91Ó°ÊÓ

The mass number of Xenon,$A_{Xe}=140$

The mass number of Strontium, $A_{Sr}=94$

The mass number of Uranium,$A_U=236$

The expression to calculate the radii of the fragment is given as follows.

$\mathit{r}\mathbf{=}{\mathit{r}}_{\mathbf{0}}{\mathit{A}}^{\mathbf{1}\mathbf{/}\mathbf{3}}$ ...(i)

Here, A is the mass number.

The expression to calculate the electrical potential energy is given as follows.

$\mathit{W}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{Ï€}{\mathbf{Î\mu }}_{\mathbf{0}}}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{r}}$ ...(ii)

Here, ${\mathbf{q}}_{\mathbf{1}}\mathbf{.}{\mathbf{q}}_{\mathbf{2}}$are the charges and r is the distance between the charges.

The atomic number of Strontium,$Z_{Sr}=38$

The atomic number of Xenon,$Z_{Xe}=54$

Calculate the radii of the Xenon,

Substitute 1.2 fm for $r_0$and 140 for A into equation (i).

$$\begin{gathered} r_{Xe}=1.2\mathrm{fm}×{\left(140\right)}^{1/3} \\ {r}_{Xe}=1.2\mathrm{fm}×5.1924 \\ {r}_{Xe}=6.2\mathrm{fm} \end{gathered}$$

Calculate the radii of the strontium,

Substitute 1.2 fm for$r_0$and 94 for A into equation (i).

$$\begin{gathered} r_{Sr}=1.2\mathrm{fm}×{\left(94\right)}^{1/3} \\ {r}_{Sr}=1.2\mathrm{fm}×4.546 \\ {r}_{Sr}=5.48\mathrm{fm} \end{gathered}$$

Calculate the electrical potential energy.

Substitute$Z_{Xe}$for $q_1$,$Z_{Sr}$for $q_2$,$r_{Sr}+r_{Xe}$for r into above equation (i).

$W=\frac{1}{4\mathrm{Ï€}×{\mathrm{Î\mu }}_0}×\frac{Z_{Xe}{Z}_{Sr}}{\left(r_{Xe}+r_{Sr}\right)}$

Substitute 6.24 for $r_{Xe}$, 5.48 for $r_{Sr}$, 54e for$Z_{Xe}$and 38e for$Z_{Xe}$into above equation.

localid="1661927633337" $\begin{array}{r}W=\frac{1}{4\mathrm{π}×\left(8.85×{10}^{−12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right)}×\frac{54\mathrm{e}×38\mathrm{e}}{(6.24+5.48)×{10}^{−15}\mathrm{m}}\\ W=8.99×{10}^9\frac{{\mathrm{Nm}}^2}{{\mathrm{C}}^2}×\frac{2052{\mathrm{e}}^2}{11.72×{10}^{−15}\mathrm{m}}\end{array}$

Substitute$1.6×{10}^{−19}\mathrm{C}$for e into above equation.

$\begin{array}{r}W=8.99×{10}^9\frac{{\mathrm{Nm}}^2}{{\mathrm{C}}^2}×\frac{2052{\left(1.6×{10}^{−19}\mathrm{C}\right)}^2}{11.72×{10}^{−15}\mathrm{m}}\\ W=\frac{18447.48×2.56×{10}^{24}×{10}^{−38}\mathrm{J}}{11.72}\\ W=\frac{47225.54×{10}^{−14}\mathrm{J}}{11.72}\\ W=4029×{10}^{−11}\mathrm{J}\end{array}$

Convert the energy into MeV,

$$\begin{gathered} W=\frac{4.033×{10}^{−11}}{1.602×{10}^{−19}}\mathrm{eV} \\ \mathrm{W}=2.515×{10}^8\mathrm{eV} \\ \mathrm{W}=251.5\mathrm{MeV} \end{gathered}$$

By rounding down the value of the energy is 251 MeV.

Hence the potential electrical energy is 251 MeV.

The energy released in the typical fission energy is 200 MeV and the electrical potential energy from the part (a) is 251 MeV . By comparing the two energies, it is clear that the electrical potential energy is greater than the typical fission energy. This energy appears in the form of kinetic, beat and sound energy.

Hence the electrical potential energy is greater than typical fission energy.