Main Question or Discussion Point

[SOLVED] A question about Photoelectric Effect

According to my teacher "photoelectric current is independent of frequency but dependent on intensity". This statement seems to conflict the following exercise:

"In an experiment to investigate the photoelectric effect, monochromatic light is incident on a metal surface. The photoelectric current and the maximum kinetic energy of the photoelectrons are measured.

Which one of the following correctly shows the change, if any, in the photoelectric current and in the maximum kinetic energy of the photoelectrons when light of the same intensity but higher frequency is incident on the same metal surface?"

A.
Photoelectric current is the same
Maximum kinetic energy increases

B.
Photoelectric current decreases
Maximum kinetic energy increases

The solution is B. Hence, my teacher's statement is ambiguous. Clearly, photoelectric current is NOT independent of the frequency as an increase in frequency has decreased the current. Hence, i think I don't understand the statement of independence. Please, help me!

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ZapperZ
Staff Emeritus
You are correct. It appears that you have a better understanding of the photoelectric effect phenomenon than your teacher.

Zz.

Yet, your response conflicts with other exercise in which the independence of photoelectric current about frequency is clearly stated:

State three pieces of evidence provided by the photoelectric effect that support the particle nature of electromagnetic radiation.

1. photoelectric current / rate of emission independent of frequency;
2. photoelectric current / rate of emission depends on intensity of radiation;
3. (max) kinetic energy of electron dependent on frequency;
etc.

It is the solution 1. that confirms my teacher's statement that "photoelectric current is independent of frequency". Sorry but i still do not get the meaning of independence. What does it mean here then? Please, help me.

ZapperZ
Staff Emeritus
I had to read this several times because these things are worded in a rather convoluted way.

"Intensity", it appears in this case, is the # of photons per unit time. So if intensity remains the same, then the amount of photoelectrons being emitted per unit time should remains the same.

However, if the the frequency is higher, then on average, the photoelectrons should have higher average kinetic energy as well. It also means that the max kinetic energy should be higher. Off the top of my head, I would say that the photocurrent remains the same if the intensity remains the same because you still have the same number of photoelectrons per unit time being emitted. So (1) is correct. This means that in your original post, the solution should have been (A).

Zz.

Shooting Star
Homework Helper
According to my teacher "photoelectric current is independent of frequency but dependent on intensity".
Increasing the light intensity increases the number of photons per second arriving at the metal electrode, if the frequency is kept the same, because there are more photons knocking out more electrons. So, the current should increase. This seems to be correct.

"In an experiment to investigate the photoelectric effect, monochromatic light is incident on a metal surface. The photoelectric current and the maximum kinetic energy of the photoelectrons are measured.

Which one of the following correctly shows the change, if any, in the photoelectric current and in the maximum kinetic energy of the photoelectrons when light of the same intensity but higher frequency is incident on the same metal surface?"

A.
Photoelectric current is the same
Maximum kinetic energy increases

B.
Photoelectric current decreases
Maximum kinetic energy increases
If the intensity is constant but the frequency is higher, then the total incident energy of the light is constant, and so the number of photons must be lesser because each photon carries higher energy. Then the number of electrons emitted must be less, but each electron will have a higher max KE. The total average energy of the electrons reaching the positive electrode would be the same, since the total energy supplied by the light is the same as before.

This shows that the current should remain the same if the intensity is constant, and this can be looked at as the current being independent of the frequency, provided the intensity is higher than the threshold frequency.

The correct answer seems to be (A) to me.

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Shooting Star
Homework Helper
"Intensity", it appears in this case, is the # of photons per unit time. So if intensity remains the same, then the amount of photoelectrons being emitted per unit time should remains the same.
If intensity is the number of photons, then if you increase the frequency, the current must increase. Without going into any calculations, the number of electrons with higher energies will be passing through the positive electrode, per unit time.

ZapperZ
Staff Emeritus
If intensity is the number of photons, then if you increase the frequency, the current must increase. Without going into any calculations, the number of electrons with higher energies will be passing through the positive electrode, per unit time.
Not necessarily, because the number of photon passing a unit area per unit time will still be the same. It means that the photoelectron density in a unit volume is different.

Zz.

Shooting Star
Homework Helper
Not necessarily, because the number of photon passing a unit area per unit time will still be the same. It means that the photoelectron density in a unit volume is different.
In this case, do you mean higher or lower by different?

If you mean lower, then I will have to read the "detailed" details of photoelectric effect all over again.

