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

Solution:

(a) Take the D-T nuclear reaction: ${ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+\mathrm{n}$

It is given that:

Mass of $_{1}^{2} \mathrm{H}, m_{1}=2.014102 \mathrm{u}$

Mass of ${ }_{1}^{3} \mathrm{H}, m_{2}=3.016049 \mathrm{u}$

Mass of ${ }_{2}^{4} \mathrm{He} m_{3}=4.002603 \mathrm{u}$

Mass of ${ }_{0}^{1} \mathrm{n}, m_{4}=1.008665 \mathrm{u}$

Q-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 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:

$V=\frac{e^{2}}{4 \pi \in_{0}(d)}$

Where,

 

∈0 = Permittivity of free space

$\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2}$

$\therefore V=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{4 \times 10^{-15}}=5.76 \times 10^{-14} \mathrm{~J}$

$=\frac{5.76 \times 10^{-14}}{1.6 \times 10^{-19}}=3.6 \times 10^{5} \mathrm{eV}=360 \mathrm{keV}$

Hence, $5.76 \times 10^{-14} \mathrm{~J}$ or $360 \mathrm{keV}$ of kinetic energy $(\mathrm{KE})$ is needed to overcome the Coulomb repulsion between the two nuclei.

However, it is given that:

$\mathrm{KE}=2 \times \frac{3}{2} k T$

Where,

k = Boltzmann constant = 1.38 × 10−23 m2 kg s−2 K−1

 

T = Temperature required for triggering the reaction

$\therefore T=\frac{\mathrm{KE}}{3 K}$

$=\frac{5.76 \times 10^{-14}}{3 \times 1.38 \times 10^{-23}}=1.39 \times 10^{9} \mathrm{~K}$

Hence, the gas must be heated to a temperature of 1.39 × 109 K to initiate the reaction.