Potential energy of a crystal, electrity and magnetism intermediate course

[fluidistic]

Homework Statement

Calculate the potential energy, by ion, in an infinite bi-dimensional ionic crystal.
Hint : Use the power series expansion of $$\ln (x+1)$$.

Homework Equations

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

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%.

 

[gabbagabbahey]

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

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

 

[fluidistic]

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

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

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.

 

[gabbagabbahey]

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.

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)?

 

[fluidistic]

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

Yes.

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)?

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

How about the potential energy due to just the effects of its next two nearest neighbor (the two positive charges---one on either side)?

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

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.

 

[gabbagabbahey]

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

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

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 $U=\sum_{n=1}^\infty U_n$

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.

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 $-\ln(2)$.

 

[fluidistic]

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 $-\ln(2)$.

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 $$\ln 2$$).

 

[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?>...

 

[fluidistic]

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?>...

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