Photoelectric effect and work functions

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Homework Help Overview

The discussion revolves around the photoelectric effect and the relationship between the work functions of two metals and their threshold wavelengths. The original poster is seeking clarification on how these concepts are interconnected and is struggling to find an appropriate equation to describe this relationship.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between work function and threshold frequency, as well as the inverse relationship between threshold wavelength and work function. Questions about the correct terminology and definitions are also raised.

Discussion Status

Some participants have provided insights into the equations related to work function and threshold frequency, while others have pointed out potential errors in terminology. The discussion appears to be ongoing, with various interpretations being explored.

Contextual Notes

There is a mention of a potential spelling error affecting the clarity of the discussion. The original poster's understanding of the relationship between frequency and wavelength is also questioned, indicating a need for further exploration of these concepts.

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, [itex]\phi = hf_{0}[/itex], where [itex]f_{0}[/itex] 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 [itex]\phi[/itex] is related to the wavelength as:

[itex]\phi[/itex][itex]= \frac{hc}{ \lamda }[/itex]
 
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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