Photoelectric Effect and circuits

AI Thread Summary
The discussion centers on calculating the threshold frequency of a metal when exposed to 520nm light, resulting in a stopping voltage of 1.13V. The formula E = h.f is used to derive a threshold frequency of 2.72 x 10^14 Hz. Participants clarify that the reverse voltage relates to the maximum kinetic energy of ejected electrons, which is expressed as KE = eV. The stopping voltage effectively indicates the energy needed to halt the current, linking it to the work function of the metal. Understanding the relationship between stopping voltage and kinetic energy is essential for grasping the photoelectric effect.
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Homework Statement


When 520nm light falls on a metal, the current through the circuit is brought to zero by a reverse voltage of 1.13V. What is the threshold frequency for this metal.


Homework Equations


E = h.f


The Attempt at a Solution


Using the above formula I get an threshold frequency of 2.72 x 10^14 Hz. Except here's my question, the reverse voltage doesn't have anything to do with the work function if it brings the current down to zero. Or does it, can someone please explain the process? I'm just confused by the term reverse voltage, or stopping voltage.

Thanks for any help :)
 
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The maximum kinetic energy of the ejected electron is given by the stopping potential.
According to photoelectric effect
KE = hν - φ where φ is the work function.
 
Thanks for your reply, but I'm still unclear on how the reverse voltage, or stopping voltage relates to the equarion.
 
Kinetic energy = e*V, where e is the charge on electron and V is the stopping voltage.
 
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