Energy of a photon scattered due to the Compton Effect

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SUMMARY

The discussion centers on the Compton Effect and the calculation of the maximum energy of a scattered photon. Participants emphasize the importance of understanding the relationship between the energy of the incoming photon (E0) and the scattered photon (E) as a function of the scattering angle (θ). The maximum energy of the scattered photon is established as 2m0c2, regardless of the incident photon's energy. The conversation highlights the necessity of using calculus to derive the function E(θ) and the implications of minimizing the change in wavelength (Δλ).

PREREQUISITES
  • Understanding of the Compton Effect and photon scattering
  • Familiarity with relativistic energy equations, specifically E = mc2
  • Basic knowledge of calculus, particularly in maximizing functions
  • Concept of wavelength change (Δλ) in photon interactions
NEXT STEPS
  • Study the derivation of the Compton wavelength shift formula
  • Learn how to apply calculus to optimize functions in physics problems
  • Explore the implications of relativistic energy conservation in particle interactions
  • Investigate real-world applications of the Compton Effect in medical imaging and astrophysics
USEFUL FOR

Physics students, educators, and professionals interested in quantum mechanics and particle physics, particularly those focusing on photon interactions and energy calculations.

PhysicsTruth
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Homework Statement
Show that the maximum energy of a scattered photon in Compton experiment is 2moc^2(mo=rest mass of electron). Also find the corresponding angle at which this max energy occurs.
Relevant Equations
Delta lambda=h/moc (1-cos(theta))
I don't really know how to proceed. Can someone help me out in this?
 
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I don't really know how to proceed. Can someone help me out in this?
You can't get any help from anyone here if you say this. This is just ##\textbf{put-into-the-formula}## problem
 
No it really isn't ! I'm not that dumb really. I had tried using the relativistic energies of photon and electron, and the fact that delta lambda should be minimum for the scattered photon to have max energy, but to no avail.
 
Do you know how to use calculus to maximize a function?
 
Yeah, but on which formula should I use it? I'm confused. For delta lambda to be minimum, theta had to be 0. But then, how is the energy moc^2? Really, I don't get it.
 
PhysicsTruth said:
Yeah, but on which formula should I use it? I'm confused. For delta lambda to be minimum, theta had to be 0. But then, how is the energy moc^2? Really, I don't get it.
2moc^2, my bad.
 
phyzguy said:
Do you know how to use calculus to maximize a function?
PhysicsTruth said:
Yeah, but on which formula should I use it?
From the problem statement, it should be apparent that the relevant function is the energy of the scattered photon as a function of angle. Call the energy of the incoming photon ##E_0##, and call the energy of the scattered photon ##E##. Can you come up with a function ##E(\theta)##?

Hint: what is the relationship between ##E## and ##\lambda##?
 
jtbell said:
From the problem statement, it should be apparent that the relevant function is the energy of the scattered photon as a function of angle. Call the energy of the incoming photon ##E_0##, and call the energy of the scattered photon ##E##. Can you come up with a function ##E(\theta)##?

Hint: what is the relationship between ##E## and ##\lambda##?
I get: ##1/E##= ##1/E_0## + (1-cos(theta))/moc^2
But for E to be max, theta should be 0,so how to do it now?
 
PhysicsTruth said:
I get: ##1/E##= ##1/E_0## + (1-cos(theta))/moc^2
But for E to be max, theta should be 0,so how to do it now?
Can you please guide me how should I proceed further?
 
  • #10
Please help someone!
 
  • #11
PhysicsTruth said:
But for E to be max, theta should be 0
Indeed! and what does E equal in that case?

Hmmm... are you sure you copied the problem statement correctly and completely?
 
  • #12
The question says- **Show that the maximum energy of the scattered photon will be 2moc^2, irrespective of the energy of the incident photon.**
 
  • #13
I imagine most of the helpers here think that question is obviously wrong, which may be why you're a little confused as to what to do. You may want to check with your instructor if it was worded correctly.
 

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