John Ye Posted August 30, 2018 Author Posted August 30, 2018 29 minutes ago, studiot said: Your reply has nothing whatsoever to do with my question. Here is my reference for the definition you are clearly trying to avoid. Thanks. I know electrical dipole. H atom in ground state has no dipole, because electron can drift in the balanced sphere according to the external electric fields.
swansont Posted August 30, 2018 Posted August 30, 2018 1 hour ago, John Ye said: You must have not read my script carefully. I explained it in detail. It's possible that I used poor English, not clearly explains it. If so, I feel sorry. In n=2 state, two electrons are temporarily attached to proton. No, that is not the case I am talking about. How do you explain when you have only one electron, but it is in an excited state? There is only one electron. The atom is neutral. Your model does not address this. Incidentally, if you has two (or mode) electrons, you would see a shift in the energy states, and would not replicate the Bohr energy levels, as these electrons would repel each other. Quote This paper does not mention atom's fine structure. The angular momentum is for the quantum theory to explain the fine structure. These are observed phenomena, and QM accounts for them. Your model does not. Quote the quantum theory has not defined a circle track motion of electron, the momentum is not applicable. QM includes quantized angular momentum, which is experimentally confirmed. Quote I am hoping that this can explain energy levels of exciting hydrogen gas It doesn't, because the energy levels under consideration are for neutral hydrogen. One electron is attached. 5 minutes ago, John Ye said: Thanks. I know electrical dipole. H atom in ground state has no dipole, because electron can drift in the balanced sphere according to the external electric fields. Doesn't matter of it can drift. That would just mean that the dipole moment would be "drifting" as well. But it would be measurable. Has anyone measured such a dipole moment? It would have a huge effect on two hydrogen atoms bonding to each other, but no sense in going there when we have basic elements of your conjecture is question.
John Ye Posted August 30, 2018 Author Posted August 30, 2018 (edited) 33 minutes ago, swansont said: No, that is not the case I am talking about. How do you explain when you have only one electron, but it is in an excited state? There is only one electron. The atom is neutral. Your model does not address this. Incidentally, if you has two (or mode) electrons, you would see a shift in the energy states, and would not replicate the Bohr energy levels, as these electrons would repel each other. These are observed phenomena, and QM accounts for them. Your model does not. QM includes quantized angular momentum, which is experimentally confirmed. It doesn't, because the energy levels under consideration are for neutral hydrogen. One electron is attached. Doesn't matter of it can drift. That would just mean that the dipole moment would be "drifting" as well. But it would be measurable. Has anyone measured such a dipole moment? It would have a huge effect on two hydrogen atoms bonding to each other, but no sense in going there when we have basic elements of your conjecture is question. This should be tested and proved. When H gas is in external electric field, all elections in atoms will be drifting to positive side. But the test may be difficult, need to carefully designed 23 minutes ago, John Ye said: This should be tested and proved. When H gas is in external electric field, all elections in atoms will be drifting to positive side. But the test may be difficult, need to carefully designed 23 minutes ago, John Ye said: This should be tested and proved. When H gas is in external electric field, all elections in atoms will be drifting to positive side. But the test may be difficult, need to carefully designed Atoms in ground state is in H2 molecules which are moving in a very high speed. Bullet speed. Hard to design the test 33 minutes ago, swansont said: No, that is not the case I am talking about. How do you explain when you have only one electron, but it is in an excited state? There is only one electron. The atom is neutral. Your model does not address this. Incidentally, if you has two (or mode) electrons, you would see a shift in the energy states, and would not replicate the Bohr energy levels, as these electrons would repel each other. These are observed phenomena, and QM accounts for them. Your model does not. QM includes quantized angular momentum, which is experimentally confirmed. It doesn't, because the energy levels under consideration are for neutral hydrogen. One electron is attached. Doesn't matter of it can drift. That would just mean that the dipole moment would be "drifting" as well. But it would be measurable. Has anyone measured such a dipole moment? It would have a huge effect on two hydrogen atoms bonding to each other, but no sense in going there when we have basic elements of your conjecture is question. Quote. It doesn't, because the energy levels under consideration are for neutral hydrogen. One electron is attached. /quote One election H atom has only one energy level 13.6ev, no other levels exists. 