Consider a Photoelectric effect experiment apparatus as shown in Figure 11.1

Figure 11.1

The variation of photocurrent with the voltage applied across $\mathrm{A}$ and $\mathrm{C}$ is as shown in Figure 11.3.

Figure 11.3

My question –

Why is the current non-zero even if the voltage is zero?

Consider a situation in which the voltage across the plates is zero. Photons of sufficient frequency strike the outermost surface of the cathode, emitter plate $\mathrm{C}$. Electrons become free from the attraction of nucleus as their net energy becomes positive.

Why I think that there should be no net current?

As soon as photons strike the outermost surface of the cathode(assumed to be $1$ to $10$ atomic diameters thick), electrons become free. Now, as the electrons have kinetic energy, they'll come out of that outermost surface. Now, the electrons(with kinetic energy) should have an equal probability of going in every direction of the surface. So, around half of the electrons should go in the right direction towards the anode and strike it and half should go to the left, trying to penetrate the inner surfaces of the cathode plate.

Now, the electrons which will go to the right will face some field due to the space charge present in the tube but nevertheless, some electrons will definitely reach the anode plate. After they strike with the anode plate, they'll face some resistance but still, some electrons will reach the low resistance copper wires connected to the anode plate and milliammeter.

Now, consider the electrons which go towards left. They'll definitely face resistance from cathode plate but some will still reach the copper wire.

Now, as there is no potential difference applied between $\mathrm{A}$ and $\mathrm{C}$, there is no need of commutator and voltmeter. So, the ammeter is connected in series with the anode plates and cathode plates with copper wires.

Now, if there is steady state current in the circuit, electrons emitted to any one of the sides should dominate over the other. But the number of electrons emitted on both the sides is same, and all the electrons should face the same resistance as the current(electrons) should flow in a closed loop. But if the number and velocity of electrons coming from both the sides are same, there should be no net current. But there is. Why?

Edit –

From the answers of anna v and mmainville, it can be seen that the emission of electrons is dependent on the angle of emission of light. But still, the question remains. If the electrons emitted towards anode have enough energy to complete the circuit and again reach the holes in cathode despite the resistance in between, the electrons emitted to the left should have enough energy to come to anode plate via copper wires and again get ejected through anode plate by the virtue of their kinetic energy and reach the cathode plate. In this case too, the current should be zero which is not true. Why?

  • $\begingroup$ I'm not sure if this is the answer to your question, but the metal is ionized by the incident radiation. The free electrons which move to the left are hitting the metal and are absorbed by it, i.e., the ionization process is reversed and energy is emitted but there is no current. $\endgroup$
    – Virft
    Commented Aug 15, 2017 at 7:20
  • $\begingroup$ @Virft Resistance of $\mathrm{C}$ and wires is not infinite. So, yes some energy will be disappated as heat but there will be an electric charge flow nevertheless. Also, after electrons strike also, they'll have to face resistance too to complete the circuit. $\endgroup$ Commented Aug 15, 2017 at 9:29
  • $\begingroup$ "but there is. Why"Because the electrons that went to the negative hemisphere of the scatter, into the lattice of the cathode, transfer their momentum to the whole cathode lattice ( hyperphysics.phy-astr.gsu.edu/hbase/Solids/band.html ) and are caught in the fermi level of the cathode and neutralize the charge . No current is generated because there is no electron motion outside the cathode. The electrons scattered from the cathode become a beam of charged particles, by definition a current which hitting the anode are detected as extra charge. $\endgroup$
    – anna v
    Commented Aug 16, 2017 at 16:00
  • $\begingroup$ balanced by the positive charge of the cathode left over from the extraction of the electrons from the fermi level. in the backward direction, no charge extracted,in the forwared extracted electrons $\endgroup$
    – anna v
    Commented Aug 16, 2017 at 16:02
  • $\begingroup$ I still can't understand why can't the electrons come out of the cathode plate towards the copper wires in the left. The cathode plate will catch the electrons in the Fermi level but won't some electrons still escape? How can one guarantee that the positive charge created on the outermost surface will be completely neutralized by the electrons going the left? $\endgroup$ Commented Aug 18, 2017 at 15:25

4 Answers 4


You have made an assumption that the photo-electrons will be randomly distributed in all directions. This assumption defies the Law of Conservation of Energy and the Law of Conservation of Momentum.

The incoming radiation from source S has momentum and energy that must be conserved. When the incoming radiation hits the cathode surface, some of the energy and momentum is distributed throughout the metal lattice on average inward and at opposite to the normal angle with an energy of the work function. The tensile forces in the metal lattice structure ensure this uniform distribution and the energy is transformed into heat. The remaining energy must be reflected at some angle which conserves momentum and energy. This reflected energy and momentum is carried by the photo-electrons.

