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 23592U to be about 200MeV.
By 2020 India need 2,00,000 MW of electric energy per year, out of which 10% of energy is to be obtained from nuclear power plants.
Thus, Nuclear power plants should produce,
= 20,000×106×365×24×60×60 J
= 6.3072×1017 J of energy
Given that Fission of one 
releases 200 MeV of energy.
200 MeV = 200×106×1.6×10-19 J = 3.2×10-11 J
Since, reactor is 25% efficient, total energy produced per atom,
Ea = 0.25×3.2×10-11 J = 8×10-12 J
∴ Number of atoms to be fission to produce required energy,
We know that, 235 g Uranium contains 6.023×1023 atoms.
Thus, for 7.884×1028 atoms we need,
Hence, India need 3.076×104 Kg of uranium in 2020 to produce sufficient electricity.