Compton Scattering Question

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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 sucient 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
 

Answers and Replies

  • #2
rude man
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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.
 
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Well, I'm screwed. Thanks, though.
 
  • #4
rude man
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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.
 
  • #5
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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 :/
 
  • #6
rude man
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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 ...
 
  • #7
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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!
 
  • #8
rude man
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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.
 

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