Phy 310 chapter 9

i d fi di d fi d

DrDr Ahmad Ahmad TaufekTaufek Abdul Abdul RahmanRahmanSchoolSchool of of PhysicsPhysics & Material & Material StudiesStudies,, FacultyFaculty of of AppliedApplied SciencesSciences, , UniversitiUniversiti TeknologiTeknologi MARA MARA Malaysia,Malaysia, Campus Campus of Negeri of Negeri SembilanSembilan

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

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

of an atomic of an atomic nucleusnucleus.

CHAPTER CHAPTER 9: 9: Nuclear reactionNuclear reaction(2 Hours)(2 Hours)(2 Hours)(2 Hours)

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

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L i O t

DR.ATAR @ UiTM.NS PHY310 NUCLEAR REACTION

Learning Outcome:

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: 9.1 Nuclear reaction (1 hour)

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 reactionthe equation of nuclear reactionWrite 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 reactionreactionreactionreaction

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9.1.1 Conservation of nuclear reaction9.1 Nuclear reaction

Any nuclear reaction must obeyed conservation laws stated below:

Conservation of relativistic energy (kinetic and restConservation of relativistic energy (kinetic and rest energies):

=energyicrelativist∑ energyicrelativist∑Conservation of linear momentum:

=gyreactionbefore∑ gy

reactionafter ∑

=momentumlinear reaction before∑ momentumlinear

reactionafter ∑ Conservation of angular momentum:

=momentumangular tib f∑ momentumangular

tift∑3

=reactionbefore∑ reactionafter ∑


Conservation of charge (atomic number Z):

=Znumber atomic∑ Znumber atomic∑Conservation of mass number A:

=reactionbefore∑ reactionafter ∑

=Anumber massreactionbefore∑ Anumber mass

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

N tN tNote:Note:

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

4


Energy is released (liberated) in a nuclear reaction in the form of kinetic energ of the particle emittedkinetic energ of the particle emitted the kinetic energ ofkinetic energ of

9.1.2 Reaction energy (Q)

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 gammaenergy of the gamma--ray 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 wherewhere

− nucleus of massreactionafter products∑ =defect Mass nucleus of mass

reaction before∑

The reaction energy Q is the energy equivalent to the mass

fi mmm −=∆ (9.1(9.1))

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

( ) 2∆ cmQ (9 2(9 2))

5

( )∆ cmQ = (9.2(9.2))Speed of light in vacuumSpeed of light in vacuum


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 theexothermic (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 QQ gendothermic (endoergic)endothermic (endoergic) in which the energy need to be energy need to be absorbedabsorbed for the reaction occurred.

is defined as the phenomenon in which an unstable nucleus the phenomenon in which an unstable nucleus disintegrates to acquire a more stable nucleus withoutdisintegrates to acquire a more stable nucleus without

9.1.3 Radioactivity decay

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

4involves the emission of an alpha particle (α OR ), a beta particle (β OR ) and gamma-ray (γ OR ). It also

l Q k di i t tidi i t ti

He42

γ00e0

1−

6

release an energy Q known as disintegration energydisintegration energy.


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

Example 1 :

represented by the equation below:

CalculateQ++→ HePbPo 4

220882

21284

Calculatea. the energy Q released in MeV.b. the wavelength of the gamma-ray produced.g g y p(Given mass of Po-212, mPo=211.98885 u ; mass of Pb-208, mPb=207.97664 u and mass of α particle , mα=4.0026 u)Solution :Solution : Q++→ HePbPo 4

220882

21284

before before after after

α decay

decaydecay decaydecay

7∑∑ = fi ZZ ∑∑ = fi AAand


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

fi mmm −=∆ fi mmm∆( )αmmm +−= PbPo

( )002649766420798885211 +−=

The energy released in the decay reaction can be calculated by

( )0026.497664.20798885.211 +=u 1061.9 3−×=∆m

gy y yusing two method:11stst method:method: kg 10661u 1 27−×= .( ) 2cmQ ∆=

( )( )273 1066.11061.9 −− ××=∆m

( )in kgin kg

( )( )kg 105953.1 29−×=

( )( )2829 100031059531 ××= −Q

8

( )( )1000.3105953.1 ××QJ 10436.1 12−×=Q


Solution :Solution :a. Thus the energy released in MeV is

12104361 −× J10601MeV1 13−×=

MeV988Q131060.1

10436.1−×

×=Q J 10601MeV1 ×= .

