If 200 MeV energy is released in the fission of a single nucleus of \( {}^{235}_{92}U \), how many fissions per second must occur to produce a power of 1 kW?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 3.125 × 1013
The power equals the energy per fission times the number of fissions per second. Convert the energy per fission to joules:
\( \displaystyle E = 200\,\mathrm{MeV} = 200 \times 1.6\times10^{-13}\,\mathrm{J} = 3.2\times10^{-11}\,\mathrm{J} \)
\( \displaystyle N = \frac{P}{E} = \frac{1000\,\mathrm{J\,s^{-1}}}{3.2\times10^{-11}\,\mathrm{J}} = 3.125\times10^{13}\,\mathrm{s^{-1}} \)