2. Relevant equations
From what I’ve read in my text book I believe I should be using the equations:
E(kinetic) = (Gamma-1)*m*c^2
3. The attempt at a solution
For the first part:
Ok, so first of all I wasn’t entirely sure what energy the 0.7M electron volts was referring to, but I assumed it meant the incident photon. So I took that 0.7M electron volts and worked out the initial momentum of the system, in the x direction. I also noted that the initial y momentum of the system was zero.
Then, using trig and the angles given, seperated this momentum into x and y components for both the photon and the electron. The expressions were in E’, v etc.. Basically unknown quantities. I assume I must equate these knowing that.
Px = P(photon)+P(electron) = 0.7×10^-6eV /c
Simularly, Py = P(Photon)-P(Electron) = 0
Given these both have the angles in them I assume there must be some trick to work out the angle. I’m stuck at that point, assuming I haven’t made some mistake… Which is quite possible as I’m struggling with quantum mechanics lol.
For the second part, I assume that knowing the angles the particles are scattered I can find the velocity of the electron and hence it’s kinetic energy.
Py = P(photon)-P(electron) = 0