Question

# Suppose India had a target of producing by 2020AD,200000MW of electric power, 10 % of which was to be obtained from nuclear power plants. Suppose we are given that, on an average, the efficiency of utilisation (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 235 U to be about 200 MeV .

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Solution

## Total electric power which is aimed to be produced by 2020 = 2×105 MWPower obtained from the nuclear power plant = 10% of 2×105 MW=2×104 MW=2×1010 WEnergy required from the nuclear plant in one year= Power×time=2×1010×365×24×60×60=6.3×1017 JAvailable electric energy per fission= 25% of 200 MeV=(0.25×200×1.602×10−13) J=8×10−12 JRequired no. of fission per year = 6.3×10178×10−12=0.7875×1029Now, 6.023×1023 nuclei of 23592U have mass = 235g∴ Mass required to produce 7.9×1028 nuclei = 2356.023×1023×3.95×1028g≈3.084×104 kg

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