Question

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 1

The nuclear fission raction is `._(92)U^(235) +._(0)n^(1) rarr ._(56)Ba^(141) + ._(36)Kr^(92) + 3_(0)n^(1)`
Mass defect `Deltam = [(m_(u) - m_(n)) - (m_(Ba) + m_(Kr) + 3m_(n))] = 256.0526 - 235.8373 = 0.2153u`
Energy released `Q = 0.2153 xx 931 = 200 MeV`. Number of atoms in `1g = (6.02 xx 10^(23))/(235) = 2.56 xx 10^(21)`
Energy released in fission of `1g` of `U^(235)` is `E = 200 xx 2.56 xx 10^(21) = 5.12 xx 10^(23) MeV`
`= 5.12 xx 10^(23) xx 1.6 xx 10^(-13) = 8.2 xx 10^(10) J`
`= (8.2 xx 10^(10))/(3.6 xx 10^(6)) kWh = 2.88 xx 10^(4) kWh`