[Physics] Potential energy for a crystal lattice

electromagnetismelectrostaticssolid-state-physics

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?

Best Answer

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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