If 200 MeV energy is released in the fission of a single 23592U nucleus, then the number of fissions per second required to produce a power of 10 kW are
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3.125 x 1014
The power is the energy released per second, so the number of fissions per second is the power divided by the energy released per fission, with both in joules.
\( \displaystyle E = 200\,\mathrm{MeV} = 200 \times 1.6\times10^{-13}\,\mathrm{J} = 3.2\times10^{-11}\,\mathrm{J} \)
\( \displaystyle P = 10\,\mathrm{kW} = 1\times10^{4}\,\mathrm{W} \)
\( \displaystyle N = \frac{P}{E} \)
\( \displaystyle N = \frac{1\times10^{4}\,\mathrm{J\,s^{-1}}}{3.2\times10^{-11}\,\mathrm{J}} \)
\( \displaystyle N = 3.125\times10^{14}\ \text{fissions per second} \)