Work function of a cathode

1. Nov 11, 2005

siifuthun

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?

2. Nov 11, 2005

Staff: Mentor

This one's easier than you think. Use this equation (twice; once for each case):
[tex]{KE}_{max} = hf - E_0[/itex]

3. Nov 11, 2005

siifuthun

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?

Last edited: Nov 11, 2005
4. Nov 11, 2005

siifuthun

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.

5. Nov 11, 2005

Staff: Mentor

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.

6. Nov 11, 2005

Staff: Mentor

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.