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`).

`3.125xx10^(13)`

`3.125xx10^(14)`

`3.215xx10^(15)`

`3.215xx10^(16)`

Text Solution

Verified by Experts

`E=200meV=200xx10^(6)xx1.6xx10^(-19)=3.2xx10^(-11)J`
`p=(nE)/timpliesn/t=P/E=(10^(3))/(3.2xx10^(-11))=3.125xx10^(13)`

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