13.31 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.

Hint: The energy released per fission is given by \(E=\Delta mc^2\).
Step 1: Find the amount of energy converted per fission.
Amount of electric power to be generated, P=2×105 MW
10% of this amount has to be obtained from nuclear power plants.
Amountofnuclearpower,p1=10100×2×105=2×104×106J/s=2×1010
×60×60×24×365J/y
Heatenergy releasedperfissionofaU235nucleus,E=200MeV
Efficiencyofareactor=25%
Hence,theamountofenergyconvertedintotheelectricalenergyperfissioniscalculatedas:
25100×200=50MeV
=50×1.6×10-19×106=8×10-12J
Step 2: Find the amount of uraniumm per year.
The number of atoms required for fission per year:
2×1010×60×60×24×3658×10-12=78840×1024atoms
1mole,i.e.,235gofU235contains6.023×1023atoms.
Massof6.023×1023atomsofU235=235g=235×10-3kg
Massof78840×1024atomsofU235
=235×10-36.023×1023×78840×1024
=3.076×104kg
Hence,themassofuraniumneededperyearis3.076×104kg.