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The energy levels of electrons in an atom are quantized $E_i$. A photon of a specific momentum $\vec p$ and energy $$\omega=(E_{i+1}-E_i)/\hbar$$ hits an atom and gets absorbed. Okay now say the energy of the photon is $$\omega'=(E_{i+1}-E_i)/\hbar+\Delta E/\hbar,$$ where the value $\Delta E$ is some energy, which is not even near the value of any electron state gap energy. hence the quantity $\Delta E$ cannot be absorbed by means of excitation of the other particle. What happens in this case? Does this energy-wise non-fitting photon just go through the material? Does the photon "spit" in a way that the part $E_{i+1}-E_i$ of the photon energy gets absorbed and a new photon with energy $\Delta E$ will move away? Microscopically, has this anything to do with soft radiation?

Okay well, if we model the atom as only being able to suck up energy by changing it's energy levels, this would violate momentum conservation. I guess the atom must start to move too, or if we view it as part of a bigger solid state system, the photon will make the thing wobble a little. From that perspective, we'd have to consider the kinetic energy distribution of the particles composing the material and this would be able to suck up any real energy or momentum value.

If we leave the microscopic description (in terms of which I'd like to see an answer to the first question - after all, I'm speaking of the behaviour of a single photon here) and go on and view this as a "gray body" problem, and in case the answer really is that after the photoelectric effect the rest energy gets termalized, then I'd like to know the relation (for a class of materials) between the absorption per frequency and material characterizing parameters like (electron/atom density? emissivity? ...).

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"if we view [the atom] as part of a bigger solid state system, the photon will make the thing wobble a little" You are rapidly setting up the preconditions for the Mössbauer effect. –  dmckee Jan 28 '13 at 15:52
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There is nothing that requires a reaction like $$ \gamma + A \to A^{*}$$ as opposed to $$ \gamma + A \to A^* + \gamma'$$ were the emitted photon carries off the excess energy.

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