So, there are in fact no electrons emitted to the left in your diagram. In fact, photo tubes are generally designed with a curved cathode surface such that the photo-electrons are reflected and focused towards the anode.

It should also be noted that if photo-electrons were being scattered in all directions then we would have a build up of a net positive electric charge at the location of the creation of photo-electrons. Electrons would have to come in to replace this deficiency from somewhere, but this would be inconsistent. The electric field is either pointing towards or away from this point and the average electron drift will be determined by this electric field. It cannot be both at the same time.

In reality, the photo-electrons travel towards the anode and the cathode lead supplies an inrush of electrons to replace the holes created. And thus you have a current flowing through the tube, commutator and ammeter even with the potentiometer set to 0 V.

  • 1
    $\begingroup$ Does that imply that the current will be dependent on the angle of incidence of the radiation? $\endgroup$ Commented Aug 22, 2017 at 5:53
  • 1
    $\begingroup$ @ApoorvPotnis britannica.com/science/radiation/The-photoelectric-effect slideplayer.com/slide/9561396 $\endgroup$
    – anna v
    Commented Aug 22, 2017 at 6:36
  • 1
    $\begingroup$ Yes it does. If you look at pictures of modern photo tubes and especially photo multipliers, they have focusing plates to help ensure that the reflected photo beam actually reaches the anode. Otherwise the efficiency of the photo tube is much less and the photo electric effect is very difficult to measure $\endgroup$
    – mmainville
    Commented Aug 22, 2017 at 18:10
  • $\begingroup$ I have edited the question again. Please see it. $\endgroup$ Commented Aug 24, 2017 at 6:17
  • $\begingroup$ There aren't any electrons emitted to the left. The energy that is distributed to the left is the work energy which is equivalent to the amount of energy to free the photo-electrons ( which are emitted to the right ) from their bounds. The work energy is distributed to the left mainly as heat. Holes in the cathode lattice are created from the freed electrons which creates an electric field which pulls electrons out of the anode and into the cathode via the copper wiring. Electron flow in the opposite direction must overcome this electric field. $\endgroup$
    – mmainville
    Commented Aug 24, 2017 at 20:17

The simple photo electric effect is here


The electrons appear like an elastic scatter. There are no backward ones because they are absorbed by the metal surface. Electrons in vacuum in motion in one direction are a current by definition of current. As in all scattering experiments with light, there is a beam that hits the cathode. In a photosensivive cathode part of the photons that make up the beam instead of scattering elastically in the scattering angle and rebuild the classical beam in the direction of the scattering angle, will extract an electron.

In particle scattering there is a probability for the electrons to be in all directions but there is an angle of higher probability which the majority of scattered electrons follow. Thus the photoelectrons have a dominant direction. The ones entering back into the metal lattice will neutralize with the hole they left when they scattered. The ones leaving the surface will leave a hole behind, as with normal current flow . This is for gamma rays, but they are photons after all. For low energies there are studies which also give a preferred direction.

angle of phot with electron

In your setup, the same is true, it is the electrons bouncing off the surface of the cathode and hitting the anode a current appears. There is no vacuum when the light is on, because electrons carry the current. Whether there is a voltage drop or not, the geometry of light scattering off the cathode is the same . Applying a voltage allows mapping the effect.

  • $\begingroup$ I think that current is what the Ammeter measures. The electrons emitted from both the sides experience same resistance and as the number of electrons and their speeds emitted on both the sides is same, no current should be recorded by the ammeter. $\endgroup$ Commented Aug 15, 2017 at 13:40
  • $\begingroup$ No, the side looking at the vacuum has no resistance. The electrons leave as bullets semi circularly. The ones backwards are absorbed, the ones hitting the anode create the continution of the current in your plot $\endgroup$
    – anna v
    Commented Aug 15, 2017 at 14:28
  • $\begingroup$ To complete the circuit, electrons struck on anode must again reach cathode as shown in Fig. 11.1. In a closed circuit, the current is same everywhere and all the electrons should experience same resistance. $\endgroup$ Commented Aug 16, 2017 at 1:12
  • 1
    $\begingroup$ that is the macroscopic view,please read hyperphysics.phy-astr.gsu.edu/hbase/electric/ohmmic.html . see physics.stackexchange.com/questions/123456/… Vacuum has infinite resistance so does not fall into the V=I/R classical law when there is an electron beam. Do you realize that the photoelectric effect is one of the pillars on which the discovery of quantum mechanics rests? the other is black body radiation. The photoelectric effect generates a beam of electrons, which hits the anode and closes the vaccum infinite resistance circuit $\endgroup$
    – anna v
    Commented Aug 16, 2017 at 4:21
  • $\begingroup$ . If there were no photons, there is no no current no matter what the potential difference from theother circuit elements. $\endgroup$
    – anna v
    Commented Aug 16, 2017 at 4:22

The electrons start on the cathode. If they go out into the vacuum and then right back to the cathode, well that's where they started, so the net effect is that nothing has happened. These electrons are contributing zero to the current. They are not counterbalancing the electrons that go cathode --> vacuum --> anode.

