Photoelectric effect - Frequency vs. Current

[jaumzaum]

I was solving a photoelectric effect exercise where we had to find the correct alternative. One of them was:
The increase of the light frequency results in an increase of the intensity of the current that flows trough the circuit

They were judging this wrong.
But when we increase light frequency, we are increasing the kinetic energy of the emitted electrons, as well as the initial speed of the these electrons when they go out the plate. Increasing the velocity we are decreasing the time the electron need to arrive to the other plate, thereby the current (charge divided by time) should increase.

Where is the error in this reasoning?

 

[Darwin123]

I was solving a photoelectric effect exercise where we had to find the correct alternative.

Technically, one shouldn't post a homework problem in this forum. However, your question is pretty general..

One of them was:
The increase of the light frequency results in an increase of the intensity of the current that flows trough the circuit

They were judging this wrong.

This sounds very wrong.

An increase in frequency by itself shouldn't affect the current at all. The current would be the amount of charge coming off of the metal per unit time. This would be proportional to the number of photons absorbed by the metal per unit time.

You didn't say what is set constant in this problem. The frequency increases. However, does the irradiance of the light remain the same? Does the power of the beam remain the same? Does the absorption coefficient of the metal remain the same?

Maybe they kept the irradiance of the same while increasing the frequency. Maybe the metal has a higher absorption coefficient at the higher frequency due to photoionization effects. Then, the number of photons absorbed per unit time would increase with frequency. That would result in the electric current increasing with frequency

But when we increase light frequency, we are increasing the kinetic energy of the emitted electrons, as well as the initial speed of the these electrons when they go out the plate. Increasing the velocity we are decreasing the time the electron need to arrive to the other plate, thereby the current (charge divided by time) should increase.

Where is the error in this reasoning?

Here is your error. You are referring to the time of arrival for the electrons, not the total time of collection. Increasing the velocity decreases the time of arrival. However, it would not change the amount of time that the detector is counting electrons.

You seem to be assuming the same thing I am concerning sampling time. You seem to be assuming that the photodetector system is counting electrons for the same sampling time regardless of frequency.

You may want to post the precise answer to the question on the Homework forum. What you wrote was too vague. In order to answer your question, one would have to know what other properties remained the same when the frequency of light increased.

Speculation: They meant that the amplitude was constant. However, they didn't think the consequences through.

If the amplitude of the light wave isn't changed, then the number of photons per unit time doesn't change. If the absorbance doesn't also change with frequency, the number of photons absorbed by the surface isn't changed. Then, there should be no increase or decrease with electric current as the frequency increases.

They were wrong. However, not for the reasons that you wrote.

 

[ZapperZ]

To expand further, the time it takes for the electron to go from the cathode to the anode is so short, this transit time is no longer the determining factor in the current. Rather, it is the number of electrons being emitted per second that is the predominant factor in the amount of current. That's why increasing the photo energy is not an issue for the simple photoelectron experiment.

Zz.

 

[jaumzaum]

Technically, one shouldn't post a homework problem in this forum. However, your question is pretty general..

This sounds very wrong.

An increase in frequency by itself shouldn't affect the current at all. The current would be the amount of charge coming off of the metal per unit time. This would be proportional to the number of photons absorbed by the metal per unit time.

You didn't say what is set constant in this problem. The frequency increases. However, does the irradiance of the light remain the same? Does the power of the beam remain the same? Does the absorption coefficient of the metal remain the same?

Maybe they kept the irradiance of the same while increasing the frequency. Maybe the metal has a higher absorption coefficient at the higher frequency due to photoionization effects. Then, the number of photons absorbed per unit time would increase with frequency. That would result in the electric current increasing with frequency



Here is your error. You are referring to the time of arrival for the electrons, not the total time of collection. Increasing the velocity decreases the time of arrival. However, it would not change the amount of time that the detector is counting electrons.

You seem to be assuming the same thing I am concerning sampling time. You seem to be assuming that the photodetector system is counting electrons for the same sampling time regardless of frequency.

You may want to post the precise answer to the question on the Homework forum. What you wrote was too vague. In order to answer your question, one would have to know what other properties remained the same when the frequency of light increased.

Speculation: They meant that the amplitude was constant. However, they didn't think the consequences through.

If the amplitude of the light wave isn't changed, then the number of photons per unit time doesn't change. If the absorbance doesn't also change with frequency, the number of photons absorbed by the surface isn't changed. Then, there should be no increase or decrease with electric current as the frequency increases.

They were wrong. However, not for the reasons that you wrote.

Sorry by posting the thread here :/

Thanks, I've find out what I was considering wrong, but I'm still confused about the following:
Let's say the intensity (W/m²) of the radiation remains the same, and only the frequency increases. The intensity multiplied by the Area of the plate results in the total energy that arrives the plate in each second. So IA = E/Δt, where E = nhf (n photons of frequency f)
Let's say each photon is able to pull out one electron from the plate, so the current i = ne/Δt, where e is the charge of the electron. That gives n/Δt = i/e, and
IAe=hfi -> i = IAe/hf, so when we increase the frequency, if the intensity of the radiation ramains the same, the current decreases. Is this right?

Second possibility: the spectral radiance remains the same (the number of photons per unit of time remains the same). Thereby the number of electrons ejected per unit of time remains the same too, as well as the current. So this time the current is constant.

