QM Compton scattering energy loss; check of derivation

[Ingwe]

Okay, so the question is an electron of energy 100 MeV collides with a photon of wavelength 3x10^(-3) m (ie: the CMB). What is the maximum energy loss of the electron?
After doing a few derivations for formulae, I came up with this one at work...could someone please let me know if there is anything is wrong with it?

Momentum Conservation:

Pe + Pγ = Pe' + Pγ'
Pe + hν/c = Pe' + hν'/c
c[Pe' - Pe] = h(ν-v')
c²Pe'² + c²Pe² - 2c²PePe'[cosθ] = (hv-hv')²
c²Pe'² = (hv-hv')² - c²Pe² + 2c²Pe'Pe[cosθ]
Where cosθ would be the angle between Pe and Pe'

Energy Conservation:

Ee + Eγ = Ee' + Eγ'
Ee + hv = [(mc²)² + (Pe'c)²]^½ + hv'
[Ee + (hv-hv')]² = m²c^4 + Pe'²c²
Ee² + 2(hv-hv') + (hv-hv')² = m²c^4 + (hv-hv')² - c²Pe² + 2c²Pe'Pe[cosθ]
Ee² + 2cPe' - 2cPe = m²c^4 - Pe² + 2c²Pe'Pe[cosθ]
Pe'[2c - 2c²Pe[cosθ]] = m²c^4 - c²Pe² - Ee² + 2cPe
Pe' = [m²c^4 - c²Pe² - Ee² + 2cPe]/2c[1 - cPe[cosθ]]

So since the energy loss is given as
ΔEe = Ee - Ee' = Ee - [(mc²)² + (Pe'c)²]^½, to maximize it we need to make Pe' as small as we can, which occurs when cosθ = (-1) ==> θ = 180

Does this make sense?

 

[member 553083]

If so, the maximum energy loss of the electron will be: ΔEe = Ee - [(mc²)² + [([m²c^4 - c²Pe² - Ee² + 2cPe]/2c)²]^½]

 

[wais]

Yes, this derivation looks correct. The equations you have derived follow the principles of momentum and energy conservation in Compton scattering. The final step of maximizing the energy loss by setting cosθ to -1 also makes sense, as this would result in the largest possible change in momentum and energy for the electron. Well done!