vuquta Posted September 4, 2010 Share Posted September 4, 2010 Please spew out all you know about the standard model and the photon. Thanks Link to comment Share on other sites More sharing options...
ajb Posted September 5, 2010 Share Posted September 5, 2010 People write books on the subject. I suggest you ask specific questions. Link to comment Share on other sites More sharing options...
vuquta Posted September 5, 2010 Author Share Posted September 5, 2010 People write books on the subject. I suggest you ask specific questions. OK, why is it that the photon is able to move around molecules in the infrared range (heat) but only knock off electrons for the photo electric effect? Link to comment Share on other sites More sharing options...
swansont Posted September 5, 2010 Share Posted September 5, 2010 OK, why is it that the photon is able to move around molecules in the infrared range (heat) but only knock off electrons for the photo electric effect? Because ionization energies are typically a few electron-Volts or more, and IR photons are about 1 eV or less. You need photons in the blue/UV to ionize. (and IR is not the same as heat) Link to comment Share on other sites More sharing options...
vuquta Posted September 5, 2010 Author Share Posted September 5, 2010 Because ionization energies are typically a few electron-Volts or more, and IR photons are about 1 eV or less. You need photons in the blue/UV to ionize. (and IR is not the same as heat) Does the infrared photon perform more work than the photo electric photon or the compton scattering photon? Link to comment Share on other sites More sharing options...
swansont Posted September 6, 2010 Share Posted September 6, 2010 Does the infrared photon perform more work than the photo electric photon or the compton scattering photon? Work really isn't the concept to apply here, since force and distance aren't known. Link to comment Share on other sites More sharing options...
vuquta Posted September 6, 2010 Author Share Posted September 6, 2010 Work really isn't the concept to apply here, since force and distance aren't known. Well, let's see if we can think it through. If I use a match on some covalent bond like say lighter fluid, I will get much more energy out of that interaction than I would a solar cell would provide for that match. Here is another example. The steam engine uses the work attained from the infrared radiation and they are quite efficient. Now, given light of the same magnitude I get far less work done by a solar cell. If not, then our energy troubles would have been solved long ago. Link to comment Share on other sites More sharing options...
swansont Posted September 7, 2010 Share Posted September 7, 2010 Well, let's see if we can think it through. If I use a match on some covalent bond like say lighter fluid, I will get much more energy out of that interaction than I would a solar cell would provide for that match. Here is another example. The steam engine uses the work attained from the infrared radiation and they are quite efficient. Now, given light of the same magnitude I get far less work done by a solar cell. If not, then our energy troubles would have been solved long ago. I don't know where you are going with this. Most of the IR from a steam engine is waste heat; conduction and convection would account for most of the mechanical work. Similarly, combustion's heat transfer is not solely due to IR being generated. AFAIK, solar cells have lower efficiencies because the low-energy part of the spectrum does not efficiently promote electrons into the conduction band of the photocell. I don't see where work enters into this at all; it has a specific definition in physics: the energy transferred by a force acting through a displacement. Link to comment Share on other sites More sharing options...
vuquta Posted September 7, 2010 Author Share Posted September 7, 2010 I don't know where you are going with this. Most of the IR from a steam engine is waste heat; conduction and convection would account for most of the mechanical work. Similarly, combustion's heat transfer is not solely due to IR being generated. AFAIK, solar cells have lower efficiencies because the low-energy part of the spectrum does not efficiently promote electrons into the conduction band of the photocell. I don't see where work enters into this at all; it has a specific definition in physics: the energy transferred by a force acting through a displacement. I am comparing the infared photon with the normal light photon. Link to comment Share on other sites More sharing options...
swansont Posted September 7, 2010 Share Posted September 7, 2010 I am comparing the infared photon with the normal light photon. Not so much; you are using examples that involve processes other than, or in addition to, photons.. Link to comment Share on other sites More sharing options...
vuquta Posted September 8, 2010 Author Share Posted September 8, 2010 Not so much; you are using examples that involve processes other than, or in addition to, photons.. Come on, you are not going to sell to the average folks that the photo electric photon performs more work than the infrared photon. So, how do you explain this? Link to comment Share on other sites More sharing options...
