Nuclear Physics Tutorial Solutions (1)

SERANGOON JUNIOR COLLEGE PHYSICS 9646

20-1

Topic 20: Nuclear Physics

*1(a) The diagram represents a short part of the track of an -particle as it approaches a

stationary nucleus N. Complete the diagram to show the path of the -particle as it passes by, and moves away from N. [1]

Thinking Process What is the charge of the nucleus and the alpha particle? How does this affect the motion of the alpha particle?

20.1 The Nucleus

20.2 Isotopes

N

Solution

Note the significance of the dotted line joining the nucleus to the final straight-line path of the

alpha-particle.

N


20-2

(b) Classic experiments on -particle scattering were performed by Rutherford, Geiger and Marsden. State and explain the experimental observation(s) obtained from such experiments which suggested that

(i) the nucleus is small, [2]

(ii) the nucleus is massive and charged. [4]

(c) One isotope of thallium (Tl), atomic number 81, has nuclei of mass number 205.

(i) What is meant by the term isotope? [1] (ii) Write down the notation for the representation of this thallium nucleus. [1]

Solution

Most of the alpha particles passed through without deflection. [1]

This demonstrated that the atoms should contain mostly empty spaces for most of the

alpha particles to have passed through. [1]


alpha-particle.

Solution

Some alpha particles were deflected by more than 90. [1]

The alpha particles would not have deflected by a large angle if the nucleus has a small

mass. The nucleus would have been knocked out of the atom instead. [1]

Some particles deviated from their original path indicated. [1]

This showed that the nucleus must be charged such that there is Coulomb force between

the nucleus and the alpha particles. [1]


alpha-particle.

Solution

Isotopes are atoms with the same atomic number but different mass numbers. [1]

Additional points:

They differ only in the number of neutrons.

They have similar chemical properties and therefore share the same chemical symbol.

The existence of isotopes can be demonstrated by a mass spectrometer.


alpha-particle.

Solution

205

81 lT [1]


20-3

(iii) For this thallium nucleus, write down

1. the number of protons, [1]

2. the number of neutrons. [1]

Solution

Number of protons = 81 [1] Number of neutrons = 205 -81 = 124 [1]


alpha-particle.


20-4

*2 (a) Explain the terms mass defect and nuclear binding energy. [2]

(b) In a proposed fusion reactor, one possible reaction is n He H H 10

42

31

21

.

(i) How much energy might 150 kg of the appropriate mixture of isotopes of

hydrogen produce? (5.07 x 1016 J) [3] (ii) For how long, to the nearest month, would this supply a city with an average

demand of 1.5 GW, if the process were 8.0% efficient? (1 month) [2]

mass of H21 = 2.014102 u mass of H31 = 3.016050 u

mass of He42 = 4.002603 u mass of n10 = 1.0086651 u 1 month = 2.6 x 10

6 s

20.3 Mass Defect and Nuclear Binding Energy

20.4 Nuclear Processes

Solution

Mass defect is the difference between the total mass of the individual constituent nucleons and

the mass of the nucleus. [1]

Nuclear binding energy is the energy needed to completely separate a nucleus into its

constituent nucleons (protons and neutrons). [1]

Solution

(i) Mass of 1 atom of deuterium and tritium = 2.014102 u + 3.016050 u = 5.030152 u

Number of pairs of deuterium and tritium in 150 kg of mixture of isotopes of hydrogen

= 150/(5.030152 x 1.66 x 10-27) = 1.796 x 1028 [1]

Amount of energy produced when 1 atom of deuterium react with 1 atom of tritium

25 030152 4 002603 1 0086651. . . u c = 2.821 x 10-12 J [1]

Amount of energy produced by 150 kg mixture

= 1.796 x 1028 x 2.821 x 10-12 = 5.067 x 1016 J [1]

(ii) Number of months supply can last

= 16

9

energy supply 0 08 5 067 10

energy demand per month 1 5 10 30 24 60 60

. .

. [1]

= 1.04 month [1]


20-5

*3 Uranium-235 nuclei may undergo fission upon interacting with slow-moving neutrons.

