Answer
$^{1}_{0}n+^{235}_{92}U\rightarrow ^{88}_{38}Sr+^{136}_{54}Xe+12^{1}_{0}n;126.5MeV$
Work Step by Step
Firs of all, we complete the given reaction as follows:
Number of neutrons in $^{235}_{92}U=235-92=143$
Number of neutrons in $^{88}_{38}Sr=88-38=50$
Number of neutrons in $^{136}_{54}Xe=136-54=82$
The number of neutrons in the system before reaction $143+1=144$
and the number of neutrons in the system after reaction $50+82+x=132+x$
We know that
$144=132+x$
$\implies x=12$
Thus, the required reaction is $^{1}_{0}n+^{235}_{92}U\rightarrow ^{88}_{38}Sr+^{136}_{54}Xe+12^{1}_{0}n$
Now $\Delta m=m_f-m_i$
$\implies \Delta m=235.916825u-236.05259u=-0.135765u$
The required energy can be calculated as
$E=|\Delta m|c^2$
We plug in the known values to obtain:
$E=|-(0.135765u)(\frac{931.494\frac{MeV}{c^2}}{1u})|c^2$
$E=126.5MeV$
