The binding energy of the \({}^{16}_{8}\mathrm{O}\) nucleus is 127.5 MeV and that of the \({}^{4}_{2}\mathrm{He}\) nucleus is 28.3 MeV. The ratio of the binding energy per nucleon of \({}^{16}_{8}\mathrm{O}\) to that of \({}^{4}_{2}\mathrm{He}\) is about
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(1.13\)
The binding energy per nucleon is the binding energy of the nucleus divided by its mass number, \( E_{bn}=E_b/A \). Find it for each nucleus first, then take the ratio.
\( \begin{aligned} E_{bn}(\mathrm{O}) &= \frac{127.5}{16} \\ &= 7.97\ \mathrm{MeV} \end{aligned} \)
\( \begin{aligned} E_{bn}(\mathrm{He}) &= \frac{28.3}{4} \\ &= 7.08\ \mathrm{MeV} \end{aligned} \)
\( \begin{aligned} \frac{E_{bn}(\mathrm{O})}{E_{bn}(\mathrm{He})} &= \frac{7.97}{7.08} \\ &= 1.13 \end{aligned} \)
Therefore, a nucleon in oxygen-16 is bound a little more strongly than a nucleon in helium-4.
NCERT (2026-27):
Nuclear binding energy – “… A more useful measure of the binding between the constituents of the nucleus is the binding energy per nucleon, Ebn, which is the ratio of the binding energy Eb of a nucleus to the number of the nucleons, A, in that nucleus:”
\( E_{bn}=E_b/A \)\( \qquad (13.9) \)