Question

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.

Answer 1

Average energy relased per fission of ` ""_(94)^(239) Pu ` ,`E_(av)=180 MeV`
Amount of pure `""_(94) Pu^(239)` ,m=1 kg =1000 g
`N_(A)` = Avogdro number ` = 6.023 xx10^(23)`
Mass number of `""_(94)^(239) Pu =239 g`
1 mole of `""_(94)Pu^(239 ) ` Contains `N_(A)` atoms ,
` therefore m g ""_(94)Pu ^(239)` Contins `((N_(A))/("Mass number ") xxm)`atoms
`=(6.023 xx10^(23))/(239) xx1000 =2.52 xx10^(24)` atoms
`therefore` total energy during the fistion of 1 kg of `""_(94^(239)Pu` is calculated as :
`E=E_(av)xx2.52 xx10^(24)`
`=180xx2.52 xx10^(24)=4.536 xx10^(26)MeV`
Hence ,`4.536xx10^(26)MeV` is relased if all the atoms in 1 kg oure `""_(94) Pu ^(239)` undergo fissioon,