23 minutes ago, John Ye said: This should be tested and proved. When H gas is in external electric field, all elections in atoms will be drifting to positive side. But the test may be difficult, need to carefully designed Atoms in ground state is in H2 molecules which are moving in a very high speed. Bullet speed. Hard to design the test Quote. It doesn't, because the energy levels under consideration are for neutral hydrogen. One electron is attached. /quote One election H atom has only one energy level 13.6ev, no other levels exists. Energy levels exist in hydrogen gas, not in one single hydrogen atom!!! in one single hydrogen atom, there is only one energy, that is 13.6ev. the work needed to pull electron from balanced point to infinity. or an electron from infinity goes back to balanced point, emits a 92nm light ( equal to 13.6ev) So, there is no any quantum effect in one single H atom. We detect the light of a bottle of excited H gas, incorrectly thought that the light is emitted by every single H atom. This is wrong, totally wrong. This leads to later wrong atom models. People was cheated by the H gas bottle. Edited August 30, 2018 by John Ye
swansont Posted August 30, 2018 Posted August 30, 2018 1 hour ago, John Ye said: This should be tested and proved. When H gas is in external electric field, all elections in atoms will be drifting to positive side. But the test may be difficult, need to carefully designed AFAIK, no atom has been measured to have a permanent electric dipole moment. You can measure them in molecules (where they are possible) via the Stark shift that is observed. if your model is correct there should be a permanent dipole moment, and it shouldn't take much if a field at all to align them and swamp out any thermal effects. But, again, AFAIK no such thing has ever been measured, and would be a big deal, since QM predicts no EDM for atoms. Quote Atoms in ground state is in H2 molecules which are moving in a very high speed. Bullet speed. Hard to design the test And yet people do precision experiments with hydrogen and other atoms. Quote Quote. It doesn't, because the energy levels under consideration are for neutral hydrogen. One electron is attached. /quote One election H atom has only one energy level 13.6ev, no other levels exists. Bollocks. The spectrum of hydrogen has been observed. You won't simultaneously attach multiple electrons to a bunch of hydrogen atoms, because there is a finite supply of them. How do you strip the electrons off of the hydrogen atoms, since it takes 13.6 eV to do so? Quote Energy levels exist in hydrogen gas, not in one single hydrogen atom!!! Yes, they do. Quote in one single hydrogen atom, there is only one energy, that is 13.6ev. the work needed to pull electron from balanced point to infinity. or an electron from infinity goes back to balanced point, emits a 92nm light ( equal to 13.6ev) So, there is no any quantum effect in one single H atom. That would be up to you to provide evidence for this extraordinary claim. Quote We detect the light of a bottle of excited H gas, incorrectly thought that the light is emitted by every single H atom. This is wrong, totally wrong. This leads to later wrong atom models. People was cheated by the H gas bottle. That should be something that you could test. If you are getting light from hydrogen ions that have multiple electrons, you can only get light from some small fraction of the atoms, as the rest will lack electrons. How do you strip the electrons off of the hydrogen atoms to form these ions, since it takes 13.6 eV to get each electron? Also, if i am understanding your model correctly, the energy levels depends on having excess electrons, which implies that these electrons are bound with that energy, and that emission must be accompanied by the loss of an electron. But where does that energy for the photon, and the electron expulsion, come from?
