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 `U_(92)^(235)`
`= ("Avagadro number")/("At wt") = (6.023xx10^(23))/235`
Energy released per atom = 200 MeV
`= 200xx10^(6)xx1.6 xx10^(-19)J`
Total energy released per gm ,
`E=(6.023xx10^(23))/235xx200xx10^(6)xx1.6xx10^(-19)J`
`= (6.023xx200xx1.6)/(235xx36xx10^(5))xx10^(10)kWh`
`=0.2278xx10^(5)kWh`
Total energy released from 2 gm of `U_(92)^(235)` is `0.4556 xx10^(5)KWH`