What is the electric charge of an isolated copper ball bombarded with radiation?

[liorda]

Homework Statement

What will be the electric charge of an isolated ball of copper, which was projected long enough with radiation of wavelength $$\lambda = 1400 \angstrom$$?

Homework Equations

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

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.

 

[DAKONG]

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

electron volt = electron volt_unit

$$W = FS$$

$$W = qES$$

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

$$W = qV$$

if you want to change eV unit to be Joule unit

Work_Joule = charge of electron_qoulomb x 1_volt

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

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

 

[Doc Al]

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.

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.

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?

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.)