Question

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 1 kg of pure `""_94^239 "Pu"` undergo fission?

Answer 1

Average energy released per fission of `""_94^239 "Pu", "E"_"av" = 180 "MeV"`

Amount of pure `""_94"Pu"^239` m = 1 kg = 1000 g

N_A= Avogadro number = 6.023 × 10²³

Mass number of `""_94^239 "Pu"`= 239 g

1 mole of `""_94"Pu"^239`  contains N_A atoms.

∴ mg of `""_94"Pu"^239` contain `("N"_"A"/"Mass number" xx m)` atom

`= (6.023 xx 10^23)/239 xx 1000  = 2.52 xx 10^24` atoms

∴ Total energy released during the fission of 1 kg of `""_94^239"Pu"` is calculated as:

`"E" = "E"_"av" xx 2.52 xx 10^24`

`= 180 xx 2.52 xx 10^24 = 4.536 xx 10^26` MeV

Hence, `4.536 xx 10^26` is released if all the atoms in 1 kg of pure `""_94"Pu"^239` undergo fission.