Chapter 44, Problem 77CP

To determine

To show that the electrical electrostatic potential energy of the nucleus is $\frac{3Z^2{e}^2}{20\pi {\epsilon }_{0}R}=\frac{3k_e{Z}^2{e}^2}{5R}$.

Answer to Problem 77CP

The electric electrostatic potential energy is $\frac{3Z^2{e}^2}{20\pi {\epsilon }_{0}R}=\frac{3k_e{Z}^2{e}^2}{5R}$.

Explanation of Solution

Write the expression for the electric field inside the nucleus (small sphere of radius R).

  $E_{\text{in}}=\frac{\rho r}{3{\epsilon }_0}$

Here, $E_{\text{in}}$ is the electric field inside the nucleus and $\rho$ is the charge density.

Write the expression to calculate charge density.

  $\rho =\frac{Ze}{\frac{4}{3}\pi R^3}$

Here, Z is the atomic number, e is the electric charge and R is the radius of the nucleus.

Substitute the above equation in the expression for $E_{\text{in}}$ to rewrite.

  $\begin{array}{c}{E}_{\text{in}}=\frac{r}{3{\epsilon }_0}\left(\frac{Ze}{\frac{4}{3}\pi R^3}\right)\\ =\frac{Zer}{4{\epsilon }_0\pi R^3}\end{array}$                                                                                                    (I)

Write the expression to calculate the electric field outside the nucleus.

  $E_{\text{out}}=\frac{Ze}{4\pi {\epsilon }_0{r}^2}$                                                                                                           (II)

Here, $E_{\text{out}}$ is the electric field outside the nucleus.

Write the expression to calculate the electrostatic potential energy of the nucleus.

  $U={\displaystyle {\int }_0^R\frac{1}{2}{\epsilon }_0{\left(E_{\text{in}}\right)}^{2}d\tau }+{\displaystyle {\int }_R^{\infty }\frac{1}{2}{\epsilon }_0{\left(E_{\text{out}}\right)}^{2}d\tau }$                                                               (III)

Here, U is the electrostatic potential energy and $d\tau$ is the elemental volume of the sphere.

Write the expression for $d\tau$.

  $d\tau =4\pi r^{2}dr$                                                                                                           (IV)

Substitute the equations (I), (II) and (IV) in (III) to calculate U.

  $\begin{array}{c}U={\displaystyle {\int }_0^R\frac{1}{2}{\epsilon }_0{\left(\frac{Zer}{4{\epsilon }_0\pi R^3}\right)}^2\left(4\pi r^{2}dr\right)}+{\displaystyle {\int }_R^{\infty }\frac{1}{2}{\epsilon }_0{\left(\frac{Ze}{4\pi {\epsilon }_0{r}^2}\right)}^2\left(4\pi r^{2}dr\right)}\\ =\frac{1}{8\pi {\epsilon }_0}\left[{\displaystyle {\int }_0^R\frac{Z^2{e}^2}{R^6}\left(r^{4}dr\right)}+{\displaystyle {\int }_R^{\infty }{Z}^2{e}^2\left(\frac{1}{r^2}dr\right)}\right]\\ =\frac{1}{8\pi {\epsilon }_0}\left[\left(\frac{Z^2{e}^2}{R^6}\right){\left(\frac{r^5}{5}\right)}_0^R+Z^2{e}^2{\left(\frac{-1}{r}\right)}_R^{\infty }\right]\\ =\frac{1}{8\pi {\epsilon }_0}\left[\left(\frac{Z^2{e}^2}{R^6}\right)\left(\frac{R^5}{5}-0\right)+Z^2{e}^2\left(\frac{-1}{\infty }+\frac{1}{R}\right)\right]\end{array}$

Reduce the above equation to calculate U.

  $\begin{array}{c}U=\frac{Z^2{e}^2}{8\pi {\epsilon }_0}\left[\left(\frac{1}{5R}\right)+\left(\frac{1}{R}\right)\right]\\ =\frac{Z^2{e}^2}{8\pi {\epsilon }_0}\left(\frac{6}{5R}\right)\\ =\frac{Z^2{e}^2}{4\pi {\epsilon }_0}\left(\frac{3}{5R}\right)\\ =\frac{3Z^2{e}^2}{20\pi {\epsilon }_{0}R}\\ =\frac{1}{4\pi {\epsilon }_0}\left(\frac{3Z^2{e}^2}{5R}\right)\end{array}$

Write the expression for the coulomb constant.

  $k_e=\frac{1}{4\pi {\epsilon }_0}$

Substitute the above equation in the expression for U to rewrite.

  $\begin{array}{c}U=k_e\left(\frac{3Z^2{e}^2}{5R}\right)\\ =\left(\frac{3k_e{Z}^2{e}^2}{5R}\right)\end{array}$

Conclusion:

Therefore, the electric electrostatic potential energy is $\frac{3Z^2{e}^2}{20\pi {\epsilon }_{0}R}=\frac{3k_e{Z}^2{e}^2}{5R}$.