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Expert-verified Found in: Page 1331 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Question: A ${}^{{\mathbf{236}}}{\mathbf{U}}$ nucleus undergoes fission and breaks into two middle-mass fragments, ${}^{{\mathbf{140}}}{\mathbit{X}}{\mathbit{e}}$${\mathbf{and}}{\mathbf{}}{}^{{\mathbf{96}}}{\mathbf{Sr}}$ . (a) By what percentage does the surface area of the fission products differ from that of the original ${}^{{\mathbf{236}}}{\mathbf{U}}$ nucleus? (b) By what percentage does the volume change? (c) By what percentage does the electric potential energy change? The electric potential energy of a uniformly charged sphere of radius r and charge Q is given by ${\mathbit{U}}{\mathbf{=}}\frac{\mathbf{3}}{\mathbf{5}}\left(\frac{{Q}^{2}}{4\pi {\epsilon }_{0}r}\right)$

(a) The percentage change in the surface area is 25.5%.

(b)The percentage change in the volume is 0%.

(c) The percentage change in the electrical potential energy is -35.9%.

See the step by step solution

## Step 1: Write the given data from the question.

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, $U=\frac{3}{5}\left(\frac{{\mathrm{Q}}^{2}}{4{\mathrm{\pi \epsilon }}_{0}\mathrm{r}}\right)$

## Step 2: Determine the formulas:

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

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

Here, A is the mass number.

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

${\mathbit{a}}{\mathbf{=}}{\mathbf{4}}{\mathbit{\pi }}{{\mathbit{r}}}^{{\mathbf{2}}}$ …… (2)

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

${\mathbf{V}}{\mathbf{=}}\frac{\mathbf{4}}{\mathbf{3}}{{\mathbf{\pi r}}}^{{\mathbf{3}}}$ …… (3)

## Step 3: (a) Calculate the percentage change in the surface area of the sphere:

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).

$r=4\pi {r}_{0}{\left({A}^{1/3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=4\pi {r}_{0}{A}^{2/3}$

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.

${a}_{r}=4\pi {r}_{0}\left({A}_{Xe}^{2/3}+{A}_{Sr}^{2/3}\right)$

Calculate the surface area before the fission process.

${a}_{r}=4\pi {r}_{0}{A}_{U}^{2/3}$

Calculate the percentage change in the surface.

$\frac{∆a}{a}=\frac{{A}_{f}-{A}_{i}}{{A}_{i}}$

Substitute $4\pi {r}_{0}\left({A}_{Xe}^{2/3}+{A}_{Sr}^{2/3}\right)$ for ${a}_{f}$ and $4\pi {r}_{0}{A}_{U}^{2/3}$ for ${a}_{i}$ into above equation.

$\begin{array}{r}\frac{\mathrm{\Delta }a}{a}=\frac{4\pi {r}_{0}\left({A}_{{X}_{e}}^{2/\beta }+{A}_{Sr}^{2/\beta }\right)-4\pi {r}_{0}{A}_{U}^{2/3}}{4\pi {r}_{0}{A}_{U}^{2/3}}\\ =\frac{4\pi {r}_{0}\left({A}_{{\chi }_{e}}^{2/\beta }+{A}_{Sr}^{2/\beta }-{A}_{U}^{2/3}\right)}{4\pi {r}_{0}{A}_{U}^{2/\beta }}\\ =\frac{\left({A}_{Xe}^{2/3}+{A}_{sr}^{2/3}-{A}_{U}^{2/\beta }\right)}{{A}_{U}^{2/3}}\end{array}$

Substitute 236 for ${A}_{u}$,140 for ${A}_{Xe}$ and 96 for ${A}_{Sr}$ into above equation.

$\frac{∆a}{a}=\left(\frac{{140}^{2/3}+{96}^{2/3}-{236}^{2/3}}{{236}^{2/3}}\right)\phantom{\rule{0ex}{0ex}}=\frac{26.9619+20.9659-38.1892}{38.1892}\phantom{\rule{0ex}{0ex}}=\frac{9.7386}{38.1892}\phantom{\rule{0ex}{0ex}}=0.255$

$\frac{∆a}{a}%=0.255×100%\phantom{\rule{0ex}{0ex}}=25.5%$

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

## Step 4: (b) Calculate the percentage change in the volume of the sphere:

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

Substitute ${r}_{0}{A}^{1/3}$ for r into equation (3).

