massphotonswavelength
I have the formula for Compton wavelength:
$$\lambda_{c}= \frac{h}{m_{0}c}$$
In this equation, is $m_0$ the mass of the electron that the photon hit?
I got online that this might be the photon rest mass, but it is basically 0, and it is impossible to do the calculation to get $\lambda_c$ if it is equal to 0.
As all scattering process in quantum mechanics, finding the scattered particle at a particular direction is random, because the scattered wavefunction is not a plane wave.
The usual derivation of Compton scattering overlooks the wave nature of electron and light, and hence gives you false perception that the scattered electrons and photons have a definite four-momentum in any individual event. In fact, they are in superposition of momentum eigenstates after scattering.
The frequency of light is not a property of many photons but a property of a single photon. (This is also strictly inaccurate, since we should think about a field with ripples in it; we then call a little clump of waves a photon.)
Anyway, let's imagine a photon/wavepacket as having a typical wavelength $\lambda$ given by the peak-peak spacing in this diagram:
This is a single photon, with the colour given by wavelength $\lambda$. Then this whizzes past you at the speed of light, $c$. But then you see the peaks go past at a rate of 1 per $\lambda/c$. Thus there is a sensible definition of frequency as $f=c/\lambda$. Then we let $E=hf$ and carry on happily.
This is all essentially an example of wave-particle duality. If you want to try to understand this better, try reading about the photoelectric effect. The key idea is that there is a wave-like phenomena (light) which we perceive as interacting with the rest of the world in discrete, particle-like events (photons).
Best Answer
The Compton wavelength given by,
$$\lambda=\frac{\hbar}{mc}$$
is a natural length scale associated to any particle with mass $m\neq 0$. As Professor Tong, states: