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Physics Question on Nuclei

When a slow neutron is captured by a 92235U^{235}_{92} U nucleus, a fission energy releasing 200 MeV200\, MeV. If power of nuclear reactor is 100 W100\, W then rate of nuclear fission is

A

3.6×106 s−13.6 \times 10^6 \,s^{-1}

B

3.1×1012 s−13.1 \times 10^{12}\,s^{-1}

C

1.8×104 s−11.8 \times 10^4\, s^{-1}

D

4.1×106 s−14.1 \times 10^6\, s^{-1}

Answer

3.1×1012 s−13.1 \times 10^{12}\,s^{-1}

Explanation

Solution

Number of fission per second = totalpower  energy/fission =\frac{\text { totalpower }}{\text { energy/fission }} Here, total power =100 W= 100\, W energy/fission =200 MeV=200×106×1.6×10−19 J= 200\, MeV =200 \times 10^{6} \times 1.6 \times 10^{-19}\, J =3.2×10−11 J.=3.2 \times 10^{-11}\, J . ∴\therefore fission rate =1003.2×10−11=3.1×1012s−1=\frac{100}{3.2 \times 10^{-11}}=3.1 \times 10^{12} s ^{-1}