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+0n1 2He4+1H3 1H2+1H3 2He4+0n1 During the explosion the energy released is approximately [Given ; M(Li)=6.01690 amu, M (1H2 )=2.01471 amu, M (2He4 )=4.00388 amu, and 1 amu=931.5 MeV]
- 22.22 MeV
- 28.12 MeV
- 16.48 MeV
- 12.64 MeV
Answer
(1) 22.22 MeV
Solution
Add the two reactions (0n1 and 1H3 cancel):
3Li6+1H2 2 2He4
m=6.01690+2.01471-2 4.00388
=8.03161-8.00776=0.02385 amu
Q=0.02385 931.5 22.22 MeV
Answer: (1)