What Factors Affect the Emission of a Photon by an Electron?

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SUMMARY

The discussion centers on the factors affecting photon emission by an electron, specifically when an electron traveling at 1 x 106 m/s collides with a target, resulting in a halved speed and the release of a photon. The energy of the emitted photon is calculated using the equation Ephoton = (hc) / λ, where h is Planck's constant and c is the speed of light. The conservation of energy principle is applied, stating that the initial kinetic energy of the electron (Eelectron_initial) equals the sum of its final kinetic energy (Eelectron_final) and the energy of the emitted photon (Ephoton).

PREREQUISITES
  • Understanding of kinetic energy formulas, specifically 1/2 mv2
  • Familiarity with the concepts of photon energy and wavelength
  • Knowledge of conservation of energy principles
  • Basic grasp of the relationship between speed, energy, and wavelength in quantum mechanics
NEXT STEPS
  • Study the derivation of the photon energy equation E = (hc) / λ
  • Explore the implications of conservation of energy in particle physics
  • Learn about the relationship between an electron's kinetic energy and photon emission
  • Investigate the effects of speed changes on photon wavelength and energy
USEFUL FOR

Physicists, students studying quantum mechanics, and anyone interested in the principles of photon emission and energy conservation in particle interactions.

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an electron travels at 1 x 106 m/s. when it hits a target, its speed decrease in half and a photo is released. find the phton's wavelength.

so far i know E photon = (hc) / lamda

what about the speed? does the photons' energy same as kinetic energy of the electron which is just 1/2 mv ^2? if so, do i use the halved speed or the initial?
 
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Apply conservation of energy:

E_electron_inital = E_electron_final + E_phton
 

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