- #1
Renandornelles
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Is it possible to get the photoelectric effect in red color frequency?
Is it possible to get the photoelectric effect in red color frequency?
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berkeman
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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?
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?
- #3
berkeman
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What's a ddp?Renandornelles said:
The photoelectric effect has nothing to do with solar panels, IMO.Renandornelles said:
Again, what reading have you been doing on your question (please post links), and what do you really want to learn/do?
- #4
Renandornelles
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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...berkeman said:
I wanted to demonstrate the frequency limits capable of emitting electrons for children to understand solar panels.
berkeman said:
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berkeman
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berkeman said:
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Vanadium 50
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What is the work function of cesium? What is the corresponding color?
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jtbell
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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.berkeman said:
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berkeman
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So it must be something involving recursion then...jtbell said:
- #9
Gordianus
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In Spanish ddp stands for "Diferencia de potencial" . Translated to English Is Potential difference
- #10
artis
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@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
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
- #11
apostolosdt
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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.
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.
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sophiecentaur
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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.Renandornelles said: