Chapter 16 Nuclear Reaction

7/29/2019 Chapter 16 Nuclear Reaction

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TOPIC : 16. Nuclear Reaction

SUB-TOPIC : 16.1 Nuclear reaction16.2 Nuclear fission

16.3 Nuclear fusion

DURATION : 3 hours

OBJECTIVES :1. Understand the conservation of charge (Z) and nucleon number (A) in a

nuclear reaction.

2. Write and complete the equation for nuclear reaction.

3. Understand the principle of conservation of energy and calculate the energyliberated in the process of nuclear reaction.

4. Understand nuclear fission and fusion.

5. Understand the occurrence of fission and fusion from the aspect of binding

energy per nucleon.6. State the condition for chain reaction.

7. Describe the process of nuclear fusion in the sun.

DEFINITION OF KEY TERMS:

1. Nuclear reactions collisions between a nuclei with energetic particles whichchange the structure or identity of the target nuclei

2. Nuclear fission the splitting of the nucleus of a heavy atom, such as

uranium-235, into two main parts, accompanied by the

release of much energy.3. Chain reaction A self-sustaining reaction in which the products of one

reaction event stimulate further reaction events.4. Nuclear fusion The combination of the nuclei of light atoms to form heaviernuclei, with the release of much energy.

CONTENT:

16.1 Nuclear Reaction

Rutherford suggested in 1919 that a massive particle with sufficient kinetic

energy might be able to penetrate a nucleus. The result would be either a new nucleus

with great atomic number and mass number or a decay of the original nucleus.Rutherford bombarded nitrogen with particles and obtained an oxygen nucleus

and a proton. It can be described by the following equation :

N

14

7 + He

4

2

O

17

8 + H

1

1

or 42 or p1

1

nitrogen-14 particle oxygen-17 proton

Note :

The nucleon numbers (A) balance on both sides of the equation, So do the atomicnumbers (Z)

Nuclear reactions are subject to several conservation laws. The classical

conservation principles forcharge (Z), momentum, angular momentum, and energy

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( including rest energies ) are obeyed in all nuclear reactions. An additional conservation

law, is conservation of the nucleon numbers.

When two nuclei interact, charge conservationrequires thatthe sum of the

initial atomic numbers must equal the sum of the final atomic numbers.

Because of conservation of nucleon number, the sum of the initial mass

numbers must also equal the sum of the final mass numbers. In general, these are notelastic collisions, and, correspondingly, the total initial mass does not equal the total final

mass.

Nuclear reactions have the general formA + a B + b

where the uppercase letters represent the nuclei and the lowercase letters represent the

particles. Such reactions often written in a shorthand notation:

A( a, b )BFor example, in this form, the previous reaction can be rewritten more compactly as

14N( , p ) 17O

REACTION ENERGY

The difference between the masses before and after the reaction corresponds to

the reaction energy, according to the mass-energy relation2

mCE=

If initial particles A and B interact to produce final particles C and D,

A + B C + D

the reaction energy is defined as

Q = ( MA + MB MC MD ) c2

To balance the electrons, we use the neutral atomic masses in the above equation. That is,we use the mass of H

1

1 for a proton, H2

1 for a deuteron, H4

2 for an particle, and so on.

When Q is positive, the total mass decreases and the total kinetic energy increases.

Such a reaction is called an exoergic reaction. When Q is negative, the massincreases and the kinetic energy decreases, and the reaction is called an endoergic

reaction. The terms exothermal and endothermal, borrowed from chemistry, are also

used. In an endoergic reaction the reaction cannot occur at all unless the initial kinetic

energy in the center of mass reference frame is at least as great as \ Q \ . That is, there is a

threshold energy, the minimum kinetic energy to make an endoergic reaction go.

EXAMPLE 16.1 ( A possible energy source : Q value of a reaction )

Determine if the following reaction is endoergic or exoergic, and calculate its Q value.

