How Do We Derive the Compton Electron Energy Equation?

[P-Jay1]

Hi, I'm not sure how to do B1 part c, can anyone help?

http://qm-web.library.qmul.ac.uk/exams/science/physics/2010/PHY-215.pdf

 

[sgd37]

just using momentum conservation

$$\frac{h}{\lambda} = cos(\theta)\frac{h}{\lambda'} + cos(\phi)p_{\mu}$$

you can get $$p_{\mu}$$ from the other momentum conservation equation i.e.

$$0 = sin(\theta)\frac{h}{\lambda'} - sin(\phi)p_{\mu}$$

and you can get the $$\lambda'$$ in terms of $$\lambda$$ from the Compton formula now you have all the ingredients to solve for $$\lambda$$

 

[P-Jay1]

I keep getting lambda equal to lambda' - I'm at a dead end. If both angles are 30, then Pmuon is equal to h/lambda' , right? Excuse the lack of symbology

 

[frozenguy]

I can't get into that area in the link..

Cant you copy/paste?

 

[P-Jay1]

I can't get into that area in the link..

Cant you copy/paste?

I can't, i'll type it roughly

Basically the first bit asks you to derive the Compton scattering formula.

And for the second the the question to the answer I want is-

Q. Consider the case where the angle between the incident and out going light rays is 30 degrees. Calculate the value of the incident wavelength when also the angle between the incident light ray and recoiling electron is 30 degrees.

 

[P-Jay1]

Is it even possible to derive the wavelength in its entirety when you only know the angle?

 

[frozenguy]

Is it even possible to derive the wavelength in its entirety when you only know the angle?

well.. I've never derived the equation and I'll need to think about it for a second..

But the relation of the angle of recoiling electron (phi) to the angle of scattered photon (theta) is this: $$P_{p}sin\theta = P_{e}sin\phi$$

With P_p = momentum of scattered photon
P_e = momentum of recoiling electron

The momentum of the scattered photon is given by $$\frac{hc}{\lambda'}\frac{1}{c}$$ (the 1/c is for units only. The hc is in units of [nm*eV] so the momentum will be in eV/c)

The momentum of the recoiling electron is $$\frac{\sqrt{K^{2}_{e} + 2K_{e}m_{e}c^{2}}}{c}$$


With $$K_{e} = hc\left(\frac{1}{\lambda} - \frac{1}{\lambda'}\right)$$


And of course you have the compton formula which you need to learn how to derive I guess as well lol.. It will relate $$\lambda$$ and $$\lambda'$$

You have theta and phi, both are thirty degrees.


The derivation starts with $$\lambda' = \lambda + \Delta\lambda$$

 

[frozenguy]

I keep getting lambda equal to lambda' - I'm at a dead end. If both angles are 30, then Pmuon is equal to h/lambda' , right? Excuse the lack of symbology

If both angles are thirty degrees, that means that the momentum of the scattered photon = momentum of recoiling electron.

 

[P-Jay1]

Finally! The penny has dropped. Thanks for the help.

 

[gaddu13]

how can we derive the equation of energy of compton electron

Ee = hν{α (1 − cosθ )/[1 + α (1 − cosθ )]}

where

α=hν/m0 c^2
m0=mass of electron