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Assuming that energy released by the fis...

Assuming that energy released by the fission of a single `""_(92)^(235)U` nucleus is 200MeV, calculate the number of fissions per second required to produce 1 kilowatt power.

Text Solution

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Energy released by `""_(92)^(235)U` in one fission = 200MeV
Amount of `""_(92)^(235)U` = 1kg
Avogadro no. = `6.023xx10^(23)`
Energy released = ?
Energy release when fission, is
`E=200MeV=200xx10^(6)xx1.6xx10^(-19)`
= `3.2xx10^(-11)` joule
`P=(nE)/t`
`rArrn/t=P/E`
= `10^(3)/(3.2xx10^(-11))`
E = `3.125xx10^(13)`.
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