Is it possible to get the photoelectric effect in red color frequency?

Click For Summary

Discussion Overview

The discussion revolves around the possibility of observing the photoelectric effect using red color frequency light. Participants explore the implications of using red light in experiments, particularly in educational contexts, and differentiate between the photoelectric effect and the functioning of solar panels.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Experimental/applied

Main Points Raised

  • Some participants inquire about the feasibility of generating a potential difference (ddp) using red light in a photoelectric effect experiment.
  • Others question the relevance of the photoelectric effect to solar panels, suggesting that the mechanisms involved are different.
  • A participant mentions that the photoelectric effect requires photon energies from 3 eV and above, typically in the UV range, while solar panels can operate with lower energy due to their semiconductor structure.
  • One participant proposes that red light can indeed excite the PN junction in solar panels, potentially generating current, although multiple LEDs may be needed for sufficient output.
  • Another participant discusses the complexities of photoemission as a many-body phenomenon, emphasizing the differences between photoemission in metals and the behavior of electrons in semiconductors.
  • Concerns are raised about the effectiveness of homemade demonstrations for teaching the photoelectric effect compared to standard equipment.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between the photoelectric effect and solar panels, with no consensus on whether red light can effectively demonstrate the photoelectric effect. The discussion remains unresolved regarding the specific capabilities of red light in this context.

Contextual Notes

Participants mention the need for clarity on definitions, such as "ddp," and the potential limitations of homemade experiments compared to established methods. The discussion also highlights the complexity of the photoelectric effect and its distinction from semiconductor behavior.

Renandornelles
Messages
2
Reaction score
0
Is it possible to get the photoelectric effect in red color frequency?
 
Physics news on Phys.org
Welcome to PF.

What reading have you been doing about the Photoelectric effect? How does the wavelength of the incoming EM radiation affect the emission of the electrons?
 
  • Like
Likes   Reactions: vanhees71 and Renandornelles
Renandornelles said:
I wanted to do a small experiment replicating with LEDs, a solar panel, and I would like to know if a red lamp would be able to generate ddp...
What's a ddp?
Renandornelles said:
I wanted to demonstrate the frequency limits capable of emitting electrons for children to understand solar panels.it would be like a little toy, for children to learn by playing.
The photoelectric effect has nothing to do with solar panels, IMO.

Again, what reading have you been doing on your question (please post links), and what do you really want to learn/do?
 
berkeman said:
Welcome to PF.

What reading have you been doing about the Photoelectric effect? How does the wavelength of the incoming EM radiation affect the emission of the electrons?
I wanted to do a small experiment replicating with LEDs, a solar panel, and I would like to know if a red lamp would be able to generate ddp...
I wanted to demonstrate the frequency limits capable of emitting electrons for children to understand solar panels.
berkeman said:
What's a ddp?

The photoelectric effect has nothing to do with solar panels, IMO.

Again, what reading have you been doing on your question (please post links), and what do you really want to learn/do?
 
berkeman said:
What's a ddp?
 
  • Like
Likes   Reactions: Renandornelles
What is the work function of cesium? What is the corresponding color?
 
  • Like
Likes   Reactions: vanhees71
berkeman said:
What’s a ddp?
When I do a Google seatch for “ddp photoelectric effect”, this thread is near the top of the results. The other hits don’t give anything useful, as far as I can see from an admittedly quick scan.
 
  • Haha
Likes   Reactions: berkeman and vanhees71
jtbell said:
When I do a Google seatch for “ddp photoelectric effect”, this thread is near the top of the results
So it must be something involving recursion then... :wink:
 
In Spanish ddp stands for "Diferencia de potencial" . Translated to English Is Potential difference
 
  • Informative
  • Like
Likes   Reactions: dextercioby and berkeman
  • #10
@Renandornelles
the photoelectric effect is when EM radiation of sufficient energy is hitting the surface of a metal in vacuum. The photons (if they have enough energy) can give that energy to the electrons and they become free. In the presence of an external E field they then form a current.
This effect only really starts at photon energies from 3eV and upwards, so that makes it in the UV range of light and up.
Solar panels have semiconductor structure in them. There is a PN junction. You only need to provide enough energy to overcome the bandgap energy of the electrons within that semiconductor.
It happens to be that this bandgap energy in the PN junction is much lower than the energy for UV light. Roughly it starts from around 0.7 eV and up, so in solar panels visible light also works to produce current.
That is the major difference between the photoelectric effect and solar panels.

So you can demonstrate that experiment with red light because red light has enough energy to excite the PN junction in a solar panel and cause current.
Take a red LED and shine it onto a small solar panel like those that charge mobile phones you should get output, only you might need many LED's instead of one.Oh and by the way here is a good material on the topic
http://solarcellcentral.com/junction_page.html
 
  • Like
Likes   Reactions: hutchphd, jim mcnamara and DaveE
  • #11
Photoelectric effect---by the way, now called photoemission---has to be distinguished from the ionization of a bound electron. Photoemission is a many-body phenomenon and occurs with free electrons in a lattice; as a matter of fact, it occurs with the upper few layers of a metal's surface. A surface in metal is already a different physical environment, for it has been relaxed and the interactions between the various d (usually) orbitals have been modified. The entire topic can only be studied through quantum many-particle methods and, just like almost all phenomena involving many particles, is still not well understood.

In labs doing experiments with photoemission, mere energies involved in the phenomenon are considered a sort of `bookkeeping' approach. Einstein's formula is such a bookkeeping expression; $h\nu$ and ${1\over 2}mv^2$ refer to individual photoelectrons, but the voltage term is some kind of averaged description of the effect of the entire crystal lattice, which is not a single-particle effect. Already by the 1970s, there were publications in which many-body techniques were employed with photoemission (e.g. Caroli et al).

As you know, the situation is quite different with semiconductor crystals, let alone semiconductors with embedded `impurities' as in PN junctions. Although the energies required to make an electron jump into the conduction band are lower than UV energies, the phenomenon is not classified as photoemission. Especially when one distinguishes between experiments for studying the physics and devices used in applications. Usually, photoemission measurements are made in order to produce spectra from the d-orbital electrons in metal surfaces with the purpose of studying the crystal lattice there.

Also, one can produce a beam of photoelectrons and guide them toward a detector. The detector is usually capable of angular-resolving the electrons; in more sophisticated apparatuses, the beam can produce a kind of image of the metal's surface on a suitable monitor, but the interpretation of such an image is not that simple.

Hope you find it of some help.
 
  • #12
Renandornelles said:
I wanted to demonstrate the frequency limits capable of emitting electrons for children to understand solar panels.
Home made or home devised demonstrations to kids (even more mature students) are often not as good as you can get with proprietary equipment (most things tend just not to work apart from in your shed or 'prep room' after school). The standard equipment for demonstrating the photoelectric effect just uses a beam of light to discharge a plate on an electroscope and the gold leaves drop down. But that's not 'mechanical' (apart from the Coulomb force); the presence or absence of charge is all a bit abstract.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
930
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K