Suppose India had a target of producing by $2020\; AD$, $200,000\; MW$ of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilization (i.e., conversion to electric energy) of thermal energy produced in a reactor was $25\%$. How much amount of fissionable uranium would our country need per year by $2020$? Take the heat energy per fission of $^{235}_{92}U$ to be about $200\; MeV$.

(N/A) Total electric power target, $P = 2 \times 10^{5} \; MW = 2 \times 10^{11} \; W$.
Nuclear power required, $P_{n} = 10\% \text{ of } P = 0.1 \times 2 \times 10^{11} = 2 \times 10^{10} \; J/s$.
Total energy required per year, $E_{total} = P_{n} \times (365 \times 24 \times 3600) \; s = 2 \times 10^{10} \times 3.1536 \times 10^{7} = 6.3072 \times 10^{17} \; J$.
Energy per fission, $E_{f} = 200 \; MeV = 200 \times 10^{6} \times 1.6 \times 10^{-19} \; J = 3.2 \times 10^{-11} \; J$.
Efficiency of reactor, $\eta = 25\% = 0.25$.
Useful energy per fission, $E_{u} = \eta \times E_{f} = 0.25 \times 3.2 \times 10^{-11} = 8 \times 10^{-12} \; J$.
Number of fissions required per year, $N = \frac{E_{total}}{E_{u}} = \frac{6.3072 \times 10^{17}}{8 \times 10^{-12}} = 7.884 \times 10^{28} \; \text{atoms}$.
Mass of $1 \; \text{mole}$ of $^{235}U = 235 \; g = 0.235 \; kg$.
Number of atoms in $1 \; \text{mole} = 6.023 \times 10^{23}$.
Mass of uranium required, $M = \frac{N}{N_{A}} \times 0.235 = \frac{7.884 \times 10^{28}}{6.023 \times 10^{23}} \times 0.235 \approx 3.076 \times 10^{4} \; kg$.