Consider the $D-T$ reaction (deuterium-tritium fusion):
$_{1}^{2} H +_{1}^{3} H \rightarrow_{2}^{4} He + n$
$(a)$ Calculate the energy released in $MeV$ in this reaction from the data:
$m(_{1}^{2} H) = 2.014102 \; u$
$m(_{1}^{3} H) = 3.016049 \; u$
$(b)$ 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 gas be heated to initiate the reaction?

(N/A) The given $D$-$T$ nuclear reaction is:
$_{1}^{2} H + _{1}^{3} H \rightarrow _{2}^{4} He + n$
Given masses:
$m(_{1}^{2} H) = 2.014102 \; u$
$m(_{1}^{3} H) = 3.016049 \; u$
$m(_{2}^{4} He) = 4.002603 \; u$
$m(n) = 1.008665 \; u$
The $Q$-value is given by: $Q = [m(_{1}^{2} H) + m(_{1}^{3} H) - m(_{2}^{4} He) - m(n)] c^2$
$Q = [2.014102 + 3.016049 - 4.002603 - 1.008665] \; u \cdot c^2 = 0.018883 \; u \cdot c^2$
Since $1 \; u = 931.5 \; MeV/c^2$,$Q = 0.018883 \times 931.5 \; MeV = 17.59 \; MeV$.
$(b)$ Radius $r \approx 2.0 \; fm = 2.0 \times 10^{-15} \; m$. The distance at contact is $d = 2r = 4.0 \times 10^{-15} \; m$.
The Coulomb potential energy is $V = \frac{1}{4\pi\epsilon_0} \frac{e^2}{d} = (9 \times 10^9) \frac{(1.6 \times 10^{-19})^2}{4.0 \times 10^{-15}} = 5.76 \times 10^{-14} \; J$.
Converting to $eV$: $V = \frac{5.76 \times 10^{-14}}{1.6 \times 10^{-19}} = 3.6 \times 10^5 \; eV = 360 \; keV$.
For thermal initiation,the average kinetic energy $KE = \frac{3}{2} kT$ per particle. For two particles,the total energy required is $3kT = V$.
$T = \frac{V}{3k} = \frac{5.76 \times 10^{-14}}{3 \times 1.38 \times 10^{-23}} \approx 1.39 \times 10^9 \; K$.