Photons and the Photoelectric Effect

[frankene]

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!

 

[rsk]

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

 

[frankene]

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.

 

[rsk]

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