What Is the Work Function of Millikan's Hypothetical Metal?

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Homework Help Overview

The discussion revolves around the calculation of the work function for a hypothetical metal based on observations made by Robert A. Millikan regarding the photoelectric effect. Participants are analyzing the relationship between the maximum kinetic energy of ejected electrons and the wavelength of incident light.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to calculate the work function using given kinetic energy values and wavelengths. There are discussions about deriving the stopping potential and its relation to the work function, as well as the need to determine the stopping potential (Vo) from the maximum kinetic energy.

Discussion Status

Some participants have made progress in calculating frequencies and energies, while others are seeking clarification on how to derive the stopping potential and its implications for the work function. There is an ongoing exploration of different forms of the photoelectric effect equation.

Contextual Notes

Participants are working under the constraint of not knowing the value of Planck's constant and are trying to navigate the problem without complete information on the stopping potential.

TJDF
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Homework Statement



The American physicist Robert A. Millikan (1868-1953) won the 1923 Nobel Prize in physics, in part for his work on the photoelectric effect. Assume that Millikan observed for a hypothetical metal a maximum kinetic energy of 0.535 eV when electrons were ejected with 431.7 nm light. When light of 258.6 nm was used, he observed a maximum kinetic energy of 2.52 eV. Using these results, calculate the work function, W0, for the metal, without knowing the value for Planck's constant.

Homework Equations



Vo = [(h/e)*f]-(θ/e)

where θ is the work function.

The Attempt at a Solution



Frequency 1 (using wavelength 1) = 6.94927e14 Hz
Frequency 2 (using wavelength 2) = 1.1601e15 Hz

Delta Frequency = 4.65173e14

Energy 1 (using eV1) =8.571644e-20 J
Energy 2 (using eV2) =4.0375e-19 J

Delta Energy = 3.18032e-19

thus, Delta Energy over Delta Frequency gives me my experimental h...
h = 6.83685e-34 Js

That's all fine and good... but

I have NO idea how to solve for the work function.

Greatly appreciate a reply, this is my last question of the day.
 
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Hmmm... still no luck on this one yet. ideas?
 
some progress...

so e = 1.60e-19 C

f will equal 0... so...

Vo = θ/e ; θ = e*Vo = (1.60e-19J)*( ? V)

hmmmm... how to figure out Vo?
 
Hi TJDF,

TJDF said:
some progress...

so e = 1.60e-19 C

f will equal 0... so...

Vo = θ/e ; θ = e*Vo = (1.60e-19J)*( ? V)

hmmmm... how to figure out Vo?


Vo is the stopping potential, which is the potential required to stop even those electrons with the maximum kinetic energy. In that case, the maximum kinetic energy is being converted completely into potential energy. Do you see how to calculate Vo?

(You might also find some other forms of the photoelectric effect equation that have the maximum kinetic energy explicitly, instead of the stopping potential; those forms would be more straightforward for this problem.)
 

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