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 ${}^{235}U$ to be about 200MeV.

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Payal Gupta | Contributor-Level 10

8 months ago

13.31 Amount of Electric power to be generated, P = 200000 MW

10% of which to be obtained from nuclear power.

Hence, amount of nuclear power = 10% of 2 $* 10^{5}$ MW = 2 $* 10^{4}$ MW = 2 $* 10^{4} * 10^{6}$ J/s

= 2 $* 10^{4} * 10^{6} * 60 * 60 * 24 * 365$ J/y = 6.3072 $* 10^{17}$ J/y

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

Efficiency of a reactor = 25%

So the amou

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Physics Ncert Solutions Class 12th 2023

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