# Stopping Voltage

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I've been working on the photoelectric effect and I'm stuck on this question. I know you have the formula KE = e * Vs but it doesn't seem to work here. And it's the only formula that seem relevant, any help?

Thank you

Thank you

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#2

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I've been working on the photoelectric effect and I'm stuck on this question. I know you have the formula KE = e * Vs but it doesn't seem to work here. And it's the only formula that seem relevant, any help?

Thank you

**Mlopez14**)I've been working on the photoelectric effect and I'm stuck on this question. I know you have the formula KE = e * Vs but it doesn't seem to work here. And it's the only formula that seem relevant, any help?

Thank you

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I only have the frequency and voltage. I need kinetic energy here. Unless I am missing something?

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#4

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I only have the frequency and voltage. I need kinetic energy here. Unless I am missing something?

**Mlopez14**)I only have the frequency and voltage. I need kinetic energy here. Unless I am missing something?

_{max}= eV

_{s}. You know the values of V

_{s}, e is a constant, so you can work out each value of KE

_{max}.

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That's where I am confused, there is no charge. No charge that I know of.

F is frequency,

Vs is stopping voltage

Ek is?

F is frequency,

Vs is stopping voltage

Ek is?

Last edited by Mlopez14; 6 months ago

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#8

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That's where I am confused, there is no charge. No charge that I know of.

F is frequency,

Vs is stopping voltage

Ek is?

**Mlopez14**)That's where I am confused, there is no charge. No charge that I know of.

F is frequency,

Vs is stopping voltage

Ek is?

Here's a diagram showing light falling on a metal and ejecting electrons via the photoelectric effect:

(Diagram taken from this page, which also has a good explanation.)

The metal plate that the light falls onto is inside an evacuated tube. There's another metal plate at the other end of the tube, and they are connected together as part of a circuit. If you imagine the rheostat (variable resistor) set so that the "arrow" is at the very right hand end, then the circuit goes from one of the plates, through the ammeter, and back to the other plate. The battery is effectively disconnected from the circuit, so you can ignore it.

You know that the electrons emitted from the first metal plate have kinetic energy up to a maximum value, KE

_{max}. Because they've got kinetic energy, they can move through the tube and strike the other metal plate, and this is what sets up a current in the circuit. It is going to be a very tiny current, so the ammeter will have to be very sensitive. But the reading on the ammeter tells you nothing about the kinetic energy of the electrons, only how many make the journey through the tube each second.

We now change the rheostat so that the battery comes into play. Because of the way that the battery is connected, it will try to push the electrons the other way around the circuit, which will reduce the current. In particular, there will now be a potential difference across the ends of the tube, and this is what the voltmeter is there to measure.

The definition of potential difference is V = W/Q, where W is the work done or energy transferred per unit charge Q. So think about what this does to the electrons emitted from the metal that try to get to the other plate. Each electron has charge e, so V = W/e, which rearranges to W = eV. So, the work done on each electron trying to make it through the tube is eV. How is this work done? It is done by trying to pull the electrons back to the plate that they were emitted from. If we set the rheostat to just the right setting so that the current recorded by the ammeter falls to zero, then the work done by this potential difference must be just the right amount to balance the maximum kinetic energy of the emitted electrons.

So, if V

_{s}is the potential difference required to make the current fall to zero - that is, the potential that stops the current, the stopping potential - then using the idea that work done = energy transferred, we can see that eV

_{s}= KE

_{max}. This circuit gives us a way to measure the maximum kinetic energy of the emitted electrons, which isn't something we can do directly.

So in your question, you find the maximum kinetic energy of the electrons by multiplying V

_{s}by the charge on an electron, e.

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Thank you very much for all that. This is what useful.

So I multiply V by the charge of an electron (-1) right?

So for the first row, I get - 0.188?

So I multiply V by the charge of an electron (-1) right?

So for the first row, I get - 0.188?

Last edited by Mlopez14; 6 months ago

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#10

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Thank you very much for all that. This is what useful.

So I multiply V by the charge of an electron (-1) right?

**Mlopez14**)Thank you very much for all that. This is what useful.

So I multiply V by the charge of an electron (-1) right?

You also don't need the negative sign. The electrons cannot have a negative amount of kinetic energy. The work done on the electrons by the potential difference across the tube is against this KE, so keep it positive.

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Right of course, however E remains a constant correct? Since the charge of an electron doesn't just change like that?

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#12

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Right of course, however E remains a constant correct? Since the charge of an electron doesn't just change like that?

**Mlopez14**)Right of course, however E remains a constant correct? Since the charge of an electron doesn't just change like that?

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In this situation here, what does F0 stand for? I was thinking it was threshold frequency but I'm not certain on how to calculate it.

(Bear in mind I had to draw a graph earlier, maybe that's relevant)

Thank you

(Bear in mind I had to draw a graph earlier, maybe that's relevant)

Thank you

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#15

(Original post by

In this situation here, what does F0 stand for? I was thinking it was threshold frequency but I'm not certain on how to calculate it.

(Bear in mind I had to draw a graph earlier, maybe that's relevant)

Thank you

**Mlopez14**)In this situation here, what does F0 stand for? I was thinking it was threshold frequency but I'm not certain on how to calculate it.

(Bear in mind I had to draw a graph earlier, maybe that's relevant)

Thank you

*just*the right amount of energy to release electrons, there will be no energy left over to give those electrons kinetic energy.

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So basically the frequency when kinetic energy is 0? Am I suppose to draw a graph for that or use a formula? I would think a formula since a graph isn't accurate

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#17

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So basically the frequency when kinetic energy is 0? Am I suppose to draw a graph for that or use a formula? I would think a formula since a graph isn't accurate

**Mlopez14**)So basically the frequency when kinetic energy is 0? Am I suppose to draw a graph for that or use a formula? I would think a formula since a graph isn't accurate

Start with the photoelectric equation, rearrange it so that it corresponds to your graph in a y = mx + c way/

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