How Is Energy Conserved in a Photon-Electron Collision?

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SUMMARY

The discussion centers on calculating the total energy of a photon-electron system after a collision, specifically involving a photon with a frequency of 2.00 x 1019 Hz colliding with a stationary electron. The key to solving this problem lies in utilizing the known speed of light to derive the photon's wavelength and applying the principles of energy conservation in particle collisions. The total energy can be calculated using the equation E = hf for the photon and considering the rest mass energy of the electron.

PREREQUISITES
  • Understanding of photon energy calculations using E = hf
  • Familiarity with the concept of rest mass energy, E = mc2
  • Knowledge of the speed of light as a constant in physics
  • Basic principles of conservation of energy in particle physics
NEXT STEPS
  • Research the calculation of photon energy using frequency and Planck's constant
  • Study the implications of relativistic effects in particle collisions
  • Learn about the Compton effect and its relation to photon-electron interactions
  • Explore the concept of momentum conservation in collisions involving photons
USEFUL FOR

Students studying physics, particularly those focusing on particle physics and energy conservation principles, as well as educators seeking to explain photon-electron interactions in a classroom setting.

Phoenixtears
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Homework Statement



An incident photon having a frequency of 2.00 E10^19 Hz collides with an electron initially at rest.

What is the total energy of the two-particle system after the collision? (In joules


The Attempt at a Solution



I've stared at this problem for at least an hour an have no idea where to go from here. Can anyone give me a leg up on this one. It feels like there isn't enough given information to even begin this... no velocity, no wavelength.

Thanks in advance!

~Phoenix
 
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Well, to be fair... it's a photon, so you know the velocity (ie. the speed of light), and since you have the frequency, you have the wavelength - perhaps you can see it from there?
 

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