Reflection of Light off of an Atom

[evankiefl]

If a photon is not absorbed by an atom during collision, what determines the direction of the reflected photon? Explicitly...

Does the photon reflect off of the valence probability 'shell' by abiding angle of incidence = angle of reflection? This seems counter-intuitive to me because some (most?) of the amplitudes of EMR are larger than some atoms themselves. If not, how is this phenomena carried out?

 

[SpectraCat]

Light can't reflect off an atom. Light can't reflect off any feature that is much smaller than it's wavelength. It *certainly* can't undergo specular reflection as you suggest in your post.

When light interacts with an atom, it is either scattered, or absorbed. For the description of scattering phenomena, look up Rayleigh scattering on wikipedia.

 

[ardie]

the photon will undergo compton scattering should the atom be in a non-relativistic setup, the details of which you can find over here:
http://en.wikipedia.org/wiki/Compton_scattering
should the energies be more in the relativistic realm, then an application of quantom field theory leads to the Klein-Nishina formulae, describing the behaviour of the photon.
the photon can also go through photoelectric absorption which u didnt want to follow
Rayleigh scattering is a generalised version of compton scattering that assumes spherical atoms, but gives a near correct approximation to visible light ray scattering

 

[SpectraCat]

the photon will undergo compton scattering should the atom be in a non-relativistic setup, the details of which you can find over here:
http://en.wikipedia.org/wiki/Compton_scattering
should the energies be more in the relativistic realm, then an application of quantom field theory leads to the Klein-Nishina formulae, describing the behaviour of the photon.
the photon can also go through photoelectric absorption which u didnt want to follow
Rayleigh scattering is a generalised version of compton scattering that assumes spherical atoms, but gives a near correct approximation to visible light ray scattering

Unless the incoming photon is at least as energetic as an X-ray, then there is no mechanism for Compton scattering, which requires momentum transfer from the photon to an electron in the atom, usually (always?) involving ionization of the atom. This clearly seems different from what the OP was asking about.

Also, I have never heard Rayleigh scattering described as "a generalized version of Compton scattering" .. in fact, I can't see how that even makes sense .. Compton scattering is inelastic by definition, Rayleigh scattering is elastic by definition. Perhaps Rayleigh scattering describes some kind of low-energy limit of Compton scattering, but that's what I thought Thompson scattering is. I would be interested to see a reference where the connection between the derivations of Compton and Rayleigh scattering is given.

Anyway, even if there is a mathematical relationship, I think it is more useful and instructive to restrict Compton scattering to short wavelength cases where the EM radiation interacts with an electron in an atom, and use Rayleigh scattering to describe scattering of long wavelength radiation of larger scale objects, like atoms and molecules. That is certainly how I learned to distinguish the two.

 

[ardie]

as you rightly mentioned, Rayleigh scattering assumes elastic scattering. what does that mean in basic terms? that the atoms are like giant blocks of matter and the light cannot move them, so it arrives it collides and it shoots of in another direction.
the correct description of a light when incident on an atom, is that when a photon collides with an electron, should no absorption or interaction take place. the chances of the photon encountering the nucleus is minimal. hence the compton description applies to all photon energies in the non-relativistic terms. the more complicated but complete description is that given by the Klein-Nishina formulae.

 

[alxm]

Also, I have never heard Rayleigh scattering described as "a generalized version of Compton scattering" .. in fact, I can't see how that even makes sense

Well, the thing about generalizations is that everything's the same as everything else once you get general enough :) Although I agree that particular statement doesn't make much sense to me either.

On the other hand, in the most general sense, you can of course describe all scattering processes (Rayleigh, Raman, Thomson, Compton) in more or less the same framework (Which https://www.amazon.com/dp/0471625566/?tag=pfamazon01-20 does, for instance).