# 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.

Question:

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 235U to be about 200MeV.

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.