We have a crystall with $N$ atoms.

A Schottky defect is one of the atoms leaving their points in the lattice and going to the surface with $N' = cN^{2/3}$ possible atom places.

I want to compute the (Boltzmann) entropy of a crystal with $N_S$ Schottky defects, where $1 \ll N_S \ll N'$.

My first attempt:

We pick $N_S$ atoms out of the total $N$ atoms, which gives



Then we can place these $N_S$ atoms to the $N'$ points on the surface, which gives another


possibilities. In total we have

$$\frac{N'!}{N_S!(N'-N_S)!} \cdot \frac{N!}{N_S!(N-N_S)!} $$

possibilities to 'build' a crystall with $N_S$ defects.

However, in multiple sources i found the possibilities to 'build' a crystall with $N_S$ defects, to be given as:


I interpret this as picking $N$ atoms and placing them on a total of $N + N_S$ possible lattice points.

This doesn't make any sense to me. If we take $N$ atoms and place them along the total of $N + N_S$ lattice points, it includes the possibilities to just put the $N$ atoms back to their original places, which then would correspond to a crystall with $0$ defects. Also the latter formula includes the possibilities, where we place just one atom on the surface and so on...

So how can the latter formula give the correct number of possibilities to build a crystall with $N_S$ Schottky defects?

I thought maybe the latter formula is some limit of the first (using the conditions given at the beginning of the question), but i didn't manage to figure this out.

Where is the mistake?

I appreciate your help :)

  • 1
    $\begingroup$ For clarity and completeness, please cite the sources you're referring to. Otherwise, it's difficult to understand the context and assumptions of those models. $\endgroup$ Commented Feb 13 at 18:44
  • $\begingroup$ (Bounty awarded to the older of two good answers with equal upvotes.) $\endgroup$ Commented Feb 20 at 19:44

2 Answers 2


So, the perhaps disappointing/underwhelming answer is that the formula $\frac{(N + N_s)!}{N!N_s!}$ is suspiciously consistent with the conventional interpretation of substitution defects in a bulk crystal. If $N_s$ counts the number of substitutions (or "impurities"), there are $N_s + N$ sites in the underlying crystal lattice, $N_s$ of which are (undistinguished or indistinguishable) substitutions: hence, $N+N_s \choose N_s$ possible configurations. Alternatively, one could think of $N+ N_s$ as counting the number of available interstitial sites in a crystal lattice, with $N_s$ the number of occupied sites. Clearly, in either case the system is completely different from the subject of your question, and there is no physical basis to expect a correspondence between the two (one formula relates/connects systems at two distinct dimensionalities, while the other is dimension-independent.)


Actually, your approach is closer to usual approach to tackle Frenkel defects. Interpreting $N_S$ the number of defects, $N$ the number of sites and $N'$ the number of interstitial sites: $$ W = \binom N{N_S}\binom{N'}{N_S} $$

In the case of Schottky defects, you need to distinguish atoms and lattice sites. In the final formula, $N_S+N$ corresponds to your $N$. Indeed, even if you call $N$ the number of atoms, you rather use it as the number of sites in your reasoning.

You should rather view the formation of vacancies as a bulk process without referring the surface. I suspect that imagining an accumulation at the surface is not realistic. After all, the number of vacancies scale extensively $N_S\sim N$, but the surface sites scale $N'\sim N^{2/3}$ (assuming a smooth boundary). Therefore in the thermodynamic limit, there is simply not enough space for the vacancies to pile up at the surface.

For your situation, it's best to think in terms of the grand canonical ensemble. You can interpret the energy of a vacancy as a chemical potential. The switch to the grand canonical potential allows you to treat each site independently connected to a reservoir of atoms. It easily gives you the entropy in the thermodynamic limit without using combinatorics and the Stirling formula.

Hope this helps.


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