Photoelectric Effect and electrostatic potential

[ems]

Homework Statement

A silver ball is suspended by an insulating string in a vacuum chamber and light of wavelength 200nm is directed at it. The ball and the chamber are both initially at zero electrostatic potential. Electrons ejected from the ball causes the electrostatic potential to change.
Describe the magnitude of the current ejected from the ball and the electrostatic potential of the ball, both as a function of time, assuming the radiation stays constant. Assume the current changes linearly with respect to the potential and that the capacitance of the ball is C.

The work function of silver (4.7 eV) is also given.

The Attempt at a Solution

I understand physically what is happening in this problem. The radiation causes electrons to be ejected and causes charge build-up on the walls of the chamber and the ball which increases the electrostatic potential. Also, I realize that as the potential increases, it is more difficult for the charge to flow, and when the potential reaches the stopping voltage (I already calculated this to be 1.5 eV), there will be no current flow.
I just can't work out the problem on paper using the linear relationship... help!

 

[Jhero]

I can provide a solution to this problem by considering the principles of electrostatics and the photoelectric effect.

Firstly, we know that the photoelectric effect is the phenomenon where electrons are ejected from a material when light of a certain wavelength (in this case, 200nm) is incident on it. The work function of silver (4.7 eV) tells us the minimum energy required to eject an electron from the surface of the silver ball.

Next, we can consider the electrostatic potential of the silver ball. Initially, both the ball and the chamber are at zero electrostatic potential. As electrons are ejected from the ball, the potential will increase due to the accumulation of negative charge on the surface of the ball. This increase in potential will make it more difficult for electrons to be ejected, as you correctly stated. This can be described by the equation:

V = q/C

Where V is the electrostatic potential, q is the charge on the ball, and C is the capacitance of the ball. We can rearrange this equation to solve for q:

q = CV

Substituting in the stopping voltage of 1.5 eV, we can calculate the maximum charge that can accumulate on the ball before the potential reaches this value.

q = (C)(1.5 eV)

Now, we can consider the current ejected from the ball. The rate of electron ejection is directly proportional to the intensity of the light and the number of electrons per unit time that are ejected from the surface of the ball. This can be described by the equation:

I = Φ*E*A

Where I is the current, Φ is the photoelectric effect constant (which depends on the intensity of the light and the properties of the material), E is the energy of the incident light, and A is the surface area of the ball. We can also substitute in the work function of silver (4.7 eV) and the energy of the incident light (200nm = 6.2 eV) to calculate the current ejected from the ball.

I = (4.7 eV)(6.2 eV)(A)

We can also rewrite this equation in terms of potential:

I = (4.7 eV)(6.2 eV)(A) / CV

Now, we can see that the current ejected from the ball is directly proportional to the electrostatic potential. As the potential increases, the