Energy released in fusion of 1 kg of deuterium nuclei 

Fusion reaction of deuterium is
${ }_1{\mathrm{H}}^2{+}_1{\mathrm{H}}^2{\to }_2{\mathrm{He}}^3{+}_0{\mathrm{n}}^1+3.27\mathrm{MeV}$

$$\begin{gathered} \mathrm{So} \mathrm{E}=\frac{6.02\times {10}^{23}\times {10}^3\times 3.27\times 1.6\times {10}^{-13}}{2\times 2}=7.8\times {10}^{13}\mathrm{J} \\ \simeq 8\times {10}^{13}\mathrm{J} \end{gathered}$$