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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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Here, energy released/fission
`=200MeV`
`=200xx1.6xx10^(-13)J=3.2xx10^(-11)J`
Total energy required/sec `= 1kw=1000w`
`=1000J//s`.
Number of fission/sec
`=("Energy reqd. per sec")/("energy released "//"fission")`
`=1000/(3.2xx10^(-11))`
`=3.125xx10^(13)s`
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