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I would like some help understanding the following passage:

Consider a crystal lattice such that its unit cell has 27 ions arranged such that there are alternative positive and negative ions for same magnitude.

Then the electrical potential of the crystal lattice is just the potential of arrangement of any one of the ion taken with all the ions in the crystal times times the total number of ions halved.

i.e,

$$U = {1\over2}nN_0\sum^{nN_0}_{k = 2} {q_1 q_k \over r_{1k}}, $$

where $n$ is the number of moles of the solid.

I understand what author is saying but not why he is saying so. I don't understand why taking total potential of one ion multiplied by total number of ions gives the total potential. For instance take a ion at the bottom right corner of the crystal, clearly the potential of this ion wrt to a ion in middle and an ion at top left corner of lattice will differ. What is the author's reasoning here?

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  • $\begingroup$ What it seems to me the author is trying to say, or what I would expect him to say, is that total potential is equal to the potential of Coulomb interactions of atoms in the unit cells multiplied by the number of unit cells. Are you sure that you have copied out the formula correctly? $\endgroup$ – Ilya Lapan Apr 22 '17 at 20:44
  • $\begingroup$ I differ with you. Why do you think number of moles times avogadro's constant gives number of unit cells rather than number of ions ? $\endgroup$ – A---B Apr 22 '17 at 20:47
  • $\begingroup$ I confess that I made up the first paragraph because the picture in the book is not available with me as soft copy but I copied the formula correctly. $\endgroup$ – A---B Apr 22 '17 at 20:49
  • $\begingroup$ Oh, okay, I got it, let me write you an answer. $\endgroup$ – Ilya Lapan Apr 22 '17 at 20:51
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For instance take a ion at the bottom right corner of the crystal, clearly the potential of this ion wrt to a ion in middle and an ion at top left corner of lattice will differ.

The author here is treating the problem as a symmetric one and ignoring the edge effects (i.e. crystal is extremely big). Say you have N atoms in the crystal. So imagine you have an atom somewhere in the lattice. The total potential energy due to that individual atom is: $$ \frac{1}{4\pi\epsilon_0}\sum_k^N{q_1q_k/r_{1k}}$$ I.e sum of potential energies of interaction of this atom with every other atom in the crystal. Now the author is assuming, that wherever in the crystal you look, it all looks the same (i.e. crystal is infinite). Although, you are right, on the edge of the crystal this does not work, this is a common assumption. Now, because you have N atoms in the crystal, multiply energy of each atom by N to get the total potential energy. But you need to divide by 2 because if do not do that, you will be double counting.

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  • $\begingroup$ So this is a approximation at best. Thanks for the answer. $\endgroup$ – A---B Apr 22 '17 at 21:09
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    $\begingroup$ @A---B as Crimson pointed out, $1/r$ potential fall off so fast, that you can pretty much ignore the edge effects as soon as you a little inside the crystal. In your example " a ion at the bottom right corner of the crystal, clearly the potential of this ion wrt to a ion in middle and an ion at top left corner", ion at the top is soooo far away, it doesn't really feel the contribution of the ion in middle. In reality, only the neighbours matter. $\endgroup$ – Ilya Lapan Apr 22 '17 at 21:16
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You are correct in asserting that the equation gives the potential energy for a single ion and then multiplies it by the number of ions to obtain the total potential energy.

The reason the effects of the boundary can be ignored is based on two factors:

  1. Nearby ions have more influence on the potential of a given ion than far away ions. (the influence drops of with 1/r)
  2. The majority of the ions is not at the border of the crystal. If there are several moles of atoms, only a very small percentage is at the border.

Number 1 shows that for the ions that are not close to the boundary, the boundary is unimportant. Number 2 shows that this holds for almost all ions.

This approximation is of course only valid for huge numbers of ions

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  • $\begingroup$ Thanks for your answer. This answer is criminally low voted. $\endgroup$ – A---B Apr 22 '17 at 21:10

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