22ndnd method:method:

MeV98.8=Q( ) 2cmQ ∆= 2MeV/ 5.931u 1 c=

22MeV/ 5.931 ccm ⎥⎤

⎢⎡ ⎞

⎜⎜⎛

∆=

in uin u

u 1cm ⎥⎦

⎢⎣ ⎠

⎜⎜⎝

∆=

( ) 22

3 MeV/5931 c ⎤⎡⎟⎞

⎜⎛( ) 23

u 1MeV/5.931u1061.9 cc

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛×= −

9

MeV95.8=Q


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

its a elength can be calc lated b appl ing the Planck’sits wavelength can be calculated by applying the Planck’s quantum theory: QhcE ==

λλ

Qhc

=λQ( )( )

12

834 1000.31063.6 − ××= 1210436.1 −×

m1039.1 13−×=λNote:Note:The radioactive decay only occurredradioactive decay only occurred when the value of ∆∆mmOR QQ is positivepositive.

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

Example 2 :( )Ni66

28beta particle.a. Write an equation to represent the nuclear reaction.b. If the new nucleus found in part (a) has the atomic mass of p ( )

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 × 10−4 u and c =3.00 × 108 m s−1) Solution :Solution :

N l ti ti t b th ti f t ia. Nuclear reaction equation must obey the conservation of atomic number and the conservation of mass number.

Q++→ eXNi 06666 β decayQ++→ − eXNi 12928 β decay

11


Solution :Solution :b. Given

The mass defect (difference) of the reaction is given byu 9284.65;u 9291.65 XNi == mm

The mass defect (difference) of the reaction is given by fi mmm −=∆( )eXNi mmm +−=

( )( )eN

( )41049.59284.659291.65 −×+−=u 1051.1 4−×=∆m

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

QK =max

( ) 2∆( )( )( )28274 1000.31066.11051.1 ×××= −−

( ) 2cm∆=

12

( )( )( )J 1026.2 14

max−×=K

Example 3 :


Table 9.1 shows the value of masses for several nuclides.Example 3 :

Nuclide Mass (u)Nuclide Mass (u)

4.0026

22 9898

He42

Na23

Di h h i i ibl f i l

22.9898

26.9815Al2713

Na11

Table Table 9.19.1Al27Discuss whether it is possible for to emit spontaneously an

alpha particle. Solution :Solution :

Al2713

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

Al2713

Al2713 Na23

11→ + He4213 11 2

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

26.9924 u26.9924 u

13

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

26.9924 u26.9924 u


is defined as an induced nuclear reaction that does not an induced nuclear reaction that does not occ r spontaneo sl it is ca sed b a collision bet een aocc r spontaneo sl it is ca sed b a collision bet een a

9.1.4 Bombardment with energetic particles

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,

Qan emitted particle y and reaction energy Q:

sometimes this reaction is written in the more compact form:

Q++→+ YyxXsometimes this reaction is written in the more compact form:

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

bombarding bombarding particleparticle

emitted emitted particleparticle

14

The calculation of reaction energy Q has been discussed in section 9.1.2.

pp

Examples of bombardment reaction:


Examples of bombardment reaction:Q++→+ HOHeN 1

117

842

147 ( ) Op,N 17

8147 αOR

QH2HLi 417 ( )47

Q++→+ HeLinB 42

73

10

105

Q+→+ He2HLi 42

11

73 ( ) He,pLi 4

273 α

( ) Li,nB 73

105 α

OR

OR

A nitrogen nucleus is converted into an oxygen nucleus and a proton if it is bombarded by an alpha particle carrying certain

Example 4 :N14

7 O178

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 thisb. Calculate the minimum energy of the alpha particle for this reaction to take place.

