Q. 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 $1\ eV = 1.6 \times 10^{- 19}J$)
A) $3.125 \times 10^{13}$
B) $3.125 \times 10^{14}$
C) $3.125 \times 10^{15}$
D) $3.125 \times 10^{16}$
Explanation
(a) $P = n\left( \frac{E}{t} \right) \Rightarrow 1000 = \frac{n \times 200 \times 10^{6} \times 1.6 \times 10^{- 19}}{t}$
$\Rightarrow \frac{n}{t} = 3.125 \times 10^{13}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.