In the nuclear reaction \( {}^{7}_{3}\mathrm{Li}+{}^{1}_{1}\mathrm{H}\rightarrow 2\,{}^{4}_{2}\mathrm{He} \), an energy of 17.3 MeV is released. If the binding energy per nucleon of \( {}^{4}_{2}\mathrm{He} \) is 7.06 MeV, the binding energy per nucleon of \( {}^{7}_{3}\mathrm{Li} \) is about
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 5.60 MeV
The numbers of protons and neutrons stay the same in a nuclear reaction. So the energy released is the gain in total binding energy, \( Q=E_b(\text{products})-E_b(\text{reactants}) \). The total binding energy of a nucleus is its mass number times the binding energy per nucleon. The proton \( {}^{1}_{1}\mathrm{H} \) is a single nucleon, so it has no binding energy.
The total binding energy of the two helium nuclei (8 nucleons) is:
\( \begin{aligned} E_b(\mathrm{He}) &= 2\times4\times E_{bn}(\mathrm{He}) \\ &= 8\times7.06\,\mathrm{MeV} \\ &= 56.48\,\mathrm{MeV} \end{aligned} \)
With \( Q=E_b(\mathrm{He})-E_b(\mathrm{Li}) \), the binding energy of \( {}^{7}_{3}\mathrm{Li} \) is:
\( \begin{aligned} E_b(\mathrm{Li}) &= E_b(\mathrm{He})-Q \\ &= 56.48\,\mathrm{MeV}-17.3\,\mathrm{MeV} \\ &= 39.18\,\mathrm{MeV} \end{aligned} \)
Shared among its 7 nucleons, this gives:
\( \begin{aligned} E_{bn}(\mathrm{Li}) &= \frac{E_b(\mathrm{Li})}{7} \\ &= \frac{39.18\,\mathrm{MeV}}{7} \\ &\approx 5.60\,\mathrm{MeV} \end{aligned} \)