisions 11.1, 11.3 14 Elastic and Inelastic Collisions 11.2, 11.3 15 Rotation about a Fixed Axis 12.2 16 Oscillations and Simple Harmonic Motion 15.1 17 Simple Pendulum 15.4 18 Wave Superposition 16.3, 16.4 19 Doppler Effect 17.4 20 Fluid Flow 18.1, 18.2, 18.6 21 Ideal-Gas Law 19.1 22 Specific Heat and Changes of State 20.1, 20.4 23 Heat Engines 21.2 24 Coulomb’s Law 22.2 25 Electric Charge 22.1, 22.5 26 Electric Force Superposition 22.3 27 Electric Fields 23.1, 23.3 28 Electric Flux 24.1 29 Gauss’ Law 24.2, 24.3 30 Electrostatic Potential 25.1, 25.2, 25.4 31 Superconductors 27.3 32 DC Circuits 28.1, 28.2, 28.3, 28.4, 28.7 33 Motion in a Uniform Magnetic Field 30.1 34 Electromagnetic Induction 31.2, 31.3 35 AC Circuits 32.1, 32.2, 32.3, 32.5 36 Polarization 33.3 37 Huygens’ Construction 34.1, 34.2, 34.3 38 Geometric Optics and Lenses 34.4, 34.5 39 Interference and Diffraction 35.3, 35.5 40 X-Ray Diffraction 35.4 41 Special Relativity 36.1, 36.2 42 Implications of Special Relativity 36.2, 36.3 43 Bohr Model of the Atom 38.1, 38.2, 38.4 44 Quantum Numbers 39.1, 39.2 45 Radioactive Decay 40.3, 40.4

P r o b l e m - S o l v i n g Te c h n i q u e s (chapter section; page) Units and Significant Figures (1.6; 18) General Guidelines (2.6; 50) Vector Addition and Subtraction (3.3; 79) Dot Product and Cross Product of Vectors (3.4; 86) Projectile Motion (4.4; 108) “Free Body” Diagrams (5.6; 158) Friction Forces and Centripetal Forces (6.3; 190) Calculation of Work (7.3; 218) Energy Conservation in Analysis of Motion (7.4; 223) Energy Conservation (8.1; 238) Conservation of Momentum (10.1; 310) Center of Mass (10.2; 320) Conservation of Energy and Momentum in Collisions (11.4; 353) Angular Motion (12.3; 375) Torques and Rotational Motion (13.2; 405) Conservation of Angular Momentum (13.3; 410) Static Equilibrium (14.2; 437) Bernoulli’s Equation (18.6; 586) Ideal-Gas Law (19.2; 608) Temperature Units; Thermal Expansion (20.2; 636) Thermodynamic Calculations (21.2; 672) Electric Fields (23.1; 728) Electric Field of Charge Distribution (23.2; 732) Gauss’ Law for Charge Distributions with Symmetry (24.5; 777) Energy Conservation and Motion of Point Charge (25.1; 797) Electrostatic Potential and Field (25.4; 810) Combinations of Capacitors (26.2; 837) Combinations of Resistors (27.4; 878) Kirchhoff ’s Rules and Multiloop Circuits (28.4; 900) Direction of Magnetic Force (right-hand rule) (29.2; 936) Ampère’s Law for Current Distributions with Symmetry (29.4; 947) Faraday’s Law; Lenz’ Law (31.3; 1005) Sign Conventions for Mirrors (34.4; 1132) Images of Spherical Mirrors (34.4; 1134) Sign Conventions for Lenses (34.5; 1138) Images of Lenses (34.5; 1143) Thin-Film Interference (35.1; 1174) Two-slit Interference (35.4; 1181) Nuclear Reactions (1371; 40.3)

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TRIGONOMETRIC IDENTITIES (See Appendix 3 and 4 for More) In the following, u is in radians: r

90° – θ

α

y C

θ

x

sin   y r  cos(90  ) cos   xr  sin(90  ) tan   yr  sin cos  cot   x y  1tan   tan(90  ) sec   rx  1cos  csc   ry  1sin  cos2   sin2   1 sec2   1  tan2  csc2   1  cot2  sin 2  2 sin  cos  cos 2  2 cos2   1

B β

A

γ

d sin u  cos u du d cos u  sin u du d tan u  sec2 u du d sec u  tan u sec u du d csc u  cot u csc u du In the following, ku is in radians:

sin(  )  sin  cos   cos  sin  cos(  )  cos  cos   sin  sin  C 2  A 2  B 2  2 AB cos  sin  sin g sin    A B C

 sin (ku) du   k cos (ku) 1

 cos (ku) du   k sin (ku) 1

Physics in Practice (chapter section; page number)

Math Help (chapter section; page number)

Stopping Distances (2.5; 47) Vectors in Navigation (3.1; 71) Velocity Vectors (4.2; 99) Elevators (5.6; 157) Ultracentrifuges (6.3; 188) Hydroelectric Pumped Storage (8.1; 242) Communications Satellites and Weather Satellites (9.3; 281) Center of Mass and Stability (10.2; 320) Automobile Collisions (11.1; 343) Gyrocompass (13.4; 414) Efficiency of Automobiles (21.2; 674) Chaos (15.5; 492) Musical Instruments (17.3; 546) The Sphygmomanometer (18.4; 579) Xerography (22.5; 709) Electrostatic Precipitators (23.2; 735) Electric Shielding (25.3; 804) Capacitor Microphone (26.1; 833) Fuses and Circuit Breakers (28.5; 903) Magnetic Recording Media (30.4; 978) Magnetic Levitation (31.3; 1008) Frequency Filter Circuits (32.2; 1037) AM and FM Radio (33.4; 1089) Optical Fibers (34.3; 1124) Photomultiplier (37.3; 1266) Ultramicroscopes (38.5; 1310) Radioactive Dating (40.4; 1376)

See also Appendices 2, 3, 4, and 5. Trigonometry of the Right Triangle (1.6; 19) Differential Calculus; Rules for Derivatives (2.3; 38) Integrals (7.2; 213) Ellipses (9.4; 283) Derivatives of Trigonometric Functions (15.1; 473) Small-angle Approximations for Sine, Cosine, and Tangent (15.4; 486) The Exponential Function (28.7; 910)