The binding energy per nucleon of deuterium and helium atom is 1.1 MeV and 7.0 MeV. If two deuterium nuclei fuse to form helium atom, the energy released is
Answer & explanation
Correct answer: option 2
${ }_1 H^2+{ }_1 H^2 \rightarrow{ }_2 H e^4$ + energy
Binding energy of a $\left({ }_1 H^2\right)$ deuterium nuclei = 2 × 1.1 = 2.2 MeV
Total binding energy of two deuterium = 2.2 × 2 = 4.4 MeV
Binding energy of a $\left({ }_2 He^4\right)$ nuclei = 4 × 7 = 28 MeV
So, energy released in fusion = 28 − 4.4 = 23.6 MeV