Two possible reactions are

reaction 1: 235 1 139 95 1

92 0 54 38 0U n Xe Sr 2 n energy

reaction 2: 235 1 116

92 0 46U n 2 Pd c energyx

(a) For reaction 2, identify the particle c and state the number x of such particles

produced in the reaction. [2] (b) The binding energy per nucleon E for a number of nuclides is shown below.

Nuclide E / MeV 95

38Sr 8.74

139

54 Xe 8.39

235

92U 7.60

1. State what is meant by the binding energy per nucleon of a nucleus. [1] 2. Show that the energy released in reaction 1 is 210 MeV. [2]

Solution

By conservation of atomic and mass numbers,

Atomic number of xc = 92 - (2 x 46) = 0

Mass number of xc = (235 + 1) (2 x 116) = 4

Hence, c is a neutron [1]

x = 4 [1]

such that xc is 4 10n

Solution

The amount of energy required to completely separate a nucleus into its constituent nucleons divided by the total number of nucleons in the nucleus. [1]

Solution

Energy released = (total binding energies of Xe-139 and Sr-95) Binding energy of U-235

= (8.39 x 139) + (8.74 x 95) (7.60 x 235) [1]

= 210.51 MeV [1]


20-6

3. The energy released in reaction 2 is 163 MeV. Suggest, with a reason, which one of the two reactions is more likely to occur. [2]

(c) Sketch a fully labeled graph to show the variation with nucleon number of the binding

energy per nucleon. [2]

(d) Hence explain why fusion of nuclei having high nucleon numbers is not associated with

a release of energy. [3]

Solution

Reaction 1 releases more energy than reaction 2, resulting in daughter nuclei which are more stable. [1] As processes which result in more stable products are favoured, reaction 1 is more likely to occur compared to reaction 2. [1]

Solution

[1] correct shape [1] correctly labeled axis

Solution

The binding energy is equal to the product of the binding energy per nucleon and the

nucleon number. [1]

For fusion of nuclei having high nucleon numbers, the binding energy of the product is

lower than the total binding energy of the reactants. [1]

Hence energy is not released but absorbed. [1] (As Energy released in nuclear reaction =

binding energy of product binding energy of reactant, the negative value of energy

released implies that energy is actually absorbed instead of being released.)


20-7

*4 A collision takes place between an -particle ( He42 ) traveling at 3.0 x 107 m s1 and a

stationary nitrogen nucleus. It results in the following nuclear reaction.

14 4 17 1

7 2 8 1N He O H

The masses of the nuclei involved are given below.

14

7N : 13.9993 u, 4

2He : 4.0015 u, 17

8O : 16.9947 u, 1

1H : 1.0073 u

The particles move in a straight line, as shown in the diagram below. The speed of the proton after the collision is 6.0 x 107 m s1.

(a) State the number and type of particles which form an -particle. [1]

(b) Calculate the small change in mass, in kilograms, which takes place in this nuclear

reaction. (1.99 x 10-30 kg) [2] (c) Use your answer to (b) to calculate the minimum kinetic energy needed by the

-particle to cause the nuclear reaction. (1.79 x 10-13 J) [2]

Solution

2 protons and 2 neutrons [1]

Solution

Change in mass = [ (16.9947 + 1.0073) - (13.9993 + 4.0015) ] x 1.66 x 10-27 [1]

= 1.99 x 10-30 kg [1]

Solution

Assuming that all the kinetic energy of the particle is converted into the increase in mass of

the product,

Minimum kinetic energy needed by the particle to cause the nuclear reaction,

E = mc2

= (1.99 x 10-30) (3.0 x 108) 2 [1]

= 1.79 x 10-13 J [1]


20-8

(d) Calculate v, the velocity of the oxygen nucleus after the collision. State the direction of this velocity. [3]

(3.51 x 106 m s-1, to the right)

Solution

By the principle of conservation of momentum,

mHevHe = m0v + mHvH

v = (mHevHe - mHvH)/m0

= [(4.0015 x 3.0 x 107) (1.0073 x 6.0 x 107)] / 16.9947 [1]

= 3.51 x 106 m s-1 [1]

After the collision, the oxygen moves off with a velocity of 3.51 x 106 m s-1 in the direction to the

right (same direction with which the proton moves off). [1] for the correct direction stated


20-9

5 The large amount of energy released in a nuclear fission reaction, together with the emission of more than one neutron, has made it possible for neutron-induced fission to be used as a source of useful energy.

