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Question
How much mass of 235U is required to undergo fission each day to provide 3000 MW of thermal power? Average energy per fission is 202.79 MeV.
Numerical
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Solution
Data: Power = 3000 MW = 3 × 109 J/s
∴ Energy to be produced each day
= 3 × 109 × 86400 J each day
= 2.592 × 1014 J each day
Energy per fission = 202.79 MeV
= 202.79 × 106 × 1.6 × 10−19 J = 3.245 × 10−11 J
∴ Number of fissions each day = `"Total Energy"/"Enery per Fusion"`
= `(2.592 xx 10^14)/(3.245 xx 10^-11)`
= 7.988 × 1024 each day
0.235 kg of 235U contains 6.02 × 1023 atoms
∴ Mass per Atom = `"Molar Mass"/"Avogadro’s Number"`
= `(0.235)/(6.02 xx 10^23)`
= 0.039023 × 10−23kg
Total Mass = Number of Fissions × Mass per Atom
∴ M = 7.988 × 1024 × 0.039023 × 10−23
= 3.118 kg
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