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If we increase frequency of light, then energy of photons will also increase. We can clearly see this from this equation,

$E=h\nu$ ; $\nu$ is the frequency of light

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We know that electrons lose energy on their way out of the metal lattice due to collisions. Some electrons lose all the energy due to the collisions before coming out of the metal and thus they cannot leave the metal lattice. But now since the energy transferred to them by the photons is more, those can now make their way out of the atom despite all the collisions. So, considering these electrons which now manage to come out, number of photoelectrons should definitely increase on increasing the frequency of light.

If the number of photoelectrons have increased, then saturation current, (photocurrent at which electrons with zero kinetic energy are also able to leave the metal lattice) will increase. It is because it may so happen that the electron which previously lost all its energy before coming to the surface of the atom will come out of the metal lattice with zero kinetic energy.

Therefore, this photocurrent vs anode potential graph at two different frequencies is wrong. Saturation current of light with more frequency must be more since the number of electrons that reach the other plate is more. ( Saturation current is actually a measure of photoelectrons that reach the other plate).

What am I missing out?

Please don't give any explanation based on formulas.

To illustrate what I am saying better,

Consider an electron. In situation 1, it absorbed 3ev (numbers are randomly taken) energy from a particular photon. On its way out of the atom, it faced collisions. Say, to overcome those collisions, it needs to give 4ev energy. But since the electrons does not have that much energy, it will not be able to come out of the atom. In situation 2, say, the frequency of the light increases, and therefore more energy gets transferred to this electron, say 5ev. Now, on its way out of the atom, it will have enough energy to face all these collisions and come out. Therefore, it becomes a photoelectron.

enter image description here

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Kinetic energy of a photoelectron depends on the energy needed to take it out of the lattice and the energy of the incident photon that did this. Since it is acctually surface effect collisions with lattice atoms are not very important. So, when you shine a light onto the surface of the metal, number of electrons emitted depends on the NUMBER of photons. Emission itself of course, depends on the energy. So because electrons in metal need some exact amount of energy to be taken away from the lattice a photon must have at least this much energy. But one photon can kick out only one electron etc...now, in your scenario you say that there are electrons that simply could not get out of the metal...but have you considered how could a photon get to this electron? Metals are not very good trasperent materials...light interacts with the metal right on the surface. So electrons from the surface are the ones that are kicked out. So collision effects that you are talking about have almost ne effect in photoelectric effect. Experimental data shows that there is only one unique energy needed to start the effect in a given metal.

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  • $\begingroup$ All this while, in the entire theory of photoelectric effect, we have been hesitant to talk about kinetic energies of electrons. We only talked about range of kinetic energy. If your assumption was so valid, then we could have taken the maximum kinetic energies that the outer surface electrons will possess and work with that. $\endgroup$ – Arishta Jan 10 '17 at 11:06
  • $\begingroup$ What do you mean "Metals are not very good transparent materials". A metal is made up of unit cells which when repeated in all dimensions generate the entire lattice. So metal lattice at the surface has the same environment as the inner parts of lattice. Also, electrons at the metal surface do not belong to a particular nucleii, they can move around the entire lattice. So, an electron which was in the inner kernel might be at the outer surface at the next instant. After light ejects out some photons, electrons inside the metal lattice will come to take their place to overcome the deficiency. $\endgroup$ – Arishta Jan 10 '17 at 11:11
  • $\begingroup$ Now, those electrons that have come to take the place of the now photoelectrons get ejected out . So, I think it clearly explains why your argument that light ejects out electrons on the surface of the metal atom mostly is baseless. $\endgroup$ – Arishta Jan 10 '17 at 11:12
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    $\begingroup$ @Blue, No. The environment of an atom at the metal surface is not identical as that of one in the bulk. The long range order ( extending infinitely in all directions) is a theoretical description about a crystal. In reality, no crystals have infinite length along the three dimensions. $\endgroup$ – UKH Jan 10 '17 at 11:14
  • $\begingroup$ @Unnikrishnan, okay even if we consider that, it is true that when electrons from the surface are ejected out, the electrons which are in the inner part of kernel will rush to take their place. Therefore, we can say, all the electrons of the metal lattice have good chances of getting out. $\endgroup$ – Arishta Jan 10 '17 at 11:16
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The metal has a workfunction $\phi$ and associated with $\phi$ is a very particular frequency, sometimes known as the threshold frequency and as you are no doubt aware the EM radiation must have a frequency greater than or equal to this to cause photoelectron emission.

If you increase this frequency of the EM radiation the no of photons liberated from the metal will not increase as the light interacts with the electrons in the matter on a one to one basis. That is to say: Only 1 photon can provide enough energy to liberate 1 electron. Increasing the frequency will just give the electons more kinetic energy as they are expelled from the surface of the metal.

You would have to increase the intensity of the EM radiation (provided its frequency is above threshold frequency) to increase the number of photoelectrons liberated per second.

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  • $\begingroup$ Consider an electron. In situation 1, it absorbed 3ev (numbers are randomly taken) energy from a particular photon. On its way out of the atom, it faced collisions. Say, to overcome those collisions, it needs to give 4ev energy. But since the electrons does not have that much energy, it will not be able to come out of the atom. In situation 2, say, the frequency of the light increases, and therefore more energy gets transferred to this electron, say 5ev. Now, on its way out of the atom, it will have enough energy to face all these collisions and come out. Therefore, it becomes a photoelectron $\endgroup$ – Arishta Jan 10 '17 at 7:25
  • $\begingroup$ @Blue Okay good point, I need some more time to think about this. If I can't figure it out then hopefully someone else will be able to give you a better answer than this one. $\endgroup$ – BLAZE Jan 10 '17 at 7:30
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The question has got a really a simple conceptual answer.

On increasing the frequency of photons the energy of a photon increases. The photoelectric effect is a phenomena which increases only when a large number of surface electrons are knocked out, that is, a large number of surface electrons collide with collide with incoming photons.

Increasing frequency increases the energy of each photon, but increasing intensity increases the number of photon. The photoelectric effect increases only when large electrons are knocked out from surface, which is only done by high intensity of light.

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