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## Homework Statement

A photon whose energy equals the rest energy of the electron undergoes a Compton collision with an electron if the electron moves off at an angle of 40 degrees with the original photon direction, what is the energy of the scattered photon.

## Homework Equations

E

^{2}= (m

_{0}c

^{2})

^{2}+ (pc)

^{2}

[tex]\lambda'=\lambda_{c}(1-cos\phi)+\lambda[/tex]

## The Attempt at a Solution

This problem has been bothering me for a while. I have tried several methods of solution but they all seem too complicated. One successful version (yet actually solved with a calculator) is this:

[tex]\theta=-40^{o}[/tex]

[tex]P_{x}=\frac{h}{\lambda}=\frac{h}{\lambda'}cos\phi+P_{e}cos\theta[/tex]

[tex]P_{y}=0=\frac{h}{\lambda'}sin\phi+P_{e}sin\theta[/tex]

Solve for the terms involving theta and divide the equations to eliminate the momentum of the electron

[tex]tan\theta=\frac{\frac{h}{\lambda'}sin\phi}{\frac{h}{\lambda'}cos\phi-\frac{h}{\lambda}}[/tex]

Then substitute in for lambda prime using the Compton equation to get

[tex]tan\theta=\frac{\frac{h}{\lambda_{c}(1-cos\phi)+\lambda}sin\phi}{\frac{h}{\lambda_{c}(1-cos\phi)+\lambda}cos\phi-\frac{h}{\lambda}}[/tex]

Now you have an equation involving only phi as a variable and 'can' solve for it, I had to use a calculator though. Is there a better method or a way to solve this equation analytically?

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