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

19.2 MeV

${ }_1 H^2+{ }_1 H^2 \rightarrow{ }_2 H e^4$ + 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