# How many times can a photon be absorbed?wC

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.

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