I know that when light is incident on certain metallic surfaces, electrons are emitted from the surface. And I know that the maximum velocity can be calculated. But why does the textbooks use 'maximum velocity' rather than 'velocity'? My assumption is that those emitted photoelectrons don't have the same velocity. If it is the case, then what causes the phenomenon? And is it possible to derive the velocity distribution of these photoelectrons by simple experiments?
[Physics] what is the velocity distribution of photoelectrons in photoelectric effect
photoelectric-effect
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In order to answer your question you must think through the entire measurement. In this case you are measuring the photoelectric effect by placing a photodiode into a circuit. What kind of circuit element is the photodiode? It is a source. And as you have pointed out above, it is a constant voltage source (with voltage determined by the energy of the photons and the work function of the metal). It is also current limited (with no load it still cannot draw more current than there are incoming photons). I think you may be slightly confusing yourself because current is charge/time so it depends on how fast the electrons are moving (i.e. the voltage). Always recall that it is a constant voltage source.
To proceed with understanding your lab you need to understand your circuit. There is a schematic diagram of the photodiode, op-amp, and associated circuitry in your pdf. I am sure you are familiar with the basic idea: the circuit acts as a capacitor which builds up voltage until the current is 0 (or approaches 0). The 'current-limited' feature of the diode power supply effects the capacitor charging only at the start of a measurement. The trick to this equipment is understanding the op-amp. There is plenty of literature out there on op-amps. Wiki is always a good place to start.
If you have done your lab by now you will have noticed that there IS some intensity dependence. How can you explain this? (as an aside... ideally you have recorded voltage vs time for each measurement and fit it to an exponential--this assumes it is capacitor charging at constant voltage) I usually tell my students to chalk it up to the finite input impedance of the amp causing the current at the input to the amp to be non-zero and changing the feedback mechanism. I got this explanation from a professor a long time ago, and I am not sure I entirely buy it (Pasco claims there is an enormous input impedance). I never came up with a better answer while I was teaching the lab though.
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.
Best Answer
Photo-emission from a metal surface is a multistep process. When the photon hits the metal surface it excites a photoelectron with a very high quantum yield, but that photoelectron is travelling in the same direction as the photon i.e. down into the metal. For an electron to be emitted from the surface the initial photoelectron has either to ricochet back out of the metal, or more likely to transfer energy to other electrons so one of the other electrons has enough energy to leave the surface. This process is essentially random and consequently has a very low quantum yield, so the overall quantum yield for photoemission is around $10^{-5}$ to $10^{-6}$.
The scattering of the initial photoelectron occurs via inelastic collisions, so whether it's the original electron scattered backwards or other electrons scattered by the initial photoelectron, the energy of the electron leaving the surface is less that the energy of the photon. The balance of the energy goes into lattice vibrations of the metal i.e. heat.
Offhand I don't know the energy distribution of the photoelectrons, but it is a well studied phenomenon.