swansont Posted September 8, 2010 Share Posted September 8, 2010 Come on, you are not going to sell to the average folks that the photo electric photon performs more work than the infrared photon. So, how do you explain this? No, as I've explained, I am "not going to sell to the average folks that the photo electric photon performs more work than the infrared photon" since the concept of work isn't in play here. How do I explain what? An infrared photon has less energy than a visible photon. E = hv Link to comment Share on other sites More sharing options...
vuquta Posted September 8, 2010 Author Share Posted September 8, 2010 No, as I've explained, I am "not going to sell to the average folks that the photo electric photon performs more work than the infrared photon" since the concept of work isn't in play here. How do I explain what? An infrared photon has less energy than a visible photon. E = hv OK, if it has less energy, then how does an infrared photon move entire molecules and atoms and yet the light photon only move electons. For example, the average person would say if you move an entire building you do more work than moving a glass in the building and this is a fair comparison. Link to comment Share on other sites More sharing options...
swansont Posted September 9, 2010 Share Posted September 9, 2010 OK, if it has less energy, then how does an infrared photon move entire molecules and atoms and yet the light photon only move electons. What makes you think that this is a true statement? Link to comment Share on other sites More sharing options...
vuquta Posted September 12, 2010 Author Share Posted September 12, 2010 What makes you think that this is a true statement? Because the Brownian motion moves around atoms which are much more massive than elections. Hence, the infrared photon does more work than the photoelectric effect photon. If this is false, then the concept of physics work is false. Link to comment Share on other sites More sharing options...
swansont Posted September 13, 2010 Share Posted September 13, 2010 Because the Brownian motion moves around atoms which are much more massive than elections. Hence, the infrared photon does more work than the photoelectric effect photon. If this is false, then the concept of physics work is false. Brownian motion usually refers to larger particles being moved around by collisions with atoms and molecules. But why would you assume that a higher energy photon wouldn't move atoms around? Momentum is still conserved. Link to comment Share on other sites More sharing options...
darkenlighten Posted September 13, 2010 Share Posted September 13, 2010 vuquta, you are thinking about this all wrong. First there aren't specific "photo-electric effect photons", just certain materials that only release electrons due to the energy required to kick on out of the material. This has to do more with the separation and atomic bonding of the material. Any photon can release electrons, as long as it has enough energy. So has stated before it does have less energy, but you are not moving a particle that is in free space when it comes to the electron, it has a certain energy that it needs to overcome to be released from the material. While a molecule might be in "free space" needing less energy to be moved in the desired direction. Link to comment Share on other sites More sharing options...
vuquta Posted September 13, 2010 Author Share Posted September 13, 2010 vuquta, you are thinking about this all wrong. First there aren't specific "photo-electric effect photons", just certain materials that only release electrons due to the energy required to kick on out of the material. This has to do more with the separation and atomic bonding of the material. Any photon can release electrons, as long as it has enough energy. So has stated before it does have less energy, but you are not moving a particle that is in free space when it comes to the electron, it has a certain energy that it needs to overcome to be released from the material. While a molecule might be in "free space" needing less energy to be moved in the desired direction. Well, I know there aren't really "photo-electric effect photons". I used this phrase to refer to photons in the light spectrum. And yes, I know only certain materials are sensitive to the photo electric effect. As a side note, did you know if a plant did not operate in the pico second range for the photo electric effect for photosynthesis, there would be no life on this planet? We just hit this range in the last decade. http://photoscience.la.asu.edu/photosyn/education/photointro.html Do you find that amazing? The plant knows to use the photo electric effect to break water into H and O. It then reasembles them into a carbohydrate for energy storage. Naturally, higher level plants use N etc for protein construction using this same method. Anyway, this issue under consideration is the work function. Clearly, photons operating in the infrared range perform more work than do photons that operate in the higher frequency ranges. Do you agree or disagree? Link to comment Share on other sites More sharing options...
