The energy released by fission of one `U^(235)` atom is 200 MeV. Calculate the energy released in kWh, when one gram of uranium undergoes fission.

Answer

`2.278xx10^(4)kWh`

Explanation

Number of atoms in 1 gram of `._(92)U^(235)`
`=("Avogadro's number")/("Mass number")=(6.023xx10^(23))/235`
Energy released per fission =200MeV.
`:.` Total energy released in fission of 1 gram of `._(92)U^(235)`
`=(6.023xx10^(23))/235xx200MeV`
`=5.126xx10^(23)MeV`
As `1kWh=3.6xx10^(6)J`,
`:.` Total energy released
`=(5.126xx10^(23)xx1.6xx10^(-13))/(3.6xx10^(6))`
`=2.278xx10^(4)kWh`