In various sources (1, 2, 3, 4, to name a few) I have seen this graph shown below, that shows how intensity depends on the wavelength of the scattered photon $\lambda'$.

enter image description here

Now, I do understand what this graph shows conceptually: front-scattered photons preserve most of it's energy, so $\lambda'=\lambda_0$, and as the scattering angle increases from 0°to 180° (back-scattering), photon loses part of it's original energy so the energy of scattered photon is smaller (i.e. wavelength is larger $\lambda'>\lambda_0$).

What I don't understand is: how is intensity found analytically in this case? My guess is that these graphs are a depiction of experimental results, and intensity is being measured by detectors placed at certain angles.

But, also, I guess that there must be an analytical way to express this intensity, so that when it is graphed for certain $\theta$, it shows a pattern as seen in the picture above.

I tried using Planck's radiation intensity formula combined with $\Delta\lambda=\frac{h}{mc}(1-\cos\theta)$, but it didn't meet with the graphs above.

So my question is: how is intensity expressed analytically as a function of $\theta$ and $\lambda'$ in the case of Compton scattering?


1 Answer 1


Intensity depends on number of photons.

  1. lambda0 is received on collision with atom
  2. lambda` is received on collision with electron

xray photons have energies of 17 KeV and the bound state energies of carbon are about 300 eV. There are more free electrons (valence + loosely bound) as compared to number of atoms.Thats why the second peak is higher.

reference : https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2016/f928b8dce3d6a218fddda9617c5eb4f2_MIT8_04S16_LecNotes3.pdf


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