IN a nuclear reactor, fission is produced in 1 g of `.^(235)U`
`(235.0349 am u)`. In assuming that `._(53)^(92)Kr (91.8673 am u)`
and `._(36)^(141)Ba (140.9139 am u)` are produced in all reactions and no energy is lost, calculate the total energy produced in killowatt. Given: `1 am u =931 MeV`.

Answer

`2.28 xx 10^(4)kWh`

Explanation

The nuclear fission reaction in the above process is
`._(92)^(235)U +._(0) + ._(0)^(1)n rarr._(56)^(141)Ba +._(36)^(92)Kr+3 (._(0)^(1)n`
Thesum of the mass before reaction is
`235.0439 +1.007=236.0526 a.m. u.`
The sum of the masses after reaction is
`140.9139 +91.8973 +(1.0087) =235.8375 a.m.u.`
Mass defect,
`Delta m =236.0526 -235. 8373 =0.2175 a.m.n`
Energy released in the fission of `.^(235)U` nulceus is given by
`DeltaE=Deltam xx 931.2 MeV`
`=0.2153 xx 931.2 =200 MeV`
Number of atoms in `1 g` of `.^(235)U`
`N=(6.02 xx10^(23))/(235)=2.56 xx10^(21)`
Energy released in fission of `1g` of `.^(235)U`,
`E=200 xx 2.56 xx 10^(21) MeV`
`=5.12 xx10^(23) MeV =(5.12 xx 10^(23)) xx (1.6 xx10^(-13))`
`=8.2 xx 10^(10)J`
`(8.2 xx 10^(10))/(3.6 xx 10^(6) kWh =2.28 xx10^(4) kWh`