Incorrect interpretation of the photoelectric effects

[worlov]

Hi, I've discovered the following:

 

If the metal surface is illuminated, some electrons can leave the metal. This phenomenon is called "Photoelectric effect". But the frequency of the radiation must be high enough. To explain this effect Einstein postulated in 1905 the existence of light quanta - photons. Each photon corresponds to the energy E = hv (h is Planck's constant and v - frequency of light). When a photon penetrates into the metal, it could all his energy transferred to an electron. Taking into account the work function W is then the kinetic energy of an ejected electron: E_kin = hv - W. And indeed, there is in this respect a good agreement with experiment. For the interpretation of the photoelectric effect Einstein received 1922 the Nobel Prize in Physics: "for his services to Theoretical Physics, and especially for his discovery of the law of the photoelectric effect".

 

But there is an unpleasant shade: in the saturation the photocurrent depends on the intensity of the light, but not its frequency. This is in conflict with the light quantum hypothesis of Einstein. At constant light intensity I the photon flux density n must fall with frequency v: n = I / hv. Therefore, fewer electrons are knocked out (ideally, n_electron = n_photon). Higher kinetic energy of the electrons has nothing to do with it, because the photocurrent is determined as i = ne (e is the charge of an electron). The relationship is illustrated by the following animation

 

 

Therefore, it is expected that the photocurrent curves would thus run:

 

But this is not the case! - In reality, the photocurrent in the saturation for different frequencies is the same. This fact is usually not even mentioned. Therefore, I will present several sources to confirm this statement:

 

 

 

 

Remarkable is the fact that Einstein for incorrect interpretation of the photoelectric effect was even awarded with the Nobel Prize. The division is still taught in first grade of school. Why members of Nobel Prize Committee could not solve simple arithmetic task - intensity divided by the energy of the photon? Anyway, it is clear that the photo effect still needs a proper interpretation.

[swansont]

After you have ionized an atom, it is unavailable to contribute to additional photocurrent until an electron recombines with it. So you would expect the effect to saturate, and for the saturation to give you the same current regardless of the photon energy (as long as it's above the work function, of course)

 

Remarkable is the fact that Einstein for incorrect interpretation of the photoelectric effect was even awarded with the Nobel Prize. The division is still taught in first grade of school. Why members of Nobel Prize Committee could not solve simple arithmetic task - intensity divided by the energy of the photon? Anyway, it is clear that the photo effect still needs a proper interpretation.

 

Assuming that generations of physicists are stupid probably isn't the best tactic.

[worlov]

After you have ionized an atom, it is unavailable to contribute to additional photocurrent until an electron recombines with it. So you would expect the effect to saturate, and for the saturation to give you the same current regardless of the photon energy (as long as it's above the work function, of course)

 

It's about the metals. There are so-called conduction electrons, they are not bound to individual atoms. Because of the applied voltage - negative to cathode - there is always an excess of electrons.

[swansont]

The conduction electrons are still in finite supply, and it takes time to replenish them.

[worlov]

The conduction electrons are still in finite supply, and it takes time to replenish them.

 

Then, where do the electrons when the intensity of light increased?

 

[Klaynos]

My version of Serway only appears to have 39 chapters in it... Could you please provide some text as to what the applied potential difference is being applied to.

 

I'd have thought what you should be looking at is photocurrent vs. number of photons.

[worlov]

My version of Serway only appears to have 39 chapters in it...

 

The picture I have from here http://www.uwsp.edu/...300/Lect26.html

[Klaynos]

Then I fear there might not be enough information to interpret it.

[worlov]

Then I fear there might not be enough information to interpret it.

 

"Figure 27.5 Photoelectric current versus applied potential difference for two light intensities. The current increases with intensity..." http://books.google.com/books?id=CX0u0mIOZ44C&lpg=PA872&ots=X3EtNX1HWT&dq=photoelectric%20effect%20Serway&pg=PA873#v=onepage&q&f=false

[swansont]

Then, where do the electrons when the intensity of light increased?

 

 

Where do the electrons what when the intensity is increased?

 

(This is essentially the same as the top graph in the fig 4.3 picture.)

 

Photocurrent increases with intensity until you reach saturation. That does not appear to be shown here. Not really surprising for an introductory text.

 

Do you have any information on the value of the saturation intensity and/or current? That might (ahem) shed some light on all of this.

[worlov]

Perhaps the following computer simulation helps: http://phet.colorado.edu/en/simulation/photoelectric . But beware - it contains a serious error: if you in the saturation (U >> 0V) with the same intensity change the frequency (Wavelength) of light , the photocurrent also changes:

 

But that should not happen!

[swansont]

The demonstration doesn't show intensity saturation effects; the battery voltage has nothing to do with that (that's the stopping/accelerating voltage, which is a separate saturation effect).

