The explosive in a Hydrogen bomb is a mixture of and in some condensed form. The chain reaction is given by

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Solution

Q.
The explosive in a Hydrogen bomb is a mixture of ${}_{1}{H^{2}, {}_{1}H^{3}}$ and ${}_{3}{L i^{6}}$ in some condensed form. The chain reaction is given by

${}_{3}{L i^{6} + {}_{0}{n^{1}}} \to {}_{2}{H e^{4} + {}_{1}{H^{3}}}$

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

During the explosion the energy released is approximately [Given : M(Li) = 6.01690 amu, M ( ${}_{1}{H^{2}}$ ) = 2.01471 amu, M( ${}_{2}{H e^{4}}$ ) = 4.00388 amu, and 1 amu = 931.5 MeV] [2024]

1 28.12 MeV

2 12.64 MeV

3 16.48 MeV

4 22.22 MeV

Ans.
(4)

   ${}_{3}{L i^{6}} + {}_{0}{n^{1}} \to {}_{2}{H e^{4}} + {}_{1}{H^{3}}$

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

   ${}_{3}{L i^{6}} + {}_{1}{H^{2}} \to 2 ({}_{2}{H e^{4})}$

   $Q = Δ m c^{2}$

   $Q = [M (L i) + M (H_{1}^{2}) - 2 \times M (H e^{4})] \times 931.5 M e V$

   $Q = [6.01690 + 2.01471 - 2 \times 4.00388] \times 931.5 M e V$

$\Rightarrow Q \approx 22.22 M e V$

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