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I found only specialized articles on the subject, do you know of any accessible article explaining the laws governing a photon colliding with a proton?

Would the same formula apply to the phenomenon of scattering by an electron if we changed $m_p$ with mass of the electron: $$\lambda' - \lambda = \frac{h}{m_p c}(1-\cos{\theta})$$

Does the same law apply to a H atom and all mono-atomic molecules?

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  • $\begingroup$ Why would you think they are different? Does anything in the derivation of the Compton effect rely on the particle being an electron specifically? $\endgroup$ – ACuriousMind Mar 7 '16 at 12:10
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For low photon energies, the behaviour is the same, although obviously the different mass of the proton and electron will affect the precise details.

At higher photon energies things change substantially, because the unlike an electron, a proton has structure and isn't a fundamental particle. In very broad terms, once the photon energy becomes comparable to the binding energy of the proton, things start to change, because a) the fact that you have three point charges (the discerete quarks) rather than a single point charge starts to matter, and b) you are dealing with bound charges, so consideration of excited states, energy levels etc. come in to play.

As an analogy consider scattering of photons off bound electrons in atoms. In the least Compton-like scenario, a photon is absorbed raising an electron to a higher energy state, and it then radiatively de-excites emitting multiple lower energy photons. This is nothing whatsoever like Compton scattering.

(Although I can't pretend to know enough about QCD to have any idea of what energy levels of quarks in a proton might look like, or whether there are in fact any).

Leaving aside the issue of whether an electron is a fundamental particle or not - it doesn't matter as long as the energy of the photon scattering off the electron is $\ll$ the binding energy of whatever constituent particles an electron might be made of, and at the highest energy interactions we've observed there is no indication of any electron structure.

There are additional complications for very high energy photons anyway, which also apply to Compton scattering, such as the probability for particle - anti-particle pair production which means that Compton scattering isn't the only thing going on once you get above the pair production threshold.

I'd suspect that that would actually come in to effect before any proton structure effects did, so as long as the photon is below the pair production threshold ($\approx 1$ MeV), protons will behave like a single point charge, much like an electron, aside from the mass.

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  • $\begingroup$ Does the formula apply to air molecules, that is to Raleygh scattering, too? $\endgroup$ – user104372 Mar 8 '16 at 5:28
  • $\begingroup$ @user104 Not really. In one sense the formula you quote is just an expression of conservation of energy and momentum, and is independent of a lot of the details of the scattering as long as it is elastic. Photons scattering of point charges with no structure are elastic. But when it comes to molecules in a gas, we don't really have elastic scattering. An molecule is neutral overall, with a complicated distribution of charge and lots of modes of excitation in the electron energy levels and molecular bonds, and is very different from Compton scattering. $\endgroup$ – PhillS Mar 8 '16 at 9:19
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Yes,the law is the same for these particles, neglecting gravitation forces just because of the order of magnitude of the masses of the particles that we are talking about.As these particle exert no force on the photon except fermi forces, the law is valid.

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