- #1

Kelvin

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**Calculate the potential energy, per ion, for an infinite one-dimensional ionic crystal, that is, a row of equally spaced charges of magnitude e and alternating sign.**

Hint: The power series expansion of ln(1+x) may be useful.

Hint: The power series expansion of ln(1+x) may be useful.

Here's my calculation:

Let [tex]u_n[/tex] be the potential energy when there is n charges.

[tex]u_1 = 0[/tex]

[tex]u_2 = \frac{1}{4 \pi \epsilon_0} \frac{-e^2}{d}[/tex] where d is the separation of two charges

[tex]u_3 = \frac{1}{4 \pi \epsilon_0} \left( 2\times \frac{-e^2}{d} + \frac{e^2}{2d}\right)[/tex]

[tex]u_4 = \frac{1}{4 \pi \epsilon_0} \left( 3 \times \frac{-e^2}{d} + 2 \times \frac{e^2}{2d} + \frac{-e^2}{3d}\right)[/tex]

...

[tex]u_n = \frac{1}{4 \pi \epsilon_0} \left( (n-1) \times \frac{-e^2}{d} + (n-2) \times \frac{e^2}{2d} + \dots + \frac{(-1)^{n-1} e^2}{(n-1)d}\right)[/tex]

[tex] = \frac{e^2}{4 \pi \epsilon_0 d} \sum_{i=1}^{n-1} \frac{(-1)^i (n-i)}{i}[/tex]

[tex] = \frac{e^2}{4 \pi \epsilon_0 d} \sum_{i=1}^{n-1} (-1)^{i-1} \left( 1- \frac{n}{i} \right) [/tex]

[tex] = \frac{e^2}{4 \pi \epsilon_0 d} \left( \sum_{i=1}^{n-1} (-1)^{i-1} - n \sum_{i=1}^{n-1}\frac{(-1)^{i-1}}{i} \right)[/tex]

potential energy per ion

= [tex]u_n/n = \frac{e^2}{4 \pi \epsilon_0 d} \left( \frac{1}{n}\sum_{i=1}^{n-1} (-1)^{i-1} - \sum_{i=1}^{n-1}\frac{(-1)^{i-1}}{i} \right)[/tex]

as n tends to infinity,

[tex]u_n/n = \frac{e^2}{4 \pi \epsilon_0 d}\left( 0 - \ln 2 \right)[/tex]

I am not sure my last 2 steps are correct or not. The series [tex]\sum_{i=1}^{n-1}(-1)^{i-1}[/tex] diverges as n tends to infinity. If we divide the series by n, do we get really zero?

Thanks for your help!