Consider the D-T reaction (deuterium-tritium fusion)
`""_1^2H + ""_1^3H to ""_2^4He + ""_0n^1`
(ii) Consider the radius of both deuterium and tritium to be approximately 2 fm. What is the kinetic energy needed to overcome the Coulomb repulsion between 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)

Explanation

Repulsive potential energy of two nuclei when they almost touch each other
`U = 1/(4pi epsilon_0).(q^2)/(2r) = (9 xx 10^9 (1.6 xx 10^(-19))^2)/(2 xx 2 xx 10^(-15))`
`= 5.76 xx 10^(-14) J`
Also we know that,
Kinetic energy required for one fusion event = Average thermal kinetic energy available with the interacting particles.
Kinetic energy `=3/2 xx KT xx 2` [for two nuclei]
`= 3KT`
`("Kinetic energy")/(3K) = (5.76 xx 10^(-14))/(3 xx 1.38 xx 10^(-23))`
`= 1.39 xx 10^9 K`
This temperature cannot be achieved in actual behaviour.
`("Kinetic energy")/(3K) = (5.76 xx 10^(-14))/(3 xx 1.38 xx 10^(-23))`
`= 1.39 xx 10^(9) K`
This temperature cannot be achieved in actual behaviour .