Electron velocity after Compton Scattering

[Inferniac]

Homework Statement

In Compton scattering,how much energy must the photon have in order for
the scattered electron to achieve relativistic velocity?

Homework Equations

Compton scattering formula: $$λ'-λ=\frac{h}{mc}(1-cosθ)$$
$E=\frac{h}{λ}$,conservation of mass and momentum,possibly Lorentz transformations for velocity and kinetic energy?

The Attempt at a Solution

My train of thought goes like this:
Assume that θ=90°. That gives us $λ'-λ=\frac{h}{mc}$.
$λ'-λ$ can easily be turned into $E'-E$ using $E=\frac{h}{λ}$.
Using law of conservation of energy we solve for the kinetic energy of the scattered electron
Since we now know the electron's kinetic energy we can calculate its speed.

My problem lies with my first assumption.I don't know if it's correct. It was made after a hint that my professor made that we should use big angles.

Please note that I don't study physics in English so some things might require clarification.

Thank you for your time.

 

[szynkasz]

I think you should assume 180^o so that energy trasferred from photon to electron is maximal.

 

[BruceW]

Hi inferniac, welcome to physicsforums :)

The equation $E=\frac{h}{λ}$ is true in a natural system of units (where c=1), but in the rest of your post, it looks like you are keeping c not equal to 1. So maybe you forgot a c in the equation above?

Also, szynkasz has the right idea with the angle, although it is not usual in this forum to give the answer outright.

 

[Inferniac]

Thank you for the welcome.

You are absolutely correct,I misstyped $E=\frac{h}{λ}$ instead of $E=\frac{hc}{λ}$ the first time and I copy-pasted again.
Yes,assuming a 180° angle makes more sense.I'll see were it goes from here.

Thanks again for your time.