Nuclear Physics, photons, electrons and positrons

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SUMMARY

The minimum energy required for a photon to create an electron-positron pair is determined by the rest mass energy of the electron, which is 511 keV (or 8.19 x 10-14 J). Therefore, the photon must have at least double this energy, totaling 1.022 MeV (or 1.64 x 10-13 J) to satisfy conservation of charge and energy. The relevant equation used to derive this energy is E=mc2.

PREREQUISITES
  • Understanding of photon energy and mass-energy equivalence
  • Knowledge of electron and positron properties
  • Familiarity with the concept of particle-antiparticle pairs
  • Basic proficiency in unit conversions between keV and Joules
NEXT STEPS
  • Study the principles of pair production in high-energy physics
  • Learn about the conservation laws in particle interactions
  • Explore the implications of E=mc2 in various physical contexts
  • Investigate the properties of photons and their interactions with matter
USEFUL FOR

Students of nuclear physics, educators teaching particle physics, and researchers interested in high-energy particle interactions will benefit from this discussion.

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Homework Statement


A photon of sufficient energy can spontaneously create an electron and positron. This will conserve charge and must conserve energy. What is the minimum energy photon (in MeV and J) that can create an electron-positron pair?


Homework Equations



V=W/q (?)


The Attempt at a Solution



I know that the photon must have twice the energy of the electron (or of the electron and positron added together), but I have no idea how to find it. I think I am missing the formula or some simple conversion, help please!
 
Physics news on Phys.org
Rest energy of an electron: 511KeV

Does that help?

It's obtained by just using the E=mc^2 formula on the Electron's mass.
 
Looks good to me! Thank you!
 

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