# Homework Help: Finding the work function using stop potential

1. Feb 20, 2006

### Calmeir

Apologies, but I have another photoelectron question. In this one I get a result that seems reasonable, however it does not show up on my list of work functions that I have available to choose from. The problem is:

A photoelectric-effect experiment finds a stopping potential of 1.93V when light of 200nm is used to illuminate the cathode. What is the cathode made of?

I initially thought this would be an easy problem as Vstop = (hf - Eo) / e

I have all the values except Eo which is what I am trying to solve for.

Vstop = 1.93 V
h = 4.14x10^-15 eV
f = 3x10^8/200x10^-9 = 1.5x10^15
e = 1.6x10^-19

So, I plug all of the values into the equation and solve for Eo

1.93 = ((4.14x10^-15)(1.5x10^15)-Eo)/1.6x10^-19

The resulting value of Eo was 6.21
This seems valid, except the list I have to choose from is:

Potassium: 2.3
Sodium: 2.75
Aluminum: 4.28
Tungsten: 4.55
Copper: 4.65
Iron: 4.7
Gold: 5.1

Likely I am having the same problem as my previous question and not seeing a simple unit difference. Anyhow, I do appreciate your help.

2. Feb 20, 2006

### Tom Mattson

Staff Emeritus
That is not Planck's constant in SI units. In fact h is not expressed in eV in any system of units.

3. Feb 20, 2006

### topsquark

Tom is right and wrong. The problem IS units, but the Planck's constant being used by Calmeir is in eV-s, which is correct. The problem is that he is already working in eV, and so needs to drop the "1.6x10^-19" at the end of his equation.

I reworked it and got e=4.28 eV.

-Dan

4. Feb 21, 2006

### Tom Mattson

Staff Emeritus
No, Tom is right and right!

While the numerical value of h that he used corresponds to h in eV-s, he did report the units in eV, which is wrong.

That's one fix, but I was hinting at a different one. Since he put all of his other quantities in SI units I was trying to get him to put h in SI units as well. He would have obtained E0 in Joules and then convert to eV. Changing h to eV-s and leaving the electron charge as e is certainly more direct though...