Physice-nuclear Fission and Fusion

NUCLEAR FISSION The breaking of a heavy nucleus into two or more fragments of comparable masses, with the release of tremendous energy is called as nuclear fission. The most typical fission reaction occurs when slow 235 moving neutrons strike 92U . The following nuclear reaction takes place. If more than one of the neutrons produced in the above fission reaction are capable of inducing a 235 fission reaction (provided U is available), then the number of fissions taking place at successive stages goes increasing at a very brisk rate and this generates a series of fissions. This is known as 235 chain reaction. The chain reaction takes place only if the size of the fissionable material (U ) is greater than a certain size called the critical size. 92U

235

1

141

+ 0n ——> 56Ba

+ 36Kr

92

1

+ 3 0n + 200 MeV

If the number of fission in a given interval of time goes on increasing continuously, then a condition of explosion is created. In such cases, the chain reaction is known as uncontrolled chain reaction. This forms the basis of atomic bomb. In a chain reaction, the fast moving neutrons are absorbed by certain substances known as moderators (like heavy water), then the number of fissions can be controlled and the chain reaction is such cases is known as controlled chain reaction. This forms the basis of a nuclear reactor.

Illustration 1: When a beta particle is emitted from a nucleus the effect on its neutron-proton ratio is (A)

increased

(B)

decreased

(C)

remains same

(D)

first (1) then (2)

Solution: ZX

A

t 0n

A

0

——> Z+1Y + –1e (β) + v 1

0

——> 1P + –1e (β) + v

Nn / Np is decreased

Illustration 2: In which sequence, the radioactive radiations are emitted in the following nuclear reactions? ZX

A

A

A–4

—> Z+1Y —> Z–1K

A–4

—> Z–1K

(A)

β, α and γ

(B)

β, γ and α

(C)

α, β and γ

(D)

γ, α and β

Solution: We know that on emission of α-particle from the nucleus, atomic number reduces by 2 and mass number reduces by 4. And on emission of β-particle from the nucleus, the atomic number increases

by 1. Similarly on emission of γ-particle from the nucleus, there is no change in atomic number of mass number.

NUCLEAR FUSION

The process in which two or more light nuclei are combined into a single nucleus with the release of tremendous amount of energy is called as nuclear fusion. Like a fission reaction, the sum of masses before the fusion (i.e. of light nuclei) is more than the sum of masses after the fusion (i.e. of bigger nucleus) and this difference appears as the fusion energy. The most typical fusion reaction is the fusion of two deuterium nuclei into helium. 1H

1

2

4

+ 1H —> 2He + 21.6 MeV

For the fusion reaction to occur, the light nuclei are brought closer to each other (with a distance of –14 10 m). This is possible only at very high temperature to counter the repulsive force between nuclei. Due to this reason, the fusion reaction is very difficult to perform. The inner core of sun is at very high temperature, and is suitable for fusion, in fact the source of sun's and other star's energy is the nuclear fusion reaction

Solved Examples – Part I ASSIGNMENT 1.

The mass number of a nucleus is (A)

always less than atomic number

(B)

always more than atomic number

(C)

equal to atomic number

(D)

some times more than and sometimes equal to atomic number.

Solution: (D) Mass number of a nucleus represents the number of nucleons (neutrons + protons). When no neutrons are present in the nucleus, mass number equals to the atomic number. 2.

In which of the following decays, the element does not change? +

(A)

? – decay

(B)

? – decay

(C)

? – decay

(D)

y – decay

Solution: (D) γ-ray has no charge and no mass during γ-decay the element will not have any change in atomic number or mass number. 3.

In the reaction 7N

14

4

+ 2He —> 8O

17

1

+ 1H the minimum energy of the α-particle is

(A)

1.21 MeV

(B)

1.62 MeV

(C)

1.89 MeV

(D)

1.96 MeV.

(MN = 14.00307 amu, MHe = 4.00260 amu and MO = 16.99914 amu, MH = 1.00783 amu and 1 amu = 931 MeV) Solution: (A) 7N

14

4

+ 2He —> 8O

17

+ 1H

1

Total mass of reactants = 18.00567 amu Total mass of products = 18.00697 amu Mass defect = 18.00697 – 18.00567 = 0.0013 amu Energy (E) = 931 (0.0013) = 1.2103 MeV 4.

In the carbon cycle of fusion 1

4

1

4

2

4

2

4

(A)

Four 1H fuse to form 2He and two positrons

(B)

Four 1H fuse to form 2He and two electrons

(C)

Two 1H fuse to form 2He

(D)

Two 1H fuse to form 2He and two neutrons

Solution: (A) 1

1

4

0

4 H —> 2He + 2. +1e + Energy. 235

5. In each fission of U , 200 MeV of energy is released. If a reactor produces 100 MW power, the rate of fission in it will be (A)

3.125 × 10

18

per minute>

17

per second

3.125 × 10

17

per minute>

3.125 × 10

18

per second

(B)

3.125 × 10

(C) (D) Solution: (D)

We know P = nE/t 6

13

∴ n/t = P/E = 100×10 / 200[1.6×10 ] = 3.125 × 10

18

/sec

6. To generate a power of 3.2 MW, the number of fissions of U –19 per fission = 200 MeV, 1 eV = 1.6 ´ 10 J) 18

235

per minute is (Energy released

17

(A)

6×10 ¡

(B)

6×10

(C)

10

17

(D)

6×10

16

Solution: (A) Power of reactor P = nE/t Where 'n' is number of fissions 't' is time 'E' is energy released per fission 6

6

∴ 3.2 × 10 = n(200×10 )(1.6×10 => n = 6 × 10

–19

) / 60

18

235

7. If in nuclear reactor using U as fuel, the power output is 4.8 MW, the number of fissions per 235 –19 second is (Energy released per fission of U = 200 MeV watts, e eV = 1.6 ´ 10 J) 17

(A)

1.5×10

(C)

1.5×0

(B)

25

(D)

19

3×10

3×10

25

Solution: (A) 6

P = 4.8 MW = 4.8×10 W Power of a nuclear reactor P = nE/t 6

∴ n/t = P/E = 4.8×10 / (200)(1.6×10

–13

) = 1.5 × 10

17

Solved Examples – Part II 8.

In the following nuclear reaction 6C

11

11

—> 5B

(A)

A neutron

(B)

A neutrino

(C)

an electron

(D)

A proton

+ β + X. What does X stand for?

Solution: In ?-decay process a neutrino is accompanied with the emission of a positron. 235

9. If 200 MeV energy is released in the fission of a single U nucleus, the number of fissions –19 required per second to produce 1 kilowatt power shall be (given 1 eV = 1.6 × 10 J) 13

(B)

3.125×10

15

(D)

3.125×10

(A)

3.125×10

(C)

3.125×10

14

16

Solution: (A) Energy released in the fission –19

6

E = 200 MeV = 200 × 10 × 1.6 × 10

= 3.2 × 10

–11

Joule

P = nE/t 3

=> n/t = P/E = 10 / 3.2 × 10

–11

= 3.125×10

13

4

10. MP = 1.008 a.m.u., MN = 1.009 a.m.u. and 2He = 4.003 a.m.u. then the binding energy ofαparticle is (A)

21.4 MeV

(B)

8.2 MeV

(C)

34 MeV

(D)

1.35 A/m.

Solution: (D) BE = 2Mp + 2Mn – Ma = 2 × 1.008 + 2 × 1.009 – 4.003 = 0.031 amu = 0.031 × 931 = 28.8 MeV

11.

Energy released in fusion of 1 kg of deuterium nuclei 13

(B)

6×10 J

7

(D)

8×10 MeV

(A)

9×10 J

(C)

2×10 KwH

27

23

Solution: (A) Fusion reaction of deuterium 1H

2

2

3

E = 6.02 × 10

23

× 10 × 3.27 × 1.6 × 10

13

J

=> E ∼ 9 × 10 12.

1

+ 1H —> 2He + 0n +3.27 MeV 3

–13

/2×2

The energy produced in the sun is due to (A)

fission reaction

(B)

fusion reaction

(C)

chemical reaction

(D)

motion of electrons and ions

Solution: (A) We know that source of the huge solar energy is the fusion of lighter nuclei. Fusion of hydrogen nuclei into helium nuclei is continuously taking place in the plasma, with the continuous liberation of energy. Therefore energy produced in the sun is due to fusion reaction. 13. Atomic number of a nucleus is Z, while its mass number is M. What will be the number of neutrons in its nucleus? (A)

M

(B)

Z

(C)

(M–Z)

(D)

(M+Z)

Solution: (C) M=Z+N 14.

In uncontrolled chain reaction, the quantity of energy released, is (A)

very high

(B)

very low

(C)

normal

(D)

first (A) to (B)

Solution: (A) We know that in the uncontrolled chain reaction, more than one of the neutrons produced in a particular fission cause further fissions so that the number of fissions increases very rapidly. Thus this is a very fast reaction and the whole substance is fissioned in a few moments. A huge quantity of energy is released. 15. The net force between two nucleons 1 fm apart is F1 if both are protons, F2 if both are neutrons, and F3 if one is a neutron and the other is a proton. (A)

F1 < F2 < F3

(B)

F2 < F1 < F3

(C)

F1 < F2 = F3

(D)

F1 = F2 < F3

Solution: (C) The nuclear force of interaction between any pair of nucleons is identical i.e. force between two neutrons (F2) equals the force between neutron and proton (F3). However, between two protons net force is equal to the resultant of nuclear force between them (attractive in nature) and electrostatic force between them (repulsive in nature). Hence F1 is a value lesser than F2 and F3. So F1 < F2 = F3.