Stopping Voltage

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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Reply 1

Pangol

Original post by Mlopez14

Why do you think that KEmax = eVs won't work here?

Reply 2

Mlopez14

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

Reply 3

Pangol

Original post by Mlopez14

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

But as you correctly say, KE_max = eV_s. You know the values of V_s, e is a constant, so you can work out each value of KE_max.

Reply 4

Mlopez14

Doesn't e stand for charge?

Reply 5

Pangol

Original post by Mlopez14

Doesn't e stand for charge?

Yes - the charge of what?

Reply 6

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?

(edited 4 years ago)

Reply 7

Pangol

Original post by 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?

You need a clear understanding of what the stopping voltage is to see how charge comes into this.

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.

Reply 8

Mlopez14

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?

(edited 4 years ago)

Reply 9

Pangol

Original post by 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 need to use SI units. The charge on an electron is not -1 when measured in coulomb. You are using the relative charge. The unit for the KE at the top of the column should be a clue that you need to use something other than just -1.

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.

Reply 10

Mlopez14

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

Reply 11

Pangol

Original post by Mlopez14

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

If you mean e when you say E, then yes, the charge on an electron is indeed constant.

Reply 12

Mlopez14

Ok, thank you very much

Reply 13

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

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Reply 14

Pangol

Original post by Mlopez14

Yes, that's exactly what it is. If the incoming photons have just the right amount of energy to release electrons, there will be no energy left over to give those electrons kinetic energy.

Reply 15

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

Reply 16

Pangol

Original post by 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

If you use the formula, you will be relying on one particular data point from the table, that may be an outlier or otherwise inacurate. The point of using the graph is that your line of best fit is an averaging process, so that you don't have to rely on a single data point any more.

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