If the the photo electron density in a unit volume is indeed higher, would it not knock out more electrons, thus increasing the current?

ZapperZ
Staff Emeritus
In this case, do you mean higher or lower by different?

If you mean lower, then I will have to read the "detailed" details of photoelectric effect all over again.

If the the photo electron density in a unit volume is indeed higher, would it not knock out more electrons, thus increasing the current?
If the KE is higher, than the density of photoelectron per unit forum is lower. This is because they are all moving faster, but they are, on average, further apart. This is the only way to get the same number of photoelectrons passing a unit cross-sectional area per unit time with the case of photoelectrons having lower KE.

One can approximate this by assuming the photoelectrons in a cylindrical beam. The ones with a higher KE but with the same number of photoelectrons in 1 second will have a longer cylinder than the ones with a smaller KE but also having the same number of photoelectrons in 1 second.

Zz.

Shooting Star
Homework Helper
Not necessarily, because the number of photon passing a unit area per unit time will still be the same. It means that the photoelectron density in a unit volume is different.

Zz.
I think I am bit tired, because all this while I meant photon density when I was writing photoelectron density. I will clarify this whole thing afterward. The last post of mine was meaningless. Sorry.

However, my question in post #6 has not been answered properly, I believe. I'll frame it properly and come back to here.

ZapperZ
Staff Emeritus
I think I am bit tired, because all this while I meant photon density when I was writing photoelectron density. I will clarify this whole thing afterward. The last post of mine was meaningless. Sorry.

However, my question in post #6 has not been answered properly, I believe. I'll frame it properly and come back to here.
All you need to keep in mind is that the # of photons per unit time hitting the cathode is the same. This means that the number of photoelectrons emitted per unit time is the same, no matter how fast they are emitted. So if I collect the same number of electrons in 1 second, no matter what their KE are, then I have the same current.

Zz.

All you need to keep in mind is that the # of photons per unit time hitting the cathode is the same. This means that the number of photoelectrons emitted per unit time is the same, no matter how fast they are emitted. So if I collect the same number of electrons in 1 second, no matter what their KE are, then I have the same current.

Zz.
This is not correct. The intensity of light is not equivalent to the number of photons per unit time.

Intensity is the power delivered to a surface per unit area. Power is rate at which energy is delivered. So, if each photon has an energy $h\nu$, and the rate at which photons hit a unit area of the cathode is $R$, the intensity of the light will be $I=Rh\nu$. This means that, if we assume constant intensity, the rate at which photons hit a unit area of the cathode is $R=\frac{I}{h\nu}$.

Since the number of photons in is assumed to be the same as the number of electrons out (and, even if it isn't the relationship should just involve a constant factor), the photocurrent should look like $J=eRA=\frac{IeA}{h\nu}$, where A is the area (or, really, effective area) of the cathode.

Clearly, if we increase the frequency of the light without changing the intensity, the photocurrent must decrease because there are fewer photons to knock out electrons.

The intensity of light is not equivalent to the number of photons per unit time.

Intensity is the power delivered to a surface per unit area. Power is rate at which energy is delivered.
You are correct. This is the main essence of this problem. Your analysis shows clearly that photoelectric current is dependent on frequency. Yet, a problem arises. If we suppose that each electron has energy $h\nu$ and "intensity is the power delivered to a surface per unit area", then why intensity would not be defined as "the number of electrons per unit time delivered to a surface per unit area"? Clearly, the number of electrons is proportional to the energy. But then I think there would be no sense in having frequency as the definition would be the same. Hence, I reckon the best definition for intensity is yours and not to mix it with the number of electrons.

As for the independence, it is very hard to accept that there would be a mistake in IBO's final exams. Can the following sentence in post #3 mean then something else: "1. photoelectric current / rate of emission independent of frequency;"? In what sense, photoelectric current could be independent of frequency? Is there any other meaning for frequency in this context except "the number of photons striking the metal surface per unit time"?

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ZapperZ
Staff Emeritus
This is not correct. The intensity of light is not equivalent to the number of photons per unit time.

Intensity is the power delivered to a surface per unit area. Power is rate at which energy is delivered. So, if each photon has an energy $h\nu$, and the rate at which photons hit a unit area of the cathode is $R$, the intensity of the light will be $I=Rh\nu$. This means that, if we assume constant intensity, the rate at which photons hit a unit area of the cathode is $R=\frac{I}{h\nu}$.

