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Q.

The explosive in a Hydrogen bomb is a mixture of 1H2, 1H3 and 3Li6 in some condensed form. The chain reaction is given by
3Li6+0n12He4+1H31H2+1H32He4+0n1
During the explosion, the energy released is approximately ________MeV. [Round-off to the nearest integer]
[M(3Li6) = 6.01690amu, M(1H2) = 2.01471amu, M(2He4) = 4.0038amu, 1 amu = 931.5MeV]

 

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answer is 22.

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Detailed Solution

3Li6+0n12He4+1H3 1H2+1H32He4+0n1 3Li6+1H22×2He4
Energy Q = mC2
m931.5MeV
Q=MLi+MH22×M22e4×931.5
= (6.0169 + 2.01471 - 2 x 4.0038) x 931.5MeV
= 22.216 MeV
= 22 MeV

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