When a neutron is captured by a Uranium-235 (U-235) nucleus, it causes the nucleus to fission. On average 2.5 neutrons are emitted in these fission reactions. This is illustrated in the Fig. 5.1.

When the conditions are suitable, a chain reaction can occur. If this chain reaction is not controlled, an explosion is likely. However, if the chain reaction is controlled, as in a nuclear reactor, a source of continuous power may be created.

(a) (i) Explain, with the aid of a diagram, what is meant by a chain reaction, with

reference to fission. [3]

Fig. 5.1

Solution

Where X and Y are fission products [1] for clearly labeled diagram

A chain reaction refers to the state of a system containing fissile nuclei, where each

neutron which produces a fission releases more neutrons, which themselves produce

further fission, with the release of yet more neutrons. [1]

Consequently, the number of fissions per unit time increases exponentially and the reaction

continues to operate on its own. [1]


20-10

(ii) Suggest why, in an uncontrolled chain reaction where all neutrons are captured by U-235 nuclei, the majority of the energy is released during the final stages of the fission of a sample of the uranium. [1]

(b) The induced fission reaction of U-235 may be represented by a nuclear reaction of the

form: 235 1 192 0 0U n P Q 2 or 3 n 198 MeVa c

b d .

Fission products P and Q

have approximately equal masses. However, when any two U-235 nuclei undergo fission, the fission products may not be the same. Hence, in a large sample of U-235 undergoing fission, many different fission products will be produced. The percentage amount of each fission product in the fissioned materials is referred to as the percentage yield. The variation with nucleon number of the percentage yield of different fission products is referred to as a fission yield curve and is illustrated in Fig. 5.2.

(i) Suggest why the percentage yield is shown on a logarithmic scale. [1]

Fig. 5.2

Solution

An uncontrolled chain reaction will proceed at an accelerating rate. The number of fissions per

unit time will escalate as time goes on, as does the amount of energy produced. [1]

Solution

The percentage yields are shown on a logarithmic scale because the percentage yield takes

on a very wide range of values spanning a few orders of magnitude.

[1]


20-11

(ii) Use Fig. 5.2 to determine the nucleon numbers of those fission products that have the same percentage yield as the nuclide with a nucleon number of 82. [1]

(iii) By reference to the nuclear equation, and your answer to (ii), suggest the nucleon

number of the nuclide that would be produced in the same fission reaction as the nuclide of nucleon number 82. [1]

(iv) Use Fig. 5.2 to determine the percentage yield of fission products having nucleon

numbers of 95 and 139. [1]

(v) Hence show that the fission products in (iv) are about 600 times more likely to be produced than those having masses equal to each other. [2]

Solution

From graph, the nucleon numbers are 107, 127 and 152. [1] if all are correct

Solution

From equation and answer in part (ii), answer is 152. [1]

Solution

From the graph, Percentage yield = 6.4 % [1]

Solution

The nucleon number of fission products that have masses equal to one another is

approximately 118 (From 235/2). [1]

From Fig. 5.2, the percentage yield of a fission product with nucleon number 118 = 0.01

Compare to the fission products in (iv),

Ratio of percentage yield = 6.4 /0.01 = 640 [1]

Hence, the fission products in (iv) are around 600 times more likely to be produced than

those having masses equal to each other.


20-12

(c) The energy released in the reaction in (b) occurs partly as kinetic energy of the daughter nuclei (167 MeV) and of the neutrons (5 MeV).

