How to Determine the Work Function and Initial Wavelength of a Cathode?

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Homework Help Overview

The discussion revolves around determining the work function and initial wavelength of a cathode in the context of photoelectric effect problems. Participants are exploring the relationships between kinetic energy, frequency, and wavelength of light affecting photoelectrons.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using the equation for maximum kinetic energy of photoelectrons and express confusion about the implications of negative work functions. There are attempts to relate changes in wavelength to changes in energy and questions about the necessity of finding the work function before the wavelength.

Discussion Status

Some participants have successfully used the kinetic energy equation to find the work function, while others express uncertainty about the process of finding the initial wavelength without prior knowledge of the work function. There is an acknowledgment of multiple approaches to the problem, including setting work functions equal for different scenarios.

Contextual Notes

Participants note that the problem context involves limited coverage of the topic in lectures and textbooks, leading to questions about missing equations or concepts necessary for solving the problem.

siifuthun
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The maximum kinetic energy of photoelectrons is 3.30 eV. When the wavelength of the light is increased by 50%, the maximum energy decreases to 1.50 eV.

What is the work function of the cathode? What is the initial wavelength?

K= E_elec - deltaE
K_max = hf - E_0 = E_elec - E_0
f = c/(lambda)

so I set K=K_max and get:
E_elec - deltaE = E_elec - E_0
and solved for E_0 with deltaE= 1.5 eV - 3.3 eV, but then I get a negative work function (E_0)= -1.8, can work functions be negative? I also get the feeling that I'm not using the kinetic energy equations correctly here.

For the initial wavelengthI took
(3/2)(lambda_i) = hc/E
for E I used 1.5 eV and I got it wrong as well, and that was the last of my tries and the told me the answer was 230 nm, but I still can't see how they came to that because the book barely goes over these topics, and we didn't spend that much time on it during lecture. Is there an equation that I'm supposed to use that I'm missing?
 
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This one's easier than you think. Use this equation (twice; once for each case):
[tex]{KE}_{max} = hf - E_0[/itex][/tex]
 
Doc Al said:
This one's easier than you think. Use this equation (twice; once for each case):
[tex]{KE}_{max} = hf - E_0[/itex][/tex]
[tex] Thanks, I was able to solve for E_0 easily using the equation, but, I'm still confused as to how I would have been able to find the wavelength in the first place without knowing the work function. Since they asked what the work function is before they asked for the initial wavelength, does that mean it's not necessary to find the wavelength first to find E_0?[/tex]
 
Last edited:
siifuthun said:
Thanks, I was able to solve for E_0 easily using the equation, but, I'm still confused as to how I would have been able to find the wavelength in the first place without knowing the work function. Since they asked what the work function is before they asked for the initial wavelength, does that mean it's not necessary to find the wavelength first to find E_0?

Nevermind, forget I asked that. In explaining the problem to my friend I realized I could've just set the work function for both situations equal to each other and solve for lambda 1 knowing that lamda 2 is increased by 50%. Although I wonder if there's another way to do it.
 
I'm not sure what you did to solve for the work function. But once you know the work function, use the same equation to find the initial frequency (and thus the wavelength).

Why don't you show how you got the work function.
 
siifuthun said:
Nevermind, forget I asked that. In explaining the problem to my friend I realized I could've just set the work function for both situations equal to each other and solve for lambda 1 knowing that lamda 2 is increased by 50%. Although I wonder if there's another way to do it.
Right! That's the only way that I can see. You have two equations and two unknowns (Work function and initial frequency). Solve for both unknowns.
 

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