# Angle of Photon after Compton Scattering

1. ### Prologue

185
1. The problem statement, all variables and given/known data

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.

2. Relevant equations

E2 = (m0c2)2 + (pc)2

$$\lambda'=\lambda_{c}(1-cos\phi)+\lambda$$

3. 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:

$$\theta=-40^{o}$$

$$P_{x}=\frac{h}{\lambda}=\frac{h}{\lambda'}cos\phi+P_{e}cos\theta$$

$$P_{y}=0=\frac{h}{\lambda'}sin\phi+P_{e}sin\theta$$

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

$$tan\theta=\frac{\frac{h}{\lambda'}sin\phi}{\frac{h}{\lambda'}cos\phi-\frac{h}{\lambda}}$$

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

$$tan\theta=\frac{\frac{h}{\lambda_{c}(1-cos\phi)+\lambda}sin\phi}{\frac{h}{\lambda_{c}(1-cos\phi)+\lambda}cos\phi-\frac{h}{\lambda}}$$

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?

Last edited: Sep 13, 2009
2. ### kuruman

3,449
Are both angles theta and phi known? The statement of the problem leaves that ambiguous.

3. ### Prologue

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Sorry I was going to have to put some degree thing in there and just forgot to by the time I finished. It is fixed now. The angle of the electron is known, but the angle of the photon is not.

4. ### kuruman

3,449
OK, but why are you looking for lambda? The scattered photon has wavelength lambda prime. You know what lambda is because you have the energy of the initial photon. You need to eliminate angle phi from the equation.

Last edited: Sep 13, 2009
5. ### Prologue

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You are completely right. I just typed the wrong thing and didn't realize it. We need phi, then we can figure out lambda prime.

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Any insight?

7. ### kuruman

3,449
It is easier to eliminate φ rather than "finding" it. Go back to your momentum conservation equations. Isolate momentum components, i.e. write

pesinφ = ...
pecosφ = ...

Square both sides of each equation, then add them. You will be left with pe2 on the left. What can you do about that?

8. ### Prologue

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Those terms aren't in the equation. They are sin and cos of theta not phi. If I were to isolate the terms P_ecos and P_esin, then square and add it eliminates theta and leaves this

$$P_{e}^{2}=\frac{h^{2}}{\lambda'^{2}}+\frac{h^{2}}{\lambda^{2}}-\frac{2h}{\lambda'}cos\phi$$

Which gives us P_e squared as a function of phi. Basically the same type of relationship we had with the momentum equations, nothing new because we knew theta to begin with.

If I try the same approach with the phi terms we get

$$P_{e}^{2}-2P_{e}cos\theta+(\frac{h^{2}}{\lambda^{2}}-\frac{h^{2}}{\lambda'^{2}})=0$$

Which is a quadratic in P_e, if we solve that we get

$$P_{e}=cos\theta+(4cos^{2}\theta-4(\frac{h^{2}}{\lambda^{2}}-\frac{h^{2}}{\lambda'^{2}}))^{1/2},cos\theta-(4cos^{2}\theta-4(\frac{h^{2}}{\lambda^{2}}-\frac{h^{2}}{\lambda'^{2}}))^{1/2}$$

But this is a function of lambda prime which means it is a function of phi through the compton relation. It doesn't seem like we accomplished anything.

9. ### kuruman

3,449
Ignore the Compton formula. I am referring to your own equations that you posted. You can write

sinφ = ...
cosφ = ...

(I was wrong adding the pe up front), square and add. This gives one equation with two unknowns, pe and λ'. The energy conservation equation will allow you to eliminate pe and you will end up with an equation in λ' only. Actually, since you are asked to find the energy of the scattered photon, I would replace 1/λ' with E'/hc in my expressions and solve for E'.

10. ### zibaba

1
Through what angle must a 2x10^5 ev be scattered by free electron so that the photon looses 10% of its energy