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Interactions between e-m radiation and matter.

  1. May 10, 2009 #1
    Hi,
    I'm an undergraduate physics student and I'd like to know more about interactions of light with matter.

    Since I haven't studied Quantum Physics yet, I'd like to know what happen on a microscopic scale, when e-m radiation passes through a material (transmission), when it's reflected/scattered and when it's absorbed.

    For example, glass is in general transparent to visible wavelengths, but it's opaque to ultraviolet or infrared. Does that mean that visible wavelengths, unlike ultraviolet and infrared, are not in the specific range to excite electrons and bring them to another orbital?
    Hope this picture helps:
    http://img135.imageshack.us/img135/4940/unexcited.jpg [Broken]
    This is just what I thought can happen.
    Furthermore, is there any relation between polarization and transmission?

    About reflection, I supposed e-m is absorbed by atoms in the material and then are re-emitted at the same angle of incidence. Then why not all materials reflect radiation with the same angle of incidence (scattering)?

    About absorption, why a material should absorb radiation and transform it in internal energy instead of reflecting it?


    I'm sorry for the poor language, but I've done the best I could do, since English is my second language.
    I thank so much anyone that try to clarify what's going on when radiation encounters matter. :smile:
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 11, 2009 #2

    malawi_glenn

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    I must say that in solids such as glass, it is not individual atoms which participate in the photon interactions, but rather the solid as a whole. The individual levels of the atoms will "smear" when they are put close together and form a band.
     
  4. May 11, 2009 #3

    alxm

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    Yes. But electronic excitations only account for the part of the spectrum.

    There are many many kinds of 'polarization', so it depends. If you mean the circular polarization of the incident light, then yes, that can matter depending on the structure of the material, a fact which is exploited in a kind of spectroscopy called circular dichroism.

    Reflection is purely a bulk property of materials and can't really be explained in simple terms of individual atoms.

    I'll just assume you mean 're-emitting it' rather than reflecting it, as reflection is a more specific process.

    Where does that absorbed energy go? Heat. What's heat? Atomic/molecular motion/vibrations. (and the equivalent radiation, usually in the IR range). That energy can be transmitted through simple collisions ('non-radiatively')

    So that's essentially your answer right there: If the absorption of light corresponds to a change in motion, either directly (absorbing heat radiation) or indirectly (e.g. changing the electronic state such that the vibrational state changes - 'vibronic coupling' being an example) then the re-emitted photon (if any) will have less energy.

    Actually there's a third option, which is that the light is absorbed in a photochemical reaction (which is essentially another, but stable, electronic state). But that's a lot less common.
     
  5. May 11, 2009 #4
    It would be useful to examine why sky is blue, because the phenomenon is based on excitation and re-emission by individual molecules, with nearly complete polarization at 90 degrees, and with very little absorption of the incident sunlight. Here is a very brief discussion:
    http://en.wikipedia.org/wiki/Rayleigh_scattering
     
  6. May 14, 2009 #5
    That's what I meant, sorry for that!

    However, summing up, I've understood that to understand reactions between radiation and solid matter is very important to consider the "crystal lattice" formed by atoms in a solid and not atoms singularly, while in low pressure gases they can be considered singularly.

    That's why is complicated to explain perfect reflection and why each material has its own optical properties.
    For example diamond and graphite are both made of carbon atoms, but have different structure and indeed they have completely different properties.

    Anyway can anybody suggest me in general what happens when visible photon passes through a transparent material (solid or liquid)? What are the conditions that permit the photon not to be absorbed? Or I need some more "quantum knowledge" to understand? :)

    Thanks to all for the replies.
     
  7. May 14, 2009 #6
    A visible photon can pass through a "truly transparent material" ONLY IF there is NO dispersion. Review the Kramers Kronig relations (alias the dispersion relations) for proof.
     
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