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Light-wavelength calculation

  1. Feb 8, 2005 #1
    Hi.

    How do I calculate this:

    How long is the shortest light-wavelenght that can breake loose a elektron from a natrium surface?
     
  2. jcsd
  3. Feb 8, 2005 #2
  4. Feb 8, 2005 #3
    Thank you for the link, clive.
    This is what I've got now:

    [tex]eV_{0} = \frac{hc}{\lambda}-\phi[/tex]
    [tex]h = 6,6261 * 10^{-34}[/tex]
    [tex]c = 2,99792458*10^8[/tex]
    [tex]\phi =2,28eV[/tex]

    I need [tex]eV_0[/tex] to calculate [tex]\lambda[/tex]. How do I get it?

    And ofcourse I want the longest wavelength and not the shortest as I said before. :smile:
     
  5. Feb 8, 2005 #4

    Doc Al

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    Staff: Mentor

    [tex]eV_0[/tex] is the maximum KE of the ejected electron; to find the minimum energy photon (longest wavelength) needed to eject an electron, set [tex]eV_0[/tex] to 0.
     
  6. Feb 9, 2005 #5
    Okay so we got

    [tex]\frac{hc}{\lambda} = \phi[/tex]
    [tex]h = 6,6261*10^{-34}[/tex]
    [tex]c\approx 3*10^8[/tex]
    [tex]\phi = 2,28eV[/tex]

    This gives us

    [tex]\lambda\approx 8,7*10^{-26}[/tex]

    Is this correct? Maybe i should turn eV to joule first?

    Btw: Is the answer in meter? (I'm new to frequency)
     
    Last edited: Feb 9, 2005
  7. Feb 9, 2005 #6

    Doc Al

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    Staff: Mentor

    use standard units

    You'd better use consistent units throughout. Since h = 6.625 x 10-34 J-s, you'd better convert eV to Joules!

    The standard unit for length is the meter, so that equation for [itex]\lambda[/itex] will give you meters.
     
  8. Feb 10, 2005 #7
    Another question. If I'm lighting up a natriumsurface with a quicksilver-lamp electrons will "release". But which formulas should I use to calculate which energy the fastest electrons will get?

    Thank you in advance.
     
  9. Feb 10, 2005 #8

    Doc Al

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    Staff: Mentor

    See post #4 in this thread. You have the formula.
    (Photon energy) - (energy needed to eject an electron) = maximum KE of ejected electron

    This may help you: http://en.wikipedia.org/wiki/Photoelectric_effect#Explanation
     
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