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}U to be about 200MeV.

#### Solution

Amount of electric power to be generated, *P* = 2 × 10^{5} MW

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

∴Amount of nuclear power, `P_1 = 10/100 xx 2 xx 10^5`

= 2 × 10^{4} MW

= 2 × 10^{4} × 10^{6} J/s

= 2 × 10^{10} × 60 × 60 × 24 × 365 J/y

Heat energy released per fission of a ^{235}U nucleus, *E* = 200 MeV

Efficiency of a reactor = 25%

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

`25/100 xx 200 = 50 MeV`

` = 50 xx 1.6 xx 10^(-19) xx 10^(6) = 8 xx 10^(-12) J`

Number of atoms required for fission per year:

`(2 xx 10^10 xx 60 xx 60 xx 24 xx 365)/(8 xx 10^(-12)) = 78840 xx 10^(24) "atoms"`

1 mole, i.e., 235 g of U^{235} contains 6.023 × 10^{23} atoms.

∴Mass of 6.023 × 10^{23} atoms of U^{235 }= 235 g = 235 × 10^{−3} kg

∴Mass of 78840 × 10^{24} atoms of U^{235}

`= (235 xx 10^(-3))/(6.023 xx 10^23) xx 78840 xx 10^24`

`= 3.076 xx 10^4 kg`

Hence, the mass of uranium needed per year is 3.076 × 10^{4} kg.