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Question

On fission of one nucleus of U235, the amount of energy obtained is 200 MeV. The power obtained in a reactor is 1000 kW. Number of nuclei fissioned per second in the reactor is

A
3.125×1016
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B
6.25×1010
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C
3.125×1032
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D
6.25×1020
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Solution

The correct option is A 3.125×1016
Let E be the amount of energy obtained per one nucleus.

Power obtained from the reactor is P=nEt(nt)=PE

P=1000 kW=1000×1000=106 Js1

Also, 1 MeV=1.6×1013 J

Number of nuclei fissioned per second =PE=106200×1.6×1013=3.125×1016 s1

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