The energy released in the fusion of 2 ~kg of hydrogen deep in the...

In each fusion reaction, $4{ }_1^1 \mathrm{H}$ nucleus are used. Energy released per Nuclei of ${ }_1^1 \mathrm{H}=\frac{26.7}{4} \mathrm{MeV}\therefore$ Energy released by $2 \mathrm{~kg}$ hydrogen \left(\mathrm{E}_{\mathrm{H}}\right) =\frac{2000}{1} \times \mathrm{N}_{\mathrm{A}} \times \frac{26.7}{4} \mathrm{MeV} \qquad \qquad\quad \& $$\begin{aligned} & \therefore \text{ Energy released by } 2 \mathrm{~kg} \text{ Vranium }\left(\mathrm{E}_{\mathrm{V}}\right) \\ & =\frac{2000}{235} \times \mathrm{N}_{\mathrm{A}} \times 200 \mathrm{MeV}\end{aligned}$$$So,$\frac{E_H}{E_V}=235 \times \frac{26.7}{4 \times 200}=7.84\therefore$ Approximately close to 7.62