The binding energy of deuteron ${ }_1^2 H$ is 1.112 MeV per nucleon and an $ \alpha$-particle ${ }_2^4 \mathrm{He}$ has a binding energy of 7.047 MeV per nucleon. Then in the fusion reaction ${ }_1^2 H+{ }_1^2 H \rightarrow_2^4 H e+Q$, the energy Q released is
Answer & explanation
Correct answer: option 3
Mass of ${ }_1 H^2=2.01478$ a.m.u.
Mass of ${ }_2 He^4=4.00388$ a.m.u.
Mass of two deuterium = 2 × 2.01478 = 4.02956
Energy equivalent to $2_1 H^2$
= 4.02956 × 1.112 MeV = 4.48 MeV
Energy equivalent to ${ }_2 H^4$
= 4.00388 × 7.047 MeV = 28.21 MeV
Energy released = 28.21 - 4.48 = 23.73 MeV = 24 MeV