H21 + H2

1 He3

2 + n1

0

deuteron deuteron helium neutron( 2.014102 u ) ( 2.014102 u) ( 3.016029 u ) ( 1.008665 u )

SOLUTION :

Thinking it through :

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The reaction is exoergic if Q > 0, and endoergic if Q < 0. We need the mass

difference ( m ) to determine Q.

m is calculated by subtracting the final masses from the initial masses.Therefore,

m = 2mD - mHe - mn

= 2 ( 2.014102 u ) - 3.016029 u - 1.008665 u= + 0.00351 u

Thus, mass has been lost, the total kinetic energy has increased, and the reaction is

exoergic. The Q value isQ = ( 0.00351 u ) ( 931.5 MeV/u )

= + 3.27 MeV

EXAMPLE 16.2 (An exoergic reaction)

When Lithium ( Li7 ) is bombarded by a proton, two alpha particles ( He4 ) are produced.

The reaction can be written

H11 + Li73 He42 + He42( 1.007825 u ) ( 7.016004 u ) ( 4.002603 u ) ( 4.002603 u )

Find the reaction energy.

SOLUTION

Here are the initial and final masses :

A : H1

1 1.007825 u C : He4

2 4.002603 u

B : Li7

3 7.016004 u D : He4

2 4.002603 u

__________ ___________8.023829 u 8.005206 u

m = ( MA + MB MC MD ) c2

= 0.018623 u

Q = (0.018623 u)( 931.5 MeV/u)

= 17.35 MeV

This is an exoergic reaction; the final total kinetic energy of the two separating alpha

particles is 17.35 MeVgreater than the initial total kinetic energy of the proton and the

lithium nucleus.

EXAMPLE 16.3( An endoergic reaction )Calculate the reaction energy for the following nuclear reaction :

N147 + He4

2 O

17

8 + H1

1

(14.003074 u ) ( 4.002603 u ) ( 16.999132 u ) ( 1.007825 u )

SOLUTION :

The initial and final masses are :

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A: N14

7 14.003074 u C : O17

8 16.999132 u

B : He4

2 4.002603 u D : H1

1 1.007825 u

___________ ___________

18.005677 u 18.006957 u

m = 18.005677 u 18.006957 u

= - 0.001280 uQ = (-0.001280 u ) ( 931.5 MeV/u )

= -1.192 MeV

In the center-of mass system, that is, in a head-on collision with zero total momentum,the minimum total initial kinetic energy for this reaction to occur is 1.192 MeV.

16.2 Nuclear fission

During nuclear fission, a heavy nucleus ( e.g. of uranium or plutonium ) splits to

form two nuclei of roughly the same mass, plus several neutrons. Rarely, fission happensspontaneously. More usually, it occurs when a neutron hits and is captured by the

nucleus.

The fission of U235 by thermal neutrons can be represented by the equation

QneutronsYXUnU ++++ *236

92

1

0

235

92

where *236U is an intermediate excited state that lasts only for about 10 -12 s before

splitting into nuclei X and Y, which are called fission fragments. In any fission equation,

there are many combinations of X and Y that satisfy the requirements of conservation ofenergy and charge. With Uranium, for example, there are about 90 daughter nuclei that

can be formed.

Fission also results in the production of several neutrons, typically two or three.

For example, here are some typical fission reactions for uranium-235 :

i. QnKrBanU ++++ 10

90

36

144

56

1

0

235

922

ii. QnKrBanU ++++ 10

92

36

141

56

1

0

235

923

iii. QnMoSnnU ++++ 10

101

42

132

50

1

0

235

923

The reactions release energy, mostly as kinetic energy of the heavier decay

products. So fission is a source of heat.

Figure 1 is a graph of the distribution of fission products versus mass number A.

The most probable products have mass numbers A (90-100) and A (135-145), both of

which fall to the left of the stability line shown in Figure 2, meaning they contain more

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neutrons than protons. These fragments, because they are unstable owing to theirexcess of neutrons, almost instantaneously release two or three neutrons.

Figure 1

Figure 2

From Figure 3 we see that the binding energy per nucleon is about 7.2 MeV for

heavy nuclei ( A 240) and about 8.2 MeV for nuclei of intermediate mass. This means

that the nucleons in fission fragments are more tightly bound and therefore have less thanthe nucleons in a parent nucleus. This decrease in nucleon mass appears as released

energy when fission occurs.

Figure 3

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EXAMPLE 16.4 ( The energy released in the Fission of U235 )

Calculate the energy released when 1.00 kg of U235 fissions, taking the disintegration

energy per event to be Q = 208 MeV.

