Physics · Nuclear Physics

JEE Main 2024 — 9 April, Shift 2 — Question 43

The energy released in the fusion of 2 kg of hydrogen deep in the sun is $\mathrm{E}_{\mathrm{H}}$ and the energy released in the fission of 2 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 MeV per

fission nucleus and $\mathrm{N}_{\mathrm{A}}=6.023 \times 10^{23}$ )

Option A:
9.13
Option B:
15.04
Option C:
7.62
Correct
Option D:
25.6

Answer: C

Step-by-step 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 kg hydrogen $\left(\mathrm{E}_{\mathrm{H}}\right)$

\begin{gathered} =\frac{2000}{1} \times \mathrm{N}_{\mathrm{A}} \times \frac{26.7}{4} \mathrm{MeV} \& \end{gathered}

$\therefore$ Energy released by 2 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

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Nuclear Physics
Topic: Nuclear Fission and Fusion

A nucleus at rest disintegrates into two smaller nuclei with their masses in the ratio of 2: 1 . After disintegration they will move

Answer: C

Step-by-step solution

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