My question is about photo electric but it could be applied to other daily routine phenomenon. As we know rest mass of photon is zero. When a photon strikes the metal surface it transfers its energy to the electrons. Whether electron will be emitted or not, it depends upon work function. But my question is about that photon which was hit on metal surface. Does that photon vanished? Does that photon turned into nothing? where does it go after scattering? When I study this I only find the story about the emitted electrons but not about photons after collision. Am i missing some basic concept?
[Physics] where does the photon go after scattering
photoelectric-effectphotons
Related Solutions
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:
- Each electron has an effective cross section of interaction (each electron has some "size"). An average cross section of interaction may be defined.
- Electrons are distributed in some manner on the specimen. An average density of electrons per unit area may be defined.
- After an interaction with photon, each electron has some characteristic time during which a second interaction is possible (this time is very hard to estimate; in fact, I don't know if there are analytical methods for performing this estimation). An average characteristic time may be defined.
- The number of photons per unit area per unit time depends on the intensity of the light. An average number of photons per unit area per unit time may be defined.
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.
When Electron "orbits" nucleus it's trapped in potential barrier caused by nucleus:
Electron needs some energy $E_0$ to escape (overcome) that barrier ($E_0$ is same as work function $\phi$), When photon with frequency $f$ (Energy of that photon is $E=hf$) comes and hits electron, it gives it energy ($E=hf$), and if it's greater than $\phi$ then electron can escape the nucleus (overcome the potential barrier):
and it will have some kinetic energy ($KE$) too. So in order to electron escape the nucleus photon which will hit it must have greater energy than $\phi$ (it's same as greater frequency than $f_0$ (where $f_0$ is minimum frequency for photon which will hit electron so it will escape the nucleus and you can calculate it using this equation $\phi = hf_0$ and $f_0=\frac{\phi}{h}$)). But when photon which will hit electron has same energy as minimum energy required for electron to escape the nucleus ($\phi = E$ or $f=f_0$) then electron will just "go up" to the top of potential barrier and then it will "go down" back to the bottom of the potential barrier:
and it wont be able to escape the nucleus.
(sorry for my poor drawings)
Related Question
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Best Answer
In particle interactions the total number of particles is not conserved. For example in a collision in the LHC two photons collide and many hundreds of particles are created in the collision.
There are still some conserved quantities, for example lepton number is still conserved so you cannot just create an electron. You need to create an electron and positron together so the total lepton number doesn't change (the electron has number +1 and the positron -1, so they add to zero).
However the number of photons is not conserved. Photons are bosons and they are their own antiparticle so no particle number conservation law is violated when you create a photon. Specifically an accelerating electron can emit any number of photons, and the corollary of this is that an electron can absorb photons and be accelerated. This is what happens in the photoelectric effect. An electron in the metal absorbs the photon and its energy is increased by the photon energy (so energy is conserved). The electron will in turn collide with other electrons in its vicinity, and in a small percentage of cases enough energy is transferred to another electron that it can escape from the surface.
Note that not every photon falling on the metal emits a photon. Far from it in fact as the quantum yield is usually around $10^{-5}$ i.e. only one photon in 100,000 manages to eject an electron. In the other 99,999 cases the energy of the photon just ends up heating the metal.