The fission properties of `._84Pu^(239)` are very similar to those of `._92U^(235)`. The average energy released per fission is `180MeV`. How much energy in MeV is released if all the atoms in 1kg of pure `._94Pu^(239)` undergo fission.

Explanation

Average energy released per fission = 180 MeV
Number of atoms in 239 g of `""_(94)Pu^(239) = 6 xx 10^(23)`
Number of atoms in 1 kg = 1,000 g of `""_(94)Pu^(239)`
`=(6.023 xx 10^(23) xx 1,000)/239 = 0.025 xx 10^(26)`
`therefore` Energy released from the fission of 1 kg of `""_(94)Pu^(239)`
`E = 0.025 xx 10^(26) xx 180`
`=4,536 xx 10^(26)` MeV.