Matriculation Physics ( Nuclear Reaction )

1

PHYSICS CHAPTER 14

Bombardment Bombardment

with energetic with energetic

particlesparticles

Nuclear Nuclear fusionfusion

Nuclear Nuclear fissionfission

CHAPTER 14: Nuclear reactionCHAPTER 14: Nuclear reaction(2 Hours)(2 Hours)

Four types of Four types of nuclear reaction:nuclear reaction:

is defined as a is defined as a physical process physical process

in which there is a in which there is a change in identity change in identity

of an atomic of an atomic nucleusnucleus.

RadioactivRadioactiv

e decaye decay

PHYSICS CHAPTER 14

2

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: StateState the conservation of charge ( the conservation of charge (ZZ) and nucleon ) and nucleon

number (number (AA) in a nuclear reaction.) in a nuclear reaction. Write and completeWrite and complete the equation of nuclear reaction. the equation of nuclear reaction. CalculateCalculate the energy liberated in the process of nuclear the energy liberated in the process of nuclear

reactionreaction

Learning Outcome:

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14.1 Nuclear reaction (1 hour)

PHYSICS CHAPTER 14

3

14.1.1 Conservation of nuclear reaction Any nuclear reaction must obeyed conservation laws stated

below: Conservation of relativistic energy (kinetic and rest

energies):

Conservation of linear momentum:

Conservation of angular momentum:

14.1 Nuclear reaction

energy icrelativistreaction before energy icrelativist

reactionafter

momentumlinear reaction before momentumlinear

reactionafter

momentumangular reaction before momentumangular

reactionafter

PHYSICS CHAPTER 14

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Conservation of charge (atomic number Z):

Conservation of mass number A:

However, it is very hard to obey all the conservation laws.

Anumber massreaction before Anumber mass

reactionafter

Znumber atomicreaction before Znumber atomic

reactionafter

Note:Note:

The most importantimportant of conservation lawsconservation laws should be obeyedobeyed by every nuclear reaction are conservation of chargeconservation of charge (atomic atomic numbernumber )and of mass numbermass number.

PHYSICS CHAPTER 14

5

Energy is released (liberated) in a nuclear reaction in the form of kinetic energy of the particle emittedkinetic energy of the particle emitted, the kinetic energy of kinetic energy of the daughter nucleusthe daughter nucleus and the energy of the gamma-ray energy of the gamma-ray photonphoton that may accompany the reaction.

The energy is called the reaction OR disintegration energy (Q). It may be calculated by finding the mass defect of the reaction

where

The reaction energy Q is the energy equivalent to the mass defect m of the reaction, thus

14.1.2 Reaction energy (Q)

2Δ cmQ

nucleus of massreactionafter products defect Mass

nucleus of massreaction before

fi mmm (14.1)(14.1)

(14.2)(14.2)Speed of light in vacuumSpeed of light in vacuum

PHYSICS CHAPTER 14

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is defined as the phenomenon in which an unstable nucleus the phenomenon in which an unstable nucleus disintegrates to acquire a more stable nucleus without disintegrates to acquire a more stable nucleus without absorb an external energyabsorb an external energy.

The disintegration is spontaneousdisintegration is spontaneous and most commonly

involves the emission of an alpha particle ( OR ), a beta

particle ( OR ) and gamma-ray ( OR ). It also

release an energy Q known as disintegration energydisintegration energy.

14.1.3 Radioactivity decay

Note:Note:

If the value of mm OR QQ is positivepositive, the reaction is called exothermic (exoergic)exothermic (exoergic) in which the energy releasedenergy released in the form the kinetic energy of the product.

If the value of mm OR QQ is negative, the reaction is called endothermic (endoergic)endothermic (endoergic) in which the energy need to be energy need to be absorbedabsorbed for the reaction occurred.

He42

γ00e0

1

PHYSICS CHAPTER 14

7

Polonium nucleus decays by alpha emission to lead nucleus can be represented by the equation below:

Calculate

a. the energy Q released in MeV.

b. the wavelength of the gamma-ray produced.

(Given mass of Po-212, mPo=211.98885 u ; mass of Pb-208,

mPb=207.97664 u and mass of particle , m=4.0026 u)

Solution :Solution :

Example 1 :

Q HePbPo 42

20882

21284

Q HePbPo 42

20882

21284

before before decaydecay

after after decaydecay

decay

fi ZZ fi AAand

PHYSICS CHAPTER 14

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a. The mass defect (difference) of the reaction is given by

The energy released in the decay reaction can be calculated by

using two method:

11stst method: method:

fi mmm αmmm PbPo

0026.497664.20798885.211 u 1061.9 3m

273 1066.11061.9 mkg 105953.1 29

kg 10661u 1 27 .

