A thermal neutron strikes `U_(92)^(235)` nucleus to produce fission. The nuclear reaction is as given below :
`n_(0)^(1) + U_(92)^(235) to Ba_(56)^(141) + Kr_(36)^(92) +3n_(0)^(1) + E`
Calculate the energy released in MeV. Hence calculate the total energy released in the fission of 1 Kg of `U_(92)^(235)`.
Given mass of `U_(92)^(235) = 235.043933` amu
Mass of neutron `n_(0)^(1) =1.008665` amu
Mass of `Ba_(56)^(141)=140.917700` amu
Mass of `Kr_(36)^(92)=91.895400` amu

Explanation

Mass of the reactants `=235.043933+1.008665`
`=236.052598`
Mass of the product `=140.917700 +91.895400+3(1.008665)`
`=235.839095`
Energy E `= Delta m xx 931 "MeV"`
`=198.77` MeV
235 grams contain `6.0233 xx 10^(23)` atoms of `U^(235)`
1000 gram contains N number of atoms
`N=(6.0233 xx 10^(23) xx 1000)/( 235)`
`=2.5629 xx 10^(24)` atoms
Total energy released `=198.77 xx 2.5629 xx 10^(24)`
`=509.43 xx 10^(24)` MeV
or `E=8.15 xx 10^(13)` J