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
Answer & explanation
Correct answer: option 4
Energy released in the fission of one nucleus = 200 MeV
$=200 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J}=3.2 \times 10^{-11}$ J
P = 16 KW = 16 $\times 10^3$ Watt
Now, number of nuclei required per second
$n=\frac{P}{E}=\frac{16 \times 10^3}{3.2 \times 10^{-11}}=5 \times 10^{14}$.