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Work function

  1. May 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Red light of wavelength 670 nm produces photoelectrons from a certain material. Green light of wavelength 522 nm produces photoelectrons from the same material with 1.2 times the previous maximum kinetic energy. What is the material's work function? answer in units of eV

    2. Relevant equations

    hf = KEmax + Wo

    3. The attempt at a solution

    (6.63e^-34)(3e^8) / 522e-9 = 1.2 + W

    the answer i get after punching this into the calculator is 1.2 since the left side of the equation virtually equals zero. Please help me solve this. Thanks!
     
  2. jcsd
  3. May 3, 2009 #2

    dx

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    You just have to write the equation hf = KEmax + W0 for both the cases and solve them as simultaneous equations. If KEmax for red is A, then KEmax for green is (1.2)A.
     
  4. May 3, 2009 #3
    Yes I did that. But its the work function that I need. everytime i had 1.2 on the left side of the equation , the work function equalled 1.2
    Please, can you provide any further help?
     
  5. May 3, 2009 #4

    dx

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    Write out your steps and I'll tell you where you went wrong.
     
  6. May 3, 2009 #5
    6.63e^-34 (3e^8) / (522e^-9) = 1.2 + W

    7.8 e^-19 = 1.2 + W

    7.8 e ^-19 - 1.2 = W
    W= 1.2 (since 7.8 e^-19 is almost equal to zero)
    help =(
     
  7. May 3, 2009 #6

    dx

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    1.2 multiplies KEmax of red. You seem to have it by itself added to W, and I don't really get what you did.

    First write out the equations in symbols. What are the two equations in symbols?
     
  8. May 3, 2009 #7
    hf = KE + W and hf = 1.2KE + W

    ???
     
  9. May 3, 2009 #8
    I dont have the KE of red. So wht do i multiply 1.2 by?
     
  10. May 3, 2009 #9

    dx

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    The equations you just wrote are correct. Now there are two variables, KE and W in these equations, and you have two equations, so you can eliminate KE, and solve for W.
     
  11. May 3, 2009 #10
    sorry still a bit confused. Do I not take KE into account at all? wht do i do with the 1.2?
     
  12. May 3, 2009 #11

    dx

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    Do you know how to solve simultaneous equations?
     
  13. May 3, 2009 #12
    eeeek ...not sure .. can you get me started at least , please.
     
  14. May 3, 2009 #13

    dx

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    Ok. Take the first equation, and solve it for KE. You will get it in terms of fred and W. Now substitute this in the second equation, and you will get an equation without KE in it.
     
  15. May 3, 2009 #14
    So....

    KE = hf - W
    then after substitution ..
    hf = 1.2 (hf - W) + W
    hf = 1.2 hf -1.2W + W
    -0.2 hf = -0.2 W

    since f=c/lambda
    0.2 hc / lambda =0.2 W

    right?
     
  16. May 3, 2009 #15

    dx

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    Almost, but you made one mistake. The f in the first equation should be fred = 670 nm and the f in the second equation should be fgreen = 522 nm. They're different.
     
  17. May 3, 2009 #16
    ok thanks!
    but my answer is coming out to be a negative ..

    (6.63 e^-34)(670E-9)= 1.2 (6.63e^-34)(522e^-9) - 0.2W
    4.4421e^-40 = 4.15303 e^-40 -0.2W
    2.8907e^-41 = -0.2W this makes W negative..
    help please =(
     
    Last edited: May 3, 2009
  18. May 3, 2009 #17
    can someone please help?!!?
     
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