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 the explosion the energy released is approximately [Given : M ( Li ) = 6 . 01690 amu , M H 2 1 = 2 . 01471 amu , M He 4 2 = 4 . 00388 amu and 1 amu = 931 . 5 MeV ]

Step-by-step solution

Combining the reactions, it can be written that Li 6 3 + n 1 0 → He 4 2 + H 3 1 H 2 1 + H 3 1 → He 4 2 + n 1 0 _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ Li 6 3 + H 2 1 → 2 He 4 2 Hence, the energy released in process can be calculated as follows Q = Δ m c 2 = M ( Li ) + M H 2 1 - 2 × M He 4 2 × 931 . 5 MeV = [ 6 . 01690 + 2 . 01471 - 2 × 4 . 00388 ] × 931 . 5 MeV = 22 . 22 MeV