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Q.
$200\, MeV$ of energy can be obtained per fission. In a reactor generating $1000\, kW$, the number of nuclei under going the fission per second is $3.125 \times 10^{n}$. Find $n$.
Nuclei
Solution:
$P=1000\, kW$
$P=\frac{\text { Energy }}{\text { Time }}$
$P=\frac{\text { Energy released per fission } \times \text { Number of fission }}{\text { Time }}$
$1000 \times 10^{3}=\frac{3.2 \times 10^{-11} \times \text { Number of fission }}{\text { Time }}$
Number of fission per second $=3.125 \times 10^{16}$