John Ye Posted August 30, 2018 Author Posted August 30, 2018 (edited) 45 minutes ago, swansont said: AFAIK, no atom has been measured to have a permanent electric dipole moment. You can measure them in molecules (where they are possible) via the Stark shift that is observed. if your model is correct there should be a permanent dipole moment, and it shouldn't take much if a field at all to align them and swamp out any thermal effects. But, again, AFAIK no such thing has ever been measured, and would be a big deal, since QM predicts no EDM for atoms. And yet people do precision experiments with hydrogen and other atoms. Bollocks. The spectrum of hydrogen has been observed. You won't simultaneously attach multiple electrons to a bunch of hydrogen atoms, because there is a finite supply of them. How do you strip the electrons off of the hydrogen atoms, since it takes 13.6 eV to do so? Yes, they do. That would be up to you to provide evidence for this extraordinary claim. That should be something that you could test. If you are getting light from hydrogen ions that have multiple electrons, you can only get light from some small fraction of the atoms, as the rest will lack electrons. How do you strip the electrons off of the hydrogen atoms to form these ions, since it takes 13.6 eV to get each electron? to excite hydrogen gas, fire or electric ionization, just the same as people did to observe the spectrum in previous experiment. 35 minutes ago, John Ye said: to excite hydrogen gas, fire or electric ionization, just the same as people did to observe the spectrum in previous experiment. Yes, a small fraction of them. The more electrons attached, the less possible the energy level to appear. This is proved by the densities of each wavelength of H spectrum. 45 minutes ago, swansont said: AFAIK, no atom has been measured to have a permanent electric dipole moment. You can measure them in molecules (where they are possible) via the Stark shift that is observed. if your model is correct there should be a permanent dipole moment, and it shouldn't take much if a field at all to align them and swamp out any thermal effects. But, again, AFAIK no such thing has ever been measured, and would be a big deal, since QM predicts no EDM for atoms. And yet people do precision experiments with hydrogen and other atoms. Bollocks. The spectrum of hydrogen has been observed. You won't simultaneously attach multiple electrons to a bunch of hydrogen atoms, because there is a finite supply of them. How do you strip the electrons off of the hydrogen atoms, since it takes 13.6 eV to do so? Yes, they do. That would be up to you to provide evidence for this extraordinary claim. That should be something that you could test. If you are getting light from hydrogen ions that have multiple electrons, you can only get light from some small fraction of the atoms, as the rest will lack electrons. How do you strip the electrons off of the hydrogen atoms to form these ions, since it takes 13.6 eV to get each electron? It is hard to design and implement the test. 45 minutes ago, swansont said: AFAIK, no atom has been measured to have a permanent electric dipole moment. You can measure them in molecules (where they are possible) via the Stark shift that is observed. if your model is correct there should be a permanent dipole moment, and it shouldn't take much if a field at all to align them and swamp out any thermal effects. But, again, AFAIK no such thing has ever been measured, and would be a big deal, since QM predicts no EDM for atoms. And yet people do precision experiments with hydrogen and other atoms. Bollocks. The spectrum of hydrogen has been observed. You won't simultaneously attach multiple electrons to a bunch of hydrogen atoms, because there is a finite supply of them. How do you strip the electrons off of the hydrogen atoms, since it takes 13.6 eV to do so? Yes, they do. That would be up to you to provide evidence for this extraordinary claim. That should be something that you could test. If you are getting light from hydrogen ions that have multiple electrons, you can only get light from some small fraction of the atoms, as the rest will lack electrons. How do you strip the electrons off of the hydrogen atoms to form these ions, since it takes 13.6 eV to get each electron? If you agree this sentence: Energy levels exist in hydrogen gas, not in one single hydrogen atom!!! then, you have got my idea. but, the quantum model does NOT think so. It thinks that one H atom has that many energy levels and emits that many different wavelength of light. One single H atom has one electron and one proton. It can only emit a 13.6ev wavelength light, no others. and the electron has only 2 position ( so called track ) to stay, one is 52.9pm position, one is infinite far. No other positions in the middle exist. Edited August 30, 2018 by John Ye
Strange Posted August 30, 2018 Posted August 30, 2018 (edited) 42 minutes ago, John Ye said: to excite hydrogen gas, fire or electric ionization, just the same as people did to observe the spectrum in previous experiment You didn't answer the question. Is that because you can't? But I guess you have to deny that photons are quantised as well to support your idea? Edited August 30, 2018 by Strange
John Ye Posted August 30, 2018 Author Posted August 30, 2018 So there is no discrete track, no discrete energy levels in one single H atom. Quantum is a group effect, produced by mass H atoms, I called a GAS effect. This is why we want to observe the spectrum of other matters, we must first gasify them. We can't observe the spectrum of a solid matter without making it into gas.
