Q. When 92U235 undergoes fission. About 0.1% of the original mass is converted into energy. Then the amount of 92U235 should undergo fission per day in a nuclear reactor so that it provides energy of 200 mega watt electric power is
(a) 9.6 × 10-2 kg
(b) 4.8 × 10-2 kg
(c) 19.2 × 10-2 kg
(d) 1.2 × 10-2 kg
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Ans: (c)
Sol: E = mc2
= 0.001 × (3×108)2
E = 10-3×9×1016 = 9×1013 J
Total mass of U required/sec
m= M×0.1% = M/1000
Power P = 200×106 J/s
200×106 = (M/1000)c2
M = 200×106 ×1000/c2
M = 0.22×10-5
Mass required per day = 0.22×10-5 ×24×60×60 = 19.2 ×10-2 kg