(i) Suggest one other mechanism by which energy is released in the fission

reaction. [1]

(ii) In a nuclear power station, 25% of the energy of the fission products is converted into electrical energy. Calculate, for the fission of a mass of 1.0 kg of U-235, 1. the number of nuclei in 1.0 kg of U-235, (2.56 x 1024) [1]

2. the electrical energy generated, (1.76 x 1013 J) [2]

3. the average power output, in megawatts, of the power station if the uranium is fission in a time of 24 hours. (204 MW) [2]

Solution

Electromagnetic radiation [1]

Solution

Method 1: (Finding the mass of 1 single nucleus of U-235 first)

Mass of 1 single U-235 nucleus = 235u

= 235 x 1.66 x 10-27

= 3.901 x 10-25 kg

Total number of nuclei in 1.0 kg of U-235 = 1.0 / 3.901 x 10-25

= 2.56 x 1024 [1]

Method 2: (Finding the number of moles of Uranium in 1.0 kg of U-235 first)

Number of nuclei in 1.0 kg of Uranium-235 = (1000/235) x 6.02 x 1023

= 2.56 x 1024 [1]

Solution

Electrical energy generated = 25% energy of fission products

= 25% x ((167 + 5) x 106 x 1.60 x 10-19) x 2.56 x 1024 [1] = 1.76 x 1013 J [1]

Solution

Average power input = (1.76 x 1013 )/ (24 x 60 x 60) [1]

= 2.04 x 108 W

= 204 MW [1]


20-13

(d) The fission products are usually radioactive and give rise to a series of radioactive decay products. Each decay product has its own half-life, but eventually a stable nuclide is reached.

Two such fission products with their decay products and half-lives are shown below.

5 67 hours 2 x 10 years99 99 9942 43 44Mo Tc Ru

stable solid

16 seconds 1.1 minute 13 days 40 hours140 140 140 140 14054 55 56 57 58Xe Cs Ba La Ce

stable solid

Considering equal amount of these two fission products, suggest why there are very

different problems for the storage of this nuclear waste. [3]

Solution

For the first fission product Mo-42, its decay product Tc-43 has an extremely long half-life.

Hence, it will take a very long time before this fission product decays to yield the stable

Ru-44. [1]

In comparison for the second fission product, Xe-54, its decay products have much shorter

half-lives. Hence, it yields the stable Ce-58 in a comparatively much shorter time. [1]

As a result, the nuclear waste for Mo-42 has to be stored for a substantially long time

before the waste ceases to be radioactive and harmful whereas the nuclear waste of Xe-54

is likely to have a high activity initially which needs to be carefully managed. [1]


20-14

*6 The activity of a piece of radioactive material is 4.3 x 105 Bq at time t = 0 s. The

number of undecayed atoms in the material at time t = 0 s is 7.9 x 1015. Calculate

(a) (i) the activity after 4.0 half-lives have elapsed. (2.69 x 104 Bq) [1]

(ii) the number of undecayed atoms after 4.0 half-lives, (4.94 x 1014) [1]

(b) the decay constant , (5.44 x 10-11 s-1) [1]

(c) the half-life 1 2/t (1.27 x 1010 s) [1]

20.5 Radioactive Decay

Solution

Using the relationship 01

2

n

A A

, 4

5 414 3 10 2 69 10 Bq2

A . .

[1]

Solution

Using the relationship 01

2

n

N N

, 4

15 1417 9 10 4 94 10 2

N . .

[1]

Solution

Using the relationship, A = N At time t = 0 s,

4.3 x 105 = x 7.9 x 1015

= 5.44 x 10-11 s-1

[1]

Solution

Using the relationship, 1 22

/ln

t

10

1 2 11

21 27 10 s

5 44 10

/

lnt .

. [1]


20-15

7 Measurements are made of the activity of a specimen of carbon from pieces of wood

found in a fireplace at an archaeological site. The specimen is found to contain one

Carbon-14 (C-14) atom per 8.6 x 1010 Carbon-12 (C-12) atoms. In similar carbon from

a modern fire the concentration of C-14 atoms is greater at one C-14 atom per

3.3 x 1010 C-12 atoms. The difference between these two figures is because C-14 is

radioactive and some atoms have decayed over the years.

