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Question

Physics Question on Nuclei

If 200 MeV200\,MeV energy is released in the fission of a single nucleus of 92U235{ }_{92} U ^{235}, how many fissions must occur per second to produce a power of 1 kW?1 \,kW ?

A

3.125×10133.125 \times 10^{13}

B

1.52×1061.52 \times 10^{6}

C

3.125×10123.125 \times 10^{12}

D

3.125×10143.125 \times 10^{14}

Answer

3.125×10133.125 \times 10^{13}

Explanation

Solution

Total energy /s=1000 J/ s =1000\, J
Energy released/fission =200 MeV=200\, MeV
=200×1.6×10−13 J=200 \times 1.6 \times 10^{-13} \,J
=3.2×10−11 J=3.2 \times 10^{-11} \,J
∴\therefore Number of fission /s=10003.2×10−11/ s =\frac{1000}{3.2 \times 10^{-11}}
=3.12×1013=3.12 \times 10^{13}