Q. If 200 MeV energy is released in the fission of a single nucleus of $\ _{92}U^{235}$ . How many fissions must occur per second to produce a power of 1 kW?
A) $3.125\ \times \ 10^{13}$
B) $6.250 \times \ 10^{13}$
C) $1.525\ \times 10^{13}$
D) None of these
Explanation
(a) We know that 1 kW $= \ 1\ \times \ 10^{3}\ Js^{- 1}$
Also, $1.6\ \times \ 10^{- 9}\ J\ = \ 1\ eV$
$\therefore\ MeV = \ 200\ \times \ 1.6\ \times \ 10^{- 19}\ \times \ 10^{6}\ J$
Number of fissions = $\frac{Power}{Energy\ released}$
= $\frac{10^{3}}{200\ \times \ 1.6\ \times 10^{- 13}} = \ 3.125\ \times \ 10^{13}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.