1
$\begingroup$

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?

$\endgroup$
1

1 Answer 1

0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.