Photoelectric Effect - Reflected or Absorbed?

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SUMMARY

The discussion centers on the behavior of photons when they interact with a metal surface, particularly regarding their energy relative to the metal's work function. When a photon has insufficient energy, it is reflected without any loss of energy, contrary to the teacher's assertion that it is absorbed and emitted as infrared radiation. A smooth metal surface, such as that of a mirror, reflects light without altering its frequency, while a blackbody absorbs all incident light and reemits energy at a constant rate, potentially changing the frequency.

PREREQUISITES
  • Understanding of photon energy and work function in metals
  • Knowledge of the photoelectric effect
  • Familiarity with the concepts of reflection and absorption in physics
  • Basic principles of blackbody radiation
NEXT STEPS
  • Research the photoelectric effect and its implications in quantum physics
  • Study the properties of blackbody radiation and its applications
  • Explore the relationship between photon energy and metal work functions
  • Investigate the behavior of different materials in terms of light reflection and absorption
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Students of physics, educators explaining the photoelectric effect, and anyone interested in the principles of light interaction with materials.

wavingerwin
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Hi

When a photon of insufficient energy (lower that the metal's work function) arrive on the metal's surface, what will happen?

One part of my notes says that it will be "reflected with no loss of energy", but my teacher
has recently stated that it will be absorbed and emitted back as infrared radiation/heat.

Please help :-) thanks!
 
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A smooth metal surface reflects light without changing the frequency. The "backing" of a mirror is a metal coating. You can even see the reflection of your own face in a shiny spoon, etc.

A "blackbody" absorbs all of the light incident on it, and reemits energy at the same rate if the temperature remains constant, but the frequency can be changed.
 

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