The binding energy per nucleon is 1.1 MeV for \( {}_{1}^{2}\mathrm{H} \) and 7.0 MeV for \( {}_{2}^{4}\mathrm{He} \). The energy released in the given fusion reaction will be \( {}_{1}^{2}\mathrm{H} + {}_{1}^{2}\mathrm{H} \rightarrow {}_{2}^{4}\mathrm{He} \)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 23.6 MeV
The energy released in a nuclear reaction equals the total binding energy of the products minus that of the reactants. The total binding energy of a nucleus is its binding energy per nucleon times its mass number.
\( \text{Each } {}_{1}^{2}\mathrm{H}: 2 \times 1.1 = 2.2\,\mathrm{MeV} \)
\( \text{Two } {}_{1}^{2}\mathrm{H} \text{ nuclei}: 2 \times 2.2 = 4.4\,\mathrm{MeV} \)
\( {}_{2}^{4}\mathrm{He}: 4 \times 7.0 = 28.0\,\mathrm{MeV} \)
\( Q = 28.0 - 4.4 = 23.6\,\mathrm{MeV} \)