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Photoelectric effect question

  1. Nov 12, 2014 #1
    1. The problem statement, all variables and given/known data
    If I know the work function of various metals in eV and need to find the largest wavelenghts required to emit photoelectrons from the metals would i just use ɸ=hf0=hc/λ0?

    2. Relevant equations
    ɸ=hf0=hc/λ0

    3. The attempt at a solution
    The formula only gives the threshold wavelenght and not the largest right?

    Thanks
     
  2. jcsd
  3. Nov 12, 2014 #2

    BvU

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    Hello M and welcome to PF :)

    Bit of a problem to help without answering (as we aren't allowed to do). Look at it this way: if you find the largest wavelength, the energy with which you kick the electron is just enough to overcome the work function. Work function is a kind of binding energy. Shorter wavelengths mean higher energy. What do you think happens to the excess energy ? Where could it go ? Check it out

    3. No !
     
  4. Nov 12, 2014 #3
    .."the energy of the photoelectrons emitted is exactly the energy of the incident photon minus the material's work function or binding energy.."
    The only thing i can think of is to calculate the initial energy of the photon (or the Ek of the electron + work func) and run it throught the forumla again with known E.

    You said "no" to "3", however it states; "and the second term is the work function (ϕ) of the surface with threshold frequency (ƒ0) or threshold wavelength (λ0)" here; http://physics.info/photoelectric/
     
  5. Nov 12, 2014 #4

    BvU

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    .."the energy of the photoelectrons emitted is exactly the energy of the incident photon minus the material's work function or binding energy.."
    And that energy of the photoelectrons emitted is kinetic energy. The exercise wording points at zero kinetic energy.

    3. No ! meaning "... threshold wavelength and not which is the largest.. "
     
  6. Nov 12, 2014 #5
    If Ek=0 dosnt that mean the electron in the metal is only exited not actually emitted?
     
  7. Nov 12, 2014 #6
    If it did point to that dosnt that mean the following: Kmax = E − ϕ where Kmax=0 E=ϕ hence λmax= hc/ ɸ?
    another question, are eV ever used in calculations or do you always convert to joules?
     
  8. Nov 12, 2014 #7

    BvU

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    It means it comes out of the metal with zero speed. But it does "come out".
    The experiment described in the link (under Stopping potential) shows that the maximum kinetic energy at the threshold frequency is zero.

    550px-Photoelectric_effect_diagram.svg.png


    So finding the threshold wavelength is a matter of plotting the stopping potential versus frequency (a bit more linear than versus wavelength). Where the line crosses the V axis is the threshold.
    Usually it's not as straight as the picture suggests and you have to draw a straight line through some points with higher frequencies to find the intercept on the frequency axis.


    eV are very useful in particle physics, nuclear physics, and so on. Those folks express (almost) everything in eV (or multiples like keV, MeV, GeV,TeV).

    (almost) meaning mostly mass and energy, but also momentum. You never put a proton on a balance to weigh it anyway :)
     
    Last edited: Nov 12, 2014
  9. Nov 12, 2014 #8

    BvU

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    Correct. I think you understand it.
     
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