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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: Assume that immediately after the fission of ${}^{{\mathbf{236}}}{\mathbf{U}}$ according to Eq. 43-1, the resulting ${}^{{\mathbf{140}}}{\mathbf{Xe}}{\mathbf{}}{\mathbf{and}}{\mathbf{}}{}^{{\mathbf{94}}}{\mathbf{Sr}}$ nuclei are just touching at their surfaces. (a) Assuming the nuclei to be spherical, calculate the electric potential energy associated with the repulsion between the two fragments. (Hint: Use Eq. 42-3 to calculate the radii of the fragments.) (b) Compare this energy with the energy released in a typical fission event.

(a) The potential electrical energy is 251 MeV.

(b) The electrical potential energy is greater than typical fission energy.

See the step by step solution

## Step 1: Given data

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

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

The mass number of Uranium, ${A}_{U}=236$

## Step 2: Determine the formulas to calculate the electric potential energy

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

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

Here, A is the mass number.

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

${\mathbit{W}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{r}}$ ...(ii)

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

## Step 3: (a) Calculate the electrical potential energy.

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).

${r}_{Xe}=1.2\mathrm{fm}×{\left(140\right)}^{1/3}\phantom{\rule{0ex}{0ex}}{r}_{Xe}=1.2\mathrm{fm}×5.1924\phantom{\rule{0ex}{0ex}}{r}_{Xe}=6.2\mathrm{fm}$

Calculate the radii of the strontium,

Substitute 1.2 fm for ${r}_{0}$ and 94 for A into equation (i).

${r}_{Sr}=1.2\mathrm{fm}×{\left(94\right)}^{1/3}\phantom{\rule{0ex}{0ex}}{r}_{Sr}=1.2\mathrm{fm}×4.546\phantom{\rule{0ex}{0ex}}{r}_{Sr}=5.48\mathrm{fm}$

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).

$W=\frac{1}{4\pi ×{\epsilon }_{0}}×\frac{{Z}_{Xe}{Z}_{Sr}}{\left({r}_{Xe}+{r}_{Sr}\right)}$

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

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,

$W=\frac{4.033×{10}^{-11}}{1.602×{10}^{-19}}\mathrm{eV}\phantom{\rule{0ex}{0ex}}\mathrm{W}=2.515×{10}^{8}\mathrm{eV}\phantom{\rule{0ex}{0ex}}\mathrm{W}=251.5\mathrm{MeV}$

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

Hence the potential electrical energy is 251 MeV.

## Step 4: (b) Compare the electrical potential energy with typical fission energy.

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. ### Want to see more solutions like these? 