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Maximum Wavelength?

  1. Jul 13, 2008 #1
    1. The problem statement, all variables and given/known data
    The work function of zinc is 4.31eV. What is the maximum wavelength of light that will cause electrons to be emitted from a zinc surface?

    2. Relevant equations
    I'm not sure because, we are given the Work Function (W) which is 4.31eV and we need to find the max. wavelength (λ) so we will need a couple equations)

    v = fλ (v = speed of light)
    W = hf0 (f0 = threshold frequency) But we already know the Work Function

    3. The attempt at a solution

    I'm completely stumped. I have looked through the protons section in the book "Physics Key Ideas Part 2" but have yet to find anything like this. I would be extremely happy if someone could point me in the right direction on what equations I would need to use.

    If you would like to see the equation sheet we are given, it can be downloaded here:
    http://www.ssabsa.sa.edu.au/science/phys-formula-sheet.pdf

    Cheers for any help,
    Nathan
     
  2. jcsd
  3. Jul 13, 2008 #2

    alphysicist

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    Homework Helper

    Hi nathan17,


    Did you use the equation you listed in your post? What did you get?
     
  4. Jul 13, 2008 #3
    Ok, I think I got it :D

    W = hf0
    W/h = hf0/h
    Cancel out the h on the right side
    f0 = W/h
    where W = 4.31eV = 4.31 * 1.6x10-19 = 6.9x10-19J
    f0 = 6.9x10-19 / 6.63x10-34
    = 1.04x1015Hz

    So,

    v = fλ
    v/f = fλ / f
    Cancel out the f on the right side
    λ = v/f
    λ = 3.0x108 / 1.04x1015
    = 2.88x10-7m

    Yay, I got it :D

    Checked the back of the book and that's the answer, assuming the book is correct!

    [Edit] Also, I may need some help on another question in a minute, is it ok to post here instead of making a new thread?
     
    Last edited: Jul 13, 2008
  5. Jul 13, 2008 #4

    alphysicist

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    Your procedure looks right to me!

    (About another question: to make it more likely to get help, I think it's best to start a new thread if it is a separate problem.)
     
  6. Jul 13, 2008 #5
    Cool, thanks!

    Yeah, it's a separate question, I will have a crack at it and see how I go and start a new thread if needed.

    Cheers
     
  7. Jul 13, 2008 #6
    A good technique to practice is to use algebraic manipulation to define the variable you're solving for in terms of the variables given in the problem: for example, we have

    c=fλ
    W=hf

    So, f=c/λ, which we substitute into the second equation and get

    W=(hc)/λ

    Multiply both sides by λ/W and you have your answer

    λ = hc/W

    hc is a pretty common combination of constants, so it's good to remember that it's about 1240 eV*nm

    then, λ is about 1240 eVnm / 4.31 eV, which is about 288 nm; exactly what you found.
     
  8. Jul 13, 2008 #7
    Ah, Cool. Thankyou!
    I only recently got good at rearranging equations (I have a physics tutor) so I didn't really think of that, makes it a lot easier.

    Cheers
     
  9. Jul 13, 2008 #8

    Redbelly98

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    Staff Emeritus
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    Homework Helper

    It's a good skill to keep practicing and maintain at every opportunity. I wish more students would use it :smile:

    Regards,

    Mark
     
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