What Is the Cathode Metal in This Photoelectric Experiment?

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SUMMARY

The discussion centers on determining the cathode metal used in a photoelectric-effect experiment based on stopping potentials at two different wavelengths: 400 nm and 300 nm. The stopping potential at 400 nm is established as 25.7% of that at 300 nm. The equations utilized include V(stop) = Kmax/e and Kmax = hf - Eo, where Eo represents the work function. The participants conclude that the same metal is used for both wavelengths, necessitating the setup of two equations to solve for the work function at 300 nm.

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In a photoelectric-effect experiment, the stopping potential at a wavelength of 400 nm
is 25.7% of the stopping potential at a wavelength of 300 nm. Of what metal is the
cathode made?

So I need to find the stopping potential of the metal with a wave length of 400 nm, but first I need to find the stopping potential of the metal with a wave length of 300 nm.


Using:

V(stop)=Kmax/e

Kmax=hf-Eo(work function)

How do I find the work function for the 300 nm?


Thanks.
 
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I believe that the problem states that it is the same metal for both cases. That is, the same cathode has 400 nm light, then 300 nm light shone on it, and the result was that the stopping potential was less for the longer wavelength light (which is expected), and in fact was only 25.7% of the stopping potential for the shorter wavelength light.

Sounds like a case of setting up two equations in two unknowns and solving. Write expressions for both stopping potentials, then use the given information about their numerical relationship.
 

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