Compton scattering problem - how much did wavelength change by?

Click For Summary
The discussion focuses on calculating the change in wavelength of an X-ray photon scattered by an electron at a 30-degree angle using the Compton scattering formula. The equation applied is λ' - λ(0) = (h)/(m(e)c)(1 - cos θ), where the user initially calculated a value of 2.424 x 10^-12 m but was unsure about its correctness. Feedback indicated that the user neglected to multiply by the (1 - cos θ) term, and the initial value represented the Compton wavelength rather than the change in wavelength. It was confirmed that λ' - λ(0) does represent the increase in wavelength after scattering, with the increase ranging from 0 to twice the Compton wavelength. The user acknowledged the correction and expressed gratitude for the assistance.
daleklama
Messages
33
Reaction score
0

Homework Statement



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?

Homework 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.

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 :)
 
Physics news on Phys.org
daleklama said:

Homework Statement



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?

Homework 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.

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. :approve:

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.
 
Thank you very much, I corrected that :)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 13 ·
Replies
13
Views
7K
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K
Replies
1
Views
2K