2829 1000.3105953.1 QJ 10436.1 12Q

2cmQ in kgin kg

PHYSICS CHAPTER 14

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a. Thus the energy released in MeV is

22ndnd method: method:

MeV 98.8Q

13

12

1060.1

10436.1

QJ 10601MeV 1 13 .

2cmQ

MeV 95.8Q

22

u 1

MeV/ 5.931c

cm

2MeV/ 5.931u 1 cin uin u

22

3

u 1

MeV/ 5.931u 1061.9 c

c

PHYSICS CHAPTER 14

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b. The reaction energy Q is released in form of gamma-ray where

its wavelength can be calculated by applying the Planck’s

quantum theory:Q

hcE

Q

hc

12

834

10436.1

1000.31063.6

m 1039.1 13Note:Note:

The radioactive decay only occurredradioactive decay only occurred when the value of mm OR QQ is positivepositive.

PHYSICS CHAPTER 14

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A nickel-66 nucleus decays to a new nucleus by emitting a beta particle.

a. Write an equation to represent the nuclear reaction.

b. If the new nucleus found in part (a) has the atomic mass of

65.9284 u and the atomic mass for nickel-66 is 65.9291 u, what

is the maximum kinetic energy of the emitted electron?

(Given mass of electron, me =5.49 104 u and c =3.00 108 m

s1)


a. Nuclear reaction equation must obey the conservation of atomic

number and the conservation of mass number.

Example 2 :

Q eXNi 01

6629

6628 decay

Ni6628

PHYSICS CHAPTER 14

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b. Given

The mass defect (difference) of the reaction is given by

If the reaction energy is completely convert into the kinetic

energy of emitted electron, therefore the maximum kinetic energy

of the emitted electron is given by

fi mmm eXNi mmm

41049.59284.659291.65 u 1051.1 4m

u 9284.65;u 9291.65 XNi mm

28274 1000.31066.11051.1

J 1026.2 14max

K

QK max

2cm

PHYSICS CHAPTER 14

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Table 14.1 shows the value of masses for several nuclides.

Discuss whether it is possible for to emit spontaneously an alpha particle.


If emits an alpha particle, the decay would be represented by

Since the total mass after the reaction is greater than that before the reaction, therefore the reaction does not occur.

Example 3 :

Nuclide Mass (u)

4.0026

22.9898

26.9815

He42

Al2713

Na2311

Table 14.1Table 14.1

Al2713

Al2713

Al2713 Na23

11 He42

26.9815 u26.9815 u 22.9898 u22.9898 u 4.0026 u4.0026 u

26.9924 u26.9924 u

PHYSICS CHAPTER 14

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is defined as an induced nuclear reaction that does not an induced nuclear reaction that does not occur spontaneously; it is caused by a collision between a occur spontaneously; it is caused by a collision between a nucleus and an energetic particles such as proton, neutron, nucleus and an energetic particles such as proton, neutron, alpha particle or photonalpha particle or photon.

Consider a bombardment reaction in which a target nucleus X

is bombarded by a particle x, resulting in a daughter nucleus Y,

an emitted particle y and reaction energy Q:

sometimes this reaction is written in the more compact form:

The calculation of reaction energy Q has been discussed in section 14.1.2.

14.1.4 Bombardment with energetic particles

Q YyxX

Yyx,X daughter nucleusdaughter nucleustarget (parent) target (parent) nucleusnucleus

bombarding bombarding particleparticle

emitted emitted particleparticle

PHYSICS CHAPTER 14

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Examples of bombardment reaction:

Q HOHeN 11

178

42

147 Op,N 17

8147

Q HeLinB 42

73

10

105

OR

Q He2HLi 42

11

73 He,pLi 4

273

Li,nB 73

105

OR

OR

A nitrogen nucleus is converted into an oxygen nucleus and a proton if it is bombarded by an alpha particle carrying certain amount of energy.a. Write down an expression for this nuclear reaction, showing the atomic number and the mass number of each nuclide and particle emitted.b. Calculate the minimum energy of the alpha particle for this reaction to take place.

