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Q.

If 200 MeV energy is released in the fission of a single U235 nucleus, the number of fissions required per second to produce 1 kilowatt power shall be (Given 1eV=1.6×10−19J )

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a

3.125×1013

b

3.125×1014

c

3.125×1015

d

3.125×1016

answer is A.

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Detailed Solution

P=nEt⇒1000=n×200×106×1.6×10−19t ⇒nt=3.125×1013
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