Fatal error in computer simulation (photoelectric)

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The following computer simulation http://phet.colorado.edu/en/simulation/photoelectric contains a serious error: if you in the saturation (U >> 0V) with the same intensity change the wavelength of light, the photocurrent also changes:

http://www.uplolit.com/media/201116/d923b2bdc1817e40e9be42fcfd63e3e3.JPG

http://www.uplolit.com/media/201116/ce434b80978f2583435a805a24d697ac.JPG

But that should not happen! About cut-off frequency the photocurrent in the saturation depends only from the intensity of light:

[PLAIN]http://www.uwsp.edu/physastr/kmenning/images/sb5.40.f7.gif [Broken]
Source: http://www.uwsp.edu/physastr/kmenning/Phys300/Lect26.html [Broken]
 
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This is the answer:

"Thanks for contacting PhET. This is a common question. The physics behind the behavior you are observing is described in our teacher guide for this simulation. Please read through this document - linked below.
http://phet.colorado.edu/files/teachers-guide/photoelectric-guide.pdf

If you have any follow-up questions or concerns, please don't hesitate to be in touch again. If you could post this link on your forum too - that would be great. "
 
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  • #5
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You can see that the developers do not understand what I mean. Their Description applies only to the region U ~ 0 (Figure below). In the Saturation, i.e. at U>> 0, the photocurrent independent from the frequency.

http://www.uplolit.com/media/201117/721cbdf2b10cf96a83836547fedb81fb.GIF
Source of the original: http://iitphysics.org/photoelectric_effect.htm [Broken]
 
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  • #6
ZapperZ
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This thing is ambiguous, yes. It also can create some confusion if one is just learning about the photoelectric effect. So maybe the program needs to think about this a bit more.

However, is it "fatally wrong". It isn't.

If you have a particular light source, with a set power, and you are able to select various frequencies out of that light source, then changing frequency implies changing energy per photon. But since the light source has a set power, it means that, via simple conservation of energy, the number of photons per unit time emitted is also going to change! So a blue light from this light source will emit LESS photon per unit time than a yellow light, since blue light has more energy per photon than yellow. This means that there will be more yellow light photons than blue light photons emitted per second, leading to a larger photocurrent for yellow light than for blue light.

But now, notice that this is not what is happening in the scenario above, because the shorter wavelength (high energy photon) has more photocurrent than the longer wavelength. What is going on here?

My only guess is that they must have tested this "in real life" and decided to transpose the result in this simulation. It is true, in general (but not all the time), that high energy photons (assuming that the intensity remains the same) will liberate more photoelectrons. This is because photons with higher energy can go deeper into the conduction band and produce more photoelectrons. This corresponds to a higher "quantum efficiency" at shorter wavelength.

So the simulation here is not fatally wrong. I just think that they've included unnecessary pieces of information that will confuse very perceptive students, such as you. Still, well done in spotting such issues, and you should consider this as part of your learning process in understanding this phenomenon. I'd hire you as my summer intern in a heartbeat! :)

Zz.
 
  • #7
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This thing is ambiguous, yes. It also can create some confusion if one is just learning about the photoelectric effect. So maybe the program needs to think about this a bit more.

However, is it "fatally wrong". It isn't.
Harmless is not either. PhET simulation distort the curves:

http://www.uplolit.com/media/201117/11b77df62b0c5257801c96ab1996faa1.GIF

Is this just a little thing?
 
  • #8
ZapperZ
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Harmless is not either. PhET simulation distort the curves:

http://www.uplolit.com/media/201117/11b77df62b0c5257801c96ab1996faa1.GIF

Is this just a little thing?
BOTH are correct depending on the CONTEXT! I had tried to explain to you why I think they did what they did. I can't be sure if that is what they were trying to do, because I can't speak for them. But in my own experience (and I study photoemission phenomena intimately and as part of my job), I'm saying that it isn't wrong!

If you don't believe me, look at this Mirzoyan paper on bialkali photocathode.

http://psec.uchicago.edu/library/photocathodes/pmtquanteff.pdf

Look at Fig. 1, 2, and 3. From very long wavelength to about, say 375 nm, the QE increases with decreasing wavelength. It means that for the SAME NUMBER OF PHOTON FLUX, you get MORE PHOTOELECTRONS as the wavelength decreases. So in terms of frequency, the highest frequency will liberate more photoelectrons than the smaller frequency. Isn't this consistent with what the PhET figure showed?

This is why I said it isn't fatally flawed. It isn't the best way to do this since now, one is invoking a lot of other external factors not included in Einstein's photoelectric description. I had to invoke the QE dependance on the photon energy, which requires a lot more knowledge of the photoemission phenomenon than typically taught at the undergraduate level. So the figure isn't the best thing to present at this level. Still, it isn't wrong.

