Calculate the energy released by fission from 2g of `._(92)^(235)U` in kWh. Given that the energy released per fission is 200 MeV.

Explanation

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