I Photoelectric effect: Saturation voltage

AI Thread Summary
The discussion centers on the photoelectric effect, specifically regarding saturation current and voltage. It clarifies that saturation current is achieved when light intensity is constant and the potential difference is at its maximum, allowing for a steady emission rate of photoelectrons. However, there is confusion about the implication that all photoelectrons are collected, as graphs indicate that not all are captured, leading to a saturation point. The journey of emitted electrons is also explored, noting that they flow to the anode and return to the cathode via the voltage source. The relationship between photon wavelength and the maximum voltage of released electrons is emphasized, highlighting the proportionality of photon energy to frequency.
mgcgian
Messages
3
Reaction score
0
TL;DR Summary
Idk, what level, but this is A-levels, so AP physics?
Hi, I just got two inquiries:

My understanding is that the saturation current is reached when the intensity of light is fixed, when the potential difference is at its positive maximum magnitude. Subsequently, whatever kinetic energy store of any photoelectrons, however deep within the metal's surface, will be collected at the anode. Hence, a saturation point is reached, as the rate of emission is now at a constant rate, and now being collected at anode.

What I don't understand is: when a constant current reached, in my textbook, it states that "all the photoelectrons are collected". But, clearly from the graph, if they were to be all collected, wouldn't the voltage decrease.

But in all graphs, they asymptote to a specific value, depending on the intensity of light, showing that not all the photoelectrons are collected.

This leads onto my second, the circuit is complete when photoelectric emission occurs and the electrons are "collected",

so what happens to the electrons, do they flow around the circuit? Do they go to to the anode then go back to the photocathode. The journey of the electron is not really specified.
 
Physics news on Phys.org
Welcome to PF.

Your title "saturation voltage" suggests a maximum voltage. The wavelength of the photons decides the energy and therefore the maximum voltage of a released electron, since the energy of a photon is proportional to frequency.
Energy in eV = 1239.84 / λ nm = voltage.
https://en.wikipedia.org/wiki/Photoelectric_effect#Theoretical_explanation

Saturation current is determined by the intensity of the light, i.e the number of photons that can cause emission of an electron.
https://en.wikipedia.org/wiki/Photo...imental_observation_of_photoelectric_emission

The electrons emitted by the cathode will be collected by an external anode, then will return to the cathode via the voltage source used to set the anode voltage more positive than the photoemissive cathode.
 
Last edited:
  • Like
  • Informative
Likes mgcgian, vanhees71 and berkeman
mgcgian said:
TL;DR Summary: Idk, what level, but this is A-levels, so AP physics?

But, clearly from the graph, if they were to be all collected, wouldn't the voltage decrease.
I'd look at it as a fully charged capacitor - voltage is constant and no current in.
 
sophiecentaur said:
I'd look at it as a fully charged capacitor - voltage is constant and no current in.
Thanks for the analogy,
 
Baluncore said:
Welcome to PF.

Your title "saturation voltage" suggests a maximum voltage. The wavelength of the photons decides the energy and therefore the maximum voltage of a released electron, since the energy of a photon is proportional to frequency.
Energy in eV = 1239.84 / λ nm = voltage.
https://en.wikipedia.org/wiki/Photoelectric_effect#Theoretical_explanation

Saturation current is determined by the intensity of the light, i.e the number of photons that can cause emission of an electron.
https://en.wikipedia.org/wiki/Photo...imental_observation_of_photoelectric_emission

The electrons emitted by the cathode will be collected by an external anode, then will return to the cathode via the voltage source used to set the anode voltage more positive than the photoemissive cathode.
Thank you for the explanation
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top