$V=\frac{4}{3}\pi {r}_{0}{\left({A}^{1/3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4}{3}\pi {r}_{0}A$

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.

${V}_{r}=\frac{4}{3}\pi {r}_{0}\left({A}_{Xe}+{A}_{Sr}\right)$

Calculate the volume before the fission process.

${V}_{i}=\frac{4}{3}\pi {r}_{0}{A}_{U}$

Calculate the percentage change in the surface.

$\frac{∆V}{V}=\frac{{V}_{f}-{V}_{i}}{{V}_{i}}$

Substitute role="math" localid="1661930807586" $\frac{4}{3}\pi {r}_{0}\left({A}_{Xe}+{A}_{Sr}\right)$ for ${V}_{f}$ and $\frac{4}{3}\pi {r}_{0}{A}_{U}$ for ${V}_{i}$ into above equation.

role="math" localid="1661931138966" $\frac{∆V}{V}=\frac{\frac{4}{3}\pi {r}_{0}\left({A}_{Xe}+{A}_{Sr}\right)-\frac{4}{3}\pi {r}_{0}{A}_{U}}{\frac{4}{3}\pi {r}_{0}{A}_{U}}\phantom{\rule{0ex}{0ex}}=\frac{\frac{4}{3}\pi {r}_{0}\left({A}_{Xe}+{A}_{Sr}-{A}_{U}\right)}{\frac{4}{3}\pi {r}_{0}{A}_{U}}\phantom{\rule{0ex}{0ex}}=\frac{\left({A}_{Xe}+{A}_{Sr}-{A}_{U}\right)}{{A}_{U}}$

Substitute 236 for ${A}_{U}$, 140 for ${A}_{Xe}$ and 96 for ${A}_{Sr}$ into above equation.

$\frac{∆v}{v}=\frac{\left(140+96-236\right)}{236}\phantom{\rule{0ex}{0ex}}=\frac{236-236}{236}\phantom{\rule{0ex}{0ex}}=0$

Hence the percentage change in the volume is 0%.

## Step 5: (c) Calculate the percentage change in electrical potential energy:

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

Therefore,

${Q}_{u}={Z}_{u}e\phantom{\rule{0ex}{0ex}}{Q}_{Xe}={Z}_{Xe}e\phantom{\rule{0ex}{0ex}}Q={Z}_{Sr}e$

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

role="math" localid="1661931639748" $U=\frac{3}{5}\frac{1}{4\pi {\epsilon }_{0}{r}_{0}}\frac{{\left(Ze\right)}^{2}}{{r}_{0}{A}^{1/3}}\phantom{\rule{0ex}{0ex}}=\frac{3}{5}\frac{1}{4\pi {\epsilon }_{0}{r}_{0}}\frac{{\left(Ze\right)}^{2}}{{A}^{1/3}}$

Calculate the electrical potential energy after the fission process.

role="math" localid="1661931904914" $\begin{array}{r}{U}_{f}=\frac{3}{5}\frac{1}{4\pi {\epsilon }_{0}{r}_{0}}\left[\frac{{\left({Z}_{{\chi }_{e}e}\right)}^{2}}{{A}_{{\chi }_{e}}^{1/3}}+\frac{{\left({Z}_{Sr}e\right)}^{2}}{{A}_{Sr}^{1/3}}\right]\\ =\frac{3}{5}\frac{{e}^{2}}{4\pi {\epsilon }_{0}{r}_{0}}\left[\frac{{\left({Z}_{{X}_{e}}\right)}^{2}}{{A}_{{\chi }_{e}}^{1/3}}+\frac{{\left({Z}_{Sr}\right)}^{2}}{{A}_{Sr}^{1/3}}\right]\end{array}$

Calculate the volume before the fission process.

$\begin{array}{r}{U}_{f}=\frac{3}{5}\frac{1}{4\pi {\epsilon }_{0}{r}_{0}}\frac{{\left({Z}_{{}_{U}e}\right)}^{2}}{{A}_{U}^{1/3}}\\ =\frac{3}{5}\frac{{e}^{2}}{4\pi {\epsilon }_{0}{r}_{0}}\frac{{\left({Z}_{{X}_{U}}\right)}^{2}}{{A}_{U}^{1/3}}\end{array}$

Calculate the percentage change in the surface.