(Given mp =0.16735×10−26 kg; mα =0.66466 ×10−26 kg ; mass of

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(Given mp 0.16735 10 kg; mα 0.66466 10 kg ; mass of nitrogen nucleus , mN =2.32530×10−26 kg; mass of oxygen nucleus, mO =2.82282×10−26 kg ; c =3.00×108 m s−1)


Solution :Solution :a. The expression represents the nuclear reaction is

N14 O17→ + Q+ H1+ H4

b. The mass defect of the reaction isN14

7 O178→ + Q+ H1

1+ He42

fi mmm −=∆ fi mmm∆( ) ( )HOHeN mmmm +−+=( )2626 1066466010325302 −− +

k30

( )2626 1066466.01032530.2 ×+×=( )2626 1016735.01082282.2 −− ×+×−

Therefore the minimum energy of the alpha particle for this reaction to take place is

kg101.2 30−×−=∆m

reaction to take place is QK =min ( ) 2

min cmK ∆=( )( )2830 100031012 ×× −

16

( )( )1000.3101.2 ××=J 1089.1 13

min−×=K

E i 9 1


Exercise 9.1 :Given c =3.00×108 m s−1, mn=1.00867 u, mp=1.00782 u, 1 Complete the following radioactive decay equations :1. Complete the following radioactive decay equations :

a. [ ] HeBe 42

84 +→

[ ]b.

c.

[ ] BaSrPo 13956

9738

24094 ++→

( ) [ ]n3IU 1131236 ++→c.

d.

( ) [ ] n3IU 05392 ++→

[ ] eNa 01

2911 +→−

e.

f

[ ] ScSc 4721

4721 +→∗

[ ]CK 4040 +f. [ ] CaK 4020

4019 +→

17

E i 9 1


Exercise 9.1 :

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

(Gi f U 238 238 050786 f Th 234

Q++→ HeThU 42

23490

23892

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

mTh=234.043583 u and mass of α particle , mα=4.002603 u)ANS :ANS : 6 876 87××1010−−1313 JJANS. :ANS. : 6.876.87××1010−−1313 JJ3. The following nuclear reaction is obtained :

MeV550HCnN 114114 ++→+Determine the mass of in atomic mass unit (u).(Gi th f it l i 14 003074 )

MeV55.0HCnN 1607 ++→+

C146

(Given the mass of nitrogen nucleus is 14.003074 u)ANS. :ANS. : 14.003872 u14.003872 u

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L i O t


Learning Outcome:

At the end of this chapter, students should be able to: At the end of this chapter, students should be able to: Di ti i hDi ti i h th f l fi i d f ith f l fi i d f i

9.2 Nuclear fission and fusion (1 hour)

DistinguishDistinguish the processes of nuclear fission and fusion.the processes of nuclear fission and 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.g p g gy pg p g gy pExplainExplain chain reaction in nuclear fission of a nuclear chain reaction in nuclear fission of a nuclear reactor.reactor.D ibD ib th f l f i i thth f l f i i thDescribeDescribe the process of nuclear fusion in the sun.the process of nuclear fusion in the sun.

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9 2 1 Nuclear fission9.2 Nuclear fission and fusion9.2.1 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 p g qp g qwith 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 reactionthanthan 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 occurspontaneous fissionspontaneous fission – very rarely occur.induced fissioninduced fission – bombarding a heavy nucleus with slow neutrons or thermal neutrons of low energy (about 10−2 eV). Thi f fi i i h i i h

20

This type of fission is the important process in the energy production.

F l id th b b d t f b lU235


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

U23592

Q+++→→+ ∗ n3LaBrUnU 10

14857

8535

23692

10

23592

The reaction can also be represented by the diagram in Figure 9 1

Q+++→→+ n3LaBrUnU 0573592092Nucleus in the excited state.Nucleus in the excited state.

Br8535 n1

0

9.1.

n10

35

235

n10

n0

∗U23692

La148

U23592 n1

0

FigureFigure 9 19 1 La57Other possible reactions are:

Figure Figure 9.19.1

Q+++→→+ ∗ n3BaKrUnU 10

14456

8936

23692

10

23592

21Q+++→→+ ∗ n3XeSrUnU 1

013954

9438

23692

10

23592

Q0563692092

M t f th fi i f t (d ht l i) f th i


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 9.2.