swansont Posted August 30, 2018 Posted August 30, 2018 55 minutes ago, John Ye said: to excite hydrogen gas, fire or electric ionization, just the same as people did to observe the spectrum in previous experiment. And what are the energies involved in these various methods? Are the sufficient to ionize Hydrogen? Quote Yes, a small fraction of them. The more electrons attached, the less possible the energy level to appear. This is proved by the densities of each wavelength of H spectrum. It is hard to design and implement the test. If you agree this sentence: Energy levels exist in hydrogen gas, not in one single hydrogen atom!!! No, I do not agree. Quote then, you have got my idea. but, the quantum model does NOT think so. It thinks that one H atom has that many energy levels and emits that many different wavelength of light. Yes, that's what it predicts, and experiment agrees. Your model has serious problems, some of which have been outlined. Please address them. Such as, your acknowledgement that H has an energy of -13.6 eV (which, AFAICT, you do not predict, but use to develop your model), and you say that H- will have an energy of -3.4 eV. The difference is 10.2 eV. But why would we see a 10.2 eV photon? What process gives us that photon? The 3.4 eV represents the amount you must add to remove that electron (and I seriously doubt H- has that large of a binding energy). And your formula implies that the photon would come with the ejection of that electron. So conservation of energy appears to be grossly violated in your model. As far as your idea that you would have more than two electrons in hydrogen: "Like most negative atomic ions, H- has only one bound stable state." They also note that the electron affinity of hydrogen is 0.7542 eV — the energy needed to release that second electron — not 3.4 eV http://www-bd.fnal.gov/pdriver/laser/Bryant.pdf
John Ye Posted August 30, 2018 Author Posted August 30, 2018 9 minutes ago, Strange said: You didn't answer the question. Is that because you can't? But I guess you have to deny that photons are quantised as well to support your idea? put the H2 gas into a ionizing tube with high voltage, is it OK? Just the same as common experiment of H spectrum. Photons emitted from a H gas is quantised, I don't deny this. This is fact. H atoms emit many different and discrete wavelength photons. I said, the discrete photons are not emitted by very single H atom, but by mass atoms in H gas. 6 minutes ago, swansont said: And what are the energies involved in these various methods? Are the sufficient to ionize Hydrogen? No, I do not agree. Yes, that's what it predicts, and experiment agrees. Your model has serious problems, some of which have been outlined. Please address them. Such as, your acknowledgement that H has an energy of -13.6 eV (which, AFAICT, you do not predict, but use to develop your model), and you say that H- will have an energy of -3.4 eV. The difference is 10.2 eV. But why would we see a 10.2 eV photon? What process gives us that photon? The 3.4 eV represents the amount you must add to remove that electron (and I seriously doubt H- has that large of a binding energy). And your formula implies that the photon would come with the ejection of that electron. So conservation of energy appears to be grossly violated in your model. As far as your idea that you would have more than two electrons in hydrogen: "Like most negative atomic ions, H- has only one bound stable state." They also note that the electron affinity of hydrogen is 0.7542 eV — the energy needed to release that second electron — not 3.4 eV http://www-bd.fnal.gov/pdriver/laser/Bryant.pdf Ok, thans a lot. I will explain your questions, later. Transit from to energy in eV photon in nm 2 -> 1 10.2043 121.5003 3 -> 1 12.0940 102.5159 10.2ev photon is the transition from level 2(H2 in my paper) to level 1(H1). wavelength is 121.5nm. There are 2 jumps can do so. 1) level 2 atom ( H2, with 2 electrons attached ), lost one electron, attracted by H0 ( bare proton, with no electron ). 2) this H2 itself. when it lost one electron, the other electron will transite from energy leve 2 to level 1. in both case, a 10.2 photon will emitted. totally 2 photons. These 10.2ev photons are the main part of spectrum, with high emission density ( much brighter, easy to be observed )
swansont Posted August 30, 2018 Posted August 30, 2018 46 minutes ago, John Ye said: Transit from to energy in eV photon in nm 2 -> 1 10.2043 121.5003 3 -> 1 12.0940 102.5159 10.2ev photon is the transition from level 2(H2 in my paper) to level 1(H1). wavelength is 121.5nm. There are 2 jumps can do so. 1) level 2 atom ( H2, with 2 electrons attached ), lost one electron, attracted by H0 ( bare proton, with no electron ). 2) this H2 itself. when it lost one electron, the other electron will transite from energy leve 2 to level 1. in both case, a 10.2 photon will emitted. totally 2 photons. These 10.2ev photons are the main part of spectrum, with high emission density ( much brighter, easy to be observed ) But how does that happen? You have an atom with 2 electrons, and it transitions to be an atom with one electron. This "lost" electron was bound to the hydrogen. Where does the energy come from to get rid of that electron? And to get a 10.2 eV photon, you would have to remove the ground state electron, but not the n=2 electron, which only takes 3.4 eV (according to you. According to actual experiment, it's 0.75 eV) Why would it do that? Keep in mind, I can induce these transitions with light, with less than 13.6 eV of energy. There is no way that the electron has been ionized from some other source. If your model is correct, there should be no way for a neutral hydrogen sample to absorb 10.2 eV photons. But they do. 1
