You are given that mass  ${ }_3^{7}Li=7.0160u,$ Mass of   ${ }_2^{4}He=4.0026u,$ and mass of   ${ }_1^{1}H=1.0079u$
When 20g of  ${ }_3^{7}Li$ is converted into ${ }_2^{4}He$  by proton capture, the energy liberated, (in kWh), is :
$\left[Massofnucleon=1GeV/c^2\right]$ 
 

$\begin{array}{l}{ }_3^7\mathrm{Li}+\mathrm{P}\to 2_2^4\mathrm{He}\\ \mathrm{Mass} \mathrm{defect}=△\mathrm{M}=\mathrm{m}\left(\mathrm{Product}\right)-\mathrm{m}\left(\mathrm{Reactant}\right)=(7.0160\mathrm{u})+(1.0079\mathrm{u})-(2\times 4.0026\mathrm{u})\\ \Rightarrow △\mathrm{M}=0.0187\mathrm{u}\\ \text{ for }20\mathrm{g}{\text{ of }}_3^7\mathrm{Li}\\ \mathrm{No}. \mathrm{of} \mathrm{nucleus}=\frac{20}{7}\times 6\times {10}^{23}\\ \mathrm{Total} \mathrm{mass} \mathrm{defect} =\mathrm{\Delta M}=0.0187\times \frac{20}{7}\times 6\times {10}^{23} \mathrm{u}\\ \mathrm{Total} \mathrm{energy} \mathrm{liberated}=△\mathrm{E}={\mathrm{\Delta Mc}}^2\\ \Rightarrow △\mathrm{E}={\mathrm{\Delta Mc}}^2=0.0187\times \frac{120}{7}\times {10}^{23}\times {\mathrm{c}}^2\times \frac{{10}^9\times 1.6\times {10}^{-19}}{{\mathrm{c}}^2}\mathrm{J}\times \frac{1\mathrm{kWh}}{3.6\times {10}^6\mathrm{J}}=1.4\times {10}^6\mathrm{kWh}\\ \end{array}$