# Saturation voltage for photoelectric current

• FranzDiCoccio
In summary, the conversation discusses the concept of saturation current in the photoelectric effect. It is explained that when the current is saturated, all electrons are gathered by the collecting electrode and increasing the accelerating potential does not change the current. However, when the current is not saturated, some electrons may not reach the electrode due to factors such as their speed and direction. The conversation also explores the effects of geometric factors on the current and the role of collisions with residual atoms in the bulb. There are two explanations for why the current may be less than its saturation value, with one focusing on electrons not reaching the collector and the other on the varying kinetic energy of the emitted electrons.
FranzDiCoccio
Hi,

I was wondering about saturation current in the photoelectric effect. It is clear to me that for a sufficiently large accelerating potential all of the electrons are gathered by the collecting electrode. Since it is all of them, there cannot be more, and the current won't change if the accelerating potential is increased further.

What I'm not entirely clear about is what happens when the current is not saturated. My intuition is that some of the electrons simply won't make to the collecting electrode, but I find answers to related questions that invoke the speed of the electron (kinetic energy) and the fact that the current is the quantity of charge per unit time.

The latter explanation sort of confuses me. The perhaps naive way I'm picturing this phenomenon makes me think that electron speed does not really matter, unless it is in the wrong direction (so velocity rather than speed matters).
I guess that not all electrons are necessarily emitted towards the collecting electrode, so some of them could simply "miss the target". If an accelerating potential is switched on, the electrons trajectories are bent towards the electrode. If the potential is too weak some electrons might still escape, though.
At some point the acceleration towards the electrode is so strong that basically all of the electrons are pushed to the collecting electrode, no matter how unfortunate the direction of their initial velocity.

Does this make sense? Or else one should really take into account the different initial velocities of the electrons, and the current being "quantity of charge per unit time"? In that case the saturation would come about because the final velocities of the accelerated electrons would be much larger than any of their initial values, and so the latter would not matter. But for some reason this does not feel right.I expect that the "slope" of the ramp towards saturation is affected by geometric factors, such as the area and shape of the collector and its distance from the target. For instance, the farther apart the electrodes are, the more "stretched" the saturation is, because only electrons emitted within a small solid angle would reach the collector.

So which one is it? The current is less than its saturation value because
1) some electrons won't reach the collector
2) all electron reach the collector, but with different velocities?

ZapperZ's replies in this post and in this other post seem to point to 1), if I get them right.
If it was possible to "wrap" the collector around the target I guess that saturation would be achieved for an arbitrarily small positive potential.

Also, do collisions with residual atoms in the bulb matter? Simplified explanation claim that there is a vacuum in the bulb, but I guess that this is true only up to some degree, in reality.

Thanks a lot for any insight
Franz

FranzDiCoccio said:
Hi,

I was wondering about saturation current in the photoelectric effect. It is clear to me that for a sufficiently large accelerating potential all of the electrons are gathered by the collecting electrode. Since it is all of them, there cannot be more, and the current won't change if the accelerating potential is increased further.

What I'm not entirely clear about is what happens when the current is not saturated. My intuition is that some of the electrons simply won't make to the collecting electrode, but I find answers to related questions that invoke the speed of the electron (kinetic energy) and the fact that the current is the quantity of charge per unit time.

The latter explanation sort of confuses me. The perhaps naive way I'm picturing this phenomenon makes me think that electron speed does not really matter, unless it is in the wrong direction (so velocity rather than speed matters).
I guess that not all electrons are necessarily emitted towards the collecting electrode, so some of them could simply "miss the target". If an accelerating potential is switched on, the electrons trajectories are bent towards the electrode. If the potential is too weak some electrons might still escape, though.
At some point the acceleration towards the electrode is so strong that basically all of the electrons are pushed to the collecting electrode, no matter how unfortunate the direction of their initial velocity.

Does this make sense? Or else one should really take into account the different initial velocities of the electrons, and the current being "quantity of charge per unit time"? In that case the saturation would come about because the final velocities of the accelerated electrons would be much larger than any of their initial values, and so the latter would not matter. But for some reason this does not feel right.I expect that the "slope" of the ramp towards saturation is affected by geometric factors, such as the area and shape of the collector and its distance from the target. For instance, the farther apart the electrodes are, the more "stretched" the saturation is, because only electrons emitted within a small solid angle would reach the collector.

So which one is it? The current is less than its saturation value because
1) some electrons won't reach the collector
2) all electron reach the collector, but with different velocities?

ZapperZ's replies in this post and in this other post seem to point to 1), if I get them right.
If it was possible to "wrap" the collector around the target I guess that saturation would be achieved for an arbitrarily small positive potential.

