Practicing Success
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 |
$0.5 × 10^{14}$ $0.5 × 10^{12}$ $5 × 10^{12}$ $5 × 10^{14}$ |
$5 × 10^{14}$ |
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 $n=\frac{P}{E}=\frac{16 \times 10^3}{3.2 \times 10^{-11}}=5 \times 10^{14}$. |