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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`).

Text Solution

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Energy released `=200 MeV`
`=200xx10^(6)xx1.6xx10^(-19)=3.2xx10^(-11)J`
`P=1` mega watt `=10^(6)` watts.
`"No. of fission per second"(n)=("Total energy")/("Energy per fission")`
`n=(10^(6))/(3.2xx10^(-11))=3.125xx10^(16)` Fissions
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