Question -
Answer -
(a) Take the D-T nuclear reaction 
It is given that:
Mass of
, m1= 2.014102 u
, m2 = 3.016049 uMass of 
m3 = 4.002603 uMass of
, m4 = 1.008665 uQ-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 ofdeuterium and tritium, r ≈ 2.0 fm = 2 × 10−15 m
Distance between the two nuclei at the momentwhen 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 betweenthe two nuclei is given as:
Where,
∈0 = Permittivityof free space
Hence, 5.76 × 10−14 J or 
of kinetic energy (KE) is needed to overcomethe Coulomb repulsion between the two nuclei.
However, it is given that:
KE
Where,
k = Boltzmann constant = 1.38 × 10−23 m2 kgs−2 K−1
T = Temperature required for triggering thereaction
Hence, the gas must be heated to a temperatureof 1.39 × 109 K to initiate the reaction.