(a) Calculate the age of the wood from the ancient fire. The half-life of C-14 is 1.8 x 1011 s.

C-12 is stable. (2.49 x 1011 s) [2]

(b) The technique of dating described above

is difficult to carry out accurately. This difficulty can be minimized by using all the C-14 atoms rather than just those which happen to undergo radioactive decay when the dating is being carried out. Carbon atoms from the wood can be ionized by removing one electron from each atom. They are then formed into a beam which is passed through a magnetic field as shown in Fig. 8.1.

(i) Explain why the paths of the two types of ion are different. [2]

Fig. 8.1

Solution

Let P represent the portion of Carbon-14 atoms to Carbon-12 atoms found in the wood.

1 2

0

1

2

/

ttP

P

1110 1 8 10

10

113 3 10

1 28 6 10

t..

.

[1]

t = 7.89 x 103 years [1]

Solution

The mass of a Carbon-14 ion is different from that of a Carbon-12 atom. [1]

The same magnetic force exerted on each atom is the same as they have the same speed

and charge. [1]

Hence, the radius of curvature of the path of each type of ion through the magnetic field will

be different.


20-16

(ii) Explain why the three sections of each path followed by the ions are straight, circular, straight, and why there is no change in the speed of an ion throughout. [4]

(iii) Suggest why this method of measuring the ratio of C-14 to C-12 atoms is more reliable. [2]

*8 (a) Explain how, when making measurements of radioactivity, practical steps can be

taken to overcome problems caused by,

(i) the random nature of radioactivity, [2]

(ii) background radiation levels. [2]

Solution

The ions travel in a straight path in the vacuum as there is no electric field or magnetic field

present. There is no net force on, and thus no acceleration of, the ions. [1]

In the magnetic field, they travel in a circular path because the magnetic force exerted on

the ions is always perpendicular to its velocity, providing the necessary centripetal force. [1]

Since the force is perpendicular to the velocity, there is no work done by the force and

hence no change in kinetic energy of the ion and therefore, no change in the speed of each

ion.

[1]

The ions travel in a straight line after emerging from the magnetic field, back into a vacuum

with no electric nor magnetic field. [1]

Solution

This method enables one to find out the amount of Carbon-14 exactly [1]

instead of measuring the radioactivity which is probabilistic in nature. [1]

Solution

The radioactivity of a single atom is random. [1]

However, when taking a sufficiently large sample size, the radioactive decay follows a

predictable pattern described by the decay constant. [1]

Solution

The radioactivity level can be measured in the absence of the sample first, followed by the

radioactivity level in the presence of the sample. [1]

The difference in the readings will be the activity caused by the sample and not

background radiation. [1]


20-17

(b) Radioactive iron, 59Fe, is a radioactive nuclide with a half-life of 46 days. It is used medically in the diagnosis of blood disorders. Measurements are complicated by the fact that iron is secreted, i.e. removed, from the body at a rate such that 69 days after administering a dose, half of the iron atoms in the dose have been excreted. That is to say, iron has a biological half-life of 69 days.

(i) If the count rate from a blood sample is 960 counts per minute, what will it be

from a similar blood sample taken 138 days later? (30 counts per minute) [2]

(ii) After taking the first sample, how long further would it take for a similar sample give a count rate of 480 counts per minute? (27.6 days) [3]

(iii) Derive the relation between T, the effective half-life of the radioactive nuclide in the body, Tb, the biological half-life, and Tr, the radioactive half-life. [3]

Solution

The number of 59Fe decreases via two processes radioactive decay and excretion such

that the ratio of count rate to initial count rate is the product of two factors.

1 2 1 21 1

2 2

/ radio / bio

t tt t

o

C

C

138 46 138 691 1

960 2 2

/ /C

[1]

C = 30 counts per min [1]

Solution

0count rate [1]

480 960

ln 2 ln 2 480ln [1]

69 46 960

27 6 days [1]

B R

B R

t t

t

C C e e

e

t

t .

