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Homework Help: Photons and the Photoelectric Effect

  1. Oct 28, 2006 #1
    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!!
     
  2. jcsd
  3. Oct 28, 2006 #2

    rsk

    User Avatar

    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
    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
     
  4. Oct 28, 2006 #3
    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.
     
  5. Oct 29, 2006 #4

    rsk

    User Avatar

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