# Finding the work function of a material

1. Jul 20, 2011

### mrsvonnegut

1. The problem statement, all variables and given/known data
So today in class we did a lab where we calculated the stopping voltage of lights of different wavelengths. We did this experimentally. I have the wavelength, frequency, and stopping voltage of four different lights. I need to calculate the work function of the material, and Planck's constant. 1) Every time I calculate the work function, it comes out different for each wavelength. Am I supposed to be getting the same work function for each light? 2) How could I calculate Planck's constant, isn't it a constant? Please help! Thank you.

2. Relevant equations

Work Function: hc/lambda minus the stopping voltage of the electron
Planck's constant: 4.1 x 10^-15
c=3.0 x 10^8

3. The attempt at a solution

Various work functions I've gotten have included 1.46 eV, 1.53 eV, and 1.76 eV.

2. Jul 20, 2011

### Pi-Bond

The work function is the same for a material, regardless of the type of light used. You might want to introduce uncertainties into your readings and see if the work functions you obtains overlap for different wavelengths. The point of calculating Planck's constant would be to assess the accuracy of your experiment. You can tell from the equation of the photoelectric effect that Planck's constant should be the slope of the kinetic energy (eV) versus frequency (f) graph.

Also the value of the constant is 6.63 x 10^-34 Js

3. Jul 20, 2011

### mrsvonnegut

Thank you for your help! I'm not quite sure what you mean by introducing uncertainties. Also, the value I gave was in eV, but that's totally right for J/s!

4. Jul 20, 2011

### Pi-Bond

Oh ok. By uncertainties I mean that you can't establish an exact value of your readings - you have to give a range. For example if you got a stopping voltage of 1.5 V, you can't just record that without introducing an uncertainty; like 1.5 ± 0.1 V. It's up to you to decide how uncertain your experiment was.

5. Jul 20, 2011

### mrsvonnegut

Oh I get it! Thank you so much, I think I'll get the right answer now.