Compton Scattering formula for 180 degree scattering.

1. Jan 20, 2010

durand

Hi,
I'm trying to derive a simple formula for 180° scattering.

I've got to this stage and I really can't figure out how to simplify it further.

$$$\frac{1}{\lambda}-\frac{1}{\lambda'} = \frac{2m_ec}{h}$$$

What I actually need is:
$$$\lambda' - \lambda = \frac{2h}{m_ec}$$$

I'm pretty sure the first formula is right but I can't seem to simplify it into the second!

2. Jan 20, 2010

tiny-tim

Hi durand!

(have a lambda: λ )

If you multiply them together, you get (λ' - λ)2 = 4λ'λ, or λ'/λ = 3 ± 2√2

How did you get your equation?

3. Jan 20, 2010

durand

Uhm, I derived the first using conservation of momentum and energy at non relativistic speeds, when the photon bounces back. The second comes from the standard compton scattering formula.

4. Jan 20, 2010

durand

CoM: h/λ = mv + h/λ'
CoE: hc/λ = hc/λ' + 0.5mv²

By substituting one into the other, I reach the formula I mentioned in my first post.

5. Jan 20, 2010

Bob S

6. Jan 21, 2010

durand

Yeah, I did find that, however, it uses a relativistic derivation so I can't really see how to do the last step as it's totally different to mine :/ Thanks anyway.

My exam's in an hour so it doesn't really matter now. Thanks everyone for your help :)