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While studying photoelectric in my school, my teacher drew a graph of current versus the potential difference across the two electrodes:

I am not able to understand why do we get saturation current. I know that at saturation current all the electrons emitted from the metal surface are able to reach the other electrode. My question is suppose we increase the potential difference between the two electrodes further. I think that one of the following two things should happen:

  1. current may increase as now the electrons in the wire will be under greater potential difference thereby reducing the time in which the electrons complete one round to reach the same point again and hence increasing current. ($V=IR$)

  2. current may not be continuous (steady current might not be achieved) as the time taken by the photoelectrons to reach the positive electrode may be more than the time taken by the electrons inside the wire to go from end of the wire nearer to the positive electrode to the other end.

So in no case should we get a graph as shown in the link.

It may be that I have not understood the photoelectric effect completely. If that is the case please tell me where am i wrong?

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In the photoelectric effect, photons incident on the cathode cause the emission of electrons. Assuming there is a sufficient electric field, these electrons will make their way across to the anode, contributing current.

For simplicity, let's assume every photon generates a photo-electron. Then if $N$ photons per second hit the cathode, the current will be carried by a total of $N$ electrons per second. We always assume there are "infinitely many" electrons waiting for their turn, and the thing limiting the current is how many electrons get "released" from the cathode (i.e. how many photons hit the cathode).

Current is charge per unit time. If the electron has a charge $q_e$, then $N$ electrons per second carry a current

$$I = N\; q_e$$

There is nothing here about the velocity of the electrons... not about the time it takes them to cross the gap. If they went 100 times faster, it would not change the number of electrons crossing the gap per second. That number is determined by "how many start the trip per second" and "how many don't make it". The second of these explains that the curve starts out not completely flat: very slow electrons may not make it, especially with a small retarding potential. But once they go fast enough to fully escape, their final speed really doesn't matter. And neither does the transit time in the wire.

Did you ever calculate how slowly electrons move in a current carrying wire (for example, copper wire)? While the electrical signal is very fast, the drift velocity of the electrons themselves is very very slow... because there are so many electrons per unit volume. But that is only tangentially relevant here.

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  • $\begingroup$ Does this mean that if you attained saturation current for a particular intensity then definitely if you increased the intensity of incident radiation the saturation current would change as well? Also, I doubt the validity of the reference of importance of drift velocity of electrons can be applied here. In an electric circuit, electrons are present throughout the metal conductor. However, in the evacuated tube considered here, the electrons are only generated by the radiation and the time they take to reach the other plate would definitely depend on their speed? $\endgroup$ – LeroyJD Nov 15 '16 at 16:47
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I want to add to the answer by Floris, which explains the phenomenon correctly. Ask yourself the meanings of stopping potential and the saturation current.

The stopping potential is determined by the energy of the photons minus the work function of the material in question. Let's idealize the situation: For example, the work function of copper is 4.7 eV. Say if you have a light source and shine the copper with light of energy 5.7 eV then the stopping potential will be roughly -1 V. That is because, a photon with energy 5.7 eV hits an electron, electron uses 4.7 eV to free itself from the copper metal and gains 1 eV as kinetic energy. But it can never make it to the other electrode because it is stopped by the voltage. Increase the potential (that is go towards 0 and positive values) then the electron can make it to the other electrode and you get current.

In reality, however some of the electrons looses their kinetic energy even before they can escape from copper. That is, they end up with kinetic energies less than 1 eV and therefore they need a "push" to be able to reach the other electrode. As you increase the voltage you give that push to these poor electrons and make them reach the electrode and hence increase the current.

However at some point the electrons which can make to the other electrode will be saturated because the light source can produce certain amount of photons and therefore certain amount of photoelectrons. The scientific term for "amount of photons" is the intensity of the light source. If you have one more equivalent light source and lit it then you will get the saturation current roughly doubled.

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There nothing much to add to the beautifully scripted answers above.

Basically the number of photo electrons emitted depends on the intensity of incident radiation, while the maximum kinetic energy that an emitted photo electron can posses is a function of the potential across the collector plates. Besides, it is obvious that the maximum number of photo electrons which can be reach the collector plate depends on the intensity, therefore for a given intensity (i.e there is a given maximum number of photo electrons which are emitted), whether these reach the collector (overcoming repulsions by space charge) is depends on the potential.

Therefore the number of photoelectrons reaching the collector plate reaches a unique constant value (equaling the maximum number of photoelectrons based on the intensity of incident radiation).

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What is Current ? Current is flow of electrons in a given time i.e if the intensity of the light is Constant then Photons incident on Cathode (ejector plate) is also constant. As we know that minimum K.E, that photo-electron can have is 0 and max. is energy of the photons minus the work function of the material. And saturation current means flow of electron in a given time is constant i.e no. of electrons ejected from Cathode is constant as no. incident photons are constant.

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The number of photoelectrons emitted from metal surface will depend on the energy of light wave incident on metal surface. Now let the maximum no. Of electrons that can be emitted by the given light wave be 'x' or let this be an integer for e.g. 100 electrons are emitted, now only these 100 electrons are responsible for carrying current.

Now, the main point comes further, so read carefully.

  • the maximum kinetic energy of electron be k°

  • not all the ejected electrons will be able to reach the other plate, this is because few electrons will suffer collision with other electrons or will suffer collision with the gas molecules(basically inert gas atmosphere is taken, for which inert gases like helium or argon is taken).

    Due to collision few electrons will not be able to reach the other plate. And for few electrons, the minimum K.E. required to reach the other plate will be less than what is required.

  • let 60 electrons dont reach the other plate. Now we apply potential difference in the circuit so that electrons now have ease to reach the other plate efficiently.

Let 10V be the pot. Diff. Applied, now let 10more electrons reach the other plate in the fluence of the pot. Diff.

Further increasing the pot. Difference to 20V let 10 more electrons reach the other plate, now remaining electrons is 40

Now in this way increasing the por diff. So that all the 100 electrons are able to reach the other plate, now, the remaining electrons which are unable to reach other plate would be zero. As all the electrons are now able to reach the other plate, now the current which is directly proportional to applied potential reaches its maximum value and after reaching this maximum value, on further increasing the potential difference, current should have also increased(hypothetically) but the fact is that number of photoelectrons reaching the other plate is now 100, which means all the ejected photoelectrons are now able to reach the other plate. On further increasing pot. Diff. Would not increase the no. Of photoelectrons that reach the other plate(i.e. the carriers of the current) the no. Of photoelectrons reaching will now be 100 only, and will not increase beacuse of increase in pot. Diff thus current after reaching its saturation value cannot be further increased.

Regarding the second point, The current flowing depends on the intensity of light. The electrons flow in the area between the plates in almost same speed as in the wire.

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