Solution

0 0[1]

[1]

1 1 1[1]

eff B R

eff B R

t t t

t ( )t

eff B R

B R

A e A e e

e e

T T T


20-18

9(a) A student is provided with a freshly prepared sample of a radioactive material. The count rate C from the source is found to vary with time t as shown in Fig. 9.1 (a).

A second similar sample of the radioactive material is then prepared and the student repeats the experiment, but with the sample at a higher temperature. The variation with time of the count rate for the second sample is shown in Fig. 9.1 (b).

State the evidence that is provided by these two experiments for

(i) the random nature of radioactive decay, [1]

(ii) the spontaneous nature of radioactive decay. [1]

Fig. 9.1 (a) Fig. 9.1 (b)

Solution

There are small variations in count rates, as shown by the jagged lines on the two graphs. [1]

(This shows that the radioactive decay is random in nature, as one cannot predict exactly how

many nuclei will disintegrate the next moment.)

Solution

The two graphs have the same trend despite the samples decaying under different

temperatures. [1]

(This shows the spontaneous nature of radioactive decay, as the rate of disintegration cannot

be controlled by a change in physical conditions such as temperature.)


20-19

(b) The radioactive source in (a) is an isotope of radon ( 22086Rn ) that emits -radiation to

become polonium (Po).

(i) State the number of neutrons in one nucleus of Radon-220. [1] (ii) Write down a nuclear equation to represent the radioactive decay of a nucleus of

radon. [1] (iii) Assuming that the radon nuclide is initially stationary, determine the fraction of

the total kinetic energy produced in the radioactive decay which is carried away

by the -particle. (0.982) [4]

Solution

Number of neutrons = mass number atomic number =134 [1]

Solution

220 216 486 84 2Rn Po [1]

Solution

By conservation of linear momentum,

0 Po Pom v m v

Po Pom vvm

[1]

Kinetic energy of alpha particle,

2 21 1

2 2

Po Pom vKE m v m ( )m

As a fraction of the total energy,

2

2 2

2

2 2

1

2 [1]1 1

2 2

1

totalPo Po

Po Po Po

Po Po PoPo Po

m vKE

KEm v m v

m v mm ( )

m m

m v mm ( ) m v

m m

[1]Po

Po

m

m m

216 2160 982

216 4 220.

[1]


20-20

10 After the sudden release of radioactivity from the Chernobyl nuclear reactor accident in 1986, the radioactivity of milk in Poland rose to 2000 Bq per litre due to Iodine-131, with a half-life of 8.04 days. Radioactive iodine is particularly hazardous, because the thyroid gland concentrates iodine. The Chernobyl accident caused a measurable increase in thyroid cancers among children in Belarus.

(a) For comparison, calculate the activity of milk due to potassium. Assume that 1 litre of

milk contains 2.00 g of potassium, of which 0.0117% is the isotope 40K which has a half-life of 1.28 x 109 yr. (60.5 Bq) [3]

(b) Calculate the number of days it will take for the activity due to iodine fall below that due to potassium. (40.6 days) [2]

Solution

Activity of potassium, AK = K NK

Nmilk = 23 182 00 0 01176 02 10 3 52 10

40 100

. .. . particles [1]

17

9

ln 21 72 10

1 28 10 365 24 3600

-K .

.

[1]

AK = 17 181 72 10 3 52 10 60 5 Bq-. . . [1]

Solution

As the half-life of potassium is much longer than that of the iodine, it can be assumed that the

activity of the potassium remain constant for the period of concern in this calculation.

0 tA A e I I

ln 2

8 04 2000 60 5t

.KA e A .

I [1]

60 5ln 200040 6 days

ln 2

8 04

.

t .

.

[1]


20-21

11 A certain radioactive source with a half-life 4 years, is used in a cancer therapy treatment. When the source is first used, a particular treatment requires 10 minutes of irradiation.

If the same source is used 2 years later, what is the time required for this treatment? [2] (14.1 min)

Solution

Let the initial activity be A0.

After two years, the activity A1 will be

2

41

0

1

2

A

A

0 5

1 0

1

2

.

