# Photoelectric Effect Problem

1. May 10, 2007

### liorda

1. The problem statement, all variables and given/known data
What will be the electric charge of an isolated ball of copper, which was projected long enough with radiation of wavelength $$\lambda = 1400 \angstrom$$?

2. Relevant equations
The radius of the ball is R=1cm and the work function of copper is $$\Phi = 4.47 eV$$.

3. The attempt at a solution
$$E_{k}^{max} = \frac{hc}{\lambda} - \Phi_{Cu}$$
I want to say that the last electron which will be "released" will have kinetic energy that is equal to the potential energy on the surface of the ball.
The potential on the surface of the ball is Q/R, but how do I represent Q? Isn't it changes with the emission of the electrons?

Thanks.

2. May 10, 2007

### DAKONG

$$E_{k}^{max} = \frac{hc}{\lambda} - \Phi_{Cu}$$

electron volt = electron voltunit

$$W = FS$$

$$W = qES$$

$$W = q\frac{V}{S}S$$

$$W = qV$$

if you want to change eV unit to be Joule unit

WorkJoule = charge of electronqoulomb x 1volt

eV = another kind work unit (please review difinition of electron volt)

and $$W = \frac{1}{2}mv^2$$
m = electron mass

Last edited: May 10, 2007
3. May 10, 2007

### Staff: Mentor

Sounds good. The idea is that the charge builds up until the photoelectrons do not have enough energy to escape the field of charged ball.
Of course the charge changes--the ball starts out with no charge. Q is the maximum charge--which is what you are trying to find. (Take care with your units.)