Consider the following.

$n+{?}^{235}U{?}^{92}Kr+{?}^{140}Ba+4n$

$($a$)$ Calculate the energy $($in MeV $)$ released in the neutron$-$induced fission reaction above, given $m{(}^{92}Kr)=91\mathrm{.}926153u$ and $m{(}^{140}Ba)=139\mathrm{.}910599u.($Assume $1u=931\mathrm{.}5Me\frac{V}{c^2}.)$

$|$ MeV

$($b$)$ Confirm that the total number of nucleons and total charge are conserved in this reaction. $($The electron family number is abreviated "efn".$)$

$A_i=A_f=$

$ef{n}_i=ef{n}_f=$