I would tend to argue that the areal density of atoms cannot be well-defined for a polycrystalline material. You have a kind of average over different crystal orientations (crystallites) and their micro-surfaces might not be aligned with what you define as the surface plane.

If your material is dense, then you might proceed by considering the volume density and multiply that by  a characteristic length which approximately represents an atomic layer thickness.

Otherwise you could give an interval defined by the areal density of more open single crystal surfaces (100) on the lower end and the high density (111) surface as the upper limit. These are easily computed from the cubic lattice constant (0.408 nm). Keep in mind that the crystal structure is an fcc lattice.

Hi  samuel, In addition to Kai answer, in order to know the value of  atomic dense in the polycrystalline silver, I think you have to consider your sample homogeneity  first by running SEM. Then run TEM and AFM analysis, in order to quantify atomic dense in certain area (x nm to y nm), the depth (z nm) then compare to single silver atomic volume. I hope this will aid you. Also you can read this article (attached), thanks.   

The solution is rather simple, as far as you don't need the number at highest precision: The atomic mass of silver is 107.868 u . Remember that the value of u (or Dalton) is 1.660539 x 10^-24 g. Thus one atom of silver, irrespective of the crystaline state, has a mass of 1.79119  x 10^-22 g. Next we need he density of silver (and this is the only point where there is a - negligible, for most purposes - difference between polycrystalline and monocrystalline samples). The density is 10.49 g cm^-3. Dividing this density by the mass of a single atom gives 5.856 x 10^22 atoms in a centimeter cubed.

All values used here can be found in numerous books, and also in Wikipedia. So it is advisable to read first, and ask only when you do not find anything in the literature.

Good luck

Max

Perhaps I should have been more clear. I am looking for the approximate number of atoms per cm2 on the polycrystalline surface, not the number of atoms per unit volume. I have found values for the number of atoms per cm2 in the 111, 110 and 100 crystal faces, but would need to know the relative proportions of each crystal in the polycrystalline surface to take an average. I know this has been found for platinum, where polycrystalline Pt is said to have 1.3 x 10^15 atoms cm-2, so I was looking for an equivalent for silver. It would not need to be precise just an approximation is ok. 

Samuel, you would have to scrutinize how that number was evaluated and what it actually means. A polycrystalline film may still be textured (depending on substrate and preparation protocol) and the number vary accordingly.

Also, in addition to what you just wrote, how about rough and flat surfaces? What would you count as "the surface" or surface area?

(100): 1.2 x 1015/cm2

(111): 1.39 x 1015/cm2

They're within 10% of their average. Wouldn't that be good enough if you're just looking for an approximate number? (hope I did my math right...)

Sorry, I misunderstood your question.

Greetings, Max

calculating the cube root of Max' atoms per cm³ and sqaring the result should give us the wanted atoms per cm².

(5.856*10^22)^(2/3) = 1.508*10^15

This is also in the magnitude of Kai's atoms per cm².

These 10^15cm-² is also a number our nanomaterial science prof. wanted us to keep in mind as a role of thumb.