Compton Scattering Problem

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SUMMARY

The discussion focuses on the Compton scattering problem, specifically demonstrating that a photon cannot scatter at an angle greater than 60 degrees and still produce an electron-positron pair. The Compton equation, λ' - λ = λc(1 - cos(φ)), is utilized, where the Compton wavelength of the electron is calculated as λc = 2.426 x 10-12 m. The rest energy of the electron and positron is given as 0.511 MeV each, leading to a necessary maximum wavelength for pair production that must be derived from the energy considerations. The discussion emphasizes the importance of using the sum of rest energies rather than their product in calculations.

PREREQUISITES
  • Understanding of Compton scattering and the Compton equation
  • Knowledge of photon energy and wavelength relationships
  • Familiarity with electron-positron pair production and rest energy concepts
  • Basic proficiency in unit conversions and dimensional analysis
NEXT STEPS
  • Study the derivation of the Compton wavelength formula
  • Learn about the conditions for electron-positron pair production
  • Investigate the implications of scattering angles in photon interactions
  • Explore energy conservation principles in high-energy physics
USEFUL FOR

Physics students, educators, and researchers interested in particle physics, particularly those studying photon interactions and quantum mechanics.

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Homework Statement



Show that, regardless of its initial energy, a photon cannot undergo compton scattering through an angle of more than 60 degrees and still be able to produce an electron positron pair. (Hint: Start by expressing the Compton wavelength of the electron in terms of the maximum photon wavelength needed for pair production.)

Electron/Positron rest energy: 0.511 MeV



Homework Equations



Compton equation: [tex]\lambda' - \lambda = \lambda_{c}(1 - cos(\phi))[/tex]


The Attempt at a Solution



[tex]\lambda_{c} = \frac{h}{mc} = 2.426*10^{-12}[/tex]

[tex]\frac{Mass of e^{-}*Mass of e^{+}}{h} = \frac{2(0.511)*10^{6}eV}{3.97*10^{-15}eV*s} = 2.57 * 10^{20}Hz[/tex]


Umm yeah, that is about it. In other words I have no idea how to express it in terms of the maximum wavelength needed for pair production. I'll keep working on it though.
 
Last edited:
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The units of lambda_c are meters. Don't leave off units. For the second calculation you mean sum of the rest energies, not product of the rest masses (at least that's what you actually calculated). Now turn that Hz into a wavelength. That's the maximum value of lambda'. Now put theta=60 degrees in. What do you notice?
 
I did notice what was happening, but I didn't know how to show it. You are correct about the multiplying of the masses, I was really just trying to illustrate what I was doing in the next step, you are correct that it should be a sum. I ended up going with the 'explanation in english' approach rather than a more mathematical way. Thanks for the reply.
 

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