Also, do collisions with residual atoms in the bulb matter? Simplified explanation claim that there is a vacuum in the bulb, but I guess that this is true only up to some degree, in reality.

Thanks a lot for any insight
Franz

Both are technically correct, but it depends on the varying degree of accuracy that you want.

Here's what we know occurs:

1. Photoelectrons are emitted with a range of kinetic energy (or speed) and momentum when they are emitted.

2. However, the range of energy or speed depends very much on the energy of the incident photons. If the photons have energy just barely above the work function, then the range of energy of the photoelectrons may not matter that much. If that is the case, then the difference in KE plays a very small role in the current below saturation.

3. For photon energy well above the work function, then below saturation, both the difference in KE/speed and the missed electrons may effect photocurrent being measured.

4. The caveat in all of this is the geometry and location of the anode. If the anode is big and encompassing the cathode, then the "missed electrons" no longer matter. I've done this when I reverse-biased the cathode and ground the vacuum chamber that the whole contraption sat in. Essentially, the stainless steel chamber is the anode, meaning I captured all the emitted photoelectrons. So in that case, clearly, below saturation, the variation in current is due to photoelectrons being emitted at different KE.

Zz.

FranzDiCoccio
The knocked off electrons go to the anode or to the cathode. There is no other place to go.

So when the current is smaller than the saturation current, some electrons are returning to the cathode.

A formation of electrons tends to expand, so if we have a moving formation of electrons, the front of the formation gains kinetic energy, and the rear of the formation loses kinetic energy. I guess that is the main thing that makes some electron to return to the cathode.

If we double the speed of the electron gas, then the repulsive force decreases as the electron density decreases. The time that an electron feels the force decreases too.

It seems to me that halving the distance of the electrodes would have about the same effect as doubling the speed of electrons. I mean when the electrodes are large and impossible to miss. With small electrodes electrons spend some time wandering around the anode, and all that time they are repelling electrons near the cathode towards the cathode.

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jartsa said:
The knocked off electrons go to the anode or to the cathode. There is no other place to go.

So when the current is smaller than the saturation current, some electrons are returning to the cathode.

Actually, there is.

Most simple photoelectric effect demo or experiments are done inside evacuated glass vacuum tubes. So the electrons that do not make it to the anode can hit the walls of the tube and be absorbed there.

In photoemission experiments, this is essentially what we do, i.e. we collect electrons in all different angles that have left the samples and simply drift away due to its emitted momentum without any external field. So if a photoelectron can escape the sample (including overcoming the image potential), then it won't come back to the cathode simply because it will drift away from it.

Zz.

Thank you ZapperZ,

Both are technically correct, but it depends on the varying degree of accuracy that you want.

Here's what we know occurs: [snip]

I hoped that only the "missed target effect" really mattered. The role of the different initial velocities is much less intuitive for me. I'll need to find an intuitive way of explaining it.

4. The caveat in all of this is the geometry and location of the anode. If the anode is big and encompassing the cathode, then the "missed electrons" no longer matter.

Ok... that is what I was thinking about when I mentioned a collecting electrode "wrapped around" the target.

I got curious about this because initially I could not explain the reason why an arbitrarily small accelerating potential wasn't sufficient to saturate the current.
Then I thought that some electrons might have very "unfortunate" initial direction that cannot be corrected by a small accelerating potential.
The effect of the energy distribution remains somewhat less intuitive for me.
I'm thinking of electron extraction as a sort of "stationary" process... but perhaps is my intuition about accelerating particle that is failing me.

Thanks again.
Franz

jartsa said:
The knocked off electrons go to the anode or to the cathode. There is no other place to go.

So when the current is smaller than the saturation current, some electrons are returning to the cathode.

A formation of electrons tends to expand, so if we have a moving formation of electrons, the front of the formation gains kinetic energy, and the rear of the formation loses kinetic energy. I guess that is the main thing that makes some electron to return to the cathode.

If we double the speed of the electron gas, then the repulsive force decreases as the electron density decreases. The time that an electron feels the force decreases too.

It seems to me that halving the distance of the electrodes would have about the same effect as doubling the speed of electrons. I mean when the electrodes are large and impossible to miss. With small electrodes electrons spend some time wandering around the anode, and all that time they are repelling electrons near the cathode towards the cathode.

@jartsa,

it seems to me that you are not referring to "conventional" experiments on the photoelectric effect. In the paper you refer the light source is a synchrotron. It is a pulsed light and, as far as I understand, that's the origin of the space-charge effect you mention.

