# I Is the Compton scattering angle related to the incident energy?

1. Dec 1, 2017

The Compton equation determines the shift of the wave-length of the scattered photon as a function of the $cos\theta$ of the scattered photon. It does not depend on the energy of the incident photon. With some manipulation using the equations of conservation of the momentum and energy, one gets another equation describing the relation between the scattering angles, energy of the incident photon and that of the scattered electron.
Still, how does the scattering angle vary with the energy of the incident photon? Intuitively, the higher the energy the smaller the angle, but is it really true? I mean what makes the nature prefer a small angle when we increase the incident photon energy.
I thought of the following as a possible solution: At any angle, the higher the angle value, the lower the energy of the scattered photon by Compton equation. The difference between the incident and the scattered energy is equal to the kinetic energy of the scattered electrone by law of energy conservation. But, because the kinetric energy of the electron can not increase infinitely, this means a higher left-over energy of the scattered photon when the incident energy is high. But from Compton equation, the higher energy the lower wavelength and this can be only achieved at lower scattered angle.

2. Dec 2, 2017

### Staff: Mentor

For a given incoming photon energy, the outgoing photon can emerge at different angles, with correspondingly different energies. The familiar formula for the scattering angle doesn't give any information about which angle is most likely.

To get the probability distribution for the scattering angle, you need to use the differential cross section for Compton scattering, a.k.a. the Klein-Nishina formula. The scattering-angle formula is purely kinematic, and comes from energy and momentum conservation. The Klein-Nishina formula comes from quantum electrodynamics.

3. Dec 3, 2017