Zz.
 
  • #9
zonde
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You can see that the developers do not understand what I mean. Their Description applies only to the region U ~ 0 (Figure below). In the Saturation, i.e. at U>> 0, the photocurrent independent from the frequency.
Have you looked in the link they provided?

There it says:
"Not every photon emits an electron, even if the photons have enough energy to emit
electrons. If a photon is absorbed by an electron with binding energy greater than the
photon energy, the electron will not be released. Photons with higher energies are
more likely to release electrons because a greater proportion of the electrons in the
metal have binding energy less than the photon energy. Therefore, as you increase
the frequency, the number of emitted electrons (and therefore the current) will
increase until all photons are emitting electrons. Note that this behavior is different
from the simplified model used by many textbooks, in which every photon with
frequency greater than the threshold frequency releases an electron, so the current is
constant above the threshold frequency.
"
Basically they say that quantum efficiency increases with frequency.
And simulation behaves accordingly. Nothing wrong.
 
  • #10
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BOTH are correct depending on the CONTEXT!

If you don't believe me, look at this Mirzoyan paper on bialkali photocathode.

http://psec.uchicago.edu/library/photocathodes/pmtquanteff.pdf

Look at Fig. 1, 2, and 3. From very long wavelength to about, say 375 nm, the QE increases with decreasing wavelength. It means that for the SAME NUMBER OF PHOTON FLUX, you get MORE PHOTOELECTRONS as the wavelength decreases. So in terms of frequency, the highest frequency will liberate more photoelectrons than the smaller frequency. Isn't this consistent with what the PhET figure showed?
In the simulation are available only simple metals, i.e. no bialkali photocathode. In addition, QE is not the same as photo current.

QE = i/i_max = n*e/i_max
i_max = n_max*e, n_max = I/hv
=> QE = n/(I/hv), n is the number of emitted electron
 
  • #11
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Have you looked in the link they provided?

There it says:
"...Note that this behavior is different
from the simplified model used by many textbooks, in which every photon with
frequency greater than the threshold frequency releases an electron, so the current is
constant above the threshold frequency.
"
Basically they say that quantum efficiency increases with frequency.
And simulation behaves accordingly. Nothing wrong.
I have also not only books: http://www.cobalt.chem.ucalgary.ca/ziegler/educmat/chm386/rudiment/tourexp/photelec.htm

[URL]http://www.cobalt.chem.ucalgary.ca/ziegler/educmat/chm386/rudiment/tourexp/photef1.gif[/URL]
"This graph shows the typical results of an experiment."

Simulation does not meet this requirement, therefore it violates "the typical results of an experiment".
 
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  • #12
ZapperZ
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In the simulation are available only simple metals, i.e. no bialkali photocathode. In addition, QE is not the same as photo current.

QE = i/i_max = n*e/i_max
i_max = n_max*e, n_max = I/hv
=> QE = n/(I/hv), n is the number of emitted electron
Actually, QE IS the same as the number of emitted electrons. It is proportional to it. Have you done such measurement? Try it. Same photon intensity, but material with different QE. Tell me if you don't get different amount of photocurrrent. I did such measurement just last week!

BTW, look at the definition of "quantum efficiency". It is defined as the number of electrons emitted per photon! So by definition alone, one can already tell that the higher the QE, the more photoelectrons that will be produced!

I have also not only books: http://www.cobalt.chem.ucalgary.ca/ziegler/educmat/chm386/rudiment/tourexp/photelec.htm

[URL]http://www.cobalt.chem.ucalgary.ca/ziegler/educmat/chm386/rudiment/tourexp/photef1.gif[/URL]
"This graph shows the typical results of an experiment."

Simulation does not meet this requirement, therefore it violates "the typical results of an experiment".
You seem to be forgetting an important point that I've been making. I explicitly mentioned that both graphs were correct, within the CONTEXT! I didn't say one of them is wrong. YOU DID! I was trying to correct you into thinking that the one you think was "fatally" wrong, isn't fatal at all, because there is a perfectly valid explanation for such a curve. And yes, it can also come from metals, because different metals have different work function/QE (compare Cu with Mg, for example, both of which I know very well since I use both of them as photocathodes). Try it with photon energy of 5 eV, and then try it with photon energy of 20 eV.

If, after all this, you are STILL insisting that there is this "fatal error" in that figure, then there's nothing else I can say anymore, since I've tried. At some point, my effort will become a waste of my time.

Zz.
 
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