(a) Calculate the disintegration energy Q for the fission of the molybdenum isotope ${}^{\mathit{98}}\mathit{M}\mathit{o}$into two equal parts. The masses you will need are 97.90541u for${}^{\mathbf{98}}\mathit{M}\mathit{o}$ and 48.95002u for ${}^{\mathbf{49}}\mathit{S}\mathit{c}$. (b) If Q turns out to be positive, discuss why this process does not occur spontaneously.

The mass of ${}^{98}Mo,m_{Mo}=97.90541u$,

The mass of ${}^{49}Sc,m_{SC}=48.95002u$,

The expression to calculate the disintegrated energy is given as follows.

$$\begin{gathered} \mathit{Q}\mathbf{=}\mathbf{-}\mathbf{∆}\mathit{m}{\mathit{c}}^{\mathbf{2}} \\ \mathit{Q}\mathbf{=}\left(m_{Mo}-2m_{SC}\right){\mathit{c}}^{\mathbf{2}} \end{gathered}$$ ...(i)

Consider the fission reaction as given follow.

${}^{98}Mo\to {}^{49}Sc+{}^{49}Sc$

Calculate the disintegrated energy.

Substitute 97.90514u for $m_{Mo}$,48.95002u for$m_{Sc}$ and 931.5 MeV/u for$c^2$ into equation (i).

$$\begin{gathered} \mathrm{Q}=(97.90541\mathrm{u}-2\times 48.95002\mathrm{u})931.5\mathrm{MeV}/\mathrm{u} \\ \mathrm{Q}=0.00537\times 931.5\mathrm{MeV} \\ \mathrm{Q}=5\hspace{0.33em}\mathrm{MeV} \end{gathered}$$

Hence the disintegrated energy is 5 MeV.

From the part (a), the disintegrated energy is +5Mev. The nucleus must overcome the barrier of the energy which is larger than the +5MeV. Therefore, the positive sign of disintegrated energy does not indicate that the process is spontaneously. Therefore, due to the energy barrier the process is not spontaneously.