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Question

Physics Question on Nuclear physics

On bombardment of U235U^{235} by slow neutrons, 200 MeV200\, MeV energy is released. If the power output of atomic reactor is 1.6 M W1.6\, M\, W, then the rate of fission will be

A

5×1016/s5\times 10^{16}/s

B

10×1016/s10\times 10^{16}/s

C

15×1016/s15\times 10^{16}/s

D

20×10−16/s20\times 10^{-16}/s

Answer

5×1016/s5\times 10^{16}/s

Explanation

Solution

Rate of fission =P energy for fission =\frac{P}{\text { energy for fission }}
=1.6×106200×1.6×10−13=\frac{1.6 \times 10^{6}}{200 \times 1.6 \times 10^{-13}}
=5×1016s−1=5 \times 10^{16} s ^{-1}