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Calculate the energy released by fission from 2 gm of ""_(92)U^(235) in KWH. Given that the energy released per fission is 200 Mev. |
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Answer» Solution :No. of atoms in 1 GM of `""_(92)U^(235) = ("Avagadro number")/("mass number") = (6.023xx10^(23))/(235)` Energy released per fission = 200 MEV = `200xx10^(6)xx1.6 xx10^(-19)J` Total energy released per gm, `E=(6.023xx10^(23))/235xx200xx10^(6)xx1.6 xx10^(-19)J` `=(6.023xx200xx1.6)/(235xx36xx10^5)xx10^(10)KWH = 0.2278xx10^5` KWH Total energy released from 2 gm of `""_(92)U^(235)` is `0.4556xx10^5` KWH |
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