Saved Bookmarks
| 1. |
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» Solution :Number of atoms in 1 kg of `" "_(94)^(239)PU = N_(A)/A =(6.023xx10^(23))/0.239 =2.52 xx10^(24)` Energy RELEASED = Number of atoms `xx` Energy released PER fission` = (2.52 xx 10^(24)) xx (180) MeV = 4.536 xx 10^(26) MeV`. |
|