Two deuterons undergo nuclear fusion to form a Helium nucleus. The energy released in this process is (given…

$\mathrm{Q}=\sum \mathrm{B}.\mathrm{E}.\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{products}-\mathrm{ }\sum \mathrm{B}.\mathrm{E}.\mathrm{ }\mathrm{of}\mathrm{ }\mathrm{reactants}$

The equation for two deuterons combining to form Helium nucleus is given as ${\mathrm{H}}^2{+}_1{\mathrm{H}}^2{\to }_2{\mathrm{H}}^4\mathrm{ }$

$\mathrm{Energy}\mathrm{ }\mathrm{released}=\mathrm{Q}=4\left[\mathrm{B}.\mathrm{E}.\left({}_2{\mathrm{H}}^4\right)\right]\mathrm{ }-4\left[\mathrm{B}.\mathrm{E}.\left({}_1{\mathrm{H}}^2\right)\right]$
 $=4\times 7-4\times \text{1.1}$

$=28-\text{4.4}$

The energy released in this process is $=\text{23.6} \mathrm{MeV}$