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Threshold Frequency

  1. Jun 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the threshold frequency for a calcium surface whose work function is 3.33 eV.

    2. Relevant equations
    Ek = hf - W

    3. The attempt at a solution
    My textbook doesn't have any examples on this, but I reasoned that threshold frequency would mean that the kinetic energy of the photoelectron is zero.

    W = hf
    f = W / h

    Since the work is given in eV,
    W = 3.33 eV x 1.6x10^-16 J/eV
    W = 5.34 x 10^-19 J

    Since h = 6.63 x 10^-34,

    W = hf
    W = 8.05 x 10^14 Hz.

    However, my textbook says 5.02 x 10^14 Hz. Where did I go wrong?
  2. jcsd
  3. Jun 8, 2008 #2


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    Hm, for some reason I got the same number as you. Maybe wait for someone with more knowledge on this topic to reply?
  4. Jun 8, 2008 #3


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    Homework Helper

    I get your answer as well.

    But if you use W=hf and use their answer (5.02x10^14 Hz),you'll get the work function is 2.08eV, which isn't the work function of calcium.
  5. Jun 8, 2008 #4


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    Staff Emeritus
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    Weird, I get the same thing you do.
  6. Jun 8, 2008 #5
    I got the same answer. The answer in your book might be wrong.
  7. Jun 8, 2008 #6
    Thank you so much, guys. I really appreciate your help. =D
  8. Jun 8, 2008 #7
    No problem DMac.
  9. Jun 8, 2008 #8


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    Can someone please explain why we have to convert the Work function into jules?
  10. Jun 8, 2008 #9


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    Homework Helper

    because f=W/h

    the units of W is J and the units of h is Js

    so the units of f would be [itex]\frac{J}{Js}=s^{-1}=Hz[/itex]

    if you didn't want to convert from eV to J, you'd need to convert the units of h to eVs
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