The atomic masses of \({}^{1}_{1}\mathrm{H}\), \({}^{3}_{1}\mathrm{H}\) and \({}^{3}_{2}\mathrm{He}\) are 1.007825 u, 3.016049 u and 3.016029 u, respectively. The mass of a neutron is 1.008665 u. The binding energy of \({}^{3}_{1}\mathrm{H}\) is greater than that of \({}^{3}_{2}\mathrm{He}\) by about
(Take \(1\ \mathrm{u} = 931.5\ \mathrm{MeV}/c^{2}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(0.76\ \mathrm{MeV}\)
The binding energy of a nucleus is its mass defect times \( c^{2} \). Tritium has one proton and two neutrons. Helium-3 has two protons and one neutron. Atomic masses are used on both sides, with the mass of the \( {}^{1}_{1}\mathrm{H} \) atom for each proton. So the electron masses need no separate correction.
Mass defect of \( {}^{3}_{1}\mathrm{H} \):
\( \begin{aligned} \Delta M_{1} &= m({}^{1}_{1}\mathrm{H})+2m_n-m({}^{3}_{1}\mathrm{H}) \\ &= 1.007825+2\times1.008665-3.016049 \\ &= 0.009106\ \mathrm{u} \end{aligned} \)
Mass defect of \( {}^{3}_{2}\mathrm{He} \):
\( \begin{aligned} \Delta M_{2} &= 2m({}^{1}_{1}\mathrm{H})+m_n-m({}^{3}_{2}\mathrm{He}) \\ &= 2\times1.007825+1.008665-3.016029 \\ &= 0.008286\ \mathrm{u} \end{aligned} \)
Difference in binding energy:
\( \begin{aligned} \Delta E &= (\Delta M_{1}-\Delta M_{2})\,c^{2} \\ &= (0.009106-0.008286)\times931.5\ \mathrm{MeV} \\ &= 0.000820\times931.5\ \mathrm{MeV} \\ &\approx 0.76\ \mathrm{MeV} \end{aligned} \)
From the above, the tritium nucleus is bound more strongly than the helium-3 nucleus, by about 0.76 MeV.
NCERT (2026-27):
Summary – “7. The nuclear mass M is always less than the total mass, Σm, of its constituents. The difference in mass of a nucleus and its constituents is called the mass defect, ΔM = (Z mp + (A – Z )mn) – M Using Einstein’s mass energy relation, we express this mass difference in terms of energy as ΔEb = ΔM c2”