91Ó°ÊÓ

The Uranium ${}^{236}\mathrm{U}$fission ${}^{140}Xe$and ${}^{96}Sr$breaks into two fragments and .

The radius of the sphere is r.

The charge on the sphere is Q.

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 electric potential of the uniformly charged sphere, $U=\frac{3}{5}\left(\frac{{\mathrm{Q}}^2}{4{\mathrm{Ï€Î\mu }}_0\mathrm{r}}\right)$

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

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

Here, A is the mass number.

The expression for the surface area of the sphere is given as follows.

$\mathit{a}\mathbf{=}\mathbf{4}\mathit{π}{\mathit{r}}^{\mathbf{2}}$ …… (2)

The expression to calculate the volume of the sphere is given as follows.

$\mathbf{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{Ï€°ù}}^{\mathbf{3}}$ …… (3)

Derive the expression for the surface area in terms of the mass number.

Substitute $r=r_0{A}^{1/3}$for r into equation (2).

$$\begin{gathered} r=4\mathrm{Ï€}{r}_0{\left(A^{1/3}\right)}^2 \\ =4\mathrm{Ï€}{r}_0{A}^{2/3} \end{gathered}$$

The surface area after the fission process would be the sum of the surface area of the Xenon and Strontium.

Calculate the surface area after the fission process.

$a_r=4\mathrm{Ï€}{r}_0\left(A_{Xe}^{2/3}+A_{Sr}^{2/3}\right)$

Calculate the surface area before the fission process.

$a_r=4\mathrm{Ï€}{r}_0{A}_U^{2/3}$

Calculate the percentage change in the surface.

$\frac{∆a}{a}=\frac{A_f-A_i}{A_i}$

Substitute $4\mathrm{Ï€}{r}_0\left(A_{Xe}^{2/3}+A_{Sr}^{2/3}\right)$for $a_f$and $4\mathrm{Ï€}{r}_0{A}_U^{2/3}$for $a_i$into above equation.

$\begin{array}{r}\frac{\mathrm{Δ}a}{a}=\frac{4\mathrm{π}{r}_0\left(A_{X_e}^{2/\mathrm{β}}+A_{Sr}^{2/\mathrm{β}}\right)−4\mathrm{π}{r}_0{A}_U^{2/3}}{4\mathrm{π}{r}_0{A}_U^{2/3}}\\ =\frac{4\mathrm{π}{r}_0\left(A_{{\mathrm{χ}}_e}^{2/\mathrm{β}}+A_{Sr}^{2/\mathrm{β}}−A_U^{2/3}\right)}{4\mathrm{π}{r}_0{A}_U^{2/\mathrm{β}}}\\ =\frac{\left(A_{Xe}^{2/3}+A_{sr}^{2/3}−A_U^{2/\mathrm{β}}\right)}{A_U^{2/3}}\end{array}$

Substitute 236 for $A_u$,140 for $A_{Xe}$and 96 for $A_{Sr}$into above equation.

$$\begin{gathered} \frac{∆a}{a}=\left(\frac{{140}^{2/3}+{96}^{2/3}-{236}^{2/3}}{{236}^{2/3}}\right) \\ =\frac{26.9619+20.9659-38.1892}{38.1892} \\ =\frac{9.7386}{38.1892} \\ =0.255 \end{gathered}$$

$$\begin{gathered} \frac{∆a}{a}\%=0.255×100\% \\ =25.5\% \end{gathered}$$

Hence, the percentage change in the surface area is 25.5%.

Derive the expression for the volume in terms of the mass number.

Substitute $r_0{A}^{1/3}$for r into equation (3).

$$\begin{gathered} V=\frac{4}{3}\mathrm{Ï€}{r}_0{\left(A^{1/3}\right)}^2 \\ =\frac{4}{3}\mathrm{Ï€}{r}_{0}A \end{gathered}$$

The volume after the fission process would be the sum of the volume of the Xenon and Strontium.

Calculate the volume after the fission process.

$V_r=\frac{4}{3}\mathrm{Ï€}{r}_0\left(A_{Xe}+A_{Sr}\right)$

Calculate the volume before the fission process.

$V_i=\frac{4}{3}\mathrm{Ï€}{r}_0{A}_U$

Calculate the percentage change in the surface.

$\frac{∆V}{V}=\frac{V_f-V_i}{V_i}$

Substitute role="math" localid="1661930807586" $\frac{4}{3}\mathrm{Ï€}{r}_0\left(A_{Xe}+A_{Sr}\right)$for $V_f$and $\frac{4}{3}\mathrm{Ï€}{r}_0{A}_U$for $V_i$into above equation.

role="math" localid="1661931138966" $$\begin{gathered} \frac{∆V}{V}=\frac{\frac{4}{3}\mathrm{π}{r}_0\left(A_{Xe}+A_{Sr}\right)-\frac{4}{3}\mathrm{π}{r}_0{A}_U}{\frac{4}{3}\mathrm{π}{r}_0{A}_U} \\ =\frac{\frac{4}{3}\mathrm{π}{r}_0\left(A_{Xe}+A_{Sr}-A_U\right)}{\frac{4}{3}\mathrm{π}{r}_0{A}_U} \\ =\frac{\left(A_{Xe}+A_{Sr}-A_U\right)}{A_U} \end{gathered}$$

Substitute 236 for $A_U$, 140 for $A_{Xe}$and 96 for $A_{Sr}$into above equation.

$$\begin{gathered} \frac{∆v}{v}=\frac{\left(140+96-236\right)}{236} \\ =\frac{236-236}{236} \\ =0 \end{gathered}$$

Hence the percentage change in the volume is 0%.

