Chapter 24, Problem 62CP

(a)

To determine

The proof that the electron would be in equilibrium at the centre and if displaced from the centre with distance $r<R$ would experience the restoring force of $-Kr$.

Answer to Problem 62CP

The electron is in equilibrium at the centre of the Gaussian sphere and when the electron is at $r<R$, it would experience the restoring force of $-Kr$.

Explanation of Solution

Write the expression to obtain the charge density.

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

Here, $\rho$ is the charge density, $Z$ is the atomic number, $e$ is the charge on the electron, $R$ is the atomic radius and $\frac{4}{3}\pi R^3$ is the volume of the sphere.

Substitute $1$ for $Z$ in case of hydrogen atom in the above equation.

    $\begin{array}{c}\rho =\frac{\left(1\right)e}{\frac{4}{3}\pi R^3}\\ =\frac{e}{\frac{4}{3}\pi R^3}\end{array}$

Write the expression for the charge enclosed in the Gaussian sphere.

    $q=\rho \left(\frac{4}{3}\pi r^3\right)$

Here, $q$ is the charge enclosed in the Gaussian sphere, $\rho$ is the charge density in the Gaussian sphere, $r$ is the radius of the Gaussian sphere and $\left(\frac{4}{3}\pi r^3\right)$ is the volume of the Gaussian sphere.

Substitute, $\frac{e}{\frac{4}{3}\pi R^3}$ for $\rho$ in the above equation to calculate $q$.

    $\begin{array}{c}q=\left(\frac{e}{\frac{4}{3}\pi R^3}\right)\left(\frac{4}{3}\pi r^3\right)\\ =\frac{e{r}^3}{R^3}\end{array}$

Write the expression on the basis of Gauss law.

    ${\displaystyle \oint \overrightarrow{E}}d\overrightarrow{s}=\frac{q}{{\epsilon }_0}$

Re-write the above equation.

    $E{\displaystyle \oint d\overrightarrow{s}}=\frac{q}{{\epsilon }_0}$

Here, $E$ is the electric field in the Gaussian surface, $d\overrightarrow{s}$ is the small elemental area of the Gaussian surface, $q$ is the charged enclosed in the Gaussian surface and ${\epsilon }_0$ is the dielectric constant.

Substitute $4\pi r^2$ for ${\displaystyle \oint d\overrightarrow{s}}$ and $\frac{e{r}^3}{R^3}$ for $q$ in the above equation to find $E$.

    $\begin{array}{c}E\left(4\pi r^2\right)=\frac{\frac{e{r}^3}{R^3}}{{\epsilon }_0}\\ E=\frac{1}{4\pi {\epsilon }_0}\frac{e}{R^3}r\end{array}$

Write the expression to obtain the force in the Gaussian surface.

    $F=eE$

Here, $F$ is the force in the Gaussian surface.

Substitute $\frac{1}{4\pi {\epsilon }_0}\frac{e}{R^3}r$ for $E$ in the above equation.

    $\begin{array}{c}F=e\left(\frac{1}{4\pi {\epsilon }_0}\frac{e}{R^3}r\right)\\ =\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{R^3}r\end{array}$

Substitute $k_{\text{e}}$ for $\frac{1}{4\pi {\epsilon }_0}$ in the above equation.

    $F=k_{\text{e}}\frac{e^2}{R^3}r$                                                                                                         (I)

Here, $k_{\text{e}}$ is the electric constant.

Further substitute, $K$ for $k_{\text{e}}\frac{e^2}{R^3}$ in equation (I)..

    $F=Kr$                                                                                                              (II)

Substitute, $0$ for $r$ in equation (II) to find $F$.

    $\begin{array}{c}F=K\left(0\right)\\ =0\end{array}$

Therefore, the electron is in equilibrium at the centre of the Gaussian sphere and when the electron is at $r<R$, it would experience the restoring force of $-Kr$.

(b)

To determine

The proof that the value of $K$ is $k_{\text{e}}\frac{e^2}{R^3}$.

Answer to Problem 62CP

The value of $K$ is $k_{\text{e}}\frac{e^2}{R^3}$.

Explanation of Solution

Compare equation (I) and (II).

    $\begin{array}{c}Kr=k_{\text{e}}\frac{e^2}{R^3}r\\ K=k_{\text{e}}\frac{e^2}{R^3}\end{array}$

Therefore, the value of $K$ is $k_{\text{e}}\frac{e^2}{R^3}$.

(c)

To determine

The expression for the frequency of a simple harmonic oscillator.

Answer to Problem 62CP

The frequency of a simple harmonic oscillator is $\frac{1}{2\pi }\sqrt{\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{m{R}^3}}$.

Explanation of Solution

Write the expression of force based on Newton’s $2^{\text{nd}}$ law of motion.

    $F=ma$

Here, $F$ is the force, $m$ is the mass of the electron and $a$ is the centripetal acceleration of electron.

Substitute, $\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{R^3}r$ for $F$ in the above equation to find $a$.

    $\begin{array}{c}\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{R^3}r=ma\\ a=\frac{\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{R^3}r}{m}\\ a=\left(\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{m{R}^3}\right)r\end{array}$                                                                                   (III)

Write the expression to simple harmonic wave equation.

    $a=-{\omega }^{2}r$                                                                                                              (IV)

Here, $\omega$ is the angular frequency.

Compare equation (III) and (IV).

    $\begin{array}{l}{\omega }^2=\left(\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{m{R}^3}\right)\\ \omega =\sqrt{\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{m{R}^3}}\end{array}$

Write the expression for the frequency of the simple harmonic motion.

    $f=\frac{\omega }{2\pi }$

Here, $f$ is the frequency of the simple harmonic motion.

Substitute, $\sqrt{\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{m{R}^3}}$ for $\omega$ in the above equation to find $f$.

    $f=\frac{1}{2\pi }\sqrt{\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{m{R}^3}}$                                                                                                 (V)

Therefore, the frequency of the simple harmonic motion is $\frac{1}{2\pi }\sqrt{\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{m{R}^3}}$.

(d)

To determine

The radius of the orbit.

Answer to Problem 62CP

The radius of the orbit is $1.02\stackrel{0}{\text{A}}$.

Explanation of Solution

Re-write the equation (V).

    $f=\frac{1}{2\pi }\sqrt{\frac{1}{4\pi {\epsilon }_0}\frac{e^2}{m{R}^3}}$

Substitute $k_{\text{e}}$ for $\frac{1}{4\pi {\epsilon }_0}$ in the above equation.

    $f=\frac{1}{2\pi }\sqrt{k_{\text{e}}\frac{e^2}{m{R}^3}}$

Solve the above equation $R$.

    $\sqrt{k_{\text{e}}\frac{e^2}{m{R}^3}}=2\pi f$

Take square both the sides.

    $\begin{array}{c}{\left(\sqrt{k_{\text{e}}\frac{e^2}{m{R}^3}}\right)}^2={\left(2\pi f\right)}^2\\ k_{\text{e}}\frac{e^2}{m{R}^3}=4{\pi }^2{f}^2\\ R^3=\frac{k_{\text{e}}{e}^2}{4{\pi }^2{f}^{2}m}\\ R=\sqrt[3]{\frac{k_{\text{e}}{e}^2}{4{\pi }^2{f}^{2}m}}\end{array}$

Conclusion:

Substitute, $2.47\times {10}^{15}\text{Hz}$ for $f$, $8.987\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for $k_{\text{e}}$, $9.11\times {10}^{-31}\text{kg}$ for $m$ and $1.6\times {10}^{-19}\text{C}$ for $e$ in the above equation to calculate $R$.

    $\begin{array}{c}R=\sqrt[3]{\frac{\left(8.987\times {10}^9\text{N}{\text{.m}}^{\text{2}}/{\text{C}}^{\text{2}}\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2}{4{\pi }^2{\left(2.47\times {10}^{15}\text{Hz}\right)}^2\left(9.11\times {10}^{-31}\text{kg}\right)}}\\ =1.02\times {10}^{-10}\text{m}\\ =1.02\times {10}^{-10}\text{m}\times \left(\frac{1\stackrel{0}{\text{A}}}{{10}^{-10}\text{m}}\right)\\ =1.02\stackrel{0}{\text{A}}\end{array}$

Therefore, the radius of the orbit is $1.02\stackrel{0}{\text{A}}$.