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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?
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Solution
Average energy released per fission of `""_94^239 "Pu", "E"_"av" = 180 "MeV"`
Amount of pure `""_94"Pu"^239` m = 1 kg = 1000 g
NA= Avogadro number = 6.023 × 1023
Mass number of `""_94^239 "Pu"`= 239 g
1 mole of `""_94"Pu"^239` contains NA 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.
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