Minimum wavelength of electrons ejected from metal

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SUMMARY

The discussion focuses on calculating the minimum wavelength of electrons ejected from a metal when exposed to a laser beam. The laser produces photons with an energy of 3.5 eV, resulting in a wavelength of 221 nm. When this laser is directed at a metal with a work function of 1.5 eV in a vacuum, the minimum wavelength of the ejected electrons is determined to be 0.867 nm. The calculations utilize the equations E = hv/λ and KEmax = hf - φ to derive these values.

PREREQUISITES
  • Understanding of photon energy and wavelength relationships (E = hv/λ)
  • Knowledge of work function in photoelectric effect (φ)
  • Familiarity with the concept of kinetic energy in relation to photons and electrons (KEmax = hf - φ)
  • Basic principles of quantum mechanics, specifically de Broglie wavelength
NEXT STEPS
  • Study the photoelectric effect and its implications in quantum mechanics
  • Learn about de Broglie wavelength calculations for particles
  • Explore the relationship between energy, frequency, and wavelength in electromagnetic radiation
  • Investigate the effects of work function on electron emission in different metals
USEFUL FOR

Students in physics, particularly those studying quantum mechanics, as well as educators and professionals involved in photonics and semiconductor research.

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Homework Statement



This problem has two parts.

A laser produces photons having an energy, E = 3.5eV.
a) What is the wavelength of photons produced, assuming that the index of refraction is 1.6? ->221nm
b) If this laser beam is focused on the clean surface of a metal having a work function of φ=1.5eV in vacuum, what is the minimum wavelength λmin of the electrons that will be ejected from the metal? Assume that the experiment is done in a vacuum. -> .867nm

Homework Equations



E = hv/λ
KEmax = hf - φ

The Attempt at a Solution


So I understand part a. It's simply λ = hv/E, where v is calculated using index of refraction. However I am stuck on part B, where the answer is .867 nm. At first I was trying to do λ = hc/φ, since v = c since it is in a vacuum, and since the minimum frequency will be when KE = 0 then hf = φ, however that gives me 826nm which is not correct, or even close to being correct.
 
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The question is asking for the minimum wavelength of the ejected electrons (de Broglie wavelength).
 

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