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If I blast a material with photons at a frequency strictly greater than the difference between any two energy levels of all electrons existing in that material, will any of these electrons jump to a higher energy state? Some say the energy match must be exact. Some say the balance moves on as a lower energy photon, which might be eternal if lower than any bottom level Intuitively, the material should heat up. How is this heat manifested? Thank you

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  • $\begingroup$ Is this frequency greater than the ionization energy? $\endgroup$ – probably_someone Jun 20 '18 at 19:14
  • $\begingroup$ Is the material an atom/molecule, with discrete electronic energy states, or a solid with a continuous band? $\endgroup$ – Gilbert Jun 20 '18 at 19:21
  • $\begingroup$ Gilbert, I am an electrical engineer asking a basic question here, not one pertaining to a new state of matter. I am trying to understand how photons interact with traditional solids. I understand that the electron energies in solids are modeled as discrete. You mention a continuous band. Are you thinking of the traditional conduction band where perhaps electrons might be accelerated by an electric field an take on a continuum of energies? $\endgroup$ – Silliq Jun 20 '18 at 19:33
  • $\begingroup$ Sorry, i guess this would habe to be greater than the ionization energy. Seems like it should do something. I pose it in order to understand those who say that the photon energy hf must exactly match an energy shell difference in order for it to be absorbed. $\endgroup$ – Silliq Jun 20 '18 at 20:45
  • $\begingroup$ The ionization energy pertains to gases. I was thinking of solids here but there must be something similar for solids, probably larger $\endgroup$ – Silliq Jun 20 '18 at 20:52
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There is Compton scattering

the X-rays are scattered through an angle $θ$ and emerge at a different wavelength related to $θ$ . Although classical electromagnetism predicted that the wavelength of scattered rays should be equal to the initial wavelength, multiple experiments had found that the wavelength of the scattered rays was longer (corresponding to lower energy) than the initial wavelength.

compt

These Feynman diagrams can be used for modeling the interaction of photons with particles and/or fields.

The models are complicated, example :

Direct determinations of electron momentum densities in solids by measuring Compton scattered photons in coincidence with recoil electrons will allow more precise testing of electronic structure calculations.

In the case of solid matter, the electrons in the diagram may be virtual, representing the interaction of the photon with the field and the loss of energy of the photon. How the energy is dissipated will depend on the solid, its lattice rotation and vibration levels as the excess energy turns into kinetic energy and heat.

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Given a particular material, one would find that the material has a particular absorption spectrum. This represents a function with the probability density for a photon to be absorbed by that material as a function of the frequency (wavelength) of the photon. So, if your photon has a frequency that does not match the energy differences between the energy bands of the electrons in that material, then the probability for absorption would be low. Therefore, the material would be transparrent for the photon at that frequency.

However, one would find that the absorption peaks in the spectrum have a finite width. So, even if the frequency does not match the energy difference exactly, there is still a nonzero probability that the material can absorb the photon. What happens with the difference in energy? Well, that can go into (or be borrowed from) other forms of energy, such as thermal vibrations in the material.

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