Homework Help: Potential energy of a crystal, electrity and magnetism intermediate course

1. Aug 30, 2009

fluidistic

1. The problem statement, all variables and given/known data
Calculate the potential energy, by ion, in an infinite bi-dimensional ionic crystal.
Hint : Use the power series expansion of $$\ln (x+1)$$.

2. Relevant equations

$$\ln (x+1)=\sum _{i=1}^{\infty} \frac{(-1)^{i+1}x^i}{i}$$.

3. The attempt at a solution

$$W=U=k\sum _{j=1}^{\infty} \sum _{i=1}^{j} (-1)^i \frac{q_1^2}{a \cdot i}$$ where $$q_1$$ is the charge of an electron and $$a$$ is the distance between charges.

Here I'm stuck, I don't see how I can use the hint so I suspect I made an error.

The sketch looks like this I believe : ...$$------e^+------e^{-}\overbrace{------}^ {a}e^{+}------e^{-}$$...

When there are 2 ions, the energy needed to form the crystal is $$k\frac{q_1q_2}{a}$$ but $$q_2=-q_1$$.
For 3 ions : $$k \left ( \frac{q_1q_2}{a}+\frac{q_1q_3}{2a}+\frac{q_2q_3}{a} \right)$$ but $$q_3=q_1$$. And so on. Hence my result.

Can you help me? The series must converge I believe : I'm sure at 100%.

2. Aug 30, 2009

gabbagabbahey

Hmmm....wouldn't a bi-dimensional crystal look more like this:

$$\begin{array}{cccc}--e^{+}-- & --e^{-}-- & --e^{+}-- & --e^{-}-- \\ | & | & | & | \\ | & | & | & | \\ --e^{-}-- & --e^{+}-- & --e^{-}-- & --e^{+}-- \\ | & | & | & | \\ | & | & | & | \\ --e^{+}-- & --e^{-}-- & --e^{+}-- & --e^{-}-- \\ | & | & | & | \\ | & | & | & | \\ --e^{-}-- & --e^{+}-- & --e^{-}-- & --e^{+}-- \end{array}$$

3. Aug 30, 2009

fluidistic

Oh... silly me! You're right.
However, suppose the problem stated a unidimensional crystal, would have I solved it right? Up till I got stuck I mean.
Thanks a lot for the response.
I'll try it as you suggested.

Edit : Now I know why I did this mistake of interpretation : the problem states what I wrote here followed by "i.e. a line of equidistant charges with values e with alternate sign". So I guess I have to solve the problem I tried to solve, that is what my sketch shows.
Thanks anyway, really appreciated.

Last edited: Aug 30, 2009
4. Aug 30, 2009

gabbagabbahey

Hmmm....I guess the questioner really means a uniform 1D crystal lattice....

Anyways, break the problem into pieces....pick any charge (might as well choose one of the positive ions), what is the potential energy of that charge due to the effects of its two nearest neighbors (the two negative charges on either side)? How about the potential energy due to just the effects of its next two nearest neighbor (the two positive charges---one on either side)?

5. Aug 30, 2009

fluidistic

Yes.

Correct me if I'm wrong : $$U=-\frac{2ke^+}{a} \right )$$.

$$U=\frac{2ke^+}{2a}=\frac{ke^+}{a}$$.

Is that right? It's a very surprising way to me to approach the problem. I hope I'm not wrong though, it's all very new to me.
If this is right then I'll try to finish it tomorrow. (I have to sleep now).
Oh wait... I don't have time to check, but is the result $$\frac{2k}{a} \sum _{i=1}^{\infty} \frac{(-1)^ie^+}{i}$$?
Where $$e^+$$ is the charge of the positron and $$e^-$$ the charge of the electron.

6. Aug 30, 2009

gabbagabbahey

Assuming you mean $U_1=-\frac{2ke^2}{a}$ and $U_2=+\frac{2ke^2}{2a}$, then yes.

Now, compute $U_3$, the potential energy due to the next nearest neighbors (the charges a distance 3a away) and so on. The total potential energy of the ion will then be $U=\sum_{n=1}^\infty U_n$

Close, you should end up with $$U=\frac{2ke^2}{a}\sum_{n=1}^\infty \frac{(-1)^n}{n}$$ and when you compare that to the series for $\ln(1+x)$ you should see clearly that the sum is equal to $-\ln(2)$.

7. Aug 31, 2009

fluidistic

Ok, thank you very much. Indeed I meant $$e^2$$. I guess I was tired.
So the answer is $$- \ln 2$$, does that mean that the crystal is stable? In order to separate all the ions I'd have to do some work (related to $$\ln 2$$).

8. Aug 31, 2009

gabbagabbahey

Looks pretty stable to me...the net force on an ion due to its two nearest neighbors is?...Due to its next nearest neighbors? Due to its next nearest neighbors?>...

9. Aug 31, 2009

fluidistic

It's null, I see.
Thanks a lot for all your help!