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If 200 MeV energy is released in the fis...

If `200 MeV` energy is released in the fission of a single `U^235` nucleus, the number of fissions required per second to produce `1` kilowatt power shall be (Given `1 eV = 1.6 xx 10^-19 J`).

A

`3.125xx10^13`

B

`1.52xx10^6`

C

`3.125xx10^12`

D

`3.125xx10^14`

Text Solution

Verified by Experts

The correct Answer is:
A
Let the number of fussions per second be n.
Energy released per second = n x 200 MeV = n x 200 x 1.6 x `10^(-13)` J
Energy required per second = power x time
=1 kW x 1s = 1000 J
`therefore n xx 200 xx 1.6 xx 10^(-13)`=1000
or `n=1000/(3.2xx10^(-11)) =10/3.2 xx 10^13 = 3.125xx10^13`
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