(Given mp =0.167351026 kg; m =0.66466 1026 kg ; mass of

nitrogen nucleus , mN =2.325301026 kg; mass of oxygen nucleus,

mO =2.822821026 kg ; c =3.00108 m s1)

Example 4 :N14

7 O178

PHYSICS CHAPTER 14

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a. The expression represents the nuclear reaction is

b. The mass defect of the reaction is

Therefore the minimum energy of the alpha particle for this

reaction to take place is

N147 O17

8 Q H11

He42

fi mmm HOHeN mmmm

kg 101.2 30m

2626 1066466.01032530.2 2626 1016735.01082282.2

QK min 2min cmK

2830 1000.3101.2

J 1089.1 13min

K

PHYSICS CHAPTER 14

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Exercise 14.1 :Given c =3.00108 m s1, mn=1.00867 u, mp=1.00782 u,

1. Complete the following radioactive decay equations :

a.

b.

c.

d.

e.

f.

HeBe 42

84

BaSrPo 13956

9738

24094

n3IU 10

13153

23692

eNa 01

2911

ScSc 4721

4721

CaK 4020

4019

PHYSICS CHAPTER 14

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Exercise 14.1 :

2. Calculate the energy released in the alpha decay below:

(Given mass of U-238, mU=238.050786 u ; mass of Th-234,

mTh=234.043583 u and mass of particle , m=4.002603 u)

ANS. :ANS. : 6.876.8710101313 J J3. The following nuclear reaction is obtained :

Determine the mass of in atomic mass unit (u).(Given the mass of nitrogen nucleus is 14.003074 u)

ANS. :ANS. : 14.003872 u14.003872 u

Q HeThU 42

23490

23892

MeV 55.0HCnN 11

146

10

147

C146

PHYSICS CHAPTER 14

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At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: DistinguishDistinguish the processes of nuclear fission and the processes of nuclear fission and

fusion.fusion. ExplainExplain the occurrence of fission and fusion in the form the occurrence of fission and fusion in the form

of graph of binding energy per nucleon.of graph of binding energy per nucleon. ExplainExplain chain reaction in nuclear fission of a nuclear chain reaction in nuclear fission of a nuclear

reactor.reactor. DescribeDescribe the process of nuclear fusion in the sun. the process of nuclear fusion in the sun.

Learning Outcome:

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14.2 Nuclear fission and fusion (1 hour)

PHYSICS CHAPTER 14

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14.2.1 Nuclear fission is defined as a nuclear reaction in which a heavy nucleus a nuclear reaction in which a heavy nucleus

splits into two lighter nuclei that are almost equal in mass splits into two lighter nuclei that are almost equal in mass with the emission of neutrons and energywith the emission of neutrons and energy.

Nuclear fissionNuclear fission releases an amount of energy that is greater greater thanthan the energy released in chemical reactionchemical reaction.

Energy is released because the average binding energy per average binding energy per nucleon of the fission products is greater than that of the nucleon of the fission products is greater than that of the parentparent.

It can be divided into two types: spontaneous fissionspontaneous fission – very rarely occur. induced fissioninduced fission – bombarding a heavy nucleus with slow

neutrons or thermal neutrons of low energy (about 102 eV). This type of fission is the important process in the energy production.

14.2 Nuclear fission and fusion

PHYSICS CHAPTER 14

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n10

U23692

La14857

Br8535

U23592

n10

n10

n10

For example, consider the bombardment of by slow neutrons. One of the possible reaction is

The reaction can also be represented by the diagram in Figure 14.1.

Other possible reactions are:

U23592

Q n3LaBrUnU 10

14857

8535

23692

10

23592

Nucleus in the excited state.Nucleus in the excited state.

Figure 14.1Figure 14.1

Q n3XeSrUnU 10

13954

9438

23692

10

23592

Q n3BaKrUnU 10

14456

8936

23692

10

23592

PHYSICS CHAPTER 14

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Most of the fission fragments (daughter nuclei) of the uranium-235 have mass numbers from 90 to 10090 to 100 and from 135 to 145135 to 145 as shown in Figure 14.2.


