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Units regarding a photoelectric question

  1. Feb 28, 2005 #1
    Alright, lets be real here

    Electrons are ejected from a metal surface with speeds ranging up to 4.71×105 m/s when light with a wavelength of lambda = 628 nm is used. What is the work function (in eV) of the surface?

    Alright, money, using the formula: Max Kinetic energy = hf - Eo
    where Eo is the work function. You have all the givens, the answer is supposed to be in eV. Now, im not sure whats happening, but im not geting the right answer. To see if im doing something wrong, i got Kinetic energy=1.010485755x10^-19, and hf=-2.5999x10^-21
    try it yourself, maybe im screwing up the units
    Cheers
     
  2. jcsd
  3. Feb 28, 2005 #2
    i dont know how you got your kinetic energy but use this formula

    [tex] \Phi = \frac{1}{e} (\frac{hc}{\lambda} - m_{e} c^2 (\gamma - 1}}))[/tex]

    i turst you know what the formulas are about. The 1/e factor is to convert to electron volts.
     
  4. Feb 28, 2005 #3
    units

    Hey brah, thats cool, im also from Toronto, well, im in Hamilton right now..
    Anyways, im not famaliar with the formula, what am i solving for, the work function? and what is the greek symbol in the brackets at the the end?
    Cheers
     
  5. Feb 28, 2005 #4
    yes i simply used a different symbol for work function

    the Phi in the beginning represents the work function and
    [tex] \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} [/tex]

    Max kinetic energy = hc/lambda - Eo thats right and Eo represents the work function

    the kinetic energy of the electron is going to be so great (maybe) that you'll have to use the relativistic formula, else you can use the 1/2 mv^2 ither way you get the same answer. mc^2(gamma -1) is equivalent (for low speeds v<<c) to 1/2 mv^2.

    But im babbling, if you want to convert to electron volts just do this, solve it like you normally would, keep all units in metric SI units and only after you're done then divide by the elctron change of 1.6 x 10^-19 C
     
  6. Feb 28, 2005 #5
    nevermind, got it chief, thanks
     
  7. Mar 1, 2005 #6

    dextercioby

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    There's no need for the relativistic formula,simply because the electron's speed is roughly [tex] 1.5 \cdot 10^{-3}c [/tex],so you can use the newtonian dynamics without any problem...

    Daniel.
     
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