Calculating X-ray Photon Scattering: Change in Wavelength Explained

  • Thread starter Thread starter Noirchat
  • Start date Start date
  • Tags Tags
    X-rays
AI Thread Summary
The discussion focuses on calculating the change in wavelength of X-ray photons during Compton scattering. The relevant equation for this calculation is Δλ = h/mc(1 - cosθ), where θ is the scattering angle. For Photon 1, scattered backwards, the change in wavelength is calculated but results in an incorrect answer. Photon 2, scattered at right angles, yields a change of zero, while Photon 3, scattered at 45°, results in a calculated wavelength change of 8.666 x 10^-13. There is confusion regarding the accuracy of these calculations, particularly concerning the Compton constant.
Noirchat
Messages
15
Reaction score
0
I'd really like some explanations please, just looking at part (a) at the moment :)

Homework Statement



Suppose three 1.02 MeV X-ray photons are Compton scattered by single collisions with nearly stationary electrons.
Photon 1 is scattered backwards, in the direction opposite to its original path.
Photon 2 is scattered at right angles to its original path.
Photon 3 is scattered in a direction 45° away from the forward direction.

(a) Calculate the change of wavelength for each case. Show your working.
(b) Suppose instead, the original photons energies were 0.51 MeV. What effect does this have on the values calculated above? Explain very briefly.

Homework Equations


None were given

The Attempt at a Solution



(a) I think i use Δλ=h/mc(1-cosθ)

Photon 1:

Photon 2:

6.626 X 10^-34/9.1094x10^-31 x 3x10^8 x (1-cos90

= 0

Photon 3:

6.626 X 10^-34/9.1094x10^-31 x 3x10^8 x (1-cos45)

= 8.666x10^-13
 
Last edited by a moderator:
Physics news on Phys.org


Noirchat said:
I'd really like some explanations please, just looking at part (a) at the moment :)

Homework Statement



Suppose three 1.02 MeV X-ray photons are Compton scattered by single collisions with nearly stationary electrons.
Photon 1 is scattered backwards, in the direction opposite to its original path.
Photon 2 is scattered at right angles to its original path.
Photon 3 is scattered in a direction 45° away from the forward direction.

(a) Calculate the change of wavelength for each case. Show your working.
(b) Suppose instead, the original photons energies were 0.51 MeV. What effect does this have on the values calculated above? Explain very briefly.


Homework Equations


None were given


The Attempt at a Solution



(a) I think i use Δλ=h/mc(1-cosθ)
This would be a relevant equation. :wink:

Photon 1:

Photon 2:

6.626 X 10^-34/9.1094x10^-31 x 3x10^8 x (1-cos90

= 0

Photon 3:

6.626 X 10^-34/9.1094x10^-31 x 3x10^8 x (1-cos45)

= 8.666x10^-13
Did you have a specific question? Your answers are not correct, by the way.
 
vela said:
This would be a relevant equation. :wink:


Did you have a specific question? Your answers are not correct, by the way.

Great I'm on the right track then. Hmm I'm not sure why they're incorrect. Is it to do with the Compton constant?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top