The binding energy per nucleon of \({}^{4}_{2}\mathrm{He}\) is 7.07 MeV. The atomic mass of \({}^{1}_{1}\mathrm{H}\) is 1.007825 u and the mass of a neutron is 1.008665 u. Taking \(1\ \mathrm{u}=931.5\ \mathrm{MeV}/c^{2}\), the atomic mass of \({}^{4}_{2}\mathrm{He}\) is about
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(4.0026\ \mathrm{u}\)
A nucleus is lighter than its free protons and neutrons. The missing mass is the mass defect \( \Delta M \), and it equals \( E_b/c^{2} \). Helium-4 has 2 protons and 2 neutrons. Its atomic mass is the mass of 2 hydrogen atoms and 2 neutrons minus \( \Delta M \). The two electrons are counted on both sides.
\( \begin{aligned} E_b &= A\times E_{bn} \\ &= 4\times7.07 \\ &= 28.28\ \mathrm{MeV} \end{aligned} \)
\( \begin{aligned} \Delta M &= \frac{28.28}{931.5}\ \mathrm{u} \\ &= 0.03036\ \mathrm{u} \end{aligned} \)
\( \begin{aligned} m({}^{4}\mathrm{He}) &= [2\times1.007825+2\times1.008665]-\Delta M \\ &= 4.03298-0.03036 \\ &= 4.0026\ \mathrm{u} \end{aligned} \)
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 – “… The difference in mass of a nucleus and its constituents, ΔM, is called the mass defect, and is given by”
\( \Delta M=[Zm_p+(A-Z)m_n]-M \)\( \qquad (13.7) \)
Nuclear binding energy – “… This energy required Eb is related to the mass defect by”
\( E_b=\Delta M\,c^{2} \)\( \qquad (13.8) \)