Photoelectric efect on a silver ball

In summary, the conversation discusses a problem involving a silver ball with a radius of 1cm hanging on a string and being shone with light of wavelength 200nm. The goal is to calculate the charge on the ball and determine the necessary voltage to knock out electrons. The conversation also highlights the importance of the location of charge in a conductor and how it affects the voltage. Eventually, the conversation reaches a solution by considering the positive charge gathered in the outer shell of the ball and using the potential due to a point charge to calculate the necessary voltage.
  • #1
71GA
208
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 ?
 
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  • #2
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?
 
  • #3
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:
  • #4
Since the ball is insulated, it becomes charged when you knock out electrons!
 
  • #5
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 comming 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##.
 
  • #6
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 comming 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?
 
  • #7
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:
  • #8
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.
 
  • #9
Ok thank you guys i did it :)
 

What is the photoelectric effect?

The photoelectric effect is a phenomenon in which electromagnetic radiation, such as light, causes electrons to be emitted from a material.

How does the photoelectric effect work on a silver ball?

When light of a certain frequency, or wavelength, is shone onto a silver ball, it causes the electrons on the surface of the ball to gain enough energy to break free from the metal's surface. This results in the emission of electrons, creating an electrical current.

What factors affect the photoelectric effect on a silver ball?

The intensity of the light, the frequency of the light, and the type of material used can all affect the photoelectric effect on a silver ball. Higher intensity and frequency of light can cause more electrons to be emitted, and different materials have different energy levels required for electron emission.

What is the significance of the photoelectric effect on a silver ball?

The photoelectric effect on a silver ball has many practical applications, such as in solar panels where it is used to convert light energy into electrical energy. It also helped to provide evidence for the particle nature of light and the concept of photons.

How can the photoelectric effect on a silver ball be utilized in technology?

The photoelectric effect on a silver ball can be utilized in various technologies such as photodiodes, photomultiplier tubes, and image sensors. It is also used in photoelectron spectroscopy to study the electronic properties of materials.

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