Answer
$5.002\;MeV$
Work Step by Step
The fission of of the molybdenum isotope ${}^{98}\mathbf {Mo}$ is given by
${}^{98}\mathbf {Mo}\rightarrow {}^{49}\mathbf {Sc}+{}^{49}\mathbf {Sc}+\mathbf Q$
The mass difference for the reaction is given by
$\Delta m=(2m_{Sc}-m_{MO})$
or, $\Delta m=(2\times48.95002\;u-97.90541\;u)$
or, $\Delta m=(2\times48.95002\;u-97.90541\;u)$
or, $\Delta m=-5.37\times10^{-3}\;u$
Therefore, the disintegration energy is given by
$Q=-\Delta mc^2$
or, $Q=5.37\times10^{-3}\;u \times 931.494013\;MeV/cm$
or, $Q=5.002\;MeV$
