Question
(a) Calculate the energy released in the neutron-induced fission reaction
$n + {}^{239}\textrm{Pu} \to {}^{96}\textrm{Sr} + {}^{140}\textrm{Ba} + 4n$
given that $m({}^{96}\textrm{Sr} = 95.921750 \textrm{ u}$ and $m({}^{140}\textrm{Ba} = 139.910581 \textrm{ u}$. (b) Confirm that the total number of nucleons and total charge are conserved in this reaction. Note: At 00:39 I mis-spoke with "strontium-36" when I should have said "strontium-96".
1. $180.6 \textrm{ MeV}$
2. Please see the solution video.
Solution Video