Q. On bombarding $U^{235}$ by slow neutrons, 200MeV energy is released. If the power output of atomic reactor is 1.6 MW, then the rate of fission will be
A) $5\ \times \ 10^{22}\ s^{- 1}$
B) $5 \times \ 10^{16}s^{- 1}$
C) $8\ \times \ 10^{16}\ s^{- 1}$
D) $20\ \times \ 10^{16}\ s^{- 1}$
Explanation
(b) Energy released on bombarding $U^{235}$ by neutron=200 MeV
Power output of atomic reactor = 1.6 MW
$\therefore$ Rate of fission= $\frac{1.6\ \times \ 10^{6}}{200\ \times \ 10^{6} \times 1.6\ \times \ 10^{- 19}}$
$\ \ \ = \ 5\ \times \ 10^{16}\ s^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.