# Show Backscattered Photons have E = 250keV

1. Mar 4, 2010

### Oijl

1. The problem statement, all variables and given/known data
Gamma-rays can reach a radiation detector by Compton scattering from the surroundings. This effect is known as "backscattering." Show that, when E>>mc^2, the backscattered photon has an energy of approximately 250 keV, independent of the energy of the original photon, when the scattering angle is 180 degrees.

2. Relevant equations

3. The attempt at a solution
E>>mc^2 means that the kinetic energy of the electron is much greater than the rest energy, which means you can approximate the total energy E as the kinetic energy K, so that E = K, and also that the electron will be moving very fast, so that you can approximate v = c, which means you can approximate E = pc.

Now, if the scattering angle of the photon is 180 degrees, then the scattering angle of the electron is zero. That means

p = p$$_{e}$$ - p'

where p is the momentum of the incident photon, p' of the scattered photon, and p$$_{e}$$ of the electron.

Since for all these cases I can write p = E/c, I can say

E = Ee - E'

I don't know, but this did look like a good path, but now I'm stuck. Is it obvious to anyone how to show this, and if so, can you tell me where I should start, what I should be looking at? Thanks.