Mass of ₃Li⁷ = 7.0160 u, mass of ₂He⁴ = 4.0026 u and mass ₁H¹ = 1.0079 u. When 20 gm of ₃Li⁷ is converted into ₂He⁴ by proton capture, the energy liberated (in kWh) is [1 u = 1 GeV/c²]

$\begin{array}{l}{3}^{{\mathrm{Li}}^7}+1^{{\mathrm{H}}^1}\to 2\left(2^{{\mathrm{He}}^4}\right)\\ \mathrm{\Delta m}={\mathrm{m}}_{\mathrm{Li}}+{\mathrm{m}}_{\mathrm{H}}-2{\mathrm{M}}_{\mathrm{He}}\end{array}$
Energy released E = $∆$Mc²
In used 1gm of $\mathrm{LiE}=\frac{{\mathrm{\Delta mC}}^2}{{\mathrm{MLi}}^0}$
In used 20gm of  $\mathrm{LiE}=\frac{{\mathrm{\Delta mC}}^2}{{\mathrm{MLi}}^0}\times 20\mathrm{gm}$
$\begin{array}{l}\mathrm{E}=\frac{\left(\right(7.016+1.0079)-2\times 4.0026)}{7.016}{\mathrm{C}}^2\times 20\mathrm{g}\\ =\frac{0.0187\times {\left(3\times {10}^8\right)}^2\times 20\times {10}^{-3}}{7.016}\end{array}$
E = 480 x 10¹⁰ J
= 1.33 x 10⁶ kWh