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Photoelectric efect on a silver ball

  1. Jul 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Lets say we have a silver ball with radius ##r=1cm## hanging on a string which is an isolator. What is a charge on the ball if we shine on it with a light with ##\lambda = 200nm##? The work function for silver is ##A_0 = 4.7eV##.

    2. Relevant equations
    &W - A_0 = W_k = eU\\
    &W = h\nu = \frac{hc}{\lambda}

    3. The attempt at a solution
    First i can calculate the maximum kinetic energy that the photoelectrons have when knocked out of the metal.
    W_{k1}&=W - A_0\\
    W_{k1}&= \tfrac{hc}{\lambda} - A_0\\
    W_{k1}&= \tfrac{6.626\cdot 10^{-34}Js \cdot 2.99\cdot 10^{8}\tfrac{m}{s}}{200\cdot 10^9 m} - 4.7\cdot 1.602\cdot 10^{-19} J\\
    W_{k1} &= 2.38\cdot10^{-19}J\\
    W_{k1} &= 1.48eV\\

    So now i know the kinetic energy of the photoelectrons knocked out of the metal, but where is my voltage ##U##? I know that in clasic photoeffect we need to provide enough voltage to lower the kinetic energy to ##0##, but here i really dont know where i can get the voltage from.

    Can anyone tell me how to finish this problem ?
  2. jcsd
  3. Jul 7, 2013 #2


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    Think about what happens with the silver ball when you shine light for a long time on it knocking out more and more electrons. What does this imply for the further electrons to leave the surface?
  4. Jul 7, 2013 #3
    I don't think i understand. I only know that if there are already electrons emitted they wont stay on the surface of the ball (because they have a kinetic energy ##W_{k1}## which i calculated )... they will fly off i think.

    But lets say they stay on the surface. If they stay there further electrons will be emitted harder. But i still don't know what is the voltage ##U##.
    Last edited: Jul 7, 2013
  5. Jul 7, 2013 #4


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    Since the ball is insulated, it becomes charged when you knock out electrons!
  6. Jul 7, 2013 #5
    I know, but where do i get the voltage from? I presume that when an electron is knocked out the ball gets positively charged in the far other side because of he influence. So would a ball become positively charged in the center if we knocked out infinitely large amount of electrons (photons are afterall comming from all directions)?

    If knocked electrons are gathered in the outer shell, how can i then calculate voltage ##U##? There must be something with the radius of the ball which is ##r=1cm##.
  7. Jul 7, 2013 #6


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    You are saying here that the positive charge moves as far away from the negative electron as possible. Does this agree with the first things we are taught about charges and the force between them?

    The ball is a metal, therefore it's a conductor. What do you know about the location of charge on or in a conductor?
  8. Jul 12, 2013 #7

    Charge in a conductor can move around or can even be transfered to another objects.

    How do this help me get my voltage?

    I did rethink this situation and it is clear to me now that if electrons leave a ball the later becomes positively charged. But the positive charge repells itself and is therefore gathered in the outer shell of the ball.

    So i have a positively charged shell. How can i now get voltage?
    Last edited: Jul 12, 2013
  9. Jul 14, 2013 #8


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    Yes, that's right.

    At locations outside of the sphere, the E-field and potential for a positively charged spherical shell are the same as those of a single point charge located at the sphere's center.

    Look up the potential (i.e. voltage) due to a point charge.
  10. Jul 14, 2013 #9
    Ok thank you guys i did it :)
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