The energy released in the fusion of 2 kg of hydrogen deep in the sun is EH and the energy released

Question

The energy released in the fusion of \(2 \mathrm{~kg}\) of hydrogen deep in the sun is \(\mathrm{E}_{\mathrm{H}}\) and the energy released in the fission of \(2 \mathrm{~kg}\) of \({ }^{235} \mathrm{U}\) is \(\mathrm{E}_{\mathrm{U}}\). The ratio \(\frac{E_{H}}{E_{U}}\) is approximately :

(Consider the fusion reaction as \(4{ }_{1}^{1} \ \mathrm{H}+2 \mathrm{e}^{-} \rightarrow{ }_{2}^{4} \ \mathrm{He}+2 \mathrm{v}+6 \gamma+26.7 \ \mathrm{MeV}\), energy released in the fission reaction of \({ }^{235} \mathrm{U}\) is \(200 \ \mathrm{MeV}\) per fission nucleus and \(\mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23}\)) 

(1) 9.13

(2) 15.04

(3) 7.62

(4) 25.6

Answer 1

Correct option is :  (3) 7.62  

No. of atoms of hydrogen

\( \frac{2000}{1}=\frac{2000 \times \mathrm{N}_{\mathrm{A}}}{4} \times 26.7=\mathrm{E}_{\mathrm{H}} \)  

No. of atoms of Uranium 

\( \begin{aligned} & \frac{2000}{225} \times \mathrm{N}_{\mathrm{A}} \times 200=\mathrm{E}_{\mathrm{U}} \end{aligned} \) 

\( \begin{aligned} & \therefore \frac{\mathrm{E}_{\mathrm{H}}}{\mathrm{E}_{\mathrm{U}}}=\frac{267}{2000} \times \frac{235}{4} \approx 7.62 \end{aligned} \)