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## Homework Statement

A 100KeV collides with an electron at rest. It is scattered through 90 degrees. What is the energy after the collision? What is the kinetic energy in eV of the electron after the collision and what is the direction of its recoil.

## Homework Equations

Compton Scattering

[tex]λ_{f}-λ_{i}=\frac{h}{m_{e}c}(1-cos(θ))[/tex]

Energy of a Photon

[tex]E=hf[/tex]

## The Attempt at a Solution

What confuses me the most is the wording. What does "Scattered THROUGH 90 degrees mean?" Does that mean that the scattering angle of the photon is 90 degrees? Or does it mean that the scattering angle is between 0 and 90? I cannot confidently do this problem unless I fully understand the wording.

I took it as a scattering angle of 90 degrees.

So first I found the energy of the photon before the collision in Joules, and plugged this and theta=90 degrees into the Compton Scattering formula to get the final wave length, (from which I can get the energy of the photon.) I got about 84 KeV. Seems reasonable. Then I did an energy balance to get the energy of the electron.

[tex]E_{pi}+E_{ei}=E_{pf}+E_{ef}[/tex]

[tex]100keV+\frac{m_{e}c^{2}}{1.602×10^{-19}}=84keV+E_{ef}[/tex]

Solving this I get the final energy of the electron to be 527.89keV. Again seems reasonable considering its rest energy is on the order of 500 keV.

The last part is what I am having confusion with. If the scattering angle of the photon was indeed 90 degrees, that would mean the angle of the electron scatter would be 0 degrees. I did a momentum balance and I got an angle that was extremely close to zero. My concern is that I misunderstood the meaning of the words "through 90 degrees" and did the problem incorrectly.