John Ye Posted August 30, 2018 Author Posted August 30, 2018 20 minutes ago, swansont said: But how does that happen? You have an atom with 2 electrons, and it transitions to be an atom with one electron. This "lost" electron was bound to the hydrogen. Where does the energy come from to get rid of that electron? And to get a 10.2 eV photon, you would have to remove the ground state electron, but not the n=2 electron, which only takes 3.4 eV (according to you. According to actual experiment, it's 0.75 eV) Why would it do that? Keep in mind, I can induce these transitions with light, with less than 13.6 eV of energy. There is no way that the electron has been ionized from some other source. If your model is correct, there should be no way for a neutral hydrogen sample to absorb 10.2 eV photons. But they do. Excited atom is not stable. Extra electrons tend to be lost, ionized atom will tend to abtain one. The energy is not even distributed in the gas. 3 minutes ago, John Ye said: Excited atom is not stable. Extra electrons tend to be lost, ionized atom will tend to abtain one. The energy is not even distributed in the gas. As for the last part of question, how did people observe 10.2ev photon in past experiments? I don't know. I only compared my math value with the official spectrum database. In my script, I give the database Web site link, you can reference it.
swansont Posted August 30, 2018 Posted August 30, 2018 39 minutes ago, John Ye said: Excited atom is not stable. Extra electrons tend to be lost, ionized atom will tend to abtain one. The energy is not even distributed in the gas. Thermal energy of a gas is much smaller than 10.2 eV. To lose a bound electron, energy must be added. You need to account for this energy, and you also need to explain why your numbers do not match with experiment for the binding energy value of a second electron in hydrogen. Or why nothing beyond H- is observed. Quote As for the last part of question, how did people observe 10.2ev photon in past experiments? I don't know. I only compared my math value with the official spectrum database. In my script, I give the database Web site link, you can reference it. Then your work is incomplete. You need to have experimental evidence to support your claim. Hydrogen has an absorption spectrum — you shine light on it and it absorbs the wavelengths that QM predicts. But I don't see how your model would predict an absorption spectrum, as it would require the atom to acquire another electron (which would have to emit a photon of the appropriate energy) at precisely the same time as the photon is absorbed, and in a situation where you would not have free electrons available. The value of 10.2 eV is not in question. It's the mechanism that you are proposing.
studiot Posted August 30, 2018 Posted August 30, 2018 (edited) 4 hours ago, John Ye said: This should be tested and proved. When H gas is in external electric field, all elections in atoms will be drifting to positive side. But the test may be difficult, need to carefully designed Atoms in ground state is in H2 molecules which are moving in a very high speed. Bullet speed. Hard to design the test Quote. It doesn't, because the energy levels under consideration are for neutral hydrogen. One electron is attached. /quote One election H atom has only one energy level 13.6ev, no other levels exists. Energy levels exist in hydrogen gas, not in one single hydrogen atom!!! in one single hydrogen atom, there is only one energy, that is 13.6ev. the work needed to pull electron from balanced point to infinity. or an electron from infinity goes back to balanced point, emits a 92nm light ( equal to 13.6ev) So, there is no any quantum effect in one single H atom. We detect the light of a bottle of excited H gas, incorrectly thought that the light is emitted by every single H atom. This is wrong, totally wrong. This leads to later wrong atom models. People was cheated by the H gas bottle. Gas, gas and more gas. In the words of Mick Jagger It's a gas, gas , gas Or the popular TV program It's all gas and gaiters. What about solids? In particular ionic solids. Your explanation is in direct conflict with both observation and existing theory. That is why I asked you to a) Examine the forces acting on an electron accoring to your equation b) Review Madelung constants. You have refused to do either. I think the discussion subsequent to your reply to my definition of dipoles and dipole moments shows conclusively that you don't know what you are talking about. It is a pity that your obstinacy will not accept any helpful comments as that might let you improve your proposals to at least equal those of Lennard Jones, which I also asked you to look at. If you actually managed to achieve this your equation is much simpler than the LJ one so would be a definite improvement. Edited August 30, 2018 by studiot
swansont Posted August 30, 2018 Posted August 30, 2018 2 hours ago, John Ye said: This is why we want to observe the spectrum of other matters, we must first gasify them. We can't observe the spectrum of a solid matter without making it into gas. This is pretty ridiculous. Solid-state lasers exist, for example. Googling "spectroscopy of solids" yielded >20 million hits
Strange Posted August 30, 2018 Posted August 30, 2018 2 hours ago, John Ye said: put the H2 gas into a ionizing tube with high voltage, is it OK? No.
