# Are the atoms of a crystalline solid distinguishable from each other (canonical assembly)?

I was doing an exercise about a crystalline solid composed by $$N_s$$ atoms in the surface and $$N_b$$ inside. My question is when calculating the fugacity should I take into account that the atoms are distinguishable from each other? I think so because being in a crystalline lattice they are in certain positions, so it would not be like a gas.

My doubt comes because further on it is said: The solid is in contact with an insulated container of volume V initially empty. Thus, a number $$N_g$$ << $$N_b$$ of atoms of the solid will escape from its surface into the container until equilibrium is reached. Bearing in mind that in this simple model the number of atoms in contact with the gas will always be of the order of gas will always be of the order of $$N_s$$, and assuming the ideal gas phase, calculate the number of atoms of the gas with the data given. But after calculating it, $$N_g$$ > $$N_s$$ which I do not understand at all because the atoms inside the solid are supposed to have a higher binding energy and so far we have not taken them into account. Or is it that by removing, so to speak, all the atoms from the surface, there would be a new layer of atoms on the next surface with the same binding energy as those that were there before?

If the atoms in the solid were distinguishable the fugacity term will be $$10^{15}$$ orders of magnitude greater and $$N_g$$<$$N_s$$. That's why I am asking

• The atoms in a crystal can be labelled by their position even if the atoms themselves are indistingishable. Interchanging two atoms while keeping the same set of atomic sites occupied does not change the physical state. Commented May 3 at 19:34