Photoelectric effect, retarding potential

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Homework Help Overview

The discussion revolves around the photoelectric effect and the calculation of the work function of a material based on the retarding potential and wavelength of monochromatic radiation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the retarding potential and the work function, referencing relevant equations. There is an exploration of the significance of the wavelength provided and its conversion to meters. Some participants question how the given values fit into the equation for calculating the work function.

Discussion Status

Some guidance has been offered regarding the use of the equation to find the work function, and participants are actively engaging with the problem by clarifying units and values. Multiple interpretations of the problem setup are being explored, particularly concerning the wavelength and its implications.

Contextual Notes

One participant notes their background in a different field, which may contribute to their initial confusion regarding the physics concepts involved. There is an emphasis on the need to understand the units and their conversions in the context of the problem.

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[SOLVED] Photoelectric effect, retarding potential

Homework Statement



If the photocurrent of a photocell is cut off by a retarding potential of 0.92 volts for monochromatic radiation of 2500 A (A with the little circle above it), what is the work function of the material.

Homework Equations



My book and professor use different symbols so I am going to define the symbols I am using:
Ø means retarding potential / stopping potential
W means work function

e Ø = hc/lambda - W

hc = 1240 eV

The Attempt at a Solution



radiance is invariant with lambda b/c monochromatic
According to the graph of photocurrent i vs. anode voltage V, it is when V = -Ø that i goes to 0. I don't see where 2500A fits in, except that that must give me lambda so that I can solve for W in the equation above.
 
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You've got everything you need. Just plug them all into your equation and solve for W!
 
2500 \AA is the wavelength.

1 \AA = 10^{-10}m
 
Wow, that helps a lot!

Oh okay, having A be a measure of length helps a lot...I was an econ/poli sci major so missed a lot of this stuff undergrad = )

2500A = 2.5 * 10^-7m = 250 nm

So W = (1240 eV nm / 250 nm) - .92 eV = 4.04 eV which is the work function.

Thanks!
 

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