High frequency limit of permitivity function

Question

I am taking a course in solid state physics, and I have a question about the behavior of the (relative) permitivity function $\epsilon_r (\omega) $ for dielectric materials.

In the Drude model, we discover that electromagnetic radiation is attenuated in metals below a critical frequency, the plasma frequency, but that above this metal, they become transparent to electromagnetic radiation, and so $\epsilon_r = 1 $.

My question is, what happens to the relative permitivity function for dielectric materials in this high frequency limit? Do dielectric materials become transparent like metals for sufficiently high frequencies of light?

I know that it is equal to the "optical permitivity" given by

$\epsilon_{r,opt}$ = $\frac{N \alpha}{V \epsilon_0}$,

where $N/V$ is the density of the dielectric and $\alpha$ is the atomic polarizability, but I cannot figure out if $\epsilon_{r,opt}$ actually approaches 1 for high frequencies, which would mean that dielectrics become transparent. I would have to know the atomic polarizability for the dielectric - is there a general expression for $\alpha$ valid for all dielectrics?

Answer 1

At high frequency metals and insulators behave similarly, as the effect of valence electrons becomes small or negligible. It is the inner atomic shells that dominate. In general n deviates little from 1 and the absorptivity k is not far from 0. For precise information consult the x-ray database at http://www.cxro.lbl.gov/.