The explosive in a Hydrogen bomb is a mixture of $^2_1\text{H}$, $^3_1\text{H}$ and $^6_3\text{Li}$ in some condensed form. The chain reaction is given by
$$^6_3\text{Li} + {}^1_0n \to {}^4_2\text{He} + {}^3_1\text{H};\qquad {}^2_1\text{H} + {}^3_1\text{H} \to {}^4_2\text{He} + {}^1_0n$$
During the explosion the energy released is approximately:
[Given: $M(\text{Li}) = 6.01690\ \text{amu}$, $M(^2_1\text{H}) = 2.01471\ \text{amu}$, $M(^4_2\text{He}) = 4.00388\ \text{amu}$ and $1\ \text{amu} = 931.5\ \text{MeV}$]
Answer: (D) $22.22\ \text{MeV}$
Adding the two reactions, the neutron and $^3_1\text{H}$ cancel:
$$^6_3\text{Li} + {}^2_1\text{H} \to 2\,{}^4_2\text{He}$$
$$\Delta m = 6.01690 + 2.01471 - 2\times4.00388 = 0.02385\ \text{amu}$$
$$E = 0.02385\times931.5 \approx 22.22\ \text{MeV}$$
Solution by Sreeraj P, M.Sc Physics