# Homework Help: Light stikes metal surface; determine work function

1. Sep 15, 2009

### bw1990

1. The problem statement, all variables and given/known data

Monochromatic light with a wavelength 415 nm strikes a metal surface. Photoelectrons escape with a kinetic energy of 95 kJ/mol. What is the work function (in kJ/mol, no decimals required) of the metal surface?

2. Relevant equations

I have found various equations online relating to work function, and I have tried most of the ones that seemed like they would work, but I am still at a loss as to which equation is correct, and if I am completing the problem correctly. Some of the equations I found are:

Φ = hc/λ, where h is Planck's constant.
Φ = hfo, where fo is the minimum frequency of a photon
hc/λ = wf + KE, where wf is work function and KE is kinetic energy

3. The attempt at a solution

All of my attempts at finding a solution have resulted in -95. I have no idea if this is correct, but my best guess at solving the problem was using the last equation because I had wavelength and kinetic energy. I converted the wavelength to 4.15e-7 m..divided hc by wavelength.. subtracted kinetic energy... still get -95.

Thanks in advance to any help =)

2. Sep 16, 2009

### Kalvarin

Yep -95 KJ/mol is the right answer.

The logic behind it is:

The incident light has energy E = hf or E = h/wavelength

This energy is then transfered to an electron, in this case we are dealing with a mole of electrons so multiply the incident energy by avogadros number, because we will need a mole of incident photons to liberate a mole of electrons.

We now got the total incident energy of a mole of photons.

This energy is transfered to to a mole of electrons, some of the energy is used up in freeing the electrons from the surface, this energy is the work function.

Any left over energy becomes kinetic energy, so we can say:

Energy of 1 mole of incident photons = work function + Kinetic energy of 1 mole of liberated electrons

3. Aug 11, 2011

### thebiggerbang

I have NEVER seen a negative work function! I guess the fact why you are going wrong is that you are accounting for a mole of electrons. If I'm not wrong you'll get the correct answer accounting for a single electron. That is if you deicide the given K.E by the Avagadro number

4. Aug 11, 2011

### vela

Staff Emeritus
The first two equations apply to the case where KE=0. They allow you to solve for the cutoff frequency and wavelength. The third equation applies to the general case, and it's the one you need for this problem since you know if the incident light isn't at the cutoff wavelength.
-95 what? You need to show more details of your calculation for us to see where your mistake is. As thebiggerbang notes, you should not get a negative answer.