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We're told that semiconductors have a bandgap and photons of an energy greater than the bandgap can be absorbed, exciting electrons from the valence band to conduction band. This therefore defines their absorption spectrum.

However, metals do not have a bandgap as the uppermost energy band is half-filled. What, therefore, defines their absorption spectrum please? I've read about free-carrier absorption - is this related?

Thanks

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  • $\begingroup$ To absorb a photon, you need an occupied electron state (to get the electron from), and an empty state at the right energy and momentum to put the excited electron in to. So, you have an overlap integral of the occupied states, and the higher energy unoccupied states (which metals absolutely have). No band gap needed - that just makes it harder to figure out semiconductors. $\endgroup$ – Jon Custer Sep 24 '15 at 22:24
  • $\begingroup$ Thanks, so what determines the absorption spectrum in metals then? $\endgroup$ – email Sep 24 '15 at 22:24
  • $\begingroup$ For state-to-state absorption you have the overlap integral as above. Since the bands have structure in energy-momentum space you get some structure there. But then, you have other absorptive multi particle resonances such as plasmons, polaritons, etc. So, the gold color of gold is brought about by the plasma frequency of the free electrons being in the blue, leading to absorption in that range of frequencies. $\endgroup$ – Jon Custer Sep 25 '15 at 0:59
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In a metal light will interact with both the electrons in the conduction band and with the valance electrons of the metal lattice.

If the frequency is above the so-called plasma frequency the conduction band electrons can be considered free (very few collisions between oscillations). The electrons simply re-emit the light which makes the metal transparent. However, at lower frequencies like in the visible range the conduction band electrons will collide with the lattice much faster than the period of the wave and will absorb the energy without being able to re-emit energy. Basically, due to the lattice collisions the electrons can not follow the incoming light. This absorption prevents the light to penetrate the metal.

The effect of strong absorption somewhat counter intuitively leads to reflection. Because the waves do not penetrate deeply not many of the conduction electrons see the incoming wave and not that much overall energy is absorbed. Most of the energy is reflected by the first few layers of metal atoms in the lattice with a spectrum that depends on the scattering properties of these particular metals atoms.

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  • $\begingroup$ "...which makes the metal transparent" $\endgroup$ – boyfarrell Sep 25 '15 at 5:29
  • $\begingroup$ Thanks for the detailed reply. So is the reflectivity of metals due to scattering from these surface electrons. I see how that makes a diffuse surface reflection, but sometimes metals can give shiny specular reflections too? Or is the reflectivity due to absorption of light by these surface electrons, the re-emission in the backwards direction? $\endgroup$ – email Sep 25 '15 at 8:39
  • $\begingroup$ Yes the reflectivity is due absorption and re-emission by the atoms in the first surface layers. The specular nature of reflection is due to interference: under the right conditions (a large and flat enough surface - which can be quantified with respect to the wave length) the reflected waves only add constructively in the specular direction. $\endgroup$ – Jan Bos Sep 25 '15 at 11:29

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