Lets say we have a silver ball hanging on an isolator string. The work function $A_0$ of a silver and radius $r$ of the ball are known.

Now we shine light of known $\lambda$ on it from all the directions.

Question: Is there a way to calculate the charge gathered on the ball?

What I know: I know that light will knock photoelectrons out of the ball and the ball will become positively charged - because of the influence the positive charge distributes itself on an outer shell of the ball.

I know that i can calculate the maximum kinetic energy of the photoelectrons like this:

\begin{align} W &= W_{k0} + A_0\\ W_{k0} &= W - A_0\\ W_{k0} &= \tfrac{hc}{\lambda} - A_0 \end{align}

I am not sure how i am supposed to continue. Does there any stationary state occur here? Am i supposed to calculate some sort of the stopping voltage $U_0$ out of $W_{k0}=U_0 e$?

Please give me some advice.


1 Answer 1


Suppose the potential at the surface of the ball is $+V$, then the work required to remove an electron from the surface to infinity is simply $+V$ eV. The kinetic energy of the electrons is (using your notation) $W_{k0}$ eV, so the photoelectrons will be unable to escape the ball when $V = W_{k0}$. All you have to do is calculate the voltage of the sphere as a function of it's charge, and this is simply given by:

$$ V = \frac{Q}{4 \pi \epsilon_0 r} $$

where $Q$ is the total charge on the sphere and $r$ is the radius of the sphere. Therefore the maximum charge is:

$$ Q = (W - A_0) \space 4\pi\epsilon_0 r $$

where you need to express the photon energy, $W$, and the work function, $A_0$, in electron volts.

  • $\begingroup$ Is there a mistake in the first equation? Should it be $V = Qq/4\pi\varepsilon_0r$ ? $\endgroup$
    – 71GA
    Jul 13, 2013 at 11:01
  • $\begingroup$ One more question related to my 1st coment. Can we equate a potential and an energy. I mean shouldnt we equate potential energy and kinetic energy like this: $W_p = Qq/4\pi\varepsilon_0 r = W_k$ ? $\endgroup$
    – 71GA
    Jul 13, 2013 at 11:50
  • $\begingroup$ The potential at a distance $r$ is the energy required to move a unit charge from $r$ to infinity. So the potential is an energy and can be directly equated to kinetic energy. However the energy required to move an electron from $r$ to infinity is indeed $Qe/4\pi \epsilon_o r$. We can leave out the $e$ if we use the electron volt as the unit of energy, which is convenient as photon energies and work functions are commonly given in units of eV. $\endgroup$ Jul 13, 2013 at 14:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.