Chapter 26 Particle and Nuclear Physics (A2)

CAMBRIDGE A – LEVEL PHYSICS

PARTICLE AND NUCLEAR PHYSICS (A2)

L E A R N I N G O U TC O M E S No.

LEARNING OUTCOME

i

Apply the mass – energy conservation principle to calculate the reaction energy in a nuclear process.

ii

Understand what is meant by mass excess and the relationship and mass excess and nuclear processes.

iii

Relate the concept of binding energy with the mass defect.

iv

Relate binding energy and binding energy per nucleon.

v

Graph the relationship between binding energy p e r n u c l e o n a g a i n s t n u c l e o n n u m b e r.

LEARNING OUTCOMES No.

LEARNING OUTCOME

vi

Differentiate between nuclear fission and fusion.

vii

Understand the term activity and decay constant.

viii

Use the exponential decay method to calculate the n u m b e r o f u n d e c a y e d n u c l e i a n d a c t i v i t y.

ix

Understand what is meant by half life and calculate the half life of a radioactive sample.

UNITS OF MA SS • As we have learned at AS Level, the SI unit for mass is the kilogram. • Physicists have also developed an alternative unit for mass, the atomic mass unit (symbol = u).

UNITS OF MA SS • 1 u is equivalent to

 the mass of 1 atom of the
  isotope, in kilograms.  • The mass of one atom of  =  kg.

UNITS OF MA SS 

• Hence, kg. • The table on the next slide shows the masses of some light nuclides in atomic mass units.

UNITS OF MA SS Table 43.2, page 1441, Section 43.1: Properties of Nuclei; Chapter 43: Nuclear Physics; Sear’s and Zemansky’s University Physics, Young and Freedman, 13th edition, Pearson Education, San Francisco, 2012.

UNITS OF MA SS • The table below lists the masses of the nucleons and the electron in atomic mass units: Particle

Mass (u)

Proton

1.007276

Neutron

1.008665

Electron

0.000549

MASS - ENERGY C O N S E R VAT I O N

• As we have learned at AS Level, mass – energy is conserved in a nuclear process, but, however, mass itself is not conserved.

MASS - ENERGY C O N S E R VAT I O N • The difference between the total mass after the reaction and the total mass before the reaction gives us an idea of the amount of reaction energy involved.

MASS - ENERGY C O N S E R VAT I O N • We may use Einstein’s mass – energy  equation ; where:
  reaction energy, in J;
  ∑   ∑  , in kg, and
 speed of light, .    

   .

MASS - ENERGY C O N S E R VAT I O N • What if we have masses in atomic mass units (u)? • We now use , where:
E = energy, MeV, and
m = mass, in u.
931.494 MeV/u is a conversion constant.

MASS - ENERGY C O N S E R VAT I O N • We can also use a different version of Einstein’s equation: 

• This version relates the change in energy with the change in mass.

EXAMPLE

Table 43.1, page 1441, Section 43.1: Properties of Nuclei; Chapter 43: Nuclear Physics; Sear’s and Zemansky’s University Physics, Young and Freedman, 13th edition, Pearson Education, San Francisco, 2012.

MA SS E XC ESS • The mass excess of a nuclide is the difference between its actual mass, in atomic mass units (u) and its mass number. • Mathematically: mass mass excess excess  mass mass #in #in u' u'  neutron neutron number

MA SS E XC ESS

Figure 31.3, page 493, Chapter 31: Nuclear Physics; Cambridge International AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

MA SS E XC ESS • The table on the previous slide gives some nuclides and their masses.

Question 7, page 494, Chapter 31: Nuclear Physics; Cambridge International AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

MA SS E XC ESS • We can use mass excess values to find out whether a nuclear reaction is feasible. • We can do this by calculating: I. II. III.

the total mass excess of the reactants, the total mass excess of the products, and comparing both values.

MA SS E XC ESS • If the total mass excess of the products is smaller than the total mass excess of the reactants, the reaction is feasible. • On the other hand, if the converse is true, then the reaction will not be feasible.

MA SS E XC ESS • Let us have a look at an example: -

, →

Nuclide

1

/0 2 534 2 37

Mass Excess (u)

-

,

+0.045563

1

/0

−0.073843

5

34

7

−0.085588 +0.008665

MA SS E XC ESS • Using the values from the table in the previous slide, we can obtain: I. Total mass excess of reactants u II. Total mass excess of products  0.073843 2 0.085588 2 3  0.008665  0.133436 u

MA SS E XC ESS • In this example, the total mass excess of the reactants is larger than the total mass excess of the products. • This means that the products will be more stable than the reactants. Hence, this reaction will occur.

MA SS DE F EC T • If we were to measure the mass  of one nucleus of the  isotope, it will be different from the total mass of the 6 neutrons and the 6 protons make up the nucleus.

