The energy released in the fusion of $2 \mathrm{~kg}$ of hydrogen deep in the sun is $E_H$ and the energy…
The energy released in the fusion of $2 \mathrm{~kg}$ of hydrogen deep in the sun is $E_H$ and the energy released in the fission of $2 \mathrm{~kg}$ of ${ }^{235} \mathrm{U}$ is $E_U$. The ratio $\frac{E_H}{E_U}$ is approximately:
(Consider the fusion reaction as $4 \mid H+2 \mathrm{e}^{-} \rightarrow{ }_2^4 \mathrm{He}+2 v+6 \gamma+26.7 \mathrm{MeV}$, energy released in the fission reaction of ${ }^{235} \mathrm{U}$ is $200 \mathrm{MeV}$ per fission nucleus and $\mathrm{N}_{\mathrm{A}}=$ $\left.6.023 \times 10^{23}\right)$
7.62
25.6
15.04
9.13
Solution
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