The charges are equal to the product of the atomic number and electron charge.

Therefore,

$$\begin{gathered} Q_u=Z_{u}e \\ {Q}_{Xe}=Z_{Xe}e \\ Q=Z_{Sr}e \end{gathered}$$

Derive the expression for electrical potential energy in terms of the mass number.

role="math" localid="1661931639748" $$\begin{gathered} U=\frac{3}{5}\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}_0}\frac{{\left(Ze\right)}^2}{r_0{A}^{1/3}} \\ =\frac{3}{5}\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}_0}\frac{{\left(Ze\right)}^2}{A^{1/3}} \end{gathered}$$

Calculate the electrical potential energy after the fission process.

role="math" localid="1661931904914" $\begin{array}{r}{U}_f=\frac{3}{5}\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}_0}\left[\frac{{\left(Z_{{\mathrm{χ}}_{e}e}\right)}^2}{A_{{\mathrm{χ}}_e}^{1/3}}+\frac{{\left(Z_{Sr}e\right)}^2}{A_{Sr}^{1/3}}\right]\\ =\frac{3}{5}\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}_0}\left[\frac{{\left(Z_{X_e}\right)}^2}{A_{{\mathrm{χ}}_e}^{1/3}}+\frac{{\left(Z_{Sr}\right)}^2}{A_{Sr}^{1/3}}\right]\end{array}$

Calculate the volume before the fission process.

$\begin{array}{r}{U}_f=\frac{3}{5}\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}_0}\frac{{\left(Z_{{}_{U}e}\right)}^2}{A_U^{1/3}}\\ =\frac{3}{5}\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}_0}\frac{{\left(Z_{X_U}\right)}^2}{A_U^{1/3}}\end{array}$

Calculate the percentage change in the surface.

$\frac{∆U}{U}=\frac{U_f-U_i}{U_i}$

Substitute $\begin{array}{r}\frac{3}{5}\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}_0}\left[\frac{{\left(Z_{{\mathrm{χ}}_e}\right)}^2}{A_{{\mathrm{χ}}_e}^{1/3}}+\frac{{\left(Z_{Sr}\right)}^2}{A_{Sr}^{1/3}}\right]\\ \end{array}$for $U_f$ and role="math" localid="1661932204802" $\frac{3}{5}\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}_0}\frac{{\left(Z_U\right)}^2}{A_U^{1/3}}$for $U_f$into above equation.

role="math" localid="1661932416965" $$\begin{gathered} \frac{∆U}{U}=\frac{\frac{3}{5}\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}_0}\left[\frac{{\left(Z_{{\mathrm{χ}}_e}\right)}^2}{A_{{\mathrm{χ}}_e}^{1/3}}+\frac{{\left(Z_{Sr}\right)}^2}{A_{Sr}^{1/3}}\right]-\frac{3}{5}\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}_0}\frac{{\left(Z_U\right)}^2}{A_U^{1/3}}}{\frac{3}{5}\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}_0}\frac{{\left(Z_U\right)}^2}{A_U^{1/3}}} \\ \frac{∆U}{U}=\frac{\frac{3}{5}\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}_0}\left[\frac{{\left(Z_{{\mathrm{χ}}_e}\right)}^2}{A_{{\mathrm{χ}}_e}^{1/3}}+\frac{{\left(Z_{Sr}\right)}^2}{A_{Sr}^{1/3}}-\frac{{\left(Z_U\right)}^2}{A_U^{1/3}}\right]}{\frac{3}{5}\frac{e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}_0}\frac{{\left(Z_U\right)}^2}{A_U^{1/3}}} \\ =\frac{\frac{{\left(Z_{{\mathrm{χ}}_e}\right)}^2}{A_{{\mathrm{χ}}_e}^{1/3}}+\frac{{\left(Z_{Sr}\right)}^2}{A_{Sr}^{1/3}}-\frac{{\left(Z_U\right)}^2}{A_U^{1/3}}}{\frac{{\left(Z_U\right)}^2}{A_U^{1/3}}} \end{gathered}$$

Substitute 236 for $A_U$, 140 for $A_{Xe}$, 96 for $A_{Sr}$,54 for $Z_{Xe}$,38 for $Z_{Sr}$and 92 for $Z_U$into above equation.

$$\begin{gathered} \frac{∆U}{U}=\frac{\frac{{\left(54\right)}^2}{{140}^{1/3}}+\frac{{\left(38\right)}^2}{{96}^{1/3}}+\frac{{\left(92\right)}^2}{{236}^{1/3}}}{\frac{{\left(92\right)}^2}{{236}^{1/3}}} \\ =\frac{561.5798+315.3625+1369.6354}{1369.6354} \\ =\frac{-492.6931}{1369.6354} \\ =-0.359 \\ \frac{∆U}{U}\%=-0.359×100\% \\ =-35.9\% \end{gathered}$$

Hence, the percentage change in the electrical potential energy is -35.9%.