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Basic Fluorescence Question

  1. Sep 29, 2015 #1
    Fluorescence occurs when an atom or molecule absorbs a photon, promoting a ground state electron to an excited state, and the electron returns very quickly to the ground state, emitting a new photon. The new photon is usually of a longer wavelength because some of the energy is lost to vibrational decay.

    The quantum yield of the fluorescence process is the ratio of the number of photons emitted to the number absorbed. I have read several times now that the quantum yield can be less than 1, but not greater than 1.

    I'm not really understanding this. First, I know that fractions of photons can't be emitted, so I'm assuming that to get a quantum yield of 0.5, for example, there must be multiple decay pathways back to the ground state. Say one is a decay that produces no photon, while another decay path is through photon emission, and they are equally likely. Is that right?

    But then I run into another problem. Would it be impossible for an electron to decay by two successive radiation-producing drops in energy? That seems possible to me. The absorbed high-energy photon would then produce two lower energy photons. But that would give a quantum yield of two, which I've read is impossible.

    In summary, I'm afraid that I have some fundamental misunderstanding of fluorescence that is confusing me. I haven't learned many details behind the process, so I'm not looking for a very technical explanation yet, although I do have an undergraduate background in quantum mechanics to work with. But it would be very much appreciated if someone could explain where my misunderstanding of the basics is leading me astray.

    Thanks!
     
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  3. Sep 29, 2015 #2

    Andy Resnick

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    Remember that there's not a 100% chance that the atom/molecule will absorb a particular photon- the probability of absorption can be phrased in different ways ('transition strength' is another), but in essence the less likely a photon is absorbed, the lower the quantum yield.
     
  4. Sep 30, 2015 #3
    That makes perfect sense. But in that case wouldn't the quantum yield be "emitted photons/incident photon" rather than "emitted photons/absorbed photon"?

    I have always seen it defined the latter way.
     
  5. Sep 30, 2015 #4

    Andy Resnick

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    Ah- I was confusing quantum yield with quantum efficiency. For your application, remember that the absorption spectrum (and emission spectrum) are broad- illuminating a fluorophore 'off resonance' will yield less fluorescence, and fluorophores can decay non-radiatively. Non-radiative decay rates are strongly influenced by the solvent (chemical environment). There are many non-radiative decay processes, two examples are collisions and FRET.

    Lakowicz's book 'Principles of Fluorescence Spectroscopy' is excellent.
     
  6. Sep 30, 2015 #5
    So it sounds like I was correct in my original post: quantum yields less than 1 occur because there are additional non-radiative decay pathways that occur with some probability. Thank you for confirming that.

    In that case, I'd like to return to the second question from the original post. Why is it impossible to get a quantum yield greater than 1? After absorbing a high energy photon, why can't an electron emit two lower energy photons on its way back to the ground state?

    I'll have to check out that book! Thanks for the reference.
     
  7. Sep 30, 2015 #6

    davenn

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  8. Sep 30, 2015 #7

    Andy Resnick

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    The process you describe is exactly backwards- two photons can be absorbed, resulting in 1 fluorescence photon. This is the basis of 2nd harmonic generation and two-photon microscopy.

    What you describe is somewhat similar to parametric oscillation or 4-wave mixing, with one 'channel being the vacuum. Weird things happen- don't have a good reference on hand right now.
     
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