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Reverse Potential Difference

  1. Jul 13, 2008 #1
    1. The problem statement, all variables and given/known data
    The work function of caesium is 1.35eV. A photoelectric cell has a caesium surface and is illuminated with monochromatic light of wavelength 400nm. What reverse potential difference must be applied to the tube to just stop a current passing through it?

    2. Relevant equations
    Kmax = Vs * e
    E = hf = hc / λ

    3. The attempt at a solution
    λ = 400nm = 4.00x10-7
    E = hf = hc / λ
    = 6.63x10-34 * 3x108 / 4.00x10-7
    = 4.97x10-19J
    = 4.97x10-19 / 1.6x10-19
    = 3.11 eV

    Kmax = E - W
    = 3.11 - 1.35
    = 1.76

    Kmax = Vs * e
    Divide both sides by e
    Vs = K / e
    = 1.76 / 1.6x10-19
    = 1.1x1019

    Other Information

    Now, the answer in the back of the book is :
    Vs = 1.76 V

    I got that in the second step, but I'm not sure if I worked it out correctly. Can someone go over this and check it out for me.

    Last edited: Jul 13, 2008
  2. jcsd
  3. Jul 13, 2008 #2


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    Homework Helper

    No, that's not quite what you got in the second step; you got the numerical value but it was a different quantity. In the second step, you found the maximum kinetic energy was 1.76 eV. Now the reason that electron volts is such a useful unit is exactly because of what you observed. If an electron (or anything with charge e) goes through a potential difference of 1 V, its energy changes by 1 eV.

    In your third step, if you want to divide by [itex]e=1.6\times 10^{-19}[/itex] coulombs, you have to first convert the energy to joules. But to convert from eV to J, you have to multiply by [itex]1.6\times 10^{-19}[/itex], and so you end up with the same number as what you had in step 2.
  4. Jul 13, 2008 #3
    Oh. Heh, thanks. so it should be:

    K = 1.76 * 1.6x10-19
    = 2.8x10-19

    Vs = K / e
    = 2.8x10-19 / 1.6x10-19
    = 1.76 V

    Cheers for the help,
    really appreciate it!

    Last edited: Jul 13, 2008
  5. Jul 13, 2008 #4
    Think of the physics of the situation:

    An electron volt is the amount of energy that an electron receives when accelerated by a potential of 1 volt. When you subtract the work function from the energy for an incoming photon, you get the energy of the emitted electron, in electron volts. This difference, then is exactly the energy that you need to decelerate the electron to rest; since you have a single electron, the choice of units is convenient! 1.76 volts is the potential difference that can stop an electron with energy 1.76 eV.

    To this end, your work is correct, but you need to be careful with the formula in the third step. A volt is a joule per coulomb. If you want to use that formula, you need to put the electron energy in joules first.
  6. Jul 13, 2008 #5
    Or, what alphysicist said.

    That'll teach me to get a snack while writing a post!
  7. Jul 13, 2008 #6
    Hmm, thanks for the explanation guys!

    Yeah, it's the things like that, that I really need to be careful of when working these out when it comes exam time.
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