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Calculate the energy released by fission from 2 gm of ._92U^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 `._92U^235 ="Avagadro NUMBER"/"mass number" =(6.023xx10^23)/235` Energy RELEASED per fission = 200 Mev `= 200xx10^6xx1.6xx10^(-19)` J Total energy released per gm, `E=(6.023xx10^23)/235xx200xx10^6xx1.6xx10^(-19)` J `=(6.023xx200xx1.6)/(235xx36xx10^5)xx10^10` KWH=`0.2278xx10^5` KWH Total energy released from 2 gm of `._92U^235` is `0.4556xx10^5` KWH |
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