In the previous question, if the energy released from the fission of one $$\text{U}^{235}$$ nucleus is $$200\text{ MeV}$$, assuming $$100\%$$ efficiency the power output of the reactor is

Solution

Energy per fission $$= 200 \times 1.6 \times 10^{-13} = 3.2 \times 10^{-11}\text{ J}$$. Fissions per second $$= \frac{1.5 \times 10^{21}}{60} = 2.5 \times 10^{19}$$. Power $$= 2.5 \times 10^{19} \times 3.2 \times 10^{-11} = 8 \times 10^8\text{ W} = 800\text{ MW}$$.

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