Photoelectric effect and Saturation Current

[fog37]

Hello,
I understand the photoelectric effect, its importance, and the basic theory. But I have a few questions:

1) One photon "can" free only a single electron, correct? However, it is not certain that if we shine exactly 10 photons (frequency? $f_0$), that 10 photoelectrons will be free, correct? Why not?

2) If the incident photons frequency $f>f_0$, the freed electron will have some kinetic energy. But that kinetic energy is not the same for all photoelectrons. Some will have more than others. The maximum and possible $KE_{max}=h f - W_0$. Is that correct?

3) Saturation current: in the ideal case that 10 photons free exactly 10 electrons all having $KE_{max}=h f - W_0$ once freed from the metal. If the potential difference is such that the emitter electrode is negative and collector electrode is positive, wouldn't the 10 free electrons be accelerated by the electric field and reach the collector faster the large is the potential difference? Current is charge/time. Why wouldn't the larger $\Delta V$ not shorten the time $t$ for the electrons to reach the collector and generate a larger photocurrent $I$? I know that, in reality, the photocurrent remains the same if all 10 photoelectrons make it to the collector. It does not seem to matter how quickly the 10 photoelectrons reach the collector....Why not? When $\Delta V>=0$, the photocurrent saturates...

Thank you

 

[tech99]

It seems to me that every electron emitted is drawn to the anode, but if you increase the applied potential no more electrons are drawn off the cathode so that the current does not increase. In other words, the electrons arrive at higher velocity but their number is the same. The slope of the Va/Ia characteristic is therefore flat, so that the device has infinite dynamic resistance. The sensitivity is quoted in microamps per lumen. I think there is a minimum anode voltage of about 20 volts required so that all the electrons are drawn off and do not fall back to the cathode. In addition there must be a good vacuum, unless great sensitivity is required, when a trace of gas will give additional current due to ionisation.

 

[fog37]

It seems to me that every electron emitted is drawn to the anode, but if you increase the applied potential no more electrons are drawn off the cathode so that the current does not increase. In other words, the electrons arrive at higher velocity but their number is the same. The slope of the Va/Ia characteristic is therefore flat, so that the device has infinite dynamic resistance. The sensitivity is quoted in microamps per lumen. I think there is a minimum anode voltage of about 20 volts required so that all the electrons are drawn off and do not fall back to the cathode. In addition there must be a good vacuum, unless great sensitivity is required, when a trace of gas will give additional current due to ionisation.

Yes, the positively charged anode and negatively charged cathode help all emitted electrons get to the anode. The photocurrent $I=\frac {Q}{t}$ but the time $t$ at denominator is not the time it take the photoelectrons to travel between the electrodes through the vacuum....
I originally thought that the faster the emitted electron charges move the higher the photocurrent because, in the case of a ionized gas (plasma), we are in the presence of a current which depends on the speed of the charged particles...

 

[PeterDonis]

1) One photon "can" free only a single electron, correct? However, it is not certain that if we shine exactly 10 photons (frequency? $f_0$), that 10 photoelectrons will be free, correct?

You are running into a limitation here of the "photon" viewpoint as applied to the photoelectric effect. The EM field state in these experiments is not an eigenstate of photon number (typically it's a coherent state), so there is no such thing as shining "exactly one photon" or "exactly 10 photons" on the metal. The number of electrons freed (as measured by the current at the anode) increases as we increase the intensity of the incoming light, but there is no way to attribute the number of electrons freed to the "number of photons" in the incoming light, because there is no such thing.

2) If the incident photons frequency $f>f_0$, the freed electron will have some kinetic energy. But that kinetic energy is not the same for all photoelectrons. Some will have more than others. The maximum and possible $KE_{max}=h f - W_0$. Is that correct?

Yes.

3) Saturation current: in the ideal case that 10 photons free exactly 10 electrons

Note that this needs to be rephrased per the above. You can in principle count the electrons (based on the current measured at the anode), but you can't count incoming photons.

all having [maximum kinetic energy] once freed from the metal.

You can't just declare this by fiat. In the actual experiment, the kinetic energies will vary, so trying to analyze the experiment as though they were all the ideal maximum doesn't work.

Current is charge/time. Why wouldn't the larger $\Delta V$ not shorten the time $t$ for the electrons to reach the collector and generate a larger photocurrent $I$?

Because you've left out a key aspect: the intensity of the light is held constant as the voltage is varied, and the intensity of the light is what determines the number of photoelectrons that are freed per unit time. The saturation current is achieved when the voltage is positive enough to ensure that all of the emitted electrons reach the collector. Increasing the voltage beyond that doesn't increase the current any more because there are no more electrons to be collected; you're already collecting all of them that are emitted per unit time.

At lower voltages, when the saturation current hasn't yet been reached, the reason it hasn't been reached is that some of the emitted electrons don't reach the collector, because they don't have enough kinetic energy even counting the energy added by the voltage. Saturation occurs when the voltage is high enough that even an electron that just barely gets freed, with zero kinetic energy, still gets enough of a boost to reach the collector.

When $\Delta V>=0$, the photocurrent saturates...

No, saturation does not happen when the voltage is greater than zero. It happens only when the voltage has a large enough positive value to ensure that all emitted electrons reach the collector. At a voltage just above zero, only the electrons that are lucky enough to be emitted with the maximum possible kinetic energy will reach the collector.