John Cuthber Posted August 30, 2018 Posted August 30, 2018 3 hours ago, John Ye said: We can't observe the spectrum of a solid matter without making it into gas. Leaves are green, gold is gold. You are so wrong that it's funny. For what it's worth, sodium - as a solid is a shiny metal- it looks like aluminium. However sodium vapour is blue. so, not only do you not need to make it into a gas to see the spectrum, but if you do so, you change that spectrum enormously.
John Ye Posted August 30, 2018 Author Posted August 30, 2018 5 hours ago, John Cuthber said: Leaves are green, gold is gold. You are so wrong that it's funny. For what it's worth, sodium - as a solid is a shiny metal- it looks like aluminium. However sodium vapour is blue. so, not only do you not need to make it into a gas to see the spectrum, but if you do so, you change that spectrum enormously. Green is NOT the spectrum of leaves. It's only reflection of sun light. We can't know leave's chemical elements by green color. right? You browse definition of spectrum analysis, know its method. 6 hours ago, swansont said: This is pretty ridiculous. Solid-state lasers exist, for example. Googling "spectroscopy of solids" yielded >20 million hits Laser is not the spectrum of any solid matter. It's only one single wavelength of light generated by some matter. Spectrum is all discrete wavelength lights. 10 minutes ago, John Ye said: Green is NOT the spectrum of leaves. It's only reflection of sun light. We can't know leave's chemical elements by green color. right? You browse definition of spectrum analysis, know its method. Laser is not the spectrum of any solid matter. It's only one single wavelength of light generated by some matter. Spectrum is all discrete wavelength lights. Solid matter can emit light, but can not emit all discrete wavelengths spectrum which is normally used by spectrum analysis. To know the contents of a solid, we need its spectrum, to get spectrum,we first gasify it. Please see spectrum analysis.
John Ye Posted August 31, 2018 Author Posted August 31, 2018 2 hours ago, John Ye said: Green is NOT the spectrum of leaves. It's only reflection of sun light. We can't know leave's chemical elements by green color. right? You browse definition of spectrum analysis, know its method. Laser is not the spectrum of any solid matter. It's only one single wavelength of light generated by some matter. Spectrum is all discrete wavelength lights. Solid matter can emit light, but can not emit all discrete wavelengths spectrum which is normally used by spectrum analysis. To know the contents of a solid, we need its spectrum, to get spectrum,we first gasify it. Please see spectrum analysis. please see the device "Metal Analysis by Flame and Plasma Atomic Spectroscopy". It can detect the alloy's metal element components. to get atom spectrum of a matter, we must gasify it first. The gasification process makes sure that we obtain standalone atoms from the matter. Only in the state of atom gas, we can detect the full spectrum of the matter. 4 minutes ago, John Ye said: please see the device "Metal Analysis by Flame and Plasma Atomic Spectroscopy". It can detect the alloy's metal element components. to get atom spectrum of a matter, we must gasify it first. The gasification process makes sure that we obtain standalone atoms from the matter. Only in the state of atom gas, we can detect the full spectrum of the matter. we can NOT get H spectrum from liquid hydrogen. Hydrogen must be gasified first. It's not enough to gaisify into H2 gas, we should go further, to gasify H2 into standalone atoms( atom gas) to see the full spectrum. similarly, we want to get spectrum of the Ferrum, we must gasify Fe into Fe atom gas first. 10 hours ago, studiot said: Gas, gas and more gas. In the words of Mick Jagger It's a gas, gas , gas Or the popular TV program It's all gas and gaiters. What about solids? In particular ionic solids. Your explanation is in direct conflict with both observation and existing theory. That is why I asked you to a) Examine the forces acting on an electron accoring to your equation b) Review Madelung constants. You have refused to do either. I think the discussion subsequent to your reply to my definition of dipoles and dipole moments shows conclusively that you don't know what you are talking about. It is a pity that your obstinacy will not accept any helpful comments as that might let you improve your proposals to at least equal those of Lennard Jones, which I also asked you to look at. If you actually managed to achieve this your equation is much simpler than the LJ one so would be a definite improvement. I will explain your question later. Try my best. I should not refuse to answer any question.
