The explosive in a Hydrogen bomb is a mixture of H 2 1 , H 3 1 and Li 6 3 in some condensed form. The chain reaction is given by Li 6 3 + n 1 0 → He 4 2 + H 3 1 ; H 2 1 + H 3 1 → He 4 2 + n 1 0 During

Question

The explosive in a Hydrogen bomb is a mixture of ${}_{1}H^{2}, {}_{1}H^{3}$ and ${}_{3}{Li}^{6}$ in some condensed form. The chain reaction is given by ${}_{3}{Li}^{6} + {}_{0}n^{1} \to {}_{2}{He}^{4} + {}_{1}H^{3}$ ; ${}_{1}H^{2} + {}_{1}H^{3} \to {}_{2}{He}^{4} + {}_{0}n^{1}$

During the explosion the energy released is approximately [Given : $M (Li) = 6.01690 amu, M ({}_{1}H^{2}) = 2.01471 amu, M ({}_{2}{He}^{4}) = 4.00388 amu$ and $1 amu = 931.5 MeV$ ]

TestHub · Accepted Answer

Combining the reactions, it can be written that

${}_{3}{Li}^{6} + {}_{0}n^{1} \to {}_{2}{He}^{4} + {}_{1}H^{3}$

${}_{1}H^{2} + {}_{1}H^{3} \to {}_{2}{He}^{4} + {}_{0}n^{1}$

$_____________________{}_{3}{Li}^{6} + {}_{1}H^{2} \to 2 ({}_{2}{He}^{4})$

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

$\begin{array}{rcl} Q & = & Δ m c^{2} \\ = & [M (Li) + M ({}_{1}H^{2}) - 2 \times M ({}_{2}{He}^{4})] \times 931.5 MeV \\ = & [6.01690 + 2.01471 - 2 \times 4.00388] \times 931.5 MeV \\ = & 22.22 MeV \end{array}$

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