Energy released in fusion of 1 kg of deuterium nuclei

$8 \times 10^{23} \mathrm{MeV}$

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

So $E=\frac{6.02 \times 10^{23} \times 10^3 \times 3.27 \times 1.6 \times 10^{-13}}{2 \times 2}=7.8 \times 10^{13} \mathrm{~J}$

= 8 × 10¹³ J.