Physicsissuef
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Is in the photoelectric effect, the electrons are oppositing the electromagnetic field or they are excited from the energy?
The discussion revolves around the photoelectric effect, exploring the mechanisms by which electrons are emitted from a material when exposed to electromagnetic radiation, such as ultraviolet or visible light. Participants inquire about the role of different types of radiation, the setup of photoelectric cells, and the relationship between radiation intensity and electron emission.
Participants express varying levels of understanding and clarity regarding the photoelectric effect and related concepts. There is no consensus on some aspects, particularly concerning the initial questions posed and the specifics of how different types of radiation interact with materials.
Some participants express confusion over the terminology and setup of photoelectric cells, indicating that important information may have been omitted in earlier questions. The discussion also highlights the need for clear definitions and context when discussing the photoelectric effect.
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.