Photoelectric efect on a silver ball

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving the photoelectric effect on a silver ball. The original poster presents a scenario where a silver ball with a specified radius is illuminated by light of a certain wavelength, seeking to determine the charge on the ball as electrons are emitted due to the photoelectric effect. The work function for silver is provided, along with relevant equations.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the maximum kinetic energy of photoelectrons and expresses confusion regarding the voltage associated with the emitted electrons. Some participants question the implications of the ball becoming charged as electrons are emitted and explore the behavior of charges on a conductor.

Discussion Status

Participants are actively engaging with the problem, offering insights into the behavior of charges on conductors and the implications of the ball becoming positively charged. There is a recognition of the need to understand how to calculate the voltage resulting from the charge distribution on the ball.

Contextual Notes

There is an ongoing exploration of the assumptions regarding charge distribution on the silver ball and the effects of emitted electrons on the ball's charge state. The original poster notes the radius of the ball as a potential factor in calculating voltage.

71GA
Messages
208
Reaction score
0

Homework Statement


Lets say we have a silver ball with radius ##r=1cm## hanging on a string which is an isolator. What is a charge on the ball if we shine on it with a light with ##\lambda = 200nm##? The work function for silver is ##A_0 = 4.7eV##.

Homework Equations


\begin{align}
&W - A_0 = W_k = eU\\
&W = h\nu = \frac{hc}{\lambda}
\end{align}

The Attempt at a Solution


First i can calculate the maximum kinetic energy that the photoelectrons have when knocked out of the metal.
\begin{align}
W_{k1}&=W - A_0\\
W_{k1}&= \tfrac{hc}{\lambda} - A_0\\
W_{k1}&= \tfrac{6.626\cdot 10^{-34}Js \cdot 2.99\cdot 10^{8}\tfrac{m}{s}}{200\cdot 10^9 m} - 4.7\cdot 1.602\cdot 10^{-19} J\\
W_{k1} &= 2.38\cdot10^{-19}J\\
W_{k1} &= 1.48eV\\
\end{align}

So now i know the kinetic energy of the photoelectrons knocked out of the metal, but where is my voltage ##U##? I know that in clasic photoeffect we need to provide enough voltage to lower the kinetic energy to ##0##, but here i really don't know where i can get the voltage from.

Can anyone tell me how to finish this problem ?
 
Physics news on Phys.org
Think about what happens with the silver ball when you shine light for a long time on it knocking out more and more electrons. What does this imply for the further electrons to leave the surface?
 
vanhees71 said:
Think about what happens with the silver ball when you shine light for a long time on it knocking out more and more electrons. What does this imply for the further electrons to leave the surface?

I don't think i understand. I only know that if there are already electrons emitted they won't stay on the surface of the ball (because they have a kinetic energy ##W_{k1}## which i calculated )... they will fly off i think.

But let's say they stay on the surface. If they stay there further electrons will be emitted harder. But i still don't know what is the voltage ##U##.
 
Last edited:
Since the ball is insulated, it becomes charged when you knock out electrons!
 
vanhees71 said:
Since the ball is insulated, it becomes charged when you knock out electrons!

I know, but where do i get the voltage from? I presume that when an electron is knocked out the ball gets positively charged in the far other side because of he influence. So would a ball become positively charged in the center if we knocked out infinitely large amount of electrons (photons are afterall coming from all directions)?

If knocked electrons are gathered in the outer shell, how can i then calculate voltage ##U##? There must be something with the radius of the ball which is ##r=1cm##.
 
71GA said:
I know, but where do i get the voltage from? I presume that when an electron is knocked out the ball gets positively charged in the far other side because of he influence.

You are saying here that the positive charge moves as far away from the negative electron as possible. Does this agree with the first things we are taught about charges and the force between them?

So would a ball become positively charged in the center if we knocked out infinitely large amount of electrons (photons are afterall coming from all directions)?

The ball is a metal, therefore it's a conductor. What do you know about the location of charge on or in a conductor?
 
Redbelly98 said:
You are saying here that the positive charge moves as far away from the negative electron as possible. Does this agree with the first things we are taught about charges and the force between them?
Yes.

Redbelly98 said:
The ball is a metal, therefore it's a conductor. What do you know about the location of charge on or in a conductor?

Charge in a conductor can move around or can even be transferred to another objects.

How do this help me get my voltage?

EDIT:
I did rethink this situation and it is clear to me now that if electrons leave a ball the later becomes positively charged. But the positive charge repells itself and is therefore gathered in the outer shell of the ball.

So i have a positively charged shell. How can i now get voltage?
 
Last edited:
71GA said:
EDIT:
I did rethink this situation and it is clear to me now that if electrons leave a ball the later becomes positively charged. But the positive charge repells itself and is therefore gathered in the outer shell of the ball.
Yes, that's right.

So i have a positively charged shell. How can i now get voltage?
At locations outside of the sphere, the E-field and potential for a positively charged spherical shell are the same as those of a single point charge located at the sphere's center.

Look up the potential (i.e. voltage) due to a point charge.
 
Ok thank you guys i did it :)
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
2
Views
7K
Replies
1
Views
40K