The work function of any metal is no doubt constant for it is related to electromagnetic attraction between electrons and protons. However on increasing the intensity of any light source the kinetic energy of the emitting electrons must increase, mustn't it? Let us assume there is only 1 electron in a metal surface. Let hf be the energy required to expel it out with a velocity 'v'. Again let us increase the intensity of the source keeping frequency constant. Now ' hf ' will change to 'nhf' where n = no. of photons striking on electron at a same time. Since work function is constant the only variable must be 'v' letting it increase its K.E. This clearly shows Kinetic energy of emitting electrons is directly proportional to the intensity of light source. If kinetic energy depends upon the intensity, stopping potential for a particular frequency of light for a particular metal is a variable quantity. It is true that on increasing the intensity the no.of photo electrons will increase. But what if the no.of electrons in a metallic plate is constant. Suppose I have only two electrons in a metal, on increasing intensity i.e on increasing the no.of photons , the no.of photons colliding from different sides simultaneously may increase . So, K.E of emitted electron must increase on increasing intensity, mustn't it? But experimental data doesn't show this. What is wrong? Help me out
[Physics] In the famous Einstein’s Photoelectric effect, why does the intensity of light not raise the kinetic energy of the emitted electrons
photoelectric-effectquantum mechanics
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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.
Intensity is the total amount of energy falling (or going through) per unit area per unit time i.e, $\frac{J}{m^2.s}$.
For monochromatic radiation,
$Total\space energy = Number\space of\space photons\space \times Energy\space of\space one\space \space photon$
and $E_{photon}=h\nu$
$Intensity = \frac{Number\space of\space photons\space \times Energy\space of\space one\space \space photon}{At}$
$I=\frac{nh\nu}{At}$,
For constant area and time,
$I \propto n.\nu$
This is a very important result. You can increase the intensity of the radiation by either increasing the number of photons in it or increasing energy of each photon, or both.
How the current can be increased ? :
A single photon with energy equal or greater to the work function $(\phi)$ of the cathode plate will knock out a single photo-electron (Considering no energy is lost by the electron in collisions with other electrons and atoms of the plate).
The statement that increasing the intensity increases the number of electrons emitted by the cathode plate is indeed correct, but incomplete.
$h\nu=\phi + KE_{max}$
As stated before, intensity can also be increased by increasing the frequency of the incident radiation while the number of photons in it is kept constant and this won't change the number of photoelectrons emitted. Although, this would increase the maximum kinetic energy of the photoelectrons.
We have to be a little bit more specific.
Specifically, At constant frequency, if the intensity is increased then the number of photoelectrons will increase.
In other words, if we set the frequency (or wavelength) of the radiation to the threshold frequency (or threshold wavelength) and then increase the intensity, the number of photo-electrons will surely increase.
Best Answer
In general you're right - an electron being subject to interactions with more than a single photon may have a higher kinetic energy. However, in the vast majority of photoelectric setups you will observe that kinetic energy is independent of light's intensity.
The appropriate framework for this discussion is this of probability theory:
Now, you should ask the following question: "given the effective cross-section of interaction of electrons, the average number of electrons per unit area, the average characteristic time and the average number of photons per unit area per unit time, what is the probability for an electron to interact with more than one photon?".
The usual answer to the above question is "negligible". This happens, but so rarely that the current due to these electrons is below your measurement error.
However, in high intensity experiments (where the number of photons per unit area per unit time is enormous), multi-interaction-electrons were observed. See this for example.
Analogy:
The best analogy I can think of is this of rain. You may think about individual photons as drops of rain, about individual electrons as people in the crowd (each of whom has an effective cross-section of interaction which depends on how fat the person is :)), and about the characteristic time as of time it takes to open an umbrella over the head.
Now, if the rain is weak (usually when it just starts), each person in the crowd is hit by a single first drop. He takes his umbrella out of his bag and opens it above his head. If he does this sufficiently fast (short characteristic time), he will not be hit by more drops.
However, there are cases when the rain has no "few drops per minute" phase - it almost instantly starts and is very intensive. In this case, no matter how fast the people open their umbrellas, they will be hit by many drops.