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Question

2×108 eV energy is released due to the fission of one nucleus of 23592U, then claculate the number of fission which must take place per second in order to produce a power of 1 Kilowatt

A
1.562×1013
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B
15.62×108
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C
3.125×1013
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D
3.125×108
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Solution

The correct option is B 3.125×1013
Given,2×108ev energy released due to fission of one 23592U Power =1000 watt let number of fission take place =N the Energy release will be
N×2×108ev
or
N×2×108×1.6×1019 J 1 ev =1.6×1019Joule
Now,
N ×2×108×1.6×1019=1000
N=10003.2×1011N=3.125×1013

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