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Homework Help: Quantum Physics: Work Function

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data

    The American physicist Robert A. Millikan (1868-1953) won the 1923 Nobel Prize in physics, in part for his work on the photoelectric effect. Assume that Millikan observed for a hypothetical metal a maximum kinetic energy of 0.535 eV when electrons were ejected with 431.7 nm light. When light of 258.6 nm was used, he observed a maximum kinetic energy of 2.52 eV. Using these results, calculate the work function, W0, for the metal, without knowing the value for Planck's constant.

    2. Relevant equations

    Vo = [(h/e)*f]-(θ/e)

    where θ is the work function.

    3. The attempt at a solution

    Frequency 1 (using wavelength 1) = 6.94927e14 Hz
    Frequency 2 (using wavelength 2) = 1.1601e15 Hz

    Delta Frequency = 4.65173e14

    Energy 1 (using eV1) =8.571644e-20 J
    Energy 2 (using eV2) =4.0375e-19 J

    Delta Energy = 3.18032e-19

    thus, Delta Energy over Delta Frequency gives me my experimental h...
    h = 6.83685e-34 Js

    That's all fine and good... but

    I have NO idea how to solve for the work function.

    Greatly appreciate a reply, this is my last question of the day.
  2. jcsd
  3. Mar 18, 2009 #2
    Hmmm... still no luck on this one yet. ideas?
  4. Mar 18, 2009 #3
    some progress...

    so e = 1.60e-19 C

    f will equal 0... so...

    Vo = θ/e ; θ = e*Vo = (1.60e-19J)*( ?????? V)

    hmmmm.... how to figure out Vo?
  5. Mar 19, 2009 #4


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    Homework Helper

    Hi TJDF,

    Vo is the stopping potential, which is the potential required to stop even those electrons with the maximum kinetic energy. In that case, the maximum kinetic energy is being converted completely into potential energy. Do you see how to calculate Vo?

    (You might also find some other forms of the photoelectric effect equation that have the maximum kinetic energy explicitly, instead of the stopping potential; those forms would be more straightforward for this problem.)
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