MA SS DE F EC T

Figure 31.1, page 492, Chapter 31: Nuclear Physics; Cambridge International AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

MA SS DE F EC T • By using the values from the table on the previous slide, we will obtain a mass defect of  kg for this carbon nuclide. • What is the mass defect, in u?

MA SS DE F EC T • Definition: “The mass defect is the difference between the total mass of the individual nucleons and the mass of the nucleus.”

MA SS DE F EC T • Why is there a mass defect? • This mass defect exists because when the individual nucleons formed the nucleus, some of the mass was converted into potential energy that is used to hold the nucleons together.

BINDING E NE RGY • The strong force is responsible in producing this change in potential energy and thus binds the nucleons. • Work must be done to separate these nucleons apart to an infinite amount of separation.

BINDING E NE RGY • Definition: “The binding energy of a nucleus is the energy required to separate all the nucleons in a nucleus to an infinite amount of separation.”

BINDING E NE RGY • By using mass – energy equivalence, we can use the mass defect and convert it into the binding energy. • To achieve this, use this conversion factor: of mass defect is equivalent to of binding energy.

EXAMPLE

Exercise 43.39, page 1476; Chapter 43: Nuclear Physics; Sear’s and Zemansky’s University Physics, Young and Freedman, 13th edition, Pearson Education, San Francisco, 2012.

BINDING E NE RGY

Examples from Page 367; Section 13.7: The Mass Defect, Chapter 13: Nuclear Physics; International A/AS Level Physics, by Mee, Crundle, Arnold and Brown, Hodder Education, United Kingdom, 2008.

BINDING ENERGY PER NUCLEON • The binding energy per nucleon is equal to the binding energy of the nucleus divided by the total number of nucleons present.

BINDING ENERGY PER NUCLEON • Why is the binding energy per nucleon important? • The binding energy per nucleon value gives us the stability of that nuclide relative to its neighbours, i.e. how hard is it for that nuclei to radioactively decay into its neighbours.

BINDING ENERGY PER NUCLEON Figure 31.4, page 495, Chapter 31: Nuclear Physics; Cambridge International AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

BINDING ENERGY PER NUCLEON • The graph on the previous slide shows a sketch of the binding energy per nucleon for several elements. • The higher the value of the binding energy per nucleon compared to its neighbours, the harder it is for that nuclei to decay radioactively into one of its neighbours.

BINDING ENERGY PER NUCLEON • The nuclide with the highest binding energy per nucleon is @ .  5  • Peaks representing  ,  and  ? indicate nuclides that are relatively stable to their numbers.

EXAMPLE

Questions 9 amd 10, page 496, Chapter 31: Nuclear Physics; Cambridge International AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

NUCLEAR FISSION • Nuclear fission is a decay process in which an unstable, heavy nucleus splits into two fragments of almost the same mass.

NUCLEAR FISSION • The two fragments, known as fission fragments will have a higher binding energy per nucleon as compared to the parent nucleus. • Nuclear fission is achieved by bombarding the heavy nucleus with a neutron.

NUCLEAR FISSION • An example of a fission reaction is seen below: Page 370; Section 13.10: Nuclear Fission, Chapter 13: Nuclear Physics; International A/AS Level Physics, by Mee, Crundle, Arnold and Brown, Hodder Education, United Kingdom, 2008.

NUCLEAR FISSION • Fission reactions are accompanied by the release of energy because the binding energy per nucleon after the reaction is higher than that before the reaction.

NUCLEAR FUSION • Nuclear fusion occurs when two or more small nuclei come together (fuse) to form a larger nucleus.

NUCLEAR FUSION • Nuclear fusion only occurs under conditions of high pressure and temperature; e.g. on the Sun’s surface and in stars.

NUCLEAR FUSION • The examples below are of nuclear fusion reactions:

Page 1469, Section 43.8: Nuclear Fussion; Chapter 43: Nuclear Physics; Sear’s and Zemansky’s University Physics, Young and Freedman, 13th edition, Pearson Education, San Francisco, 2012.

F USION vs. F ISSION

Figure 31.6, page 496, Chapter 31: Nuclear Physics; Cambridge International AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

F USION vs. F ISSION • As seen on the graph on the previous slide, fusion and fission occur in order to increase the binding energy per nucleon. • Nuclei in between A and B tend to undergo fusion, while nuclei in between B and C tend to undergo fission.

EXAMPLES

Question 16; Set 45: Structure of the Nucleus and Radioactivity; page 228; PROBLEMS IN PHYSICS ; E.D GARDINER, B.L McKITTRICK; McGraw – Hill Book Company, Sydney 1985.

EXAMPLES

EXAMPLES Question 17; Set 45: Structure of the Nucleus and Radioactivity; page 228; PROBLEMS IN PHYSICS ; E.D GARDINER, B.L McKITTRICK; McGraw – Hill Book Company, Sydney 1985.

EXAMPLES Question 18; Set 45: Structure of the Nucleus and Radioactivity; page 228; PROBLEMS IN PHYSICS ; E.D GARDINER, B.L McKITTRICK; McGraw – Hill Book Company, Sydney 1985.

