Photoelectric effect - shining light on a silver ball

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 $W_{k0}=U_0 e$?

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.

• Is there a mistake in the first equation? Should it be $V = Qq/4\pi\varepsilon_0r$ ?
– 71GA
Jul 13, 2013 at 11:01
• 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: $W_p = Qq/4\pi\varepsilon_0 r = W_k$ ?
– 71GA
Jul 13, 2013 at 11:50
• 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. Jul 13, 2013 at 14:51