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Electron volts

  1. Jan 20, 2005 #1
    The photocurrent of a photocell is cut off by a retarding potential of 2.92V for radiation of wavelength 250nm. What is the work function of the photoemitter?

    I want to use the equation: PHI=hc/lambda-Ve, where V is the stopping voltage (sorrry I don't know how to do the Greek letters for PHI and lambda).

    My question: is Ve just 2.92 or is V 2.92 and e 1.602*10^-19? I have tried both ways, adjusting h as required, and I get two different answers. Using Ve=2.92 gives a answer that makes more sense, but I don't quite understand why V=2.92 is not just V. Any advice? Thanks
  2. jcsd
  3. Jan 20, 2005 #2
    if you want retarding potential

    [tex] V = \frac{h}{e} \nu - \frac{\Phi}{e} [/tex]
    and nu = c / lambda, soo

    [tex] V = \frac{h}{e} \frac{c}{\lambda} - \frac{\Phi}{e} [/tex]

    rearrange to give

    [tex] \Phi = \frac{hc}{\lambda} - eV [/tex]

    blind substitution hereafter
  4. Jan 20, 2005 #3


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    Homework Helper

    The answer is the same both ways. If you decide to use eV for energy instead of joules, you'll get phi in eV.

    1 eV = (1.6*10^-19C)(1V)= 1.6*10^-19 J


    so (2.92V)(e)=2.92 eV=2.92*1.6*19^-19 J

    Whenever you see some voltage times e, it is easier to use electron-volts, as the energy is just the same number as the voltage.

    Try getting the answer for the work function in joules and also eV (be sure to use the right number for h each time). Then convert from one to the other. They should be the same.
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