22Figure Figure 9.29.2


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

Example 5 :

undergoes fission according to

Q+++→+ n3BaKrnU 10

14456

8936

10

23592

(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 =6.02×1023 mol−1)143.92273 u and NA 6.02 10 mol )Solution :Solution :The mass defect (difference) of fission reaction for one nucleus U-235 is

fi mmm −=∆( ) ( )nBaKrnU 3mmmmm ++−+= ( ) ( )nBaKrnU

( )−+= 00867.104393.235( )00867.1392273.14391756.88 ×++

23u 1863.0=∆m

( )00867.1392273.14391756.88 ++


Solution :Solution :The energy released corresponds to the mass defect of one U-235 is ( ) 2cmQ ∆=s ( )cmQ ∆=

( ) 22MeV/ 5.931u.18630 cc⎥⎤

⎢⎡

⎟⎟⎞

⎜⎜⎛

=

MeV 174=Q

( )u1 ⎥

⎥⎦⎢

⎢⎣

⎟⎠

⎜⎝

235 × 10−3 kg of uranium-235 contains of 6.02 × 1023 nuclei20 kg of urainum-235 contains of ( )23

3 1002.610235

20×⎟

⎠⎞

⎜⎝⎛

× −

Therefore nuclei1012.5 25×=

10235 ⎠⎝ ×

Energy released by 20 kg U-235

( )( )1741012.5 25×=27

24

MeV1091.8 27×=


A uranium-235 nucleus undergoes fission reaction by bombarding it with a slow neutron The reaction produces a strontium 90 nucleus

Example 6 :

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.( )Sr90

38p p

b. If the energy released is 210 MeV, calculate the atomic mass of nucleus X.

(Gi h f U 235 235 04393 f(Given the mass of U-235 =235.04393 u, mass of neutron =1.00867 u and mass of Sr-90 =89.90775 u)Solution :Solution :a. The expression represents the fission reaction is

Q+++→+ n3XSrnU 10

14354

9038

10

23592 Q+++→+ n3XSrnU 05438092

25


Solution :Solution :The energy released of 210 MeV equivalent to the mass defect for U-235 is ( ) 2Q ∆U 35 s ( ) 2cmQ ∆=

( ) 22MeV/ 5.931210 ccm ⎥⎤

⎢⎡

⎟⎟⎞

⎜⎜⎛

∆=

u22544.0=∆m

( )u1

210 cm⎥⎥⎦⎢

⎢⎣

⎟⎟⎠

⎜⎜⎝

∆

Therefore the atomic mass of the nucleus X is given by

fi mmm −=∆( ) ( )( ) ( )nXSrnU 3mmmmmm ++−+=∆( )−+= 00867.104393.23522544.0( )

u8934.142X =m( )00867.1390775.89 X ×++m

26

X

9 2 2 Chain reaction


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

b t fi ib t fi i titi

9.2.2 Chain reaction

subsequent fission subsequent fission reactionreaction.Figure 9.3 shows a schematic diagram of the chain reaction.

27Figure Figure 9.39.3Stimulation 9.1


From Figure 9.3, 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.p pThis 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 neutronsa nuclear chain reaction where the number of neutronsa 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.

Th ll d h i i d i lThe uncontrolled chain reactions are used in nuclear weapons –atomic bomb (Figure 9.4).The controlled chain reactions take place in nuclear reactors

28

p(Figure 9.5) and release energy at a steady rate.


FigureFigure 9 49 4 FigureFigure 9 59 5Figure Figure 9.49.4 Figure Figure 9.59.5

29


is defined as a type of nuclear reaction in which two light a type of nuclear reaction in which two light n clei f se to form a hea ier n cle s ith the release ofn clei f se to form a hea ier n cle s ith the release of

9.2.3 Nuclear fusion

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 gyenergy.Examples of fusion reaction releases the energy are

Q++→+ nHeHH 1322

f

Q++→+ nHeHH 0211

Q++→+ HHHH 11

31

21

21

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 s bo uc e a e pos e y c a ged e e s a s o grepulsive 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

30

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

At these elevated temperatures however fusion reactions


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 forcessufficient energy to overcome electrostatic repulsion forces.

The nuclear fusion reaction can occur in fusion bomb and in the core of a star.D i i i f i i h l f f i iDeuterium-tritium fusion is other example of fusion reaction where it can be represented by the diagram in Figure 9.6.