PHYSICS CHAPTER 14

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Calculate the energy released in MeV when 20 kg of uranium-235 undergoes fission according to

(Given the mass of U-235 =235.04393 u, mass of neutron =1.00867 u, mass of Kr-89 =88.91756 u, mass of Ba-144

=143.92273 u and NA =6.021023 mol1)


The mass defect (difference) of fission reaction for one nucleus U-235 is

Example 5 :

Q n3BaKrnU 10

14456

8936

10

23592

fi mmm nBaKrnU 3mmmmm

u 1863.0m

00867.104393.235 00867.1392273.14391756.88

PHYSICS CHAPTER 14

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The energy released corresponds to the mass defect of one U-235 is

235 103 kg of uranium-235 contains of 6.02 1023 nuclei 20 kg of urainum-235 contains of

Therefore

Energy released

by 20 kg U-235

2cmQ

MeV 174Q

22

u 1

MeV/ 5.931u .18630 c

c

nuclei 1012.5 25

233

1002.610235

20

1741012.5 25

MeV 1091.8 27

PHYSICS CHAPTER 14

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A uranium-235 nucleus undergoes fission reaction by bombarding it with a slow neutron. The reaction produces a strontium-90 nucleus , a nucleus X and three fast neutrons.

a. Write down the expression represents the fission reaction.b. If the energy released is 210 MeV, calculate the atomic mass of nucleus X.(Given the mass of U-235 =235.04393 u, mass of neutron =1.00867 u and mass of Sr-90 =89.90775 u)


a. The expression represents the fission reaction is

Example 6 :

Sr9038

Q n3XSrnU 10

14354

9038

10

23592

PHYSICS CHAPTER 14

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The energy released of 210 MeV equivalent to the mass defect for U-235 is

Therefore the atomic mass of the nucleus X is given by

2cmQ

u 22544.0m

22

u 1

MeV/ 5.931210 c

cm

fi mmm nXSrnU 3mmmmmm

u 8934.142X m

00867.104393.23522544.0 00867.1390775.89 X m

PHYSICS CHAPTER 14

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is defined as a nuclear reaction that is self-a nuclear reaction that is self- sustaining as sustaining as a result of the products of one fission reaction initiating a a result of the products of one fission reaction initiating a subsequent fission subsequent fission reactionreaction.

Figure 14.3 shows a schematic diagram of the chain reaction.

14.2.2 Chain reaction

Figure 14.3Figure 14.3Stimulation 14.1

PHYSICS CHAPTER 14

28

From Figure 14.3, one neutron initially causes one fission of a uranium-235 nucleus, the two or three neutrons released can go on to cause additional fissions, so the process multiples.

This reaction obviously occurred in nuclear reactor. Conditions to achieve chain reactionConditions to achieve chain reaction in a nuclear reactor :

Slow neutronsSlow neutrons are better at causing fission – so uranium are mixed with a material that does not absorb neutrons but slows them down.

The fissile material must has a critical sizecritical size which is defined as the minimum mass of fissile material that will sustain the minimum mass of fissile material that will sustain a nuclear chain reaction where the number of neutrons a nuclear chain reaction where the number of neutrons produced in fission reactions should balance the produced in fission reactions should balance the number of neutron escape from the reactor corenumber of neutron escape from the reactor core.

The uncontrolled chain reactions are used in nuclear weapons – atomic bomb (Figure 14.4).

The controlled chain reactions take place in nuclear reactors (Figure 14.5) and release energy at a steady rate.

PHYSICS CHAPTER 14

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Figure 14.4Figure 14.4 Figure 14.5Figure 14.5

PHYSICS CHAPTER 14

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is defined as a type of nuclear reaction in which two light a type of nuclear reaction in which two light nuclei fuse to form a heavier nucleus with the release of nuclei fuse to form a heavier nucleus with the release of large amounts of energylarge amounts of energy.

The energy released in this reaction is called thermonuclear energy.

Examples of fusion reaction releases the energy are

The two reacting nuclei in fusion reaction above themselves have to be brought into collision.

As both nuclei are positively charged there is a strong repulsive force between them, which can only be overcome if the reacting nuclei have very high kinetic energies.

These high kinetic energies imply temperatures of the order of 108 K.

14.2.3 Nuclear fusion

Q nHeHH 10

32

21

21

Q HHHH 11

31

21

21

PHYSICS CHAPTER 14

31

At these elevated temperatures, however fusion reactions are self sustaining and the reactants are in form of a plasma (i.e. nuclei and free electron) with the nuclei possessing sufficient energy to overcome electrostatic repulsion forces.

The nuclear fusion reaction can occur in fusion bomb and in the core of a star.

Deuterium-tritium fusion is other example of fusion reaction where it can be represented by the diagram in Figure 14.6.

