Physics - Modern Physics Question with Solution | TestHub

${}_1\mathrm{H}^2 + {}_1\mathrm{H}^2 \to {}_2\mathrm{He}^4 + \mathrm{\Delta E}$

The binding energy per nucleon of a deuteron

$= 1.1 \mathrm{MeV}$

∴    Total binding energy$= 2 \times 1.1 = 2.2 \mathrm{MeV}$

The binding energy per nucleon of a helium nuclei

$= 7 \mathrm{MeV}$

∴    Total binding energy$= 4 \times 7 = 28 \mathrm{MeV}$

Hence, the energy released

$\mathrm{\Delta E} = (28 – 2 \times 2.2) = 23.6 \mathrm{MeV}$