SOLUTION :We need to know the number of nuclei in 1.00 kg of uranium. Because A = 235, Weknow that the molar mass of this isotop is 235 g/mol. Therefore, the number of nuclei in

our sample is

)1000.1)(/235

/1002.6(

323

gxmolg

molnucleixN=

= 2.56 x 1024 nuclei

Hence, the total disintegration energy is

E = NQ = ( 2.56 x 1024 nuclei) ( 208nucleusMeV )

= 5.32 x 1026 MeV

Chain Reaction

Practical amounts of energy from fission can be obtained when huge numbers of

fissions occur per second. One way of accomplishing this is by a chain reaction. For

example, suppose a 235U nucleus fissions (on its own or triggered by an external neutron)

with the release of two neutrons ( see Figure 4). Ideally, the released neutrons can theninitiate two more fission reactions, a process that, in turn, releases four neutrons. These

neutrons may initiate more reactions, and so on. Thus, the process can multiply, with the

number of neutrons doubling with each generation. When this occurs, the neutronproduction rate ( and, hence, the energy released from the sample) grows exponentially.

Figure 4

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To maintain a sustained chain reaction, there must be an adequate quantity of

fissionable material. The minimum mass required to produce a sustained chain reaction is

called critical mass. When the critical mass is attained, there is enough fissionablematerial such that at least one neutron from each fission event, on average, goes on to

fission another nucleus. Several factors determine critical mass. The conditions to

achieve chain reaction are :i) slow neutrons are better at causing fission , so that uranium are mixed with a material

that does not absorb neutrons but slows them down.

ii) the fission material must more than a critical size.Uncontrolled chain reactions are used in nuclear weapons such as atomic bomb.

Controlled chain reactions take place in nuclear reactors and release energy at a steady

rate.

16.3 Nuclear Fusion

Another type of nuclear reaction that can produce energy is fusion. In a fusion

reaction, light nuclei fuse to form a more massive nucleus, releasing energy in theprocess ( Q > 0 ). A simple fusion reaction the fusion of two deuterium nuclei ( H

2

1 ),

sometimes called a D-D reaction was examined in Example 16.1. There, it was shownthat this reaction releases 3.27 MeV of energy per fusion.

Another example is the fusion of deuterium and tritium ( a D-T reaction ) :

H21 + H3

1 He4

2 + n1

0

( 2.014102 u ) ( 3.016049 u ) ( 4.002603 u ) ( 1.008665 u )

Using the given masses, you should be able to show that this reaction involves a release

of 17.6 MeV per fusion.

A fusion reaction releases much less energy in comparison with the more than200 MeV released from a typical single fission. However, equal-mass samples of

hydrogen and uranium have many, many more hydrogen nuclei than uranium nuclei. As aresult,per kilogram, the fusion of hydrogen gives almost three times the energy released

from uranium fission.

In a sense, our lives depend crucially on nuclear fusion, because it is the source ofenergy for stars, including our Sun. One sequence of fusion reactions that is believed to

be responsible for the Suns energy output is as follows. First, we have proton-proton

fusion,

H11 + H1

1 H2

1 ++ + v

( where v represents a neutrino) Then another proton fuses with the deuteron :H11 + H

2

1 He3

2 + Finally, two of the He

3 nuclei fuse :

He3

2 + He3

2 He4

2 + H1

1 + H1

1

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The net effect of this sequence, called the proton-proton cycle, is that four protons

( H1

1 ) combine to form one helium nucleus ( He4

2 ) plus two positrons (+ ), two gamma

rays ( ), and two neutrinos ( v ) with a release of energy:

4 ( H1

1 ) He4

2 + 2+ + 2 + 2v + Q ( Q = 24.7

MeV)

In a star such as our Sun, gamma-ray photons scatter off nuclei on their way to the

surface. Each scattering results in a reduction in energy, until each photon has only a few

electron volts of energy. Thus, on reaching the surface, the photons are mostly visible-

light photons. In our Sun, fusion involves only the central 10% of the Suns mass. It hasbeen going on for about 5 billion years and should continue, approximately as is, for

another 5 billion years.

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Chapter 16 Nuclear Reaction

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