John Ye Posted August 31, 2018 Author Posted August 31, 2018 1 hour ago, John Ye said: please see the device "Metal Analysis by Flame and Plasma Atomic Spectroscopy". It can detect the alloy's metal element components. to get atom spectrum of a matter, we must gasify it first. The gasification process makes sure that we obtain standalone atoms from the matter. Only in the state of atom gas, we can detect the full spectrum of the matter. we can NOT get H spectrum from liquid hydrogen. Hydrogen must be gasified first. It's not enough to gaisify into H2 gas, we should go further, to gasify H2 into standalone atoms( atom gas) to see the full spectrum. similarly, we want to get spectrum of the Ferrum, we must gasify Fe into Fe atom gas first. I will explain your question later. Try my best. I should not refuse to answer any question. Studiot, the following is the explanation of "the forces acting on an electron according to my equation" 1 minute ago, John Ye said: Studiot, the following is the explanation of "the forces acting on an electron according to my equation" This is the formula 2. which is an extended version of formula 1 in special case. When Ne = Np = 1, it becomes formula 1 By the way, formula 2 is only applicable in a special case, that is, which I called "Equal opportunity electron configuration" If the electrons are not in the state of "equal opportunity electron configuration", I don't have a formula to describe the force. This is the reason that I can only calculated the H gas spectrum, and half He gas spectrum. In these two cases, the electrons' force can be calculated by formula 2. If I had a general formula 2 which is applicable in any case, then all atom's spectrum would be calculated. This is only a hope. it's beyond my ability. 37 minutes ago, John Ye said: Studiot, the following is the explanation of "the forces acting on an electron according to my equation" This is the formula 2. which is an extended version of formula 1 in special case. When Ne = Np = 1, it becomes formula 1 By the way, formula 2 is only applicable in a special case, that is, which I called "Equal opportunity electron configuration" If the electrons are not in the state of "equal opportunity electron configuration", I don't have a formula to describe the force. This is the reason that I can only calculated the H gas spectrum, and half He gas spectrum. In these two cases, the electrons' force can be calculated by formula 2. If I had a general formula 2 which is applicable in any case, then all atom's spectrum would be calculated. This is only a hope. it's beyond my ability. To your second question: the Madelung constants, I don't have much knowledge about crystal structure and it's math. So I can't give you answer.
