John Ye
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Posts posted by John Ye
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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
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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.
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.
revised version
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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
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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.
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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.
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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.
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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.
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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
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 )
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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.
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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.
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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.
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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.
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20 minutes ago, swansont said:
No, that's not what you did. You explain what happens when you have multiple electrons, but your model does not appear to explain the spectrum of neutral hydrogen. What happens when it gets excited to the n=2 state?
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.
29 minutes ