Suppose Nelectrons can be placed in either of two configurations. In configuration 1, they are all placed on the circumference of a narrow ring of radius Rand are uniformly distributed so that the distance between adjacent electrons is the same everywhere. In configuration 2, N-1electrons are uniformly

distributed on the ring and one electron is placed in the center of the ring. (a) What is the smallest value of Nfor which the second configuration is less energetic than the first? (b) For that value of N, consider any one circumference electron—call it e_o. How many other circumference electrons are closer to e_othan the central electron is?

4 for N and$\frac{2\mathrm{\pi }}{\mathrm{N}}$ for$\mathrm{\theta }$ in the equation${\mathrm{U}}_{1,\mathrm{N}-\mathrm{even}}=\frac{{\mathrm{Nke}}^2}{2\mathrm{R}}\left(\sum _{\mathrm{i}=1}^{\frac{\mathrm{N}}{2}-1}\frac{1}{\sin \left({\displaystyle \frac{\mathrm{i\theta }}{2}}\right)}+\frac{1}{2}\right)$

After reading the question, the distance between the two points on a circle that separated by an angle$\mathbf{\theta }$is,

$\mathbf{r}\mathbf{=}\mathbf{2}\mathbf{Rsin}\frac{\mathbf{\theta }}{\mathbf{2}}$

Here, R is the radius of the circle.

For the configuration 1 that all the N electrons are on circle, the sum of the potential energy when N = odd is as follows:

${\mathbf{U}}_{\mathbf{1}\mathbf{,}\mathbf{N}\mathbf{-}\mathbf{odd}}\mathbf{=}\frac{{\mathbf{Nke}}^{\mathbf{2}}}{\mathbf{2}\mathbf{R}}\left(\sum _{i=1}^{\frac{N-1}{2}}\frac{1}{\sin \left({\displaystyle \frac{\mathrm{i\theta }}{2}}\right)}\right)$

Here,$\mathbf{k}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{o}}}$is an electrostatic constant and e is the charge of the electron. The total angle between N electrons on the ring is as follows:

$\mathbf{N\theta }\mathbf{=}\mathbf{2}\mathbf{\pi }$

Rearrange the above equation for $\mathit{\theta }$.

$\mathbf{\theta }\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{N}}$

For the configuration 1 that all the N electrons are on circle, the sum of the potential energy when N = even is as follows:

${\mathrm{U}}_{1,\mathrm{N}=\mathrm{even}}=\frac{{\mathrm{Nke}}^2}{2\mathrm{R}}\left(\underset{\mathrm{i}=1}{\overset{\frac{\mathrm{N}}{2}-1}{}}\frac{1}{\sin \left(\frac{\mathrm{i\theta }}{2}\right)}+\frac{1}{2}\right)$

For the configuration 2 that all the N - 1 electrons are on circle and one electron at the center of the ring, the sum of the potential energy when N = odd is as follows:

${\mathbf{U}}_{\mathbf{2}\mathbf{,}\mathbf{}\mathbf{N}\mathbf{=}\mathbf{}\mathbf{odd}\mathbf{}}\mathbf{=}\frac{\mathbf{(}\mathbf{N}\mathbf{-}\mathbf{1}\mathbf{)}{\mathbf{ke}}^{\mathbf{2}}}{\mathbf{2}\mathbf{R}}\left(\underset{i=1}{\overset{\frac{N-3}{2}}{}}\frac{1}{\sin \left(\frac{{\mathrm{i\theta }}^{\text{'}}}{2}\right)}+\frac{5}{2}\right)$

The total angle between N - 1 particles on the ring is as follows:

$(N-1)\theta \text{'}=2\pi$

Rearrange the above equation for $\theta \text{'}$

$\theta \text{'}=\frac{2\pi }{N-1}$

For the configuration 2 that all the N - 1 electrons are on circle and one electron at the center of the ring, the sum of the potential energy when N = even is as follows:

$U_{2,N=even}=\frac{(N-1)k{e}^2}{2R}\left(\underset{i=1}{\overset{\frac{N}{2}-1}{}}\frac{1}{\sin \left(\frac{i{\theta }^{\text{'}}}{2}\right)}+2\right)$

$\begin{array}{c}{\mathrm{U}}_{1,4\mathrm{n}}=\frac{4{\mathrm{ke}}^2}{2\mathrm{R}}\left(\underset{\mathrm{i}=1}{\overset{\frac{4}{2}-1}{}}\frac{1}{\sin \left(\frac{\mathrm{i}2\mathrm{\pi }}{2\mathrm{N}}\right)}+\frac{1}{2}\right)\\ {\mathrm{U}}_{1,4\mathrm{n}}=\frac{4{\mathrm{ke}}^2}{2\mathrm{R}}\left(\underset{\mathrm{i}=1}{\overset{1}{}}\frac{1}{\sin \left(\frac{\mathrm{i\pi }}{4}\right)}+\frac{1}{2}\right)\\ {\mathrm{U}}_{1,4\mathrm{n}}=\frac{4{\mathrm{ke}}^2}{2\mathrm{R}}\left(\sqrt{2}+\frac{1}{2}\right)\\ {\mathrm{U}}_{1,4\mathrm{n}}=3.83\left(\frac{{\mathrm{ke}}^2}{\mathrm{R}}\right)\end{array}$

Substitute 4 for$N$and$\frac{2\pi }{N-1}$for $\theta \text{'}$in the equation

$U_{2,N-cos}=\frac{(N-1)k{e}^2}{2R}\left({\sum }_{i=1}^{\frac{N}{2}-1}\frac{1}{\sin \left(\frac{i{\theta }^{\text{'}}}{2}\right)}+2\right)$ and solve for$U_{2,4}$

$\begin{array}{c}{U}_{2,4}=\frac{(4-1)k{e}^2}{2R}\left({\sum }_{i=1}^{\frac{4}{2}-1}\frac{1}{\sin \left(\frac{i\left(\frac{2\pi }{N-1}\right))}{2}\right)}+2\right)\\ U_{2,4}=\frac{3k{e}^2}{2R}\left({\sum }_{i=1}^1\frac{1}{\sin \left(\frac{i\pi }{4-1}\right)}+2\right)\end{array}$

$\begin{array}{c}{U}_{2,4}=\frac{3k{e}^2}{2R}\left(\frac{1}{\left(\frac{\sqrt{3}}{2}\right)}+2\right)\\ U_{2,4}=4.73\left(\frac{k{e}^2}{R}\right)\end{array}$

In the similar way, we can calculate the potential energy for the other values of N as shown in the below table.

Below table shows the potential energy divided by$\frac{k{e}^2}{R}$and pure numbers. Here, the pure number depends on the value of N .

From the above table and for N <12 , the potential energy for configuration 1 is less than the configuration 2. For N$\ge$12 , the potential energy for configuration 1 is greater than the potential energy for configuration 2.

Hence, N = 12 is the smallest value when the potential energy for the configuration 1 is greater than the configuration 2 .

The configuration 2 contains 11 electrons that are distributed at equal distance around the circle and an electron is at center when N = 12 .

The distance of the one specific electron of charge$e_0$on the circle at distance R is,

$r=2Rsin\varphi$

Here,$\varphi$is the angle between the two electrons in the configuration 2. Hence, it is given as follows:

$11\varphi =\pi$

Solve the equation for$\varphi$.

$\varphi =\frac{\pi }{11}$

Su$r=2Rsin\left(\frac{\pi }{11}\right)=0.56R$bstitute $\frac{\pi }{11}$for$\varphi$ in the equation$r=2Rsin\varphi$ and solve for $r$.

Therefore, the distance away from the nearest neighbors on the circle is 0.56R .

e next nearest electron on the circle is,

$r=2Rsin\varphi \text{'}$

Here,$\varphi \text{'}$is the angle between the two electrons in the configuration 2. Hence, it is given as follows:

$11\varphi {\text{'}}^{}=2\pi$

Solve the equation for$\varphi \text{'}$.

$\varphi {\text{'}}^{}=\frac{\pi }{11}$

Substitute$\frac{2\pi }{11}$ for$\varphi$ in the equation$r=2Rsin\varphi$ and solve for$r$ .

$r=2Rsin\left(\frac{2\pi }{11}\right)=1.1R$

The distance away from the electron for the next nearest neighbor is .$1.1R$ .

Therefore, the number of electrons closer to the electron of charge than the one on thecenter is 2.