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 1kg of pure ""_(94)^(239)Pu undergo fission? |
|
Answer» Solution :Number of atoms in `239 g` of `""_(94)^(239)PU` is Avagadro number. Number of atoms in `1G = N/(239) = (6.023 xx 10^(23))/(239)` Number of atoms in 1KG = `(6.023 xx 10^(23))/(239) xx 10^(3) = 2.52 xx 10^(24) "atoms of " ""_(94)^(239)Pu` Energy released during 1 kg of `""_(94)^(239)Pu` = Energy released by `2.52 xx 10^(24)` atoms of `""_(94)^(239)Pu` = Energy released per FISSION `xx` Total number of atoms in 1kg `= 180 xx 2.52 xx 10^(24) = 4.536 xx 10^(26) MeV`. |
|