Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below: ²₁X + ²₁X = ⁴₂Y The binding energies per nucleon for ²₁X and ⁴₂Y are 1.1 MeV and 7.6 MeV respectively. The energy released in this process is _______________ MeV. Numerical answer type. Work it out, then open the answer.

Answer: 26. Q = BE(Y) - 2 BE(X). BE of Y = 4×7.6 = 30.4 MeV; BE of two X = 2×2×1.1 = 4.4 MeV. Q = 30.4 - 4.4 = 26 MeV.

Two lighter nuclei combine to form a comparatively heavier nucleus by… | JEE Main 2022 (26 Jul, Shift 2)

Q19JEE Main 2022 · 26 JUL · SHIFT 2Easy

Nuclear Fission & Fusion

Two lighter nuclei combine to form a comparatively heavier nucleus by the relation given below: ${}^{2}_{1}X + {}^{2}_{1}X = {}^{4}_{2}Y$

The binding energies per nucleon for ${}^{2}_{1}X$ and ${}^{4}_{2}Y$ are $1.1\text{ MeV}$ and $7.6\text{ MeV}$ respectively. The energy released in this process is _______________ $\text{MeV}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 26

$Q = \text{BE}(Y) - 2\,\text{BE}(X)$ .
BE of $Y = 4\times7.6 = 30.4\text{ MeV}$ ; BE of two $X = 2\times2\times1.1 = 4.4\text{ MeV}$ .
$Q = 30.4 - 4.4 = 26\text{ MeV}$ .

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