A A

01 [1]2

AA

Since intensity of radiation, I activity A Initially time required is 10 min for activity A0. For the same amount of irradiation two years later, when activity is A1, the new time required is

01 0

1

At t

A

01

0

10

2

At

A

1 1 41 10 14 1 mint . . [1]

Hence the time required is 14.1 min.


20-22

*12 Nuclei X decays to become Nuclei Y which is stable. Nuclei X is found to have a half-life of 24.0 hours A fresh sample of radioactive material contains N0 nuclei of Nuclei X and no Nuclei Y. (a) Sketch on the same axes, graphs which show the variation with time t of the

number N of nuclei in the sample which are for Nuclei X and Y. (Label the graphs clearly.) [3]

(b) Deduce the age of the sample of radioactive material when

Number of Nuclei X in sample 1

Number of Nuclei Y in sample 3

(48 hours) [3]

*** End of Tutorial ***

Solution

Correct Trend for X [1] Correct Trend for Y [1] Time when N of X and Y intersect is labeled correctly (half-life) [1]

Number of atoms

time

N

M

T

X

Y

t/hrs 24

N

N0

Solution

When the ratio of Number of Nuclei X to Number of Nuclei Y in sample is 1/3, it implies that the number of undecayed Nuclei X is N0 [1]

Using 01

2

n

N N

0 0

1 12

4 2

n

N N n

[1]

As two half-lives have elapsed, the age of the sample is 48 hours [1]


20-23

1 Fig. 1.1 below represents an experiment on Rutherford scattering in which -particles are directed at a gold foil. The detector is shown in two positions in the evacuated chamber.

(i) Why is it necessary to remove the air from the apparatus? [2] (ii) Explain why the gold foil should be very thin. [2]

Assignment

Fig. 1.1

Solution

to minimise collisions between particles and air molecules [1]

-particles have a short range in air (35 cm) [1]

Solution

the particles must only be scattered once [1]

the particles must not be absorbed by the foil [1]


20-24

(iii) 1. Explain why the count rate from the -particle detector in position 1 is much greater than that in position 2. [2]

2. What can be deduced from this observation about the structure of the atom and the properties of the nucleus of gold? [1]

(b) Fig. 1.2 illustrates the variation with nucleon number A of the binding energy

per nucleon E of nuclei.

Fig. 1.2

Solution

Most of the -particles pass straight through. [1]

Because:

- most -particles do not pass close enough to the nucleus to be deflected (or few

pass close enough to be deflected significantly)

- atoms consist mainly of open space

- the nucleus is very small in comparison to the size of an atom [1]

Solution

The nucleus is charged as it causes the charged -particles to deflect. [1]


20-25

(i) Explain what is meant by the binding energy of a nucleus. [1]

(ii) With the aid of Fig. 1.2, explain why more energy per nucleon is

released in fusion than in fission. [3]

(iii) On Fig. 1.2, mark with the letter S the region of the graph representing nuclei having the greatest stability. [1]

Solution

Binding energy of a nucleus refers to the amount of energy required to separate it into its

constituent protons and neutrons.

Solution

Energy released in a nuclear reaction is equal to the difference in binding energies between the products and the original reactants. [1]

Fusion reaction is one in which two light nuclei combine to form a heavier nucleus, while in fission a heavy nucleus splits into two lighter nuclei. [1]

From graph, the steeper slope of the binding energy curve for lighter nuclei indicates that the change in binding energy in fusion is larger, compared to that for fission reactions. [1]

Hence, more energy is released in a fusion reaction. If no reference is made to graph but generally correct, MAX [2]


20-26

(c) Uranium-235 may undergo fission when bombarded by a neutron to produce Xenon-142 and Strontium-90 as shown below.

(i) Determine the number of neutrons produced in this fission reaction. [1] (ii) The equation shown above is incomplete. State the particle which

must be included on the right side. [1]

(iii) Data for binding energy per nucleon are given in Fig. 1.3.

Calculate 1. the energy released in this fission reaction in Joules. [3]

2. the mass equivalent of this energy.

