# Homework Help: Compton Effect Question

1. Jan 26, 2012

### wolski888

1. The problem statement, all variables and given/known data
A 650-keV gamma ray Compton-scatters from an electron. Find the energy of the photon scattered at 110°, the kinetic energy of the scattered electron, and the recoil angle of the electron.
2. Relevant equations
Compton effect equation:
Δλ = $λ^{'}$ - λ = (h/mc) (1-cosθ)

Conservation of Energy:
hf + m$c^{2}$ = h$f^{'}$ + $E_{e}$

hf is initial photon energy, m$c^{2}$ is electron energy before scattering, h$f^{'}$ is the energy of scattered photon with new frequency f prime, and $E_{e}$ is the energy of the recoil electron with mc^2 and Kinetic Energy.

3. The attempt at a solution
First of all I am confused about the question. Are they saying a gamma ray is scattered? Or does a gamma ray hit the electron producing a scattering photon and the electron recoils?

For now, I am thinking that the gamma ray hits the electron (the latter mentioned).

For the recoil angle, I am thinking that it should be 80° since momentum is conserved and so the electron goes in the opposite direction, so 80°.

To find the energy of scattered photon I need to find the new frequency. My thought is that I need to find the new wavelength so that I can use f=c/λ. I can find the new wavelength using Compton Effect equation shown above. I know θ is 110°, and h,m,c are constants. But what about initial λ? Can I get that from the 650-keV? So E = 650-keV = hc/λ using h in terms of eV I can get λ. Right? Now I am all set to find the new wavelength.With that I can find the new frequency, and then find the scattering energy of the photon.

Now I should have all the information to find the Kinetic Energy using the Conservation of energy equation above.

Am I going the right way, any comments would be appreciated. Thanks! :)

2. Jan 26, 2012

### vela

Staff Emeritus
What's the difference? I don't understand the distinction you're trying to make.

You'll have to rethink this.

Good plan. You don't need to find the frequency though. The equation you already mentioned,
$E = hc/\lambda$, allows you to relate energy to wavelength directly.

3. Jan 26, 2012

### wolski888

Forget about the first question then. What do you mean rethink this? Should I use the Kinetic Energy of the recoiled electron to determine the angle?

EDIT: Or should I use conservation of momentum?

4. Jan 26, 2012

### vela

Staff Emeritus
You need to use momentum. Energy doesn't have a direction whereas momentum does.

5. Jan 26, 2012

### wolski888

Many thanks! I'll get to it then.