# Homework Help: Finding the wavelength from the work function in photoemission

1. Apr 27, 2012

### EmmaLemming

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

a) Write down the formula that relates the maximum electron energy, Emax. to the
frequency of the incident light in the photoelectric effect.

b) Calculate the maximum wavelength of light for which photoemission occurs for light
incident on a metal whose workfunction is 2.30 eV.

2. The attempt at a solution

a) E = hf - σ

where E is the maximum energy, h is planck's constant and σ is the work function.

b) E = (hc/λ) - σ

where λ is the wavelength and c is the speed of light in a vacuum

∴ λ = hc/(E+σ)

However I'm not given the energy so how am I to calculate λ?

Is there another way to calculate E?
I looked in my text book but found nothing relevant :(

2. Apr 27, 2012

### collinsmark

The problem statement asked for "maximum wavelength". Don't forget, maximum wavelength corresponds to minimum frequency; thus minimum energy.

What happens if the wavelength is so large that hc/λ = σ ? What's the wavelength that would cause photoemission to occur, but without any leftover energy?

3. Apr 28, 2012

### EmmaLemming

Okay, so I assume that minimum energy is 0 and then write,

hc/σ = λ

I get, λ = 5.4 x 10-7 m

That seems reasonable , right?

I don't know the answer to your last question.

4. Apr 30, 2012

### collinsmark

'Looks right to me.
What I mean is that if the light's wavelength is too big, there is not enough energy to overcome the work function and photo-emission doesn't take place at all. As you decrease the wavelength, the energy increases. Decrease it enough and photo-emission begins to occur. Decrease the wavelength even more and the leftover photon energy can end up becoming electron energy.

So the maximum wavelength is the wavelength where the maximum electron energy is zero.