When a photon (with energy not equal to any transition in the energy levels of a particular atom) scatters from this atom, it interacts with one of its outer electrons and then changes its direction of propagation, consequently changes its momentum. The electron itself can not change its momentum (IMO) because its on a fixed Bohr orbit on some energy level and a change in momentum leads to change in energy which is forbidden (the next level is too far). But there must be conservation of momentum of the system. Where is the lacking momentum gone? Is it transferred to the nucleus? And if yes then how?
Is this what you are looking for? If I am correct, a qauntum jump does not occur, but energy is converted to other forms:
http://en.wikipedia.org/wiki/Electron_scattering#Inelastic
Inelastic scattering
Inelastic scattering is when the collisions do not conserve kinetic energy,[23][24] and as such the internal states of one or both of the particles has changed.[22] This is due to energy being converted into vibrations which can be interpreted as heat, waves (sound), or vibrations between constituent particles of either collision party.[23] Particles may also split apart, further energy can be converted into breaking the chemical bonds between components.[23]
Furthermore, momentum is conserved in both elastic and inelastic scattering.[23] The other two results are reactions (when the structure of the interacting particles is changed producing two or more (generally complex particles)), and that new particles that are not constituent elementary particles of the interacting particles are created.[22][23]
This is very general. I am looking for the mechanism behind a very specific scattering atom - photon. Photon collides with electron which can not change its state (imo). Or are there some dense set of energy levels (vibrational) around every permited by Bohr rule level?
Ilian,
It is very difficult to think of the quantum nature of an atom if you try to perceive the constituent entities that make up an atom as separate or "separable" ... or even the atom as separate from the system in which it is located.
You describe an incoming photon. You define that photon as a photon which is not allowed to act on the *isolated* electron. The momentum of the photon is altered, but it seems you are assuming that it was because of an interaction with a specific electron isolated from the rest of the atomic system. How do you know that this change in momentum occurred because of an interaction with that electron in that "fixed Bohr orbit?"
If the specific electron in question cannot alter its energy level due to the interaction with a photon, it might be worth asking "did the photon indeed interact with just that specific electron?"
To be clear, I am not criticizing you for asking the question. Asking such questions is the only way to penetrate quantum mechanics. I asked a somewhat related question here:
https://www.researchgate.net/post/How_does_quantum_entanglement_survive_transmission_through_a_lens
I'd summarize the answer to both your question and the one I asked as "the photon does not interact as you describe it. You need to think about the system as a whole, not just as interactions between its constituent parts." (HINT: The very word "particle" is misleading when thinking about an electron in an atomic system!)
To provide a more specific answer to your question "is it transferred to the nucleus" I might say "no, it is transferred to the wavefunction of the atomic system as a whole if the atom is isolated from other atoms, but is transferred to the wavefunction of the system of atoms in which the specific atom is situated if that atom is not ideally isolated and exists, for instance in a crystal, glass, hunk of cheese or some other configuration."
[And, Ilian, don't trust my word. Keep asking the same type of questions until you can balance the answers in your own head. Even your question is not isolated from the questions and assumptions that surround it! And ... I may be wrong. ;-) ]
There are three known mechanisms:
a) Compton scattering, if the Photon energy is high enough (>1000 eV). The Photon is absorbed and a new one with longer wavelength is created. The only explanation is QM.
b) At much lower energy (< 10 eV), Thomson scattering on a charged particle like an electron may happen. This effect can not be explained with QM, you need classical electrodynamics.
c) If the energy is low and the particle is uncharged and much smaller than the wavelength, Rayleigh scattering describes what happens. The changes in wave number of the scattered wave (not photon!) are typically smaller than 50 cm−1. This can not be described with QM, you need electrodynamics.
The electrons around the nucleus are not fixed, they are "shaken" a little bit by the electric field component of the wave, but stay in the vicinity of their orbits. The nucleus is much heavier and does not move. Accelerated charges emit electromagnetic waves (like a dipole antenna) with no preferred direction. This energy is subtracted from the electromagnetic wave. Since the atom is much smaller than the wavelength, it "consumes" a very small part of the whole energy of the extended wave.
You asked for the momentum: Consider a linear y-polarized wave propagating in z-direction. The electrons wiggle in y-direction and radiate mostly in the x-z-plane. Since everything is symmetrically, there is no sideways kick in x-direction. The atom as a whole does not change position.
One last thing: Don't invent lot of photons spreading out in the x-z-plane! Everything wiggles with the same frequency f, nothing has lower frequency. If the incoming wave has the energy hf, it can not generate lot of photons with this frequency. Energy must be conserved. QM can not help to solve this problem, which is so simple in terms of classical electrodynamics. The reason: the incoming photon is not absorbed, QM can contribute nothing.
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