The only way to counterbalance the electrons that go cathode --> vacuum --> anode is to have other electrons that start in the anode and go anode --> vacuum --> cathode. But the latter doesn't exist, because light is not shining on the anode.

(UPDATE: The question was edited to suggest that electrons are also coming out of the anode into vacuum, even though there's no light shining on the anode. OK, it's true that light is not required to get out electrons---there is such a thing as "field emission", and even field emission at low or zero potential for certain anode materials (e.g. diamond can have negative electron affinity). And the number of electrons exiting the anode is not literally zero, just much much much less than the number of electrons exiting the cathode. The idea in the question edit, i.e. that electrons hit the cathode at high speed, and maintain this speed through the wire, and then come flying out the anode, is very wrong. An electron in a wire experiences something like friction, and loses its high speed probably within nanometers. Anyway, if you think electrons ought to exit the dark anode just as frequently as the illuminated cathode (they don't), maybe you should write a new question saying why you think that. It's kinda far from your original question. I think you'll get better answers that way.)

  • $\begingroup$ Won't the electrons which go cathode --> vaccum --> anode again go to cathode through the conducting wires and counterbalance the positive charge created on cathode? $\endgroup$ Commented Aug 22, 2017 at 5:56
  • $\begingroup$ Note - I've not downvoted the answer. $\endgroup$ Commented Aug 22, 2017 at 5:57
  • $\begingroup$ @ApoorvPotnis - Yes! The electrons that go cathode --> vacuum --> anode then pass through the ammeter back to the cathode to restore charge balance. Whereas the electrons that go cathode --> vacuum --> cathode do not pass through the ammeter at all. "Not passing through the ammeter at all" is very different from "passing through the ammeter in the opposite direction", which is what is required to zero out the ammeter's net current measurement. $\endgroup$ Commented Aug 22, 2017 at 13:00
  • $\begingroup$ I have edited the question again. Please see it. $\endgroup$ Commented Aug 24, 2017 at 6:20

After your edit, I'm confused. So I replace my original answer to a clarification question.

From your description it appears, as if you are not considering the situation shown in your picture, but rather the symmetric situation shown below: enter image description here Instead of labeling the "plates" as cathode and anode, I used emitter $E$ and collector $C_1$ and $C_2$. Furthermore, I grounded each plate, so that we discuss only the situation described in your question.

Now, all the points you mention make sense:

  • Half of the electrons move to the left and half to the right.
  • There should be no potential difference between the two collectors $C_1$ and $C_2$. However, this is not true for the emitter and one of the collectors. To see this, just remove the ground from all plates.
    1. Initially (=before switching on the light) all three plates $E, C_1, C_2$ will be non-changed.
    2. If we switch on the lights, electrons will be emitted by the emitter $E$ and will be collected by the two collectors $C_1, C_2$. Therefore, negative charges will accumulate at the two collectors $C_1, C_2$, while the emitter will build up a negative charge deficit.
    3. Assuming that no current other than the free electrons flows from or to the three plates, the process will eventually stop, due to build up electric potential. A equilibrium state is reached. Important: Because there is no connection between the three plates, we do not change the charge distribution. We merely capture the information, which plate charges how.

Now, we could measure the potential difference between the three plates and conclude that current will flow only from the emitter (=anode) to the collectors (cathode).

EDIT: In order to get the configuration of the original question we have to omit the plate $C_2$. This does not change the role of the emitter and collector: the collector $C_1$ still accumulates electrons (thereby becoming the cathode) and the emitter $E$ still "donates" electrons (thereby becoming the anode).

Does this situation describe your actual question, or am I completely missing it?

  • $\begingroup$ Yes. The diagram describes what I want to say. But you have changed the apparatus. I want the explanation of current recorded by the ammeter in Figure 11.1. $\endgroup$ Commented Aug 19, 2017 at 17:23
  • $\begingroup$ I changed the apparatus, because yours does not describe the situation I am describing above. Therefore, I do not understand your comment, but this might be, because I did not understand your initial question. Hope the other can help you. $\endgroup$
    – Semoi
    Commented Aug 19, 2017 at 18:20
  • $\begingroup$ By the way, how did you create that diagram? $\endgroup$ Commented Aug 20, 2017 at 10:16
  • $\begingroup$ I have edited the question again. Please see it. $\endgroup$ Commented Aug 24, 2017 at 6:20

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