1.

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`

Answer» Correct Answer - 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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