I am just being introduced to quantum physics.

I know that in order for a transition in the energy level of an electron to take place, the photon energy must be equal to the difference in energy levels.

But, I also hear great stories of the famous Compton scattering effect. If a photon must be entirely "used up" in an interaction with an atom, how could it be emitted again with a different wavelength?

I struggle to reconcile these two phenomena.

  • 5
    $\begingroup$ it's not the same photon $\endgroup$
    – user46925
    Jun 1 '15 at 14:33
  • $\begingroup$ In that case, wouldn't the quantization of energy levels mean that the absorbed and emitted photons have the same wavelength? $\endgroup$
    – Dr Coconut
    Jun 1 '15 at 14:36
  • $\begingroup$ not always; in some cases, part of the energy is kept and not output. Or, 2 photons are output like with birefringence $\endgroup$
    – user46925
    Jun 1 '15 at 14:45
  • $\begingroup$ I deleted my answer because while not incorrect, the other answer is better. $\endgroup$
    – Jimmy360
    Jun 1 '15 at 15:04

The key difference is that Compton scattering occurs on a free electron. It's all about making energy conservation work out. In an atomic process, the atom goes form having energy $E_1$ to $E_2$, and its energy is fixed by the atomic state it goes to. It can't be a $2 p ^1$ electron and also have extra energy it hangs onto. (Things are slightly more complicated than this, due to splitting of energy levels, Doppler shifting, etc. But this still just means you have a slightly broader window, rather than a true single-infinite-precision-frequency-condition.) So the photon has to have $E_2 - E_1$.

On the other hand, a Compton-scattered photon hits an electron with, say, no energy (for simplicity), which then goes off into having any number of energies. It's only limited by how much energy the photon had. Other than that, there's no reason that after collision it can't be a 20 $keV$ electron, or a 30 $keV$ electron, or anything. So there isn't the same sharp absorbance peak, because the energy condition is less stringent.

Incidentally, something like this can happen in atomic process if the photon is high enough energy to eject the electron from the atom entirely. In this case, we say the electron is excited "to the continuum"--reflecting the fact that once the electron is free, it has a whole continuum of available energies, instead of just the few bound-state energies allowed by the atomic potential.

EDIT: I just realized I ended up answering some questions raised in the comments and not the asked question. My point here is to agree with others that the photon is best viewed as having been absorbed and re-emitted in both processes, and to explain why the energetic considerations are different in each.

  • $\begingroup$ Am I right to interpret your answer as "any energy may be transferred to the electron as the electron is free (presumably after being excited by the photon)". Is that why Compton scattering is usually discussed with reference to X Rays? $\endgroup$
    – Dr Coconut
    Jun 1 '15 at 15:06
  • $\begingroup$ Yes, I think so. Basically, it makes a big difference whether you're aiming for bound states vs. free/scattering/continuum states as a final answer. Wikipedia tells me the connection to X-rays is historical (as in that's where they first noticed the effect) but I suppose the reason they saw it there first is because they were finally observing large enough energy scales. $\endgroup$
    – zeldredge
    Jun 1 '15 at 15:11

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