$\frac{∆U}{U}=\frac{{U}_{f}-{U}_{i}}{{U}_{i}}$

Substitute $\begin{array}{r}\frac{3}{5}\frac{{e}^{2}}{4\pi {\epsilon }_{0}{r}_{0}}\left[\frac{{\left({Z}_{{\chi }_{e}}\right)}^{2}}{{A}_{{\chi }_{e}}^{1/3}}+\frac{{\left({Z}_{Sr}\right)}^{2}}{{A}_{Sr}^{1/3}}\right]\\ \end{array}$ for ${U}_{f}$ and role="math" localid="1661932204802" $\frac{3}{5}\frac{{e}^{2}}{4\pi {\epsilon }_{0}{r}_{0}}\frac{{\left({Z}_{U}\right)}^{2}}{{A}_{U}^{1/3}}$ for ${U}_{f}$ into above equation.

role="math" localid="1661932416965" $\frac{∆U}{U}=\frac{\frac{3}{5}\frac{{e}^{2}}{4\pi {\epsilon }_{0}{r}_{0}}\left[\frac{{\left({Z}_{{\chi }_{e}}\right)}^{2}}{{A}_{{\chi }_{e}}^{1/3}}+\frac{{\left({Z}_{Sr}\right)}^{2}}{{A}_{Sr}^{1/3}}\right]-\frac{3}{5}\frac{{e}^{2}}{4\pi {\epsilon }_{0}{r}_{0}}\frac{{\left({Z}_{U}\right)}^{2}}{{A}_{U}^{1/3}}}{\frac{3}{5}\frac{{e}^{2}}{4\pi {\epsilon }_{0}{r}_{0}}\frac{{\left({Z}_{U}\right)}^{2}}{{A}_{U}^{1/3}}}\phantom{\rule{0ex}{0ex}}\frac{∆U}{U}=\frac{\frac{3}{5}\frac{{e}^{2}}{4\pi {\epsilon }_{0}{r}_{0}}\left[\frac{{\left({Z}_{{\chi }_{e}}\right)}^{2}}{{A}_{{\chi }_{e}}^{1/3}}+\frac{{\left({Z}_{Sr}\right)}^{2}}{{A}_{Sr}^{1/3}}-\frac{{\left({Z}_{U}\right)}^{2}}{{A}_{U}^{1/3}}\right]}{\frac{3}{5}\frac{{e}^{2}}{4\pi {\epsilon }_{0}{r}_{0}}\frac{{\left({Z}_{U}\right)}^{2}}{{A}_{U}^{1/3}}}\phantom{\rule{0ex}{0ex}}=\frac{\frac{{\left({Z}_{{\chi }_{e}}\right)}^{2}}{{A}_{{\chi }_{e}}^{1/3}}+\frac{{\left({Z}_{Sr}\right)}^{2}}{{A}_{Sr}^{1/3}}-\frac{{\left({Z}_{U}\right)}^{2}}{{A}_{U}^{1/3}}}{\frac{{\left({Z}_{U}\right)}^{2}}{{A}_{U}^{1/3}}}$

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.

$\frac{∆U}{U}=\frac{\frac{{\left(54\right)}^{2}}{{140}^{1/3}}+\frac{{\left(38\right)}^{2}}{{96}^{1/3}}+\frac{{\left(92\right)}^{2}}{{236}^{1/3}}}{\frac{{\left(92\right)}^{2}}{{236}^{1/3}}}\phantom{\rule{0ex}{0ex}}=\frac{561.5798+315.3625+1369.6354}{1369.6354}\phantom{\rule{0ex}{0ex}}=\frac{-492.6931}{1369.6354}\phantom{\rule{0ex}{0ex}}=-0.359\phantom{\rule{0ex}{0ex}}\frac{∆U}{U}%=-0.359×100%\phantom{\rule{0ex}{0ex}}=-35.9%$

Hence, the percentage change in the electrical potential energy is -35.9%. ### Want to see more solutions like these? 