darkenlighten Posted September 14, 2010 Share Posted September 14, 2010 Well, I know there aren't really "photo-electric effect photons". I used this phrase to refer to photons in the light spectrum. And yes, I know only certain materials are sensitive to the photo electric effect. As a side note, did you know if a plant did not operate in the pico second range for the photo electric effect for photosynthesis, there would be no life on this planet? We just hit this range in the last decade. http://photoscience.la.asu.edu/photosyn/education/photointro.html Do you find that amazing? The plant knows to use the photo electric effect to break water into H and O. It then reasembles them into a carbohydrate for energy storage. Naturally, higher level plants use N etc for protein construction using this same method. Yea this is interesting how they can do that and I know studies are being done in the area to figure out how the plants recombine the cells to prepare for more photosynthesis. Anyway, this issue under consideration is the work function. Clearly, photons operating in the infrared range perform more work than do photons that operate in the higher frequency ranges. Do you agree or disagree? The work function has to do with energy, not work, regardless of the name. So when you ask if the infrared does more work than higher frequencies it doesn't really make sense for this application. So we are only worried about their energies and momentums. Link to comment Share on other sites More sharing options...
vuquta Posted September 14, 2010 Author Share Posted September 14, 2010 Yea this is interesting how they can do that and I know studies are being done in the area to figure out how the plants recombine the cells to prepare for more photosynthesis. The work function has to do with energy, not work, regardless of the name. So when you ask if the infrared does more work than higher frequencies it doesn't really make sense for this application. So we are only worried about their energies and momentums. You have nice answers. But, if you recall, Einstein for the photo electric effect posited a work function as primitive. So, all logic must obey this method. ( I am amazed by the factory of photosynthesis, may I suggest you look at it. You may have, but again, it is something to me) Anyway, to analyze the photon, we must posit Einstein's work function to correctly operate on the problem. Do you agree or disagree? Link to comment Share on other sites More sharing options...
darkenlighten Posted September 14, 2010 Share Posted September 14, 2010 (edited) You have nice answers. But, if you recall, Einstein for the photo electric effect posited a work function as primitive. So, all logic must obey this method. ( I am amazed by the factory of photosynthesis, may I suggest you look at it. You may have, but again, it is something to me) Anyway, to analyze the photon, we must posit Einstein's work function to correctly operate on the problem. Do you agree or disagree? I'm not too familiar with the details of the process of photosynthesis, of course the general idea, but not in detail. But I'm not sure exactly what you are meaning by your question. If you are asking that we have to assume that the work function is correct, then yes I do agree. It has been used and calculated for a lot of materials. Edited September 14, 2010 by darkenlighten Link to comment Share on other sites More sharing options...
vuquta Posted September 14, 2010 Author Share Posted September 14, 2010 I'm not too familiar with the details of the process of photosynthesis, of course the general idea, but not in detail. But I'm not sure exactly what you are meaning by your question. If you are asking that we have to assume that the work function is correct, then yes I do agree. It has been used and calculated for a lot of materials. OK, I agree. So, 1) Where do the photons come from for fusion. 2) Why do they perform more work at the infrared level. Link to comment Share on other sites More sharing options...
darkenlighten Posted September 14, 2010 Share Posted September 14, 2010 OK, I agree. So, 1) Where do the photons come from for fusion. 2) Why do they perform more work at the infrared level. 1) I'm not sure if fusion uses specifically photons, but if it does then we either create them using a laser or the are produced from the collision of particles due to conservation of energy/momentum. 2) What example do you have for this, do you have a link to an article or what not. Its possible that what you're referencing about infrared is because its the lowest energy photon they use for this process, being the most efficient, but I'm not sure. Link to comment Share on other sites More sharing options...
swansont Posted September 14, 2010 Share Posted September 14, 2010 Clearly, photons operating in the infrared range perform more work than do photons that operate in the higher frequency ranges. Do you agree or disagree? No, this is not clear at all. 1) Where do the photons come from for fusion. 2) Why do they perform more work at the infrared level. 1) You have another thread for this 2) You haven't established that this is the case Link to comment Share on other sites More sharing options...
Klaynos Posted September 14, 2010 Share Posted September 14, 2010 The work function of a material is the energy required to free an electron, it is not classical work just poor naming. Link to comment Share on other sites More sharing options...
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