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B Few questions on light interacting with atoms

  1. May 4, 2016 #1
    Tell me if I get this straight;
    When a photon of light hits an atom and interacts with the electrons, it will only affect the electrons in the outer shell causing them to go up a state, then they will fall back down and emit that exact wavelength that was absorbed?
    And the photons of light that don't interact with the electrons, do they pass straight through?

    For emission spectra, is that electrons gaining energy from a different source and falling back down to their states and thus releasing a photon of light?

    In light spectra, couldn't an absorption spectra look like an emission spectrum and vice-versa for different atoms?
     
  2. jcsd
  3. May 4, 2016 #2

    mfb

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    That is a possible result, but not the only option. It depends on the photon energy. Also keep in mind that those excitations are always a transition of the whole atom. An isolated electron could not absorb the radiation.

    Ionization or an excitation from a deeper shell are possible as well. Stimulated emission, if the atom was in an excited state before (that's how lasers work). Simple scattering. And probably some things I forgot.
    This is again just one of multiple options.
    The photons interact with the whole atom, not with individual electrons. Apart from that: Yes, to a good approximation.
    Yes (with the same caveat as before).
    Different atoms of the same element? Yes.
     
  4. May 4, 2016 #3

    jfizzix

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    This may be surprising, but accurately describing how a single photon interacts with a single atom qualifies as cutting edge physics!

    Most treatments that you see in intro quantum physics courses treat light as a classical field interacting with a quantum atom. This is usually good enough when talking about absorption spectra.
    To describe emission spectra, you need a little more.
    If an atom in an excited state were truly isolated, it would never decay back to the ground state. The Schrodinger equation for the atom alone implies that the probabilities for the atom as a whole to be in a given energy state will be constant in time.
    The spontaneous emission and decay we see is described well by the atom interacting with the quantum electromagnetic field. Even if there's no photons in it, the interaction still drives the decay. In that case, it's the Schrodinger equation for the atom-plus-field that describes what's going on.

    When an electromagnetic field is incident on an atom, all electrons will respond to it, since the electromagnetic wave is felt by all the electrons. What transitions occur depends on the frequency of the EM wave, and how it compares with the transition frequencies of the atom. if the EM-wave is near the transition frequency of one of the levels of the valence electron, then the most likely outcome is that one photon is absorbed and the atom is excited (less likely things could happen too).

    When the valence electron decays back to where it was, it doesn't necessarily emit exactly the same light that it absorbed. In particular, if there are levels in between, it could either emit the one photon straight to the final state, or it could emit multiple photons for all the transitions in between. These cases where the light coming out has a lower energy than what came in are often lumped under the term "inelastic" photon scattering. Because of inelastic photon scattering, absorption and emission spectra are not usually identical, though they can be quite similar.

    Just because a photon doesn't get absorbed doesn't mean it isn't effected by the atom. The interaction as a whole is described by the atom and field together. Just as an atom can be effected by the field, even if it absorbs no photons from it, the field can be affected, even if it loses no photons to the atom.

    If you ever get into it, quantum optics is an incredibly rich branch of physics.

    Note: inelastic photon scattering is sometimes known as Raman scattering, though what people mean when they say Raman scattering is usually more specific.
     
  5. May 4, 2016 #4

    ChrisVer

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    In practice the absorption spectrum has a dip [lack of photons] at that particular energy as here... emission has a peak [excess of photons] at that energy as here.
     
  6. May 5, 2016 #5
    Awrepl
    awesome, you clarified pretty much everything. Thanks a lot!
     
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