Since the number of photons in is assumed to be the same as the number of electrons out (and, even if it isn't the relationship should just involve a constant factor), the photocurrent should look like $J=eRA=\frac{IeA}{h\nu}$, where A is the area (or, really, effective area) of the cathode.

Clearly, if we increase the frequency of the light without changing the intensity, the photocurrent must decrease because there are fewer photons to knock out electrons.
Yup, you are correct. I was trying to simplify the question way too much by making intensity directly proportional to # of photons per unit time. Thanks for the correction.

Zz.

Shooting Star
Homework Helper
Yup, you are correct. I was trying to simplify the question way too much by making intensity directly proportional to # of photons per unit time.
Well, that was the confusion I had yesterday from reading your post, but things seem to have been resolved in my absence.

In what sense, photoelectric current could be independent of frequency? Is there any other meaning for frequency in this context except "the number of photons striking the metal surface per unit time"?
"The number of photons striking the metal surface per unit time" has got nothing to do with the frequency of the photons. The frequency of a photon is an intrinsic property, equal to c/λ, where λ is the wavelength of the EM wave in the classical limit. (The energy of a photon is hf, where f is the frequency.)

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If the intensity is constant but the frequency is higher, then the total incident energy of the light is constant, and so the number of photons must be lesser because each photon carries higher energy. Then the number of electrons emitted must be less, but each electron will have a higher max KE. The total average energy of the electrons reaching the positive electrode would be the same, since the total energy supplied by the light is the same as before.

This shows that the current should remain the same if the intensity is constant, and this can be looked at as the current being independent of the frequency, provided the intensity is higher than the threshold frequency.

The correct answer seems to be (A) to me.
This reasoning is fallacious. Having higher frequency and constant intensity will increase the maximum kinetic energy and decrease photoelectric current. The amount of photons is directly proportional to the photoelectric current $J=eRA=\frac{IeA}{h\nu}$, where $R$ is "the rate at which photons hit a unit area of the cathode". This does not mean that the average kinetic energy of photons is directly proportional to the photoelectric current. This formula in post #12 confirms the right answer to be (B).

As it can be seen, the post #5 did not answer my question why photoelectric current / rate of emission is independent of frequency. I still have not understood the statement about independence. It is such Fuzzy Logic to me. I understand those mathematical formulae crystal clear but the use of word independence confuses me. At least, the formula shows that the photoelectric current is inversely proportional to frequency. Does this imply independence? -No, I do not think so. Perhaps, it means something else?!

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It's very simple. One photon liberates one electron, and current is equal to charge per unit time, i.e., the number of electrons per unit time. Therefore, the number of photons per unit time (i.e. intensity) determines the current. Increasing the frequency makes each electron liberated more energetic but it does not increase the number of electrons liberated per unit time, and thus, does not affect the current.

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Thanks peter0302! I understand it now. Independence means here that one photon correspond to the liberation of one electron. In that sense, photoelectric current is independent of frequency. However, the mathematical relationship was shown in post #12 that frequency is inversely proportional to the photoelectric current. In that sense, one could argue frequency to be dependent on the photoelectric current. Clearly, photoelectric effect has many dimensions -- fascinating.

ZapperZ
Staff Emeritus
Thanks peter0302! I understand it now. Independence means here that one photon correspond to the liberation of one electron. In that sense, photoelectric current is independent of frequency. However, the mathematical relationship was shown in post #12 that frequency is inversely proportional to the photoelectric current. In that sense, one could argue frequency to be dependent on the photoelectric current. Clearly, photoelectric effect has many dimensions -- fascinating.
You need to be a bit careful here in extending the simple description too far. The equation that you are dealing with here is the most naive description of the phenomenon. It assumes a quantum efficiency of 100%, i.e. one photon liberates one photoelectron. That's the assumption one has to make to talk about the amount of photocurrent. This assumption doesn't work, especially for metals, which has a quantum efficiency, at best, in the 0.01% range. When you peel off the simple picture, a number of things will pop out. For example, at higher frequency, one can get more photoelectrons even if one kept the photon flux (which I initially tied to "intensity") to be the same. This is because the higher energy photons can go deeper into the band structure of the material and thus, have a higher probability of producing photoelectrons.

So in reality, the change in photocurrent is actually quite complicated, more so than simply looking at the highest KE of the emitted photoelectrons.

Zz.

Also I think your teacher WAS treating intensity as simply the rate of photons striking the metal.

