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NCERT Class XII Chapter
Nuclei
Questions With Solutions

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Question : 31
Total: 31
Suppose India has a target of producing by 2020 AD, 200,000 MW of electric power, ten percent of which is 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? Take the heat energy per fission of 235U to be about 200 MeV.
Solution:  
10% of total power 200,000 MW to be obtained from nuclear power plant by 2020 AD.
So, power from nuclear plants
= 2 × 105 × 0.1 MW = 2 × 104 MW = 2 × 1010 W
With efficiency of power plants 25% only, the energy converted into electrical energy per fission
=
25
100
 × 200 = 50 MeV = 50 × 1.6 × 10−13 Joule = 8 × 10−3
Total energy to be produced
= 2 × 104 MW = 2 × 1010 joule/sec
= 2 × 1010 × 60 × 60 × 24 × 365 joule/year =
2×1024×36×24×265
8

Mass of
2×36×24×365
8
 × 1024 atoms
=
235×10−3
6.023×1023
 ×
2×36×24×365×1024
8
= 3.08 × 104 kg
Hence mass of uranium needed per year= 3.08 × 104 kg
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