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Here we have the standard set up to observe the compton effect. From what I understand, due to the particle like nature of light, when the x ray photons collide with the electrons in the scatterer they lose energy and thus have a larger wavelength.

I was just wondering that if we were to remove the scatterer and just shine a narrow beam of x ray photons on the crystal, would we still observe a compton effect. I am just wondering as even in that case, we would still have the x ray photons striking the electrons in the crystal, is the same loss of energy not present there as in the case of the photons striking the scatterer first.

One reason I think we will not observe the effect is because the electrons in the atom are tightly bound, so the photons colliding with the electrons have a very small change in their overall energy as the electrons have the effective mass of the entire atom, thus gain little energy during such a collision. Although, this is just a guess.

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Yes, in fact Compton scattering is often an annoyance in x-ray diffraction when you want to study the diffuse background from crystal defects and disorder. The amount of Compton scattering is highly dependent on the xray energy, and is highest when it reaches the energy scale of the electron mass of 512 keV.

Below is a plot showing the various contributions to x-ray interactions with matter as a function of energy. Note how Compton scattering begins dominating at higher energy. "Thomson scattering" is what you would call ordinary x-ray diffraction. Photoelectric contributions amount to absorption of the x-rays, so they don't show up directly in the scattered beam, but explains why the most x-rays are absorbed not scattered.

I should add though, Compton scattering can actually be useful in studying crystals, as it can measure the momentum of electrons in the crystal. In a sense you can then reconstruct both the charge density and momentum density of electrons by combining x-ray scattering in the Thomson and Compton regimes.

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