What Is the Maximum Kinetic Energy of a Photoelectron with a 400 nm Wavelength?

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SUMMARY

The maximum kinetic energy (KE max) of a photoelectron emitted from a surface with a work function of 5 eV, when illuminated by photons of 400 nm wavelength, is 0 eV. This conclusion is derived from the equation hc/(lambda) = work function + KE max. Given that the energy of the incoming photons is less than the work function, the photoelectric effect does not occur, resulting in zero kinetic energy for the emitted photoelectron.

PREREQUISITES
  • Understanding of the photoelectric effect
  • Familiarity with the work function concept
  • Knowledge of the equation relating photon energy to wavelength
  • Basic grasp of electron energy levels in physics
NEXT STEPS
  • Study the photoelectric effect in detail, focusing on the conditions for electron emission
  • Learn about the relationship between photon energy and wavelength using the equation E = hc/lambda
  • Investigate the implications of work function in various materials
  • Explore experimental setups for measuring photoelectron emission
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Students in physics, educators teaching quantum mechanics, and researchers interested in the photoelectric effect and its applications in modern technology.

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Homework Statement


What is the maximum kinetic energy (in eV) of a photoelectron emitted when a surface, whose work function is 5eV, is illuminated by photons whose wavelength is 400 nm
1. -1.89
2. 1.89
3. 0
4. 3.1

Homework Equations


hc/(lambda)=work function + KE max


The Attempt at a Solution


After plugging numbers in the formula above, I get KE max=-1.89 eV, but I guess I must have done something wrong, since energy cannot be negative.
 
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looks like your calculation is correct. Think about the physical significance.
 
thank you, I've figured it out :)
since lambda > lambda0, the photoelectric effect cannot occur, then KEmax = 0
 

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