The fission properties of `""_(94)^(239)Pu` are very similar to those of `""_(92)^(235)U`. The average energy released per fission is `180 MeV`. How much energy , in MeV, is released if all the atoms in 1kg of pure `""_(94)^(239)Pu` undergo fission?

Explanation

Number of atoms in `239 g` of `""_(94)^(239)Pu` is Avagadro number.
Number of atoms in `1g = N/(239) = (6.023 xx 10^(23))/(239)`
Number of atoms in 1kg = `(6.023 xx 10^(23))/(239) xx 10^(3) = 2.52 xx 10^(24) "atoms of " ""_(94)^(239)Pu`
Energy released during 1 kg of `""_(94)^(239)Pu` = Energy released by `2.52 xx 10^(24)` atoms of `""_(94)^(239)Pu`
= Energy released per fission `xx` Total number of atoms in 1kg
`= 180 xx 2.52 xx 10^(24) = 4.536 xx 10^(26) MeV`.