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Physics Question on Atoms

Energy released in the fission of a single 92235U^{235}_{92}U nucleus is 200 MeV. The fission rate of 92235U^{235}_{92}U filled reactor operating at a power level of 5 W is

A

1.56×1010s−11.56 \times 10^{10} s^{-1}

B

1.56×1011s−11.56 \times 10^{11} s^{-1}

C

1.56×1016s−11.56 \times 10^{16} s^{-1}

D

1.56×1017s−11.56 \times 10^{17} s^{-1}

Answer

1.56×1011s−11.56 \times 10^{11} s^{-1}

Explanation

Solution

Fission rate
=total powerenergyfission=5200×1.6×10−13=1.56×1011s−1=\frac{ total \, power}{\frac{energy}{fission}}=\frac{5}{200 \times 1.6 \times 10^{-13}}=1.56 \times 10^{11}s^{-1}