Deuterium TritiumH2

1

TritiumH3

1

Fusion reaction Q++→+ nHeHH 1

042

31

21Figure Figure 9.69.6

Alpha particleN t

0211

31

p p

He42

Neutron

n10

Stimulation 9.2


A fusion reaction is represented by the equation below:Example 7 :

Calculate

HHHH 11

31

21

21 +→+

a. the energy in MeV released from this fusion reaction,b. the energy released from fusion of 1.0 kg deuterium,(Gi f 1 007825 f i i 3 016049(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

( ) ( )fi mmm −=∆( ) ( )pTDD mmmm +−+=

3( ) ( )007825.1016049.3014102.2014102.2 +−+=

32

u1033.4 3−×=∆m


Solution :Solution :a. Therefore the energy released in MeV is

( ) 2cmQ ∆= ( )cmQ ∆=

( ) 22

3 MeV/ 5.931u 1033.4 cc⎥⎥⎤

⎢⎢⎡

⎟⎟⎞

⎜⎜⎛

×= −

MeV 03.4=Q

( )u1 ⎥

⎥⎦⎢

⎢⎣

⎟⎠

⎜⎝

b. The mass of 2 deuterium nuclei is 4 ×10−3 kg.4 × 10−3 kg of deuterium contains of 6.02 × 1023 nuclei

01 ⎞⎛1.0 kg of deuterium contains of

nuclei105051 26×=

( )233 1002.6

1040.1

×⎟⎠⎞

⎜⎝⎛

× −

ThereforeEnergy released from

nuclei10505.1 ×=

( )( )034105051 26×33

1.0 kg deuterium ( )( )03.410505.1 6×=MeV1007.6 26×=


The sun is a small star which generates energy on its own by f l f i i it i t i

9.2.4 Nuclear fusion in the sun

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 h li l i th l d l ti hi hhelium 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 9 7illustrated by Figure 9.7.

34


Qv +++→+ eHHH 01

21

11

11

neutrinoneutrinopositron (beta plus)positron (beta plus)

Qv +++→+ eHHH 1111

Q++→+ γHeHH 32

11

21 Q++→+ γHeHH 211

Q+++→+ HHHeHeHe 11433 Q+++→+ HHHeHeHe 11222

FiFi 9 79 7

The amount of energy released per cycle is about 25 MeV.N l f i i h i i f h b h

Figure Figure 9.79.7

Nuclear fusion occurs in the interior of the sun because the temperature of the sun is very high (approximately 1.5 × 107 K).

35


Table 9.2 shows the differences between fission and fusion reaction

9.2.5 Comparison between fission and fusion

reaction.

Fission FusionSplitting a heavy nucleus into two C bi t ll l i t fSplitting a heavy nucleus into two small nuclei.

Combines two small nuclei to form a larger nucleus.

It occurs at temperature can be It occurs at very high temperature controlled. (108 K).Easier to controlled and sustained

Difficult to controlled and a sustained controlled reaction has

Table Table 9.29.2

sustained. sustained controlled reaction has not yet been achieved.

The similaritysimilarity between the fission and fusion reactions is both reactions produces energyproduces energy.Graph of binding energy per nucleon against the mass number

36

Graph of binding energy per nucleon against the mass number in Figure 9.8 is used to explain the occurrence of fission and fusion reactions.


Greatest stabilityGreatest stabilityy

per

ucle

on)

g en

erg

(MeV

/nu

FissionFissionThe falling partfalling part of the binding energy curve

Bin

din

nucl

eon 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 numbern

FusionFusionThe rising partrising part of the binding energy curve

nuclei to nuclei of smaller mass numbernuclei of smaller mass number.

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.

37Mass number AFigure Figure 9.89.8

E i 9 2


Exercise 9.2 :Given c =3.00×108 m s−1, mn=1.00867 u, mp=1.00782 u, 1 Complete the following nuclear reaction equations:1. Complete the following nuclear reaction equations:

a. [ ] HeHe Li 42

32

63 +→+

[ ] 1258b.

c.

[ ] H HNi 11

21

5828 +→+

[ ]n5XenU 10

13854

10

23592 ++→+c.

d.

[ ] n5XenU 054092 ++→+

( ) C____,Be 126

94 α

e.2. Calculate the energy released in joule for the following fusion

( ) Np,n_____ 167

reaction:(Given the mass of deuterium =3.345×10−27 kg, mass of tritium =5 008×10−27 kg mass of He = 6 647×10−27 kg and

nHeHH 10

42

21

21 +→+

38

tritium =5.008×10 27 kg, mass of He = 6.647×10 27 kg and mass of neutron =1.675×10−27 kg)

ANS. :ANS. : 2.82.8××1010−−1212 JJ


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