Are these 2 statements right?

 

[Darwin123]

Sorry by posting the thread here :/

Let's say the intensity (W/m²) of the radiation remains the same, and only the frequency increases. The intensity multiplied by the Area of the plate results in the total energy that arrives the plate in each second. So IA = E/Δt, where E = nhf (n photons of frequency f)
Let's say each photon is able to pull out one electron from the plate, so the current i = ne/Δt, where e is the charge of the electron. That gives n/Δt = i/e, and
IAe=hfi -> i = IAe/hf, so when we increase the frequency, if the intensity of the radiation remains the same, the current decreases. Is this right?

Yes, this is correct. I commend you for stating units.
Just as a warning, the word "intensity" is not always used consistently even by workers in the field. The units that you gave are the "formal" units of intensity.

Second possibility: the spectral radiance remains the same (the number of photons per unit of time remains the same). Thereby the number of electrons ejected per unit of time remains the same too, as well as the current. So this time the current is constant.

I think that this time the answer is the same. The current would decrease with frequency.

I haven't found the phrase "spectral radiance" in any of my books. However, I found the word radiance. The units of radiance are (W/m²sr). However, it is still proportional to energy. So the higher the energy of each photon, the lower the number of photons.

I suspect this is a problem having more to do with units than in fundamental physics. The invariants in a radiometry or photometry problem are determined by the units of the quantities. Confusion of words like "intensity" and "radiance" are quite common. That is why I bought a very nice reference book on the subject.

As I type, I am reading from the following book:

"An Introduction to Radiometry and Photometry" by Ross McCluney (Artech House, 1994).

This is not the only text on the subject. It may not be the best for you. However, I looked long and hard for a book that would clarify the subject best. For me, this is it.

The text reviews all the standard units and standard equations for calculating light-related quantities. It has nice tables, also. They differentiate between intensity, radiance, irradiance, and so forth.


You may find to tables really useful when challenging a professor.

 

[jaumzaum]

Yes, this is correct. I commend you for stating units.
Just as a warning, the word "intensity" is not always used consistently even by workers in the field. The units that you gave are the "formal" units of intensity.

That's a thing I'm always worried about. Sometimes I get confused about the units used by intensity. I've wrote the spectral radiance (that's how we call here in Brazil), to avoid that. But I see many exercises using other units (not W/m²) to design intensity, and sometimes it can confuse you to check the right answer.

I haven't found the phrase "spectral radiance" in any of my books. However, I found the word radiance. The units of radiance are (W/m²sr). However, it is still proportional to energy. So the higher the energy of each photon, the lower the number of photons.

Here we call spectral radiance to design the amount of radiation emitted in a given frequency range. The Black-Body radiation graphic is spectral radiation vs frequency. The units are W/m²Hz, this way when we multiply the spectral radiance by the frequency (or integrate the function in a certain range) we get the real intensity at that frequency (W/m²). The problem is that many authors call spectral radiance just by intensity, what confuses me a little. What is the quantity you use to design this amount of radiation of a certain frequency?
Returning to the previous question, if the amount of photons does not change, should the current stay the same?

I suspect this is a problem having more to do with units than in fundamental physics.

I agree!

"An Introduction to Radiometry and Photometry" by Ross McCluney (Artech House, 1994).

I will really try to read your book. Don't know if it is too advanced to me (I'm beginning in physics)
Thank you

 

[Darwin123]

Returning to the previous question, if the amount of photons does not change, should the current stay the same?

Usually, but not always. The absorption or scattering efficiency also has to remain the same.

The hypothesis that you just mentioned is often used, but I am not sure what precisely to call it. Maybe it should be called the extinction effect. Every photon that creates a free electron goes extinct. This is usually a good principle, but it has exceptions. Let me start with an example where this hypothesis is completely valid.

Suppose the target is a foil so thin that most photons incident on the surface pass through it. Make it so thin that the photoexcited electrons also pass through it. Suppose that you increase the frequency but fix the number of photons. The photoelectric cross sectional area for each atom may change with frequency. If the photoelectric cross section increases, there will be more electrons emitted and fewer photons that make it through the foil.

The same type of argument can be made for a very thick foil. However, here you have to take into account the scattering of electrons from atoms.

In the photoelectric effect, there is at most one photon is destroyed for every electron that is emitted by the surface. Some of the photons may be destroyed by processes other than the photoelectric effect.

In an academic exercise or school test, I think that your hypothesis is probably fair. If the number of photons in incident radiation is constant, then the photocurrent will remain constant. This will be true under a wide range of conditions. If you are working on an actual experiment and this hypothesis doesn't seem to work, consider changes in extinction.

I will really try to read your book. Don't know if it is too advanced to me (I'm beginning in physics)

It is not an advanced book. I would think of it more like a "reference book" with detailed explanations. There is no derivation of formulas. There is no quantum mechanics or even mechanics in it.

If anything, the book is too terse. The author is probably aiming for an engineering audience.

Modern physics occasionally in this book through Planck's constant, only because there are units that incorporate Planck's constant.

One can and should learn the fundamental physics elsewhere. However, the book takes a very pragmatic and immediate approach. If the jargon of optics is confusing you, then this is the book to get.