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NCERT Class XII Chapter
Nuclei
Questions With Solutions

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Question : 28
Total: 31
Consider the D-T reaction (deuterium - tritium fusion)
2
‌
1
H+
3
‌
1
H →
4
‌
2
He + n
(a) Calculate the energy released in MeV in this reaction from the data m(
2
‌
1
H) = 2.014102 u, m(
3
‌
1
H) = 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 gases be heated to initiate the reaction?
Solution:  
(a) For the process
2
‌
1
H+
3
‌
1
H →
4
‌
2
He + n + Q
Q =
[m(
2
‌
1
H)+m(
3
‌
1
H)−m(
4
‌
2
He)−mn]
× 931 MeV
= 0.018878 × 931 = 17.58 MeV
(b) Repulsive potential energy of two nuclei when they almost touch each other is
=
q2
4πε0(2r)
 =
9×109(1.6×10−19)2
2×2×10−15
 joule = 5.76 × 10−14 joule.
Classically, K.E. atleast equal to this amount is required to overcome Coulomb repulsion. Using the relation
K.E. = 2 ×
3
2
 kT ⇒ T =
(K.E)
3k
 =
5.76×10−14
3×1.38×10−23
 = 1.39 × 109 K
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