Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below : _1^2 X+ _1^2 X= _2^4 Y The binding energies per nucleon for \frac21 X and _2^4 Y are 1.1 \,MeV and 7.6 \,MeV respectively. The energy released in this process is \underline\hspace2cm MeV .

Atoms and Nuclei PYQ — JEE Main 2022 · 26 Jul · Shift 2 · Q61 with Solution

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Correct answer: 26

The given nuclear fusion reaction is

{}^{2}_{1}X + {}^{2}_{1}X \rightarrow {}^{4}_{2}Y

So, two nuclei of mass number $2$ combine to form one nucleus of mass number $4$ .

Given:

Each ${}^{2}_{1}X$ nucleus has 2 nucleons, so total binding energy of one ${}^{2}_{1}X$ is

2 \times 1.1 = 2.2\,\text{MeV}

Since there are two such nuclei, total initial binding energy is

2 \times 2.2 = 4.4\,\text{MeV}

The nucleus ${}^{4}_{2}Y$ has 4 nucleons, so its total binding energy is

4 \times 7.6 = 30.4\,\text{MeV}

Energy released in fusion is the increase in total binding energy:

Q = \text{BE(final)} - \text{BE(initial)}

Q = 30.4 - 4.4 = 26.0\,\text{MeV}

Therefore, the energy released is

\boxed{26\,\text{MeV}}

Stored correct answer = $26$

This matches exactly.