Q. If the energy released in the fission of one nucleus is $200\ MeV$ then the number of nuclei required per second in a power plant of $16\ kW$ will be
A) $0.5 \times 10^{14}$
B) $0.5 \times 10^{12}$
C) $5 \times 10^{12}$
D) $5 \times 10^{14}$
Explanation
(d) Energy released in the fission of one nucleus $= 200\ MeV$
$= 200 \times 10^{6} \times 1.6 \times 10^{- 19}J = 3.2 \times 10^{- 11}J$
$P = 16kW = 16 \times 10^{3}watt$
Now, number of nuclei required per $second$
$n = \frac{P}{E} = \frac{16 \times 10^{3}}{3.2 \times 10^{- 11}} = 5 \times 10^{14}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.