 

In rereading your first post, I see that your first two images of graphs are not depicting the same conditions. Graph image 1 says I is constant, and graph image 2 (fig 4.3) is depicting a change in intensity. So there is no reason to expect the graphs to look the same. There is no contradiction between them.

 

So what, exactly, is the problem?

[worlov]

The demonstration doesn't show intensity saturation effects; the battery voltage has nothing to do with that (that's the stopping/accelerating voltage, which is a separate saturation effect).

 

I see that you did not understand the photoelectric effect, the debate is pointless.

 

 

[swansont]

I see that you did not understand the photoelectric effect, the debate is pointless.

 

[sarcasm]Yeah, that's it. [/sarcasm]

[imatfaal]

worlov, on 19 April 2011 - 10:47 AM, said:

 

I see that you did not understand the photoelectric effect, the debate is pointless.

 

[sarcasm]Yeah, that's it. [/sarcasm]

 

Hey you're in good company - acc the OP neither did Einstein

[worlov]

[sarcasm]Yeah, that's it. [/sarcasm]

 

Per accelerating voltage all ejected electrons independently of their kinetic energy are driven to the anode and in this way the photocurrent reached the saturation

[swansont]

Per accelerating voltage all ejected electrons independently of their kinetic energy are driven to the anode and in this way the photocurrent reached the saturation

 

But you posted a graph that showed that the maximum photocurrent is the same for different frequencies as you change the intensity, and compared it to a graph that predicted that the photocurrent would be different for different frequencies at the same intensity, and claimed an inconsistency. I did not notice this, saw the intensity variation effect, and explained the intensity saturation. The two situations are not the same. (Also note that sometimes intensity is used as number/area*time, so "same intensity" might refer to the same photon flux. You have to check the work from which you get the graph)

 

So: What, exactly, is your objection?

 

The driving voltage reaches saturation when you get all of the photoelectrons to the target. Even though their work function is the same, they can scatter or be sent off in different directions. Without the external field, not all of them reach the target. So you add the external voltage to accelerate them; this also allows you to retard the travel until you reach no current, and this lets you measure the work function. (I'm guessing, of course, since I obviously don't understand it)

[worlov]

But you posted a graph that showed that the maximum photocurrent is the same for different frequencies as you change the intensity...

 

I always assumed that the intensity of light is constant in my view. You say it's not enough of the electrons. So I ask: where are the electrons when the intensity of light increased? In this way, I just wanted to show, there are always enough electrons. But this change of intensity has nothing to do with the original idea.

[swansont]

I always assumed that the intensity of light is constant in my view. You say it's not enough of the electrons.

 

You can't assume that when the graph tells you it isn't the case.

 

So I ask: where are the electrons when the intensity of light increased? In this way, I just wanted to show, there are always enough electrons. But this change of intensity has nothing to do with the original idea.

 

 

Then what is the problem with your "original idea"? The graphs you used don't show the same conditions. Thus, they cannot be compared to each other in any meaningful way.

[worlov]

You can't assume that when the graph tells you it isn't the case.

 

The intensity of light I = const. The number of photons per second n = I / hv. The number of electrons is equal to the number of photons, i.e. also is n = I / hv. Therefore, the photocurrent i = ne = eI / hv. The relationship: I = const, v -> oo => i -> 0!

 

That's easy...

 

 

[swansont]

The intensity of light I = const. The number of photons per second n = I / hv. The number of electrons is equal to the number of photons, i.e. also is n = I / hv. Therefore, the photocurrent i = ne = eI / hv. The relationship: I = const, v -> oo => i -> 0!

 

That's easy...

 

But

 

1) you haven't shown a graph where anyone has claimed otherwise and

2) once the frequency gets high enough, you get processes other than the photoelectric effect (Compton scattering, pair production), so not only don't you have the data that shows your relationship, you can't have such data

[worlov]

The intensity of light I = const. The number of photons per second n = I / hv. The number of electrons is equal to the number of photons, i.e. also is n = I / hv. Therefore, the photocurrent i = ne = eI / hv. The relationship: I = const, v -> oo => i -> 0!

 

 

That's easy...

 

But

 

1) you haven't shown a graph where anyone has claimed otherwise...

 

I presented three graphs, where I = const, v -> oo and i = const!

[swansont]

Where? The one labeled 4.3 is not for constant I, and two of them are not in English.

[worlov]

Where? The one labeled 4.3 is not for constant I...

 

[swansont]

"Bright" is not a value; you cannot make that determination. "Bright" could (and probably is intended to) mean the same number of photons. The point of that graph is the stopping potential does not depend on the number of photons, it depends on the frequency. As I mentioned before, sometimes "intensity" is used in discussion where it should really be "flux."