The explosive in a Hydrogen bomb is a mixture of H 2 1 , H 3 1 and...

Combining the reactions, it can be written that

$Li 6 3 + n 1 0 \to He 4 2 + H 3 1$

$H 2 1 + H 3 1 \to He 4 2 + n 1 0$

_____________________ Li 6 3 + H 2 1 \to 2\left(He 4 2\right)

Hence, the energy released in process can be calculated as follows

$\begin{cases}Q & = & \Deltam{c}^{2} \\ & = & \left[M ( Li ) + M\left(H 2 1\right) - 2 \times M\left(He 4 2\right)\right] \times 931 . 5MeV \\ & = & [ 6 . 01690 + 2 . 01471 - 2 \times 4 . 00388 ] \times 931 . 5MeV \\ & = & 22 . 22MeV\end{cases}$