I am more concerned in the ideal, basic experiment on the photoelectric effect.
So when the current is smaller than the saturation current, some electrons are returning to the cathode.

apply also in the basic case? It seems strange to me that Coulomb repulsion matters. Are electrons really so closely packed in a standard experiment?

Does the Coulomb attraction from the cathode matter? I mean it should be positively charged due to some electrons leaving it.

@ZapperZ

2. However, the range of energy or speed depends very much on the energy of the incident photons. If the photons have energy just barely above the work function, then the range of energy of the photoelectrons may not matter that much. If that is the case, then the difference in KE plays a very small role in the current below saturation.

Does this mean that the "ramp" to saturation is steeper in this case? If I get it right one has initially (very?) slow electrons close to the emitting electrode, and hence a small accelerating potential should bring all of them to the collecting electrode basically "in bulk".

Based on what you say I guess that larger photon frequencies cause a less steep approach to saturation.

But then again the saturation value would be higher, so that the curve could be steep anyway... I'm not sure I'm making sense here.

FranzDiCoccio said:
@jartsa,

it seems to me that you are not referring to "conventional" experiments on the photoelectric effect. In the paper you refer the light source is a synchrotron. It is a pulsed light and, as far as I understand, that's the origin of the space-charge effect you mention.

I am more concerned in the ideal, basic experiment on the photoelectric effect.

The important thing about the paper is that in the beginning it says the same thing as I said. The rest I have not read :)

It seems strange to me that Coulomb repulsion matters. Are electrons really so closely packed in a standard experiment?

Standard experiment is conducted with many different voltages. We can be sure that Coulomb forces matter when voltage is low enough.
Does the Coulomb attraction from the cathode matter? I mean it should be positively charged due to some electrons leaving it.

Cathode is negative, because the negative electrode of the battery is connected to it. ... Except when we are doing the experiment with very low voltage. http://www.john-a-harper.com/tubes201/
See chapter 4

jartsa said:

Space-charge effects is important when the light source is "intense enough".

I worked at a particle accelerator where space-charge efffects are always present since we used high-powered laser to produce photoelectrons. However, in a typical photoelectric effect experiment, where one often used a discharge lamp, and with a bunch of optical element along the way, the photon flux is small enough that space-charge effects are not as prominent. This is especially true if one, say, uses the He line, and the work function is 5 eV or less. A high percentage of the emitted electrons will be from the Fermi energy and so, it will have quite a bit of momentum and kinetic energy when they are emitted. These will make them less effected by space-charge effects, if any.

Now, if the OP is doing this with a high-powered laser, or using photon energies just above the work function, then we certainly should consider space-charge effects as possibly affecting the varying current.

There is no one simple answer that fits every scenario.

Zz.

FranzDiCoccio said:
@ZapperZ
Does this mean that the "ramp" to saturation is steeper in this case? If I get it right one has initially (very?) slow electrons close to the emitting electrode, and hence a small accelerating potential should bring all of them to the collecting electrode basically "in bulk".

Based on what you say I guess that larger photon frequencies cause a less steep approach to saturation.

But then again the saturation value would be higher, so that the curve could be steep anyway... I'm not sure I'm making sense here.

Again, this is difficult to say because it definitely depends on the geometry of the anode. I have used just a bare wire dangling in front of the photocathode, and I can tell you that it takes a larger voltage to get all the photoelectrons when compared to using a flat disk.

As I've stated in the previous post, it is difficult to give just one answer, because depending on the situation, there could be different mechanism that dominates within a particular voltage range, and then it changes over another different range.

Zz.

FranzDiCoccio
jartsa said:
The important thing about the paper is that in the beginning it says the same thing as I said. The rest I have not read :)

I see.

Cathode is negative, because the negative electrode of the battery is connected to it. ... Except when we are doing the experiment with very low voltage.
http://www.john-a-harper.com/tubes201/
See chapter 4

Ok, but you mentioned that some electrons go back to the cathode. It being charged positively could be one reason for that. Of course this is true only for small potentials, or if you want to prevent the electrons from reaching the collecting electrode, like when performing the standard experiment on the photoelectric effect.
The collecting electrode should be at a lower potential than the cathode, right?

FranzDiCoccio said:
The collecting electrode should be at a lower potential than the cathode, right?
What do you mean by "should"? Sometimes we want the anode to be less negative than the cathode, maybe some other times we don't.In the previous post I said that the cathode may become positive if we use low voltage. That is actually wrong. Even a low voltage voltage source forces the voltage between the cathode and the anode to be the voltage of the voltage source.

