I was explained Compton effect as a collision between a photon and an electron that can be considered free. The equation is $\lambda' = \lambda + \lambda_c(1-\cos\phi)$

I've been reading more on the subject and I found the experimental curves of intensity vs wavelength. I am confused about the interpretation of these curves

For any fixed angle(except $0$ degrees) there are two picks, the first one corresponding to the wavelength of the incident radiation, and the other one corresponding to the scattered radiation. Not only that, but there's a continuous distribution of wavelengths (the black dots in the figures below) instead of only $\lambda'$ The incident radiation should not exist anymore at the detector(except for the $\phi =0$ direction where $\lambda=\lambda'$ ) .Instead I see all these wavelengths where each one should corrispond to a photon of different energy, that means that the original monochromatic radiation is not monochromatic anymore after the interaction? Where are all these additional photons coming from, if the equation says that that for a fixed $\phi$ and $\lambda$ , $\lambda'$ is uniquely determined?

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1 Answer 1


When scattering, ray of light can interact with the material in different ways:

  1. Rayleigh scattering: electrons, oscillating in EM wave, radiate wave with intensity proportional to $\frac{1}{\lambda^4}$. This happens when wavelength is greater that the size of an object (the electron is enough small and usually in the demonstrations of Compton effects we use light of visible range, so the scatterer can be thought as small)
  2. Photoelectric effect: some of the photons lose energy when they move electrons out from the atom core. The probability of this effect (and also its radiational cross-section) strongly depends on $\lambda$
  3. Ionization of atoms: some of the photons in your setup can be absorbed by atoms and then re-radiated

Finally, there are a lot of contributions in the intensity of the scattered light. Changing the wavelength, we make some of the contributions more probable and significant. But it is ok that intensity of reflected light changes continuously: when less photons scatter in Compton's processes, more photons do scatter in another kinds of processes.

  • $\begingroup$ "in the demonstrations of Compton effects we use light of visible range" ?Compton effect requires high energy radiation, typically X rays. $\endgroup$ Commented Nov 14, 2021 at 16:59
  • $\begingroup$ "Changing the wavelength....". Compton effect is done with monochromatic radiation, as far as I been reading , so I don't think this is correct, is it? $\endgroup$ Commented Nov 14, 2021 at 17:02
  • $\begingroup$ You are right about the wavelength in Compton effect. But this doesn't affect the situation much (when scatterer is in order of wavelength in size, there is Mie scattering instead of Rayleigh). On the plots in your question we have wavelength as a variable on the horizontal axis. That means that in the experiments monochromatic sources of different wavelength were used, and results for reflected intensity were then analysed $\endgroup$
    – goroshek
    Commented Nov 14, 2021 at 17:10
  • $\begingroup$ Well, all the books I have searched for say nothing about using lots of monochromatic sources to get these plots. They imply a monochromatic source is used and that's it. I find it difficult to believe that such important information is not mentioned. Do you know of any resource? $\endgroup$ Commented Nov 14, 2021 at 17:23

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