Suppose India had a target of producting by 2020 A.D., 2 × 10⁵ 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 electrical energy) of thermal energy produced in a reactor is 25%. How much amount of fissionable uranium would our country need per year by 2020? Take the heat energy per fission of U²³⁵ to be about 200 MeV.

Solution

Given,
Target of electric power to be produced = 100,000 MW.
10% power is to be obtained from nuclear power plants.
Heat energy per fission of U²³⁵ = 200 MeV

Therefore,

Required electric power from nuclear plants,

= 100000 \times \frac{10}{100} = 10, 000 MW

Therefore, required electric energy from nuclear plants per year,

= (10, 000 \times 10^{6} W) \times 365 \times 24 \times 60 \times 60 = 3.1536 \times 10^{17} J

,

Electrical energy recovered from the fission of one U²³⁵nucleus,

= 200 \times \frac{25}{100} = 50 MeV = 50 \times 1.6 \times 10^{- 13} = 8 \times 10^{- 12} J

∴

Number of fissions of

U^{235}

nucleus required is,

= \frac{3.1536 \times 10^{7}}{8 \times 10^{- 12}} = 3.942 \times 10^{28}

Number of moles of

U^{235} required per year

= \frac{3.942 \times 10^{28}}{6.023 \times 10^{23}} = 6.5449 \times 10^{4}

Therefore, mass of

U^{235}

required per year,

= 6.5449 \times 10^{4} \times 235 = 1538.054 g = 1.538054 kg .

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