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Physicsissuef
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Is in the photoelectric effect, the electrons are oppositing the electromagnetic field or they are excited from the energy?
mathman said:I don't understand your question. However, the photoelectric effect involves an electron absorbing a photon and, being more energetic, leaving the atom.
Physicsissuef said:And do ultraviolet radiation pass to the metal, or just visible light?
Ok, sorry. I just want to know, if UV radiation passes through the glass bulb, the glass will absorb that energy. So what is causing the electrons to flow? UV radiation or visible light, or maybe something else?ZapperZ said:You really need to put a bit more effort in asking your question here, expecially if you want others to put effort into responding. If not, you will continue to get responses from other people in the form of "HUH?", and you will have to keep on explaining yourself.
UV radiation can cause photoemission if the work function is below the photon energy. Now what is it exactly that you want to know here in this thread?
Zz.
Sorry for being rude, but have you ever heard about photoelectric tube or photoelectric cell? There is glass balloon or bulb or whatever... Inside there is vacuum, so it is called vacuum photocell. There are anode and cathode. So when I bring some kind of radiation to some metal, the electrons are excited and pulled off the cathode. What is that kind of radiation? UV or visible light? Also I want to know why the anode and cathode are put in vacuum photocell?ZapperZ said:This is getting utterly confusing. You WERE asking about the photoelectric effect, weren't you? Why is it optical conductivity now?
Glass bulb? When did that come in? Did you mention about glass anywhere till now? There's no "photoemission" using UV source on glass bulb, is there?
Why don't you start from the very beginning and ask your question once again. But this time, please put in as much effort and information in your question so that we know (i) what you know (ii) what exact it is the picture that you have in your mind. If you see that you have to ADD new stuff as you go along, it means clearly that you omitted important information in your original question.
Zz.
Physicsissuef said:Sorry for being rude, but have you ever heard about photoelectric tube or photoelectric cell? There is glass balloon or bulb or whatever... Inside there is vacuum, so it is called vacuum photocell. There are anode and cathode. So when I bring some kind of radiation to some metal, the electrons are excited and pulled off the cathode. What is that kind of radiation? UV or visible light? Also I want to know why the anode and cathode are put in vacuum photocell?
Ok, thanks. I understand now. I have two more questions. Why the cathode and anode are in vacuum balloon instead connected them with wire? Is this "[URL correct?ZapperZ said:Not only have I heard of them, I've used them!
What kind of radiation? Any radiation in which the photon energy is higher than the work function! I thought I mentioned this already.
Electrons do not travel very far in air, and even if they do, a lot of them get scattered off and would not reach the anode. This is not what you want when you are using a photocell to detect EM radiation. If you are trying to detect UV, the "glass" has to be either quartz or fused silica, because ordinary glass absorbs UV.
Zz.
Physicsissuef said:And why in my textbook says, that the kinetic energy of the electrons doesn't depends from the intensity of the radiation?
And what makes the current stronger? The more electrons ejected, or the speed of the electrons?ZapperZ said:Because "intensity" only increases the number of photons, not the energy within each photon.
Your textbook doesn't explain this? Please read for example, this:
http://www.colorado.edu/physics/2000/quantumzone/photoelectric.html
Zz.
Physicsissuef said:And what makes the current stronger? The more electrons ejected, or the speed of the electrons?
Zapper said:Because "intensity" only increases the number of photons
ZapperZ said:<scratching head>
This is not clear?
Zz.
Correct.Physicsissuef said:It says that the current is much stronger when I increase the intensity. So that means that more electrons are ejected and the current is getting stronger, right?
dst said:Yes. exactly so.
dst said:I find it hard to see what's at the back. A cathode ray tube? Photodiode?
If it's a photodiode or some sort of light sensitive resistor at the back, you should be able to get sound. I did a similar test with a laser beam and an amplifier and it makes for a reasonable listening tool.
The photoelectric effect is the phenomenon where electrons are emitted from a material when it is exposed to light of a certain frequency or above. This effect was first observed by Heinrich Hertz in 1887 and explained by Albert Einstein in 1905.
The photoelectric effect works by the absorption of photons, or packets of light energy, by the material. When a photon with enough energy strikes an atom in the material, it can cause an electron to be ejected from the atom. This electron is then free to move and creates an electric current.
The photoelectric effect has significant implications in the fields of physics and technology. It provided evidence for the particle nature of light and contributed to the development of quantum mechanics. It also has practical applications in various devices, such as solar cells and photomultiplier tubes.
The energy of the photons is a crucial factor in the photoelectric effect. If the photons do not have enough energy, they will not be able to eject electrons from the material. The minimum energy required is called the "work function" of the material. Additionally, increasing the energy of the photons will result in more energetic electrons being emitted.
The work function is the minimum amount of energy needed to remove an electron from the surface of a material. It is specific to each material and can vary depending on factors such as the composition and surface conditions. The work function is directly related to the energy of the photons needed to cause the photoelectric effect. If the energy of the photons is greater than the work function, electrons will be emitted from the material.