John Ye Posted August 31, 2018 Author Posted August 31, 2018 13 hours ago, swansont said: Thermal energy of a gas is much smaller than 10.2 eV. To lose a bound electron, energy must be added. You need to account for this energy, and you also need to explain why your numbers do not match with experiment for the binding energy value of a second electron in hydrogen. Or why nothing beyond H- is observed. Then your work is incomplete. You need to have experimental evidence to support your claim. Hydrogen has an absorption spectrum — you shine light on it and it absorbs the wavelengths that QM predicts. But I don't see how your model would predict an absorption spectrum, as it would require the atom to acquire another electron (which would have to emit a photon of the appropriate energy) at precisely the same time as the photon is absorbed, and in a situation where you would not have free electrons available. The value of 10.2 eV is not in question. It's the mechanism that you are proposing. In the excited H atom gas, there are many free electrons that are randomly flying everywhere. And also, we should know the energy is not even distributed in the gas. It's totally a mess
John Ye Posted August 31, 2018 Author Posted August 31, 2018 59 minutes ago, John Ye said: In the excited H atom gas, there are many free electrons that are randomly flying everywhere. And also, we should know the energy is not even distributed in the gas. It's totally a mess Many thanks for your kind suggestions. The work is not complete, further experiment is needed to prove my claim. Although I am sure the new model is correct, but only experiment can prove it. On 2017/9/8 at 11:21 AM, John Ye said: A new atom model Abstract: This paper presents a new microscopic extension to the Coulomb’s law－a formula that describes how electron and nucleus interact each other within atom world. Based on this Coulomb’s law extension, a new atom model is proposed. Compared with current electron-cloud model and with old Bohr model, this model is most like atom’s real physical structure. Using this new model and basic integral calculation, the spectrum of hydrogen gas and the spectrum of ionized helium gas are successfully derived. Key words: Energy levels, Microscopic, Coulomb’s law extension, Balance point, Balance sphere. New atom model.pdf Ye Cang 2017 rev.pdf 10 minutes ago, John Ye said: Many thanks for your kind suggestions. The work is not complete, further experiment is needed to prove my claim. Although I am sure the new model is correct, but only experiment can prove it. Ye Cang 2017 rev.pdf revised version
swansont Posted August 31, 2018 Posted August 31, 2018 3 hours ago, John Ye said: In the excited H atom gas, there are many free electrons that are randomly flying everywhere. Excited? Your claim is that there are no excited states in atoms. How do these electrons free themselves from their bound state? 7 hours ago, John Ye said: please see the device "Metal Analysis by Flame and Plasma Atomic Spectroscopy". It can detect the alloy's metal element components. to get atom spectrum of a matter, we must gasify it first. The gasification process makes sure that we obtain standalone atoms from the matter. Only in the state of atom gas, we can detect the full spectrum of the matter. we can NOT get H spectrum from liquid hydrogen. Hydrogen must be gasified first. It's not enough to gaisify into H2 gas, we should go further, to gasify H2 into standalone atoms( atom gas) to see the full spectrum. similarly, we want to get spectrum of the Ferrum, we must gasify Fe into Fe atom gas first. I will explain your question later. Try my best. I should not refuse to answer any question. If you want atomic spectra, yes, you need to have individual atoms. But there are spectra for molecules, in liquids and solids. These absorb and emit photons, too.
John Ye Posted August 31, 2018 Author Posted August 31, 2018 4 minutes ago, swansont said: Excited? Your claim is that there are no excited states in atoms. How do these electrons free themselves from their bound state? If you want atomic spectra, yes, you need to have individual atoms. But there are spectra for molecules, in liquids and solids. These absorb and emit photons, too. Yes, I see. That is not atom spectrum. To get atom spectrum, gasification is needed. 11 minutes ago, John Ye said: Yes, I see. That is not atom spectrum. To get atom spectrum, gasification is needed. Chapter 5. Hydrogen Atom Spectrum 5 Spectrum of the ... Excited hydrogen atoms are produced in an electric discharge which not only dissociates hydrogen molecules, ... 13 minutes ago, John Ye said: Yes, I see. That is not atom spectrum. To get atom spectrum, gasification is needed. Chapter 5. Hydrogen Atom Spectrum 5 Spectrum of the ... Excited hydrogen atoms are produced in an electric discharge which not only dissociates hydrogen molecules, ... In a electric discharge tube
Strange Posted August 31, 2018 Posted August 31, 2018 18 minutes ago, John Ye said: In a electric discharge tube Irrelevant. You are avoiding the question about how hydrogen atoms absorb (and emit) photons of specific energies that correspond to the energy levels of the electron in the atom and not to the ionisation energy. This, by itself, falsifies your idea.
John Cuthber Posted August 31, 2018 Posted August 31, 2018 11 hours ago, John Ye said: You browse definition of spectrum analysis, know its method. I used to work as a spectroscopist. I really don't need to browse the subject. You, on the other hand, have no idea what you are talking about. You have rehashed the bits of Bohr's work that gave the "right" answers for hydrogen, and ignored the problems where it gives the wrong answers. The "lines" in the hydrogen spectrum are actually split by spin- orbit coupling, and as far as I can see your idea doesn't deal with that. http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydfin.html
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