# In Compton effect, why is there a continuous distribution of wavelengths (including $\lambda$)instead of just $\lambda'$ for every fixed angle?

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?

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 $$\lambda$$