The energy released in fusion reaction
\( {}_1^2\mathrm{H}+{}_1^2\mathrm{H}\rightarrow{}_2^3\mathrm{He}+n \)
where Binding Energy of \( {}_1^2\mathrm{H}=2.23\,\mathrm{MeV} \) and of \( {}_2^3\mathrm{He}=7.73\,\mathrm{MeV} \), would be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3.27 MeV
The energy released in a nuclear reaction is the binding energy of the products minus the binding energy of the reactants: the more tightly bound the final nuclei, the more energy is given out.
\( B_{\mathrm{reactants}} = 2\times2.23\,\mathrm{MeV} = 4.46\,\mathrm{MeV} \)
\( B_{\mathrm{products}} = 7.73\,\mathrm{MeV} + 0 = 7.73\,\mathrm{MeV} \)
The neutron is a single free nucleon, so its binding energy is zero.
\( Q = B_{\mathrm{products}} - B_{\mathrm{reactants}} = 7.73\,\mathrm{MeV} - 4.46\,\mathrm{MeV} = 3.27\,\mathrm{MeV} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Nuclear Fusion – Energy Generation in Stars – “… When two light nuclei fuse to form a larger nucleus, energy is released, since the larger nucleus is more tightly bound, as seen from the binding energy curve in Fig.13.1. Some examples of such energy liberating nuclear fusion reactions are :
\( \displaystyle {}^{2}_{1}\mathrm{H} + {}^{2}_{1}\mathrm{H} \rightarrow {}^{3}_{2}\mathrm{He} + n + 3.27\ \mathrm{MeV} \qquad [13.13(b)] \)”