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.

Solution:

Amount of electric power to be generated, P = 2 × 105 MW

10% of this amount has to be obtained from nuclear power plants.

$\therefore$ Amount of nuclear power, $P_{1}=\frac{10}{100} \times 2 \times 10^{5}$

= 2 × 104 MW

= 2 × 104 × 106 J/s

= 2 × 1010 × 60 × 60 × 24 × 365 J/y

Heat energy released per fission of a 235U nucleus, E = 200 MeV

Efficiency of a reactor = 25%

Hence, the amount of energy converted into the electrical energy per fission is calculated as:

$\frac{25}{100} \times 200=50 \mathrm{MeV}$

$=50 \times 1.6 \times 10^{-19} \times 10^{6}=8 \times 10^{-12} \mathrm{~J}$

Number of atoms required for fission per year:

$\frac{2 \times 10^{10} \times 60 \times 60 \times 24 \times 365}{8 \times 10^{-12}}=78840 \times 10^{24}$ atoms

1 mole, i.e., 235 g of U235 contains 6.023 × 1023 atoms.

∴Mass of 6.023 × 1023 atoms of U235 = 235 g = 235 × 10−3 kg

$\therefore$ Mass of $78840 \times 10^{24}$ atoms of $U^{235}$

$=\frac{235 \times 10^{-3}}{6.023 \times 10^{23}} \times 78840 \times 10^{24}$

$=3.076 \times 10^{4} \mathrm{~kg}$

Hence, the mass of uranium needed per year is 3.076 × 104 kg.