# Compton Scattering Question

## Homework Statement

Okay, so here's the problem:

In a Compton scattering experiment, a photon is scattered through an angle of 90.0 and the
electron is set into motion in a direction at an angle of 20.0 to the original direction of the
photon.

(a) Explain how this information is sucient to determine uniquely the wavelength of the
scattered photon.

(b) Find this wavelength.

## Homework Equations

λ2 - λ1 = (h/m c)(1 - Cosθ)

## The Attempt at a Solution

So, with the angle I'm given I can find Δλ, but I won't be able to find λ2 which is what I believe the question is asking me to find. I feel like I should be using the angle of the electron but I don't really see how I can. Can anyone help me out here? :S

rude man
Homework Helper
Gold Member
Write the x and y component equations for momentum conservation:

For x, the photon has initial momentum h/λ1 and final momentum is a function of p, scattered electron angle ψ, scattered photon angle θ (actually a new photon!), and λ2. p is the scalar relativistic momentum of the electron after collision so the x component of momentum for the electron would be p*cosψ.

Similarly, for the y direction, you have 0 = a function of p, the same two angles, and λ2.

On top of that you also need to invoke energy conservation: initial photon energy plus electron rest energy = new photon energy plus relativistic electron energy which of course is E = √[(pc)2 + E02].

3 equations, 3 unknowns: λ1, λ2 and p.

Last edited:
Well, I'm screwed. Thanks, though.

rude man
Homework Helper
Gold Member
Well, I'm screwed. Thanks, though.

Why? It's just basic momentum & energy conservation, and I think I gave you most of the relativistic stuff you need.

I have a vague idea of what I'm supposed to be doing but I'm just so horrible at this.
Ok, here's my attempt:

The initial momentum of the photon in the x direction PLUS the initial momentum of the electron in the x direction (would it be zero??) has to equal the sum of both final x direction momentums.

Sooooooooo

Ppix (Momentum photon inital x direction) = h/λ1
Peix (momentum electron initial x direction) = 0?

Ppfx = I have no idea

Pefx = p Cosψ (what is p exactly? Scalar Relativistic momentum? What does that mean?)

So Ppix + Peix = Ppfx +Pefx

And likewise with the y components, although I wouldn't have the first clue where to start with that

And energy. Hmmmmmm. So....

h f1 + electron rest energy = h f2 + √[(pc)^2 + E0^2]
?
Huh. What's E0? What's electron rest energy? What exactly is this "p"?! Yeah, I'm in bad shape right now :/

rude man
Homework Helper
Gold Member
The initial momentum of the photon in the x direction PLUS the initial momentum of the electron in the x direction (would it be zero??) has to equal the sum of both final x direction momentums.
Check.
Sooooooooo

Ppix (Momentum photon inital x direction) = h/λ1 Right.
Peix (momentum electron initial x direction) = 0? Right!

Ppfx = I have no idea
C'mon - what is the magnitude of the new photon momentum - you got the right expression for the incoming one - but now take just the x component?

Pefx = p Cosψ (what is p exactly? Scalar Relativistic momentum? What does that mean?)

p = mv and m is the relativistic mass. Remember how m is corrected for relativistic motion? You should review this area.

So Ppix + Peix = Ppfx +Pefx Right.

And likewise with the y components, although I wouldn't have the first clue where to start with that

Is there a y component of initial momentum for either the electron or the photon? As for the final, for the electron, it's going to be p times the sin of ψ, right? With a - sign since ψ is negative if you draw a picture of the "after" situation. And I know you know the new photon's momentum, but again taking just the y component.

And energy. Hmmmmmm. So....

h f1 + electron rest energy = h f2 + √[(pc)^2 + E0^2]
? Exactly!!

Huh. What's E0? What's electron rest energy? What exactly is this "p"?! Yeah, I'm in bad shape right now :/

Yeah, you need to review. Relativisically, rest energy derives from the special theory which exchanges energy for mass, and you know the electron has mass ...

Ok, so would the new x component momentum for the photon be h/λ2 * Cosθ? Or is there something I'm missing?

And would the initial y components for the photon and electron be zero?

As for the final y components would they be:

Ppfy= h/λ2 * sinθ?

Pefy= p sinψ?

So, in total:

h/λ1 + 0 = h/λ2 * cosθ + p cosψ <------x components

0 = h/λ2 * sinθ + p sinψ ---> -p sinψ = h/λ2 * sinθ <--------y components

But that's assuming that I got the final x and y components right

And for energy:

hc/λ1 + mc^2 = hc/λ2 + √[(pc)^2 + E0^2] ?

I'm not totally sure how to get the last part. Is E0 rest energy?

I really do need to review this stuff. Can you let me know if this is right? And I apologize for saying that I was screwed. That was pretty immature of me. I just get so dang frustrated with this stuff!

rude man
Homework Helper
Gold Member
Ok, so would the new x component momentum for the photon be h/λ2 * Cosθ? Or is there something I'm missing? Right.

And would the initial y components for the photon and electron be zero? Right!
As for the final y components would they be:

Ppfy= h/λ2 * sinθ? Yes!

Pefy= p sinψ? Almost. What did I say about the sign having to be negative? Do you see why? Draw a picture of the trajectories of the final photon & electron motion.

So, in total:

h/λ1 + 0 = h/λ2 * cosθ + p cosψ <------x components

0 = h/λ2 * sinθ + p sinψ ---> -p sinψ = h/λ2 * sinθ <--------y components

But that's assuming that I got the final x and y components right
Fix the sign of p sinψ

And for energy:

hc/λ1 + mc^2 = hc/λ2 + √[(pc)^2 + E0^2] ?
mc^2 is not the rest energy. E0 = m0c2 is. mc^2 is the total energy = rest energy + kinetic energy. m is always the relativistic energy. Replace your mc^2 with E0 =m0c^2.

So now you see that you can solve for λ1 and λ2, right? Which is what the question seemingly asks for.