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Question
(a) Calculate the energy released in the neutron-induced fission reaction

$n + {}^{235}\textrm{U} \rightarrow {}^{92}\textrm{Kr} + {}^{142}\textrm{Ba} + 2n$

given $m({}^{92}\textrm{Kr}) = 91.926269\textrm{ u}$ and $m({}^{142}\textrm{Ba}) = 141.916361\textrm{ u}$. (b) Confirm that the total number of nucleons and total charge are conserved in this reaction.
1. $179.4\textrm{ MeV}$
2. Please see the solution video.
Solution Video