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Photoelectric Effect, identify metal given wavelength light and electric potential

What do you think the answer is?

Poll closed Oct 30, 2011.
  1. Iron

    0 vote(s)
  2. Nickel

    0 vote(s)
  3. Magnesium

    0 vote(s)
  4. Potassium

    0 vote(s)
  5. Aluminum

    0 vote(s)
  6. Zinc

    0 vote(s)
  7. Platinum

    0 vote(s)
  1. Oct 25, 2011 #1
    1. The problem statement, all variables and given/known data

    In a photoelectric effect experiment, 200.0 nm light was applied to a metal surface causing release of electrons. It was determined that a minimum 1.90V opposing electric potential was required to prevent the released electrons from striking the collection electrode. Based on the following photoelectric work functions (ionization energies, Ei, required for photoelectron release), the identity of the metal was _____.

    Iron 7.21x10-19 J
    Nickel 8.03x10-19 J
    Magnesium 5.90x10-19 J
    Potassium 3.69x10-19 J
    Aluminum 6.54x10-19 J
    Zinc 6.89x10-19 J
    Platinum 1.02x10-18 J

    2. Relevant equations
    Kmax = h * v - ∅ [maximum kinetic energy = Planck's constant*frequency - work function]
    v = c/λ [frequency = speed of light / wavelength]
    E = h * v [radiant energy = Planck's constant*frequency]
    h * v = eV + ∅ [Planck's constant * frequency = opposing electrostatic potential energy + work function] [eV = electron charge * required retarding voltage]

    3. The attempt at a solution
    I've tried this problem a dozen times. My professor has given no examples anything like this and there are no similar problems in the textbook or online, so I am completely lost. I don't even know how much of this is relevant
    I've calculated that:
    Kmax=3.0438E-19J = (6.66252E-33)(1.5E15)-∅ so ∅=9.6894E-18
    I keep coming back to ∅=9.6894E-18 but that's not an answer choice and I don't know what I'm missing. We didn't go over any of this in class.
  2. jcsd
  3. Nov 3, 2011 #2


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    Homework Helper

    Re: Photoelectric Effect, identify metal given wavelength light and electric potentia

    You calculated the photon energy with wrong h. It is h=6.66252E-34 J/s. Correct it, subtract eV and compare the result with the work functions in the table.

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