R A D I AT I O N D E T E C T I O N • Outlined below are methods of radiation detection:

R A D I AT I O N D E T E C T I O N

R A D I AT I O N D E T E C T I O N

R A D I AT I O N D E T E C T I O N

R A D I AT I O N D E T E C T I O N

Examples from Pages 355 - 357; Section 13.14: Detecting Radioactivity; Chapter 13: Nuclear Physics; International A/AS Level Physics, by Mee, Crundle, Arnold and Brown, Hodder Education, United Kingdom, 2008.

NUC LEA R DEC AY • If we were to use a GM counter to measure radioactivity by listening to the number of clicks, we will not be able to predict when the next click is heard.

NUC LEA R DEC AY • If we were to use a ratemeter, the reading on the ratemeter fluctuates up and down. • This occurs because of the random and spontaneous nature of nuclear processes.

NUC LEA R DEC AY • It is random because

we cannot predict which nucleus in a sample will decay, and ii. The probability that each of the nuclei will decay in per unit of time is constant. This probability is known as the decay constant, λ. i.

NUC LEA R DEC AY • If we plot a graph of count rate vs. time, we would obtain a graph as seen on the next slide. • The fluctuations indicate the random nature of nuclear decay.

NUC LEA R DEC AY Figure 31.8, page 497, Chapter 31: Nuclear Physics; Cambridge International AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

NUC LEA R DEC AY • Nuclear decay is spontaneous because: i.

its occurrence is independent of any external or environmental factors, and ii. not affected by the presence of other nuclei.

ACTIVITY • Radioactive nuclei undergo decay. • Therefore, the amount of parent nuclei reduces with time.

ACTIVITY AB AC

• The rate of nuclei decay, is directly proportional to the amount of undecayed nuclei present in the sample, . • Hence

AB AC

.

ACTIVITY • We can rewrite this expression AB as where: AC

decay constant that has 


units of .
the minus sign indicated this is a decay.


ACTIVITY • Definition: “The decay constant, is defined as the probability per unit time interval that the nuclei will undergo decay.” •

AB AC

is also known as the activity of the source, .

ACTIVITY • Definition: “The activity of a radioactive source is the number of nuclear decays produced per unit of time in the source”. • Activity is measured in Becquerels (Bq) , and 1 Becquerel is 1 decay per second.

ACTIVITY • By

combining the equations AB AB and , we will AC AC obtain , where:
  activity of the sample, in Bq;
B  number of undecayed nuclei, and
D  decay constant, in E  .

EXAMPLES

Questions 12 and 13, page 499, Chapter 31: Nuclear Physics; Cambridge International AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

EXAMPLES Questions 15 and 16, page 501, Chapter 31: Nuclear Physics; Cambridge International AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

HALF LIFE • The

solution

equation the form of

of

the is in .

HALF LIFE • The quantities number of undecayed nuclei, activity, and received count rate, all have the general form 

DC

.

HALF LIFE • We now have three equations that relate these three quantities with time,




DC  

DC DC

HALF LIFE • If we plot the equation B  B FDC , for three different values of λ ,we would obtain: Figure 31.9, page 497, Chapter 31: Nuclear Physics; Cambridge International AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

HALF LIFE • Definition: “The half life,
H , of a radioactive nuclide is the time taken for the number of undecayed nuclei to be reduced to half its original number”. • How do we calculate the value of IK⁄J ?

HALF LIFE K⁄ , • At J substitute into



DC

 

M.

When we the equation

, we obtain

NOKH

 

J

.

• By taking the natural logarithms on .P both sides, we get
H D

EXAMPLES Question 18, page 501, Chapter 31: Nuclear Physics; Cambridge International AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

EXAMPLES Questions 19 and 20, page 501, Chapter 31: Nuclear Physics; Cambridge International AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

EXAMPLES Question 21, page 501, Chapter 31: Nuclear Physics; Cambridge International AS and A Level Physics Coursebook, Sang, Jones, Chadha and Woodside, 2nd edition, Cambridge University Press, Cambridge, UK,2014.

HOME WOR K 1. 2. 3. 4. 5. 6. 7. 8.

Question 8, Paper 4, Summer 2008. Question 9, Paper 4, Summer 2009. Question 8, Paper 41, Winter 2009. Question 8, Paper 42, Winter 2009. Question 8, Paper 41, Summer 2010. Question 8, Paper 42, Summer 2010. Question 8, Paper 41, Winter 2010. Question 8, Paper 41, Summer 2011.

HOME WOR K 9. Question 8, Paper 42, Summer 2011. 10.Question 8, Paper 41, Winter 2011. 11.Question 8, Paper 43, Winter 2011. 12.Question 9, Paper 41, Summer 2012. 13.Question 8, Paper 42, Summer 2012. 14.Question 8, Paper 43, Winter 2012.