Fig. 1.3

Solution

Number of neutrons = 4 [1]

Solution

energy = BE of Products BE of Reactant [1]

= (8.37 142 + 8.72 90) 235 7.59 [1]

= 189.69 MeV = 189.69 x 106 x 1.6 x 10-19

= 3.04 x 10-11 J [1]

Solution

Using the concept that energy E = mc2

mass equivalent = (3.04 x 10-11) / (3.0 108)2

= 3.37 1028 kg [2]

Solution

-particle [1]


20-27

(iii) A nuclear power station supplies electrical power of 100 MW to a town

in one year at an efficiency of 35%. Calculate the mass of uranium needed in the operation of the power station in a year.

2(a) State the changes to the number of protons and of neutrons that occur within nuclei

when they emit

(i) -particles, [1]

(ii) -particles, [1]

(iii) -radiation. [1]

Solution

Useful output Power = 100 x 106 W at 35% efficiency

In a year, total power produced = 100 x 106/0.35 = 285.71 x 106 W [1]

Energy released in 1 reaction = 3.04 x 10-11 J

No of reactions required = 285.71 x 106 / 3.04 x 10-11

= 9.3985 x 1018 [1]

= N, number of uranium atoms required.

Mass of uranium required = (N/Av ) x molar mass of uranium

= 1.5612 x 10-05 x 235 g

= 3.67 x 10-6 kg [1]

Solution

The number of protons decreases by two and the number of neutrons decreases by 2 as well.

Solution

The number of protons increases by 1. The number of neutrons decreases by 1.

Solution

Both the number of protons and the number of neutrons remain unchanged.


20-28

(b) A sample of radioactive material emits a

narrow parallel beam of -particles, -particles, -radiation.as shown in Fig. 2.1

The beam passes through a region R where a uniform field may be applied at right angles to the beam and into the plane of the paper. Discuss the effect on the beam if the field in the region R is (i) an electric field [6]

(ii) a magnetic field [6]

Fig. 10.1 Fig. 2.1

Solution

The alpha particles will be deflected into the paper along a parabolic path, [1]

as alpha particles are positively charged and hence, moves in the direction of the electric

field and that the electric force is always in the same direction and hence, the path is

parabolic. [1]

The beta particles will be deflected out of the paper along a parabolic path, [1]

as beta particles are negatively charged and hence, moves opposite to the direction of the

electric field and that the electric force is always in the same direction and hence, the path

is parabolic. [1]

The gamma rays will not be deflected [1]

by the electric field as it carries no charge, hence no electric force acting on it. [1]

Solution

The alpha particles will be deflected upwards along a circular path, [1]

as alpha particles are positively charged and hence, applying Flemings left hand rule,

indicated that the magnetic force is upwards. As the force is always perpendicular to the

direction of motion of the particles, the path is circular. [1]

The beta particles will be deflected downwards along a circular path, [1]

as beta particles are negatively charged and hence, applying Flemings left hand rule,

indicated that the magnetic force is downwards. As the force is always perpendicular to the

direction of motion of the particles, the path is circular. [1]

The gamma rays will not be deflected by the magnetic field [1]

as it carries no charge and hence no magnetic force acting on it. [1]


20-29

(c) The GM counter registers the following count-rates at two different times, t. The values are corrected for background radiation

t / hour Count-rate/ min-1 0 3198

6.0 801

0.80% of the -particles leaving the source are recorded by the counter. (i) Calculate the activity of the source at t = 0. [1]

(ii) Counting is stopped when the (corrected) count-rate first falls to less than 50

min-1. Estimate the value of t at which this event occurs. [2]

Solution

Count rate C at t = 0, 3198

53 3 Bq60

C .

Activity of the source A at t = 0,

100A 53 3 66600 Bq

0 08 .

. [1]

Solution

1 2

0

1

2

/

ttC

C

1 2

6801 1

3198 2

/t

Half life = 3 hrs [1]

350 1

3198 2

t

t = 18 hrs [1]

Documents

Nuclear Physics Tutorial Solutions (1)