H21

Deuterium

Fusion reaction

TritiumH3

1

Alpha particle

He42

Neutron

n10

Q nHeHH 10

42

31

21

Stimulation 14.2


PHYSICS CHAPTER 14

32

A fusion reaction is represented by the equation below:

Calculatea. the energy in MeV released from this fusion reaction,b. the energy released from fusion of 1.0 kg deuterium,(Given mass of proton =1.007825 u, mass of tritium =3.016049 u and mass of deuterium =2.014102 u)

Solution :Solution :a. The mass defect of the fusion reaction for 2 deuterium nuclei is

Example 7 :

HHHH 11

31

21

21

pTDD mmmm

u 1033.4 3m 007825.1016049.3014102.2014102.2

fi mmm

PHYSICS CHAPTER 14

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a. Therefore the energy released in MeV is

b. The mass of 2 deuterium nuclei is 4 103 kg.

4 103 kg of deuterium contains of 6.02 1023 nuclei 1.0 kg of deuterium contains of

ThereforeEnergy released from

1.0 kg deuterium

2cmQ

MeV 03.4Q

22

3

u 1

MeV/ 5.931u 1033.4 c

c

nuclei 10505.1 26

233

1002.6104

0.1

03.410505.1 26MeV 1007.6 26

PHYSICS CHAPTER 14

34

The sun is a small star which generates energy on its own by means of nuclear fusion in its interior.

The fuel of fusion reaction comes from the protons available in the sun.

The protons undergo a set of fusion reactions, producing isotopes of hydrogen and also isotopes of helium. However, the helium nuclei themselves undergo nuclear reactions which produce protons again. This means that the protons go through a cycle which is then repeated. Because of this proton-proton cycle, nuclear fusion in the sun can be self sustaining.

The set of fusion reactions in the proton-proton cycle can be illustrated by Figure 14.7.

14.2.4 Nuclear fusion in the sun

PHYSICS CHAPTER 14

35

The amount of energy released per cycle is about 25 MeV. Nuclear fusion occurs in the interior of the sun because the

temperature of the sun is very high (approximately 1.5 107 K).

Qv eHHH 01

21

11

11

Q HeHH 32

11

21

Q HHHeHeHe 11

11

42

32

32

neutrinoneutrinopositron (beta plus)positron (beta plus)


PHYSICS CHAPTER 14

36

Table 14.2 shows the differences between fission and fusion reaction.

The similaritysimilarity between the fission and fusion reactions is both reactions produces energyproduces energy.

Graph of binding energy per nucleon against the mass number in Figure 14.8 is used to explain the occurrence of fission and fusion reactions.

14.2.5 Comparison between fission and fusion

Table 14.2Table 14.2

Fission Fusion

Splitting a heavy nucleus into two small nuclei.

Combines two small nuclei to form a larger nucleus.

It occurs at temperature can be controlled.

It occurs at very high temperature (108 K).

Easier to controlled and sustained.

Difficult to controlled and a sustained controlled reaction has not yet been achieved.

PHYSICS CHAPTER 14

37Mass number A

Bin

din

g e

ne

rgy

per

n

ucle

on

(M

eV

/nu

cle

on)

Greatest stabilityGreatest stability


FusionFusion

FissionFission

The rising partrising part of the binding energy curve shows that elements with low mass low mass number can produce energy by fusionnumber can produce energy by fusion.

The falling partfalling part of the binding energy curve shows that very heavy elementsheavy elements such as uranium can produce energy by fissionproduce energy by fission of their nuclei to nuclei of smaller mass numbernuclei of smaller mass number.

PHYSICS CHAPTER 14

38

Exercise 14.2 :Given c =3.00108 m s1, mn=1.00867 u, mp=1.00782 u,

1. Complete the following nuclear reaction equations:

a.

b.

c.

d.

e.

2. Calculate the energy released in joule for the following fusion reaction:(Given the mass of deuterium =3.3451027 kg, mass of tritium =5.0081027 kg, mass of He = 6.6471027 kg and mass of neutron =1.6751027 kg)

ANS. :ANS. : 2.82.810101212 J J

HeHe Li 42

32

63

H HNi 11

21

5828

n5XenU 10

13854

10

23592

C____,Be 126

94

Np,n_____ 167

nHeHH 10

42

21

21

39

PHYSICS CHAPTER 14

Next Chapter…CHAPTER 15 :Radioactivity

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Matriculation Physics ( Nuclear Reaction )