Physics - Modern Physics Question with Solution | TestHub

As given,

${ }_1{\mathrm{H}}^2 +{ }_1{\mathrm{H}}^2\to { }_2\mathrm{H}{\mathrm{e}}^4 +\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{r}\mathrm{g}\mathrm{y}$

The binding energy per nucleon of a deuteron${(}_1{\mathrm{H}}^2)$$=1.1 \mathrm{M}\mathrm{e}\mathrm{V}$

$\therefore$Total binding energy of one deuteron nucleus$= 2 \times 1.1=2.2 \mathrm{M}\mathrm{e}\mathrm{V}$

$\therefore$The binding energy per nucleon of Helium${(}_2\mathrm{H}{\mathrm{e}}^4)$=7 MeV

$\therefore$Total binding energy$= 4 \times 7 =28 \mathrm{M}\mathrm{e}\mathrm{V}$

Hence, energy released in the above process$=28-2 \times 2.2$

=$28 – 4.4 = 23.6 \mathrm{M}\mathrm{e}\mathrm{V}$