Compton Scattering formula for 180 degree scattering.

[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!

Thanks in advance.

 

[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?

 

[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.

 

[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.

 

[Bob S]

See
http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/compeq.html
Bob S

 

[durand]

See
http://hyperphysics.phy-astr.gsu.edu/Hbase/quantum/compeq.html
Bob S

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 :)