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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» 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` |
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