The binding energy per nucleon of \({}^{40}_{20}\mathrm{Ca}\) is 8.55 MeV. Take the mass of each nucleon as 1 u and \(1\ \mathrm{u}=931.5\ \mathrm{MeV}/c^{2}\). The mass defect of this nucleus, as a percentage of the total mass of its nucleons, is about
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.92%
The binding energy per nucleon is the total binding energy shared among the \( A \) nucleons. The mass defect is the mass equivalent of this total binding energy. The 40 nucleons of 1 u each have a total mass of 40 u.
\( \begin{aligned} E_b &= A\times E_{bn} \\ &= 40\times8.55 \\ &= 342\ \mathrm{MeV} \end{aligned} \)
\( \begin{aligned} \Delta M &= \frac{E_b}{c^{2}} \\ &= \frac{342}{931.5}\ \mathrm{u} \\ &= 0.367\ \mathrm{u} \end{aligned} \)
\( \begin{aligned} \text{Percentage} &= \frac{\Delta M}{40\ \mathrm{u}}\times100 \\ &= \frac{0.367}{40}\times100 \\ &= 0.92\,\% \end{aligned} \)
From the above, the mass defect is less than 1 per cent of the mass of the nucleons. Hence, the calcium nucleus is only a little lighter than its free nucleons.
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:”
\( E_{bn}=E_b/A \)\( \qquad (13.9) \)
Nuclear binding energy – “… This energy required Eb is related to the mass defect by”
\( E_b=\Delta M\,c^{2} \)\( \qquad (13.8) \)