If 200 MeV energy is released in the fission of a single $U^{235}$ nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (Given $1 eV = 1.6 \times 10^{-19}$ J)

$3.125 × 10^{13}$ $3.125 × 10^{14}$ $3.125 × 10^{15}$ $3.125 × 10^{16}$

$3.125 × 10^{13}$

$P =n\left(\frac{E}{t}\right) \Rightarrow 1000=\frac{n \times 200 \times 10^6 \times 1.6 \times 10^{-19}}{t}$ $\Rightarrow \frac{n}{t}=3.125 \times 10^{13}$.