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Question:

When a slow neutron is captured by a 23592U nucleus, a fission energy released is 200 MeV. If power of nuclear reactor is 100 W then rate of nuclear fission is:

1.8×104s−1;

4.1×106s−1;

3.6×106s−1;

3.1×1012s−1;

Solution:

Energy released per fission is 200 MeV = 200 × 1.6 × 10-13 J = 3.2 × 10-11 J.
Therefore rate of fission is 100/(200 × 1.6 × 10-13) = 3.125 × 1012 per second.