# Photons and the Photoelectric Effect

Light is incident on the surface of metallic sodium, whose work function is 2.3 eV. The maximum speed of the photoelectrons emitted by the surface is 1.26e6 m/s. What is the wavelength of the light?

I first converted Work Function to Joules:

2.3eV x (1.6e-19 J / 1eV) = 3.68e-19 J

The equation I know is:

hf = KEmax + Work Function

hf can also be written as:

hc/lamda

I thought I would substitute c w/ the 1.26e6 m/s and use KEmax as zero. My answer was 2.27e-9m and that is not correct.

I used the following:

f = Work Function / h = 5.55e14 Hz

Then I used:

lamda = v/f = 2.27e-9 m

If I can't substitute v for c, then what am I looking for with v??? I have looked at this problem for 3 days w/ no new ideas. Anyone else have any?

Thank you!!

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Have you worked out the ke of the electrons?

When I do this I get the E for the light different to the value you give.

And I don't understand what you're doing with this
lamda = v/f = 2.27e-9 m
Find the Ke of the electron (joules) Convert WF to joules and add. This is the energy of the light.

Then use E = hc/lamda

I get 636 nm

Thank you.

I didn't realize I could find KE by using .5 *mv^2 and use the mass of an electron. Once I did that I added WF and then divied hc by my answer. The answer I got was 1.82e-7 m and that was correct.

Thank you again for the help. I knew I was thinkging too hard and not looking at the obvious.

good - don't know why my answer's wrong....one too many glases of wine before using a calculator?