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Compton scattering problem - how much did wavelength change by?

 
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May16-12, 03:10 PM   #1
 

Compton scattering problem - how much did wavelength change by?


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

How much will the wavelength of the incident X-ray photon change by if it is scattered by 30 degrees when it interacts with an electron?

2. Relevant equations

This is Compton scattering.

This is the equation I used:

lambda ' - lambda (0) = (h)/(m(e)c) (1-cos theta)

where h is planck's constant, me is the mass of the electron, c is speed of light.

3. The attempt at a solution

Basically I just subbed in everything into the equation above, and got 2.424 x 10^-12 m.

The thing I'm not sure about is whether I'm finished or not.

Does lambda ' minus lambda (0) give me the difference? Did I use the right form of the equation here? Do I need to do anything else?

Thank you :)
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May16-12, 10:10 PM   #2
 
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Quote by daleklama View Post
1. The problem statement, all variables and given/known data

How much will the wavelength of the incident X-ray photon change by if it is scattered by 30 degrees when it interacts with an electron?

2. Relevant equations

This is Compton scattering.

This is the equation I used:

lambda ' - lambda (0) = (h)/(m(e)c) (1-cos theta)

where h is planck's constant, me is the mass of the electron, c is speed of light.

3. The attempt at a solution

Basically I just subbed in everything into the equation above, and got 2.424 x 10^-12 m.

The thing I'm not sure about is whether I'm finished or not.
You forgot to multiply by the (1 - cos θ) part of the formula. Your value of 2.424 x 10-12 m is just h/(mec), the Compton wavelength of an electron. (And you might want to check the rounding on that too.)
Does lambda ' minus lambda (0) give me the difference?
I'm pretty sure, yes.

λ' - λ0 represents the increase in wavelength of the photon after scattering, as compared to the wavelength of the original photon. This increase in wavelength is at minimum 0, and at maximum twice the Compton wavelength.
May17-12, 02:51 AM   #3
 
Thank you very much, I corrected that :)
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