With Compton scattering, does the deflection angle depend on particle size?

[FrankJ777]

I'm researching a bit on Compton Scattering, especially in relation to, backscattering scanner, like one would see in an airport. I think I understood that the angle of deflection is in relationship to the loss of energy of the photon, where λ_f - λ_i = h/mc(1-cosθ). So if you know the origional frequency and the final frequency you can calculat the angle. But there's a few things I'm trying to understand.

Is the deflection angle, θ dependent on the mass of the particle it bounces off of?

Is the h/mc(1-cosθ), the energy imparted into the particle?

Since I'm researching backscatter scanners, where in that case i'd think, since the collector is colocated with the tranmitter, meaning a 180° reflection angle, then all rays obsorved would have lost h/mc(1-cos(180)) = 2h/mc. So I'm wondering if that is just a convenient angle to measure frm an engineering standpoint, or if that provides a benifite to identify certain materials (by the fact that they cause a photon to scatter at 180° angle.

I hope my question makes some sense.

Thanks

 

[sophiecentaur]

Summary:: Is the deflection angle, θ dependent on the mass of the particle it bounces off of?

Is the h/mc(1-cosθ), the energy imparted into the particle?

Is the deflection angle, θ dependent on the mass of the particle it bounces off of?

I think the Wiki article may have the answers for you. (it's a matter of context). The angle of scattering will depend (treating the collision as with snooker balls) on the path of the incident photon and conservation of momentum on the transverse axis. It's inelastic scattering, though.
The diagram at the top of the article shows the various levels of interaction between photons and matter. Compton scattering is amongst those examples.