Consider the D−T reaction (deuterium−tritium fusion)

\[\ce{^2_1H + ^3_1H -> ^4_2He}\]

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.)

#### Solution

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:

`"V" = "e"^2/(4pi in_0("d"))`

Where

∈_{0} = Permittivity of free space

`1/(4pi in_0) = 9 xx 10^9 "Nm"^2 "C"^(-2)`

`therefore "V" = (9 xx 10^9 xx (1.6 xx 10^(-19))^2)/(4 xx 10^(-15)) = 5.76 xx 10^(-14) "J"`

`= (5.76 xx 10^(-14))/(1.6 xx 10^(-19)) = 3.6 xx 10^5 "eV" = 360 "KeV"`

Hence, 5.76 × 10^{−14} J or 360 KeV of kinetic energy (KE) is needed to overcome the Coulomb repulsion between the two nuclei.

However, it is given that:

KE = `2 xx 3/2 " kT"`

Where,

k = Boltzmann constant = 1.38 × 10^{−23} m^{2} kg s^{−2} K^{−1}

T = Temperature required for triggering the reaction

`"T" = ("KE")/("3K")`

`= (5.76 xx 10^(-14))/(3 xx 1.34 xx 10^(-23)) = 1.39 xx 10^9 "K"`

Hence, the gas must be heated to a temperature of 1.39 × 10^{9} K to initiate the reaction.