Q157 / 160 · AP EAMCET 2025 (27 MAY SHIFT 1) · Physics

If the binding energy per nucleon of deuteron $\left({ }_1 \mathrm{H}^2\right)$ is 1.15 MeV and an $\alpha$-particle has a binding energy of 7.1 MeV per nucleon, then the energy released per nucleon in the given reaction is ${ }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^2 \rightarrow{ }_2 \mathrm{He}^4+\mathrm{Q}$

A
23.8 MeV
B
26.1 MeV
Correct answer
C
5.95 MeV
D
28.9 MeV

Hence, the correct answer is option: 'C'.

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