Consider the D–T reaction (deuterium-tritium fusion)
(a) Calculate the energy released in MeV in this reaction from the data:
m (21H) = 2.014102 u
m (31H) = 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 = Boltzmann’s constant, T = absolute temperature.)
a) Given,
Mass of 
, m1 = 2.014102 u
Mass of 
, m2 = 3.016049 u
Mass of 
, m2 = 4.002603 u
Mass of 
, m3 = 1.008665 u
Where,
Δm = Mass defect (or) mass lost during reaction
c = speed of light
Given nuclear fusion reaction is,
∴ 
But, u = 931.5 MeV/c2
∴ 
b) Given,
Radius of the deuterium and tritium, r = 2fm = 2×10-15 m
Charge on deuterium and tritium nuclei = e = 1.6×10-19 C
Thus,
Distance between the two nuclei, d = r + r = 4×10-15 m
Repulsive potential energy between two nuclei is,
Where,
e = charge
ϵ0 = permittivity of the space
d = distance between charges
and, 
∴ 
(∵1.6×10-19 C = 1eV)
Hence, it needs 360 keV of kinetic energy to overcome coulomb repulsion.
But, given that Kinetic energy (KE) is,
Where,
k = Boltzmann constant = 1.38×10-23 kg m2 s-2 K-1
T = temperature required to trigger the reaction
∴ 
= 
Hence, the gas must be heat up to 1.39×109 K to start fusion.