In this question I am talking about the following situation:

enter image description here

Now, I know that the max kinetic energy of the electrons emitted is

$KE_{max} = h\nu - e\phi_{em}$

where $\phi_{em}$ is the work function of the emitter electrode (on the left in the diagram). And my lecturer agrees with that, but he tells us that the stopping potential $V_0$ can be found using

$eV_0 = h\nu - e\phi_{col}$

where $\phi_{col}$ is the work function of the collector electrode (on the right in the diagram). The emitter and collector electrodes are made from different metals.

What I don't understand is why the stopping potential doesn't depend on the kinetic energy of the emitted electrons.


I have attached the slide from the lecture course

Lecture slide

  • $\begingroup$ It may be helpful to talk in terms of total energy of the electron (kinetic plus potential energy), which is ideally conserved. Your diagram is after all drawn with a total energy quantity on the Y axis. You speak of "the kinetic energy" but the electron has a different kinetic energy at every part of its journey. $\endgroup$ – Nanite Jun 11 '14 at 19:55
  • $\begingroup$ @Nanite I always assumed it was initial kinetic energy after being liberated from the metal. Also, are you referring to my diagram in my answer or the lecturer's diagram in my question? $\endgroup$ – binaryfunt Jun 12 '14 at 16:47

Yes, the work function in that equation is in fact the collector's work function. It's a common misconception (see article: http://iopscience.iop.org/article/10.1088/0143-0807/32/4/018/pdf).

The proper way to treat this is to say: the kinetic energy of the emitted electron is $h\nu - e\phi_{emitter}$, as the electron flies towards the collector it sees an uphill potential and so decreases its energy by $e[V_{stop} + (\phi_{collector} - \phi_{emitter})]$. Now, by definition of the stopping potential, $h\nu - e\phi_{emitter} - e[V_{stop} + (\phi_{collector} - \phi_{emitter})] = 0$ (i.e. electrons arrive with zero K.E.). So we see the emitter's work function cancel out and rearranging we get Einstein's equation with the collector's work function instead!

The energy diagram you are drawing plots the total energy of the electron (K.E. + P.E.). So if you think about it, energy conservation is the same as saying that the level of $h\nu$ matches that of $e(V_{stop} + \phi_{collector})$. The emitter's work function simply does not play a role in determining the stopping potential!

There are some subtleties: we assumed that all the electrons emitted were near the Fermi level, but this is of course not a bad approximation at room temperatures, so it's fine. Now here's the thing, even in theory there are unequal numbers of electrons sitting at a given energy inside the collector (the density of states for a free particle scales as $\sqrt{\epsilon}$ in 3D), so experimentally you will find that the I-V curve does not have a sharp stopping potential, but has a so called 'Fermi tail' tending to zero current. This complaint of Einstein's theory is well documented, with articles dating back to the 1920s (Millikan).

So while Einstein was technically wrong in this aspect of the photoelectric effect, he did get the photon bit right which won him the Nobel!

Hope that clears things up!

P.S. For future generations: if you are from a certain London college, your professor is indeed correct!

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    $\begingroup$ Why is the 'uphill potential' seen $e[V_{stop} + (\phi_{collector} - \phi_{emitter})]$ and not just $eV_{stop}$? Thanks :) $\endgroup$ – SomePhysicsStudent Mar 1 '16 at 17:05
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    $\begingroup$ Towards the end of my time at said London College and I finally understand the mystery of Cullercoats! Thank you for your help with this problem! :) $\endgroup$ – SomePhysicsStudent Mar 17 '17 at 2:34
  • $\begingroup$ Finally getting around to accepting this answer, as it both makes sense when you think about it in the full detail given here, and it cites a paper on the topic. The diagram in my answer complements this picture. The other, more highly upvoted answer seems to miss the detail of the electrons facing that uphill potential in order to enter the collector. $\endgroup$ – binaryfunt Nov 26 '18 at 16:22

But the stopping potential does depend on the kinetic energy of the electrons. The stopping potential is defined as the potential necessary to stop any electron (or, in other words, to stop even the electron with the most kinetic energy) from 'reaching the other side'.

As you already stated, the maximum kinetic energy is given by $$K_\text{max}=h\nu-e\phi_\text{em}$$ In order to stop an electron with this amount of kinetic energy, you have to impose an electric field such that it will lose exactly this amount of energy while traversing it, so that it stops just slightly before reaching the other end of the setup shown in your picture.

The energy gained or lost by a charged object traversing this static electric field is given by the simple formula $$\Delta K=q(V_\text{final}-V_\text{initial})$$ In the case of an electron, $q=e$, while the difference in potential can be denoted by $\Delta V\equiv V_\text{final}-V_\text{initial}$ Now, if we want to stop the most energetic electrons, but only barely, we have to make sure that $|\Delta K|=K_\text{max}$. Let us denote the corresponding potential difference, the stopping potential, by $V_0$. Then, we obtain $$\Delta K=K_\text{max}=eV_0=h\nu-e\phi_{em} $$ As you see, we are only considering the most energetic electrons when we want to make sure all of them are stopped. This explains why the velocity does not appear as a variable.

  • $\begingroup$ But the professor who does our quantum physics lectures makes a point of saying it's $eV_0 = h\nu - e\phi_{col}$ with the collector work function in the formula, not the emitter work function. He keeps referring to his confusing energy level diagram on the slide. $\endgroup$ – binaryfunt Jun 2 '14 at 13:19
  • $\begingroup$ He always says that textbooks get this wrong or assume the emitter and collector plates are made of the same metal. I don't know who to believe! $\endgroup$ – binaryfunt Jun 2 '14 at 13:25
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    $\begingroup$ @BrianFunt ah, so that's the real question. To be completely honest, I'm not quite sure what to make of that. I don't see a clear reason why that should be the case. It seems to me that he's just plain wrong. I understand the diagram but it seems irrelevant. $\endgroup$ – Danu Jun 2 '14 at 14:54
  • $\begingroup$ He says that the photoelectrons require energy to get into the collector plate. But even if this is true, surely that would mean $eV_0 = h\nu - e\phi_{em} - e\phi_{col}$? $\endgroup$ – binaryfunt Jun 2 '14 at 17:57
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    $\begingroup$ @BrianFunt I believe your instructor has it wrong. The electrons are stopped or not stopped before they get to the collector (while they are free), so the stopping voltage can't include the collector work function. Now, the energy of collected electrons will include the collector work function, but that is not generally measured. $\endgroup$ – dmckee --- ex-moderator kitten Jun 7 '14 at 19:16

The potential difference needed to just stop the photoelectrice current is knows as stopping potential.When the electrons of highest energy are stopping the electrons of lower energy are already stopped.Using photoelectric equation. eV=Kmax and Kmax=hv-Q We clearly observe that stopping potential depends upon frequence of incident radiation and work funcgion of emitter.

  • $\begingroup$ Hi Piyush Prakriti and welcome to Physics.SE! Please see this help post to learn how to write your equations in a way nicer way i.e. in $\LaTeX$, in order to improve legibility. Thanks! $\endgroup$ – Gonenc Feb 15 '16 at 7:36

According to a fellow student, the photoelectron has to overcome the difference in work functions, $e\phi_{col}-e\phi_{em}$, before entering the collector electrode. I made a more detailed energy level diagram to fully illustrate the components involved here.

enter image description here

On the left: the full details. On the right: what the professor told us.

So the stopping potential formula is $eV_0=h\nu-e\phi_{em}-(e\phi_{col}-e\phi_{em})$ which equals $h\nu-e\phi_{col}$.


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