Relative angle after Compton scattering?

[skrat]

Homework Statement

Photon with $\lambda =10^{-12} m$ hits an electron (Compton scattering). After the interaction the photon and electron move under relative angle of 90°. Calculate the kinetic energy of electron. Hint: First find the relation between $\theta$ and $\varphi$. ($\theta +\varphi = 90°$)

Homework Equations

The Attempt at a Solution

I can't find the relation! How on Earth can I find it?

The only thing I get from components of momentum is: $\frac{hc}{\lambda ^{'}}sin\theta =cp_esin\varphi$

But this doesn't help me much... or does it?

Please help.

 

[mfb]

Use energy conservation together with momentum conservation. Alternatively, find a formula for the electron momentum and λ′ as function of the two angles.

 

[skrat]

Ok...

$cp_esin\varphi =E_{\gamma }^{'}sin\theta$ and therefore $\frac{cp_esin\varphi }{sin\theta }=E_{\gamma }^{'}$

momentum conservation along x axis:

$E_{\gamma }=cp_ecos\varphi +E_{\gamma }^{'}cos\theta$ using the last expression above:

$cp_e=\frac{E_\gamma }{cos\varphi +sin\varphi ctg\theta }$

so If I now insert this to $cp_esin\varphi =E_{\gamma }^{'}sin\theta$ than:

$E^{'}=\frac{E_\gamma }{sin(\varphi +\theta )}$

But I can't see how this helps me?

 

[mfb]

sin(φ+θ) = sin(90°) = ?

Hmm, that doesn't look right. Where does that come from?

 

[skrat]

$cp_e=\frac{E_\gamma }{cos\varphi +sin\varphi ctg\theta}$ and $cp_esin\varphi =E_{\gamma }^{'}sin\theta$

so

$\frac{E_\gamma }{cos\varphi +sin\varphi ctg\theta}sin\varphi =E_{\gamma }^{'}sin\theta$

$\frac{E_\gamma }{sin\theta cos\varphi +sin\theta sin\varphi ctg\theta}sin\varphi =E_{\gamma }^{'}$

$sin\theta cos\varphi +sin\theta sin\varphi ctg\theta=sin\theta cos\varphi+sin\varphi cos\theta =sin(\varphi + \theta )$

$\frac{E_\gamma }{sin(\theta +\varphi )}sin\varphi =E_{\gamma }^{'}$

I know it doesn't like right... :D That's why I am here asking.. :/