JEE Mains · Physics · STD 12 - 13. Nuclei
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 \(\mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23}\) )
- A \(9.13\)
- B \(15.04\)
- C \(7.62\)
- D \(25.6\)
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 \(\left(\mathrm{E}_{\mathrm{H}}\right)\)…
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