Shooting Star
Homework Helper
It's very simple. One photon liberates one electron, and current is equal to charge per unit time, i.e., the number of electrons per unit time. Therefore, the number of photons per unit time (i.e. intensity) determines the current. Increasing the frequency makes each electron liberated more energetic but it does not increase the number of electrons liberated per unit time, and thus, does not affect the current.
The current, which is the amount of charge crossing a point per unit time, depends not only on the number of electrons liberated, but also on the speed at which they reach the positive electrode and travel through the circuit, and hence on the average KE of the electrons liberated. The faster the electrons, the higher will be the value of the current. The higher the frequency, faster the electrons will be, and hence more the current.

So, one photon liberating one electron does not quite justify the constancy of the current. The intensity, i.e., the amount of energy supplied per unit time, must be taken into account.

Thanks peter0302! I understand it now. Independence means here that one photon correspond to the liberation of one electron. In that sense, photoelectric current is independent of frequency. However, the mathematical relationship was shown in post #12 that frequency is inversely proportional to the photoelectric current. In that sense, one could argue frequency to be dependent on the photoelectric current. Clearly, photoelectric effect has many dimensions -- fascinating.
Your understanding is not quite complete, irrespective of your jubilation. Please bear in mind that I have no intention of discouraging you, but I will try to start from the basics and explain step by step if I get sufficient time, which unfortunately I don’t for a few days. (Of course, that too may not be satisfactory to you.)

Have a look at http://www.play-hookey.com/optics/photoelectric_effect.html" [Broken] – it’s perhaps better than the Wikipedia page.

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The current, which is the amount of charge crossing a point per unit time, depends not only on the number of electrons liberated, but also on the speed at which they reach the positive electrode and travel through the circuit, and hence on the average KE of the electrons liberated.
I understand what you're saying, but because we all agree on the one photon per electron idea, it would be impossible to liberate more electrons simply by making the photons more energetic (leaving aside Zapper's excellent point about deeper penetration of higher frequencies). If you have 10 electrons absorb 10 photons in 1 second, and the photons are energetic enough to surpass the work function, then you're going to have a current of 10e amps. The electrons will only get from point A to B faster, and hence need a higher potential to stop.

Shooting Star
Homework Helper
I understand what you're saying, but because we all agree on the one photon per electron idea, it would be impossible to liberate more electrons simply by making the photons more energetic (leaving aside Zapper's excellent point about deeper penetration of higher frequencies).
Who is doubting that????? That's the corner stone of Photoelectric effect. But those more energetic electrons will cross a given point faster, resulting in a higher value of current, because the time taken is less.

If you have 10 electrons absorb 10 photons in 1 second, and the photons are energetic enough to surpass the work function, then you're going to have a current of 10e amps. The electrons will only get from point A to B faster, and hence need a higher potential to stop.
I am not talking of higher stopping potential here (even though I can), but that is used to measure the energy of the electrons.

To measure the current, the setup is generally a closed circuit, whereby the liberated electrons go to the cathode and again return back to the original metal. Faster electrons naturally will result in the same charge crossing a point in a shorter time, thus making the current higher.

ZapperZ
Staff Emeritus
The current, which is the amount of charge crossing a point per unit time, depends not only on the number of electrons liberated, but also on the speed at which they reach the positive electrode and travel through the circuit, and hence on the average KE of the electrons liberated. The faster the electrons, the higher will be the value of the current. The higher the frequency, faster the electrons will be, and hence more the current.
But that's is still not correct, based on what I said earlier. For example, say you apply a forward bias between the cathode and the anode to collect the photoelectron. Now, if you have collected ALL the photoelectrons being emitted, but you continue to increase the forward bias, do you think you'll get an increase in current at higher biases just because the KE of the emitted electrons are higher? Remember, per unit time, you get the same amount of photoelectrons, no matter how fast they are moving.

I am actually in the business of measuring the QE of photocathodes. When I set up the potential between the anode and the cathode, I make sure that I reach a "saturation" in the photocurrent. This tells me that I'm collecting ALL of the emitted photoelectron. So there is a plateau after a certain forward bias.

Zz.

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Shooting Star
Homework Helper
But that's is still not correct, based on what I said earlier. For example, say you apply a forward bias between the cathode and the anode to collect the photoelectron. Now, if you have collected ALL the photoelectrons being emitted, but you continue to increase the forward bias, do you think you'll get an increase in current at higher biases just because the KE of the emitted electrons are higher?
Zz.
What about when you are not collecting all the emitted electrons? And the bias between the cathode and anode is kept constant. (We are not necessarily discussing here the case when the photo current is saturated.) Just a qualitative and brief answer will do.