Assuming that about `200 MeV` of energy is released per fission of `._(92)U^(235)` nuceli, the mass of `U^(235)` consumed per day in a fission ractor of power `1` megawatt will be approximately .

Answer

`1.0575g`

Explanation

Total energy/day=`10^(6)xx24xx60xx60J`
Energy released per fission =200MeV
`=200xx1.6xx10^(-13)J`
Number of fissions/day
`=(10^(6)xx24xx60xx60)/(200xx1.6xx10^(-13))=(24xx36xx10^(19))/3.2`
`:.` Amount of `._(92)U^(235)` consumed
`=235/(6xx10^(23))xx(24xx36xx10^(19))/3.2g=1.0575g`