Photoelectric effect and work functions

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SUMMARY

The discussion centers on the relationship between the work functions of metals and the threshold wavelength in the photoelectric effect. The work function, denoted as \(\phi\), is defined by the equation \(\phi = hf_{0}\), where \(f_{0}\) represents the threshold frequency. It is established that the threshold wavelength (\(\lambda\)) is inversely proportional to the work function, expressed as \(\phi = \frac{hc}{\lambda}\). This relationship indicates that as the threshold wavelength increases, the work function of the metal also increases, highlighting the intrinsic properties of the metals involved.

PREREQUISITES
  • Understanding of the photoelectric effect
  • Familiarity with the concepts of work function and threshold frequency
  • Knowledge of the relationship between frequency and wavelength
  • Basic grasp of photon energy calculations
NEXT STEPS
  • Study the derivation of the photoelectric effect equations
  • Learn about the implications of work function variations in different metals
  • Explore the relationship between photon energy and electron kinetic energy
  • Investigate experimental methods to measure work functions of various materials
USEFUL FOR

Physics students, researchers in material science, and educators seeking to deepen their understanding of the photoelectric effect and its applications in modern technology.

bullroar_86
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what can be said about the work functions of two metals when the threshold wavelength in the photelectric effect increases.

I'm having trouble finding and equation to describe this relationship.

ANyone have any ideas?
 
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Well, the work function, \phi = hf_{0}, where f_{0} is the threshold frequency. Do you know the relationship between frequency and wavelength?
 
Threshold wavelength is inversely proportional to the work function of the metal .
Of course the threshold wavelength is decided by the metal itself.

For the critical case the work function is \phi is related to the wavelength as:

\phi= \frac{hc}{ \lamda }
 
You spelt lambda wrong, that's why it won't come out.
 
If the photon energy exceeds the work function, the excess energy appears as the kinetic energy of the electron.
 

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