Question

A photon whose energy equals the rest energy of the electron undergoes a Compton collision with an electron. If the electron moves off at an angle of 40 degrees withthe original photon direction, what is the energy of the scattered photon?

Answer 1

the compton shift equation isλ' - λ_o = (h/m_ec) * (1 -cosθ)or λ' = λ_o + (h/m_ec) * (1 -cosθ) ----------(1)the energy of the incident photon isE_o= (h * c/λ_o)or λ_o= (h * c/E_o)h = 6.63 * 10^-34 J-sc = 3 * 10^8 m/sE_o is the rest energy of the electron and is equalto 0.511 MeV = 0.511 * 10^6 eV = 0.511 * 10^6 * (1/1.6 * 10^-19)Jthe mass of electron m_e= 9.1 * 10^-31 kgθ = 40 degreessubstituting the values in equation (1) we get the shift inwavelength therefore the energy of the scattered photonE' = (h * c/λ')

Answer 2

I did that, but it's not quite correct. The theta in the aboveequation is the angle of deflection of the photon. The 40 degreesis the angle ofdeflection of the electron.There is a relation hf'sin(theta)/c = psin(phi), where phi isthe angle of deflection of the electron. But, now you have tointroduce a c...
I can't figure out a way to bring those thingstogether...