Question

Assuming the splitting of `U^235` nucleus liberates 200 MeV energy, find
(a) the energy liberated in the fission of 1 kg of `U^235` and
(b) the mass of the coal with calorific value of `30 kJ//g` which is equivalent to 1 kg of `U^235`.

Answer 1

Correct Answer - A::B::C
(a) Number of nuclei in 1 kg of `U^235`,
`N=(1/235)(6.02xx10^26)`
`:.` Total energy released
`=(Nxx200)MeV`
`=(1/235)(6.02xx10^26)(200)(1.6xx10^-13)`
`=8.19xx10^13J`
(b) `m=(8.19xx10^13)/(30xx10^3)g`
`=2.73xx10^9g`
`=2.73xx10^6kg`