 

[sophiecentaur]

At a voltage just above zero, only the electrons that are lucky enough to be emitted with the maximum possible kinetic energy will reach the collector.

It's true that the energy imparted to each photoelectron will be hf but the surface of the metal has finite thickness and the directions of all the stimulated electrons is not necessarily the same. In a simple mechanical model the position of the electron in the metal will determine how much useful ke it gets. Many will just slow down again without leaving the surface. So, at the threshold voltage, just a few electrons will make it across to the anode. As the voltage is increased, more electrons will make it but the limit is set by the flux of photons.
The descriptions of Einstein's experiments are often so brief that the above factors are ignored. The nuts and bolts description that I wrote above explains why it's not as straightforward as we learn first time through.

 

[fog37]

The photoelectric effect was discovered before Bohr came up with its Bohr model (energy levels, etc.)

When a photon of the right frequency $f> f_0$ hits a metal and ejects an electron, that electron was previously bound to a metal atom. The work function $\Phi$ of the metal is the minimum energy needed to free an electron...does that imply that the freed electron was in its ground state inside the atom? Are we freeing the electron bound to the nucleus with the least energy $E_1$ so $E_1=\Phi$?

A metal atom has multiple electrons living in stationary states different from the ground state which could be freed...Are they free too sometimes in the context of the photoelectric effect?

Thank you.

 

[PeterDonis]

The photoelectric effect was discovered before Bohr came up with its Bohr model (energy levels, etc.)

Yes, it was. So?

When a photon of the right frequency $f> f_0$ hits a metal and ejects an electron, that electron was previously bound to a metal atom.

Not really. The electrons in question are in the conduction band, which means they aren't bound to any particular atom. They are only "bound" to the overall bulk metal itself.

The work function $\Phi$ of the metal is the minimum energy needed to free an electron...does that imply that the freed electron was in its ground state inside the atom?

No. See above. The conduction band has a range of allowed energies.

Are we freeing the electron bound to the nucleus with the least energy $E_1$ so $E_1=\Phi$?

No. See above.

A metal atom has multiple electrons living in stationary states different from the ground state which could be freed...Are they free too sometimes in the context of the photoelectric effect?

No. See above.

 

[fog37]

Yes, it was. So?Not really. The electrons in question are in the conduction band, which means they aren't bound to any particular atom. They are only "bound" to the overall bulk metal itself.No. See above. The conduction band has a range of allowed energies.No. See above.No. See above.

I see. Thank you for clarifying.

When many many multiple atoms are brought together, the sharp energy levels found in the simple Bohr model of a single atom become energy bands with a range of allowed energies. The conduction energy band is the outermost band.

In the spirit of Bohr's model, electrons living in the closest energy bands have less potential energy than the electrons living in the farther out bands, like the conduction band. But the ionization energy happens to be equal to the absolute value of the ground state energy....

So it is easier to free electrons in the conduction band since that requires less energy....I believe...

 

[PeterDonis]

When many many multiple atoms are brought together, the sharp energy levels found in the simple Bohr model of a single atom become energy bands with a range of allowed energies. The conduction energy band is the outermost band.

More precisely, some of the energy levels (the outermost ones, with the smallest binding energy) that are there for single isolated atoms are "combined" into energy bands.

the ionization energy happens to be equal to the absolute value of the ground state energy....

No. The ionization energy for an element is the energy it would take to free a single electron from the outermost occupied orbital for a single isolated atom. It is not the same as the work function for a metal in the context of the photoelectric effect, which is the minimum energy needed to free an electron from the conduction band. Nor is it the same as the absolute value of the ground state energy of anything.

it is easier to free electrons in the conduction band since that requires less energy

Less energy than it would take to free an electron from an inner orbital, yes.

 

[fog37]

More precisely, some of the energy levels (the outermost ones, with the smallest binding energy) that are there for single isolated atoms are "combined" into energy bands.No. The ionization energy for an element is the energy it would take to free a single electron from the outermost occupied orbital for a single isolated atom. It is not the same as the work function for a metal in the context of the photoelectric effect, which is the minimum energy needed to free an electron from the conduction band. Nor is it the same as the absolute value of the ground state energy of anything.Less energy than it would take to free an electron from an inner orbital, yes.

I see.
To summarize, freeing electrons in the context of a single atom or molecule or low density gas would be the process of ionization.
Freeing electrons in the context of the photoelectric effect would be the result of the photoelectric effect which is different from ionization...

 

[PeterDonis]

freeing electrons in the context of a single atom or molecule or low density gas would be the process of ionization.

Yes.

Freeing electrons in the context of the photoelectric effect would be the result of the photoelectric effect which is different from ionization...

Yes.

 

[sophiecentaur]

The work function Φ of the metal is the minimum energy needed to free an electron...does that imply that the freed electron was in its ground state inside the atom?

In metals (metallic bonding) no electron is bound to any specific atom. (look this up) so the concept of 'ground state' doesn't apply as in an H atom.

 

[sophiecentaur]

So it is easier to free electrons in the conduction band since that requires less energy....I believe...

The photoelectric effect is quite hard to achieve; only a few metals work with visible light and the surface has to be really really clean, too - and in a vacuum. UV works easier as the photon energy is higher. wiki talks about it in depth. they make some practical points about it not being an easy demo to do.