ZapperZ said:
Space-charge effects is important when the light source is "intense enough".

I worked at a particle accelerator where space-charge efffects are always present since we used high-powered laser to produce photoelectrons. However, in a typical photoelectric effect experiment, where one often used a discharge lamp, and with a bunch of optical element along the way, the photon flux is small enough that space-charge effects are not as prominent. This is especially true if one, say, uses the He line, and the work function is 5 eV or less. A high percentage of the emitted electrons will be from the Fermi energy and so, it will have quite a bit of momentum and kinetic energy when they are emitted. These will make them less effected by space-charge effects, if any.
What happens to the electrons that leave the cathode but don't reach the anode?

If such electron is hanging around somewhere in the equipment, it is doing what space charge does: Exerting Coulomb force to other electrons.

jartsa said:
What happens to the electrons that leave the cathode but don't reach the anode?

If such electron is hanging around somewhere in the equipment, it is doing what space charge does: Exerting Coulomb force to other electrons.

Maybe. But perhaps the rest of the equipment is grounded to prevent charge buildup.

And anyway, the OP is about getting a sense of the saturation current in the "standard" experiment, for pedagogical reasons.
ZapperZ's replies go in that direction.

While I thank you for taking the time of addressing my question, I feel you are adding unnecessary ingredients to that, which unnecessarily complicate the answer.
Also, sometimes I'm having a hard time following your explanations, but perhaps that has to do with me not being a native English speaker.

jartsa said:
What happens to the electrons that leave the cathode but don't reach the anode?

If such electron is hanging around somewhere in the equipment, it is doing what space charge does: Exerting Coulomb force to other electrons.

Hey, I know, those electrons go to the metal vacuum vessel, where they distribute themselves so that they cause no field inside the vessel. Well okay then, the mystery of disappearing electrons is solved.

FranzDiCoccio said:
Maybe. But perhaps the rest of the equipment is grounded to prevent charge buildup.

And anyway, the OP is about getting a sense of the saturation current in the "standard" experiment, for pedagogical reasons.
ZapperZ's replies go in that direction.

While I thank you for taking the time of addressing my question, I feel you are adding unnecessary ingredients to that, which unnecessarily complicate the answer.
Also, sometimes I'm having a hard time following your explanations, but perhaps that has to do with me not being a native English speaker.

Actually, jartsa did bring in a possibly relevant point, and again, it depends on how accurately you measure your photocurrent and how sensitive your equipment are at detecting these various mechanisms. Certainly, if you do this with a light source with such a low energy above threshold and at a very low forward bias, the photoelectrons will come out "lazily" and build up space-charge above the surface and will cause subsequent electrons to be pushed back into the cathode. But as I've said, typically one uses a He or Hg discharge lamp for basic lab experiment, and the photon energy is usually several times higher than the work function.

This is why I mentioned early on that the setup ("geometry of the anode", light source, etc.) is very important in determining the dominant mechanisms contributing to the photocurrent at a particular potential difference.

Also note that this typically applies to cathode that has been grounded (as is the rest of the containment vessel), while only the anode is forward biased. This is NOT the only setup that is done. I've mentioned earlier of reverse biasing just the cathode, while keeping the rest of the vessel grounded. You might possibly notice a slightly different photocurrent curve with this one.

Oh, btw, this is off topic, but if you can crank up the forward bias so that the E-field reaches to about 1 kV/m, you might notice that the "saturation" curve is not perfectly horizontal! The value of the "saturation current" increases with increasing forward bias. :)

Zz.

## What is saturation voltage for photoelectric current?

Saturation voltage for photoelectric current is the minimum voltage required to completely stop the emission of photoelectrons from a metal surface when illuminated by a light source.

## What factors affect the saturation voltage for photoelectric current?

The saturation voltage for photoelectric current is affected by the intensity of the incident light, the work function of the metal surface, and the frequency of the incident light.

## How is saturation voltage for photoelectric current measured?

The saturation voltage for photoelectric current is measured by gradually increasing the voltage applied to the metal surface until the emission of photoelectrons stops completely.

## What is the significance of saturation voltage for photoelectric current in understanding the photoelectric effect?

The value of the saturation voltage for photoelectric current allows us to determine the maximum kinetic energy of the emitted photoelectrons, which helps us understand the relationship between the frequency of the incident light and the energy of the emitted electrons in the photoelectric effect.

## Can the saturation voltage for photoelectric current be negative?

No, the saturation voltage for photoelectric current cannot be negative. It is always a positive value as it represents the minimum voltage required to stop the emission of photoelectrons from a metal surface.

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