What is the energy of the scattered photon?

• ZachWeiner

ZachWeiner

So, while working some problems in my modern physics books, I met this 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 with the original photon direction, what is the energy of the scattered photon?

The Compton effect equation is very simple, but it feels like you're missing a variable, since there's not a simple relation between the electron's angle (given) and the photon's angle (unknown). You can solve for one in terms of the other, but to do so you have to deal with the electrons momentum.

What am I missing?

It is just a two body process which can be solved exactly.

scattered photon energy vs its scattering angle and incoming photon energy:

$$k\text{'} = \dfrac{k}{1+ k(1-\cos \phi)}$$

Electron scattering angle:
$$\cot \theta = (1+k)\tan \left( \frac{\phi}{2} \right).$$

The electron momentum is:
$$p_e = \sqrt{T^2 + 2 T m_e}$$

where T is the kinetic energy, i.e k = k' + T + m_e

So why is there not a simple relation?

The Compton effect equation is very simple, but it feels like you're missing a variable, since there's not a simple relation between the electron's angle (given) and the photon's angle (unknown). You can solve for one in terms of the other, but to do so you have to deal with the electrons momentum.

You have to take into account conservation of momentum (both x and y components) and energy.

It's just a 4-momentum conservation problem. Just equate energy (0-component of 4-momentum) and the x and y components of 4-momentum before and after collision. That will give you 3 equations in 5 unknowns. Then use two energy-momentum relations (E^2-p^2 = m^2, c=1), one for the photon and one for the electron, to reduce this to 3 unknowns. Altogether, you'll have 3 unknowns (energy of photon after, energy of electron after, angle of photon scattering) and 3 equations.