JEE Mains · Physics · STD 12 - Nuclei
- JEE Mains 2024
- Hard
- MCQ
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 \(E_U\). 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 \(\mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23}\) )
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(7.62\)
Step-by-step Solution
Detailed explanation
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 \(\left(\mathrm{E}_{\mathrm{H}}\right)\)
\(=\frac{2000}{1} \times \mathrm{N}_{\mathrm{A}} \times \frac{26.7}{4} \mathrm{MeV}\)
\(\therefore\) Energy released by \(2 \mathrm{~kg}\) Vranium \(\left(\mathrm{E}_{\mathrm{v}}\right.\) )
\(=\frac{2000}{235} \times \mathrm{N}_{\mathrm{A}} \times 200 \mathrm{MeV}\)
So,
\(\frac{E_H}{E_V}=235 \times \frac{26.7}{4 \times 200}=7.84\)
\(\therefore\) Approximately close to \(7.62\)
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