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

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The discussion focuses on determining the work function and initial wavelength of a cathode based on the kinetic energy of photoelectrons. The maximum kinetic energy of photoelectrons is initially 3.30 eV, which decreases to 1.50 eV when the wavelength is increased by 50%. Participants clarify that the work function can be calculated using the equation KE_max = hf - E_0, and it is possible to set the work functions for both scenarios equal to each other to find the initial wavelength. Confusion arises regarding the necessity of knowing the work function before calculating the wavelength, but it is ultimately resolved that both values can be derived from the equations provided. The conversation concludes with the understanding that two equations can be used to solve for the work function and initial frequency simultaneously.
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):
{KE}_{max} = hf - E_0[/itex]
 
Doc Al said:
This one's easier than you think. Use this equation (twice; once for each case):
{KE}_{max} = hf - E_0[/itex]
<br /> Thanks, I was able to solve for E_0 easily using the equation, but, I&#039;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&#039;s not necessary to find the wavelength first to find E_0?
 
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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