Page No 465: - Chapter 13 - Nuclei - Additional Exercise Solutions - Ncert Solutions class 12 - Physics
Updated On 26-08-2025 By Lithanya
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Question 13.27:
Consider the fission of 
by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are 
and
. Calculate Q for this fission process. The relevant atomic and particle masses are
m
=238.05079 u
m
=139.90543 u
m
= 98.90594 u
Answer:
In the fission of , 10 β− particles decay from the parent nucleus. The nuclear reaction can be written as:
It is given that:
Mass of a nucleus
m1 = 238.05079 u
Mass of a nucleus 
m2 = 139.90543 u
Mass of a nucleus , m3 = 98.90594 u
Mass of a neutron
m4 = 1.008665 u
Q-value of the above equation,
Where,
m’ = Represents the corresponding atomic masses of the nuclei
= m1 − 92me
= m2 − 58me
= m3 − 44me
= m4
Hence, the Q-value of the fission process is 231.007 MeV.
Question 13.28:
Consider the D−T reaction (deuterium−tritium fusion)
(a) Calculate the energy released in MeV in this reaction from the data:
= 2.014102 u
= 3.016049 u
(b)Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction? (Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)
Answer:
(a) Take the D-T nuclear reaction: 
It is given that:
Mass of
, m1= 2.014102 u
Mass of
, m2 = 3.016049 u
Mass of
m3 = 4.002603 u
Mass of
, m4 = 1.008665 u
Q-value of the given D-T reaction is:
Q = [m1 + m2− m3 − m4] c2
= [2.014102 + 3.016049 − 4.002603 − 1.008665] c2
= [0.018883 c2] u
But 1 u = 931.5 MeV/c2
∴Q = 0.018883 × 931.5 = 17.59 MeV
(b) Radius of deuterium and tritium, r ≈ 2.0 fm = 2 × 10−15 m
Distance between the two nuclei at the moment when they touch each other, d = r + r = 4 × 10−15 m
Charge on the deuterium nucleus = e
Charge on the tritium nucleus = e
Hence, the repulsive potential energy between the two nuclei is given as:
Where,
∈0 = Permittivity of free space
Hence, 5.76 × 10−14 J or 
of kinetic energy (KE) is needed to overcome the Coulomb repulsion between the two nuclei.
However, it is given that:
KE
Where,
k = Boltzmann constant = 1.38 × 10−23 m2 kg s−2 K−1
T = Temperature required for triggering the reaction
Hence, the gas must be heated to a temperature of 1.39 × 109 K to initiate the reaction.
Question 13.29:
Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decays in the decay scheme shown in Fig. 13.6. You are given that
m (198Au) = 197.968233 u
m (198Hg) =197.966760 u
Answer:
It can be observed from the given γ-decay diagram that γ1 decays from the 1.088 MeV energy level to the 0 MeV energy level.
Hence, the energy corresponding to γ1-decay is given as:
E1 = 1.088 − 0 = 1.088 MeV
hν1= 1.088 × 1.6 × 10−19 × 106 J
Where,
h = Planck’s constant = 6.6 × 10−34 Js
ν1 = Frequency of radiation radiated by γ1-decay
It can be observed from the given γ-decay diagram that γ2 decays from the 0.412 MeV energy level to the 0 MeV energy level.
Hence, the energy corresponding to γ2-decay is given as:
E2 = 0.412 − 0 = 0.412 MeV
hν2= 0.412 × 1.6 × 10−19 × 106 J
Where,
ν2 = Frequency of radiation radiated by γ2-decay
It can be observed from the given γ-decay diagram that γ3 decays from the 1.088 MeV energy level to the 0.412 MeV energy level.
Hence, the energy corresponding to γ3-decay is given as:
E3 = 1.088 − 0.412 = 0.676 MeV
hν3= 0.676 × 10−19 × 106 J
Where,
ν3 = Frequency of radiation radiated by γ3-decay
Mass of 
= 197.968233 u
Mass of 
= 197.966760 u
1 u = 931.5 MeV/c2
Energy of the highest level is given as:
β1 decays from the 1.3720995 MeV level to the 1.088 MeV level
∴Maximum kinetic energy of the β1 particle = 1.3720995 − 1.088
= 0.2840995 MeV
β2 decays from the 1.3720995 MeV level to the 0.412 MeV level
∴Maximum kinetic energy of the β2 particle = 1.3720995 − 0.412
= 0.9600995 MeV
