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AP EAMCET · PHYSICS · Nuclear 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
\({ }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^2 \rightarrow{ }_2 \mathrm{He}^4+\mathrm{Q}\)

  1. A 23.8 MeV
  2. B 26.1 MeV
  3. C 5.95 MeV
  4. D 28.9 MeV
Verified Solution

Answer & Solution

Correct Answer

(C) 5.95 MeV

Step-by-step Solution

Detailed explanation

\(BE_{\text{reactants}} = 2 \times (1.15 \text{ MeV/nucleon} \times 2) = 4.6 \text{ MeV}\) \(BE_{\text{product}} = 7.1 \text{ MeV/nucleon} \times 4 = 28.4 \text{ MeV}\) \(Q = BE_{\text{product}} - BE_{\text{reactants}} = 28.4 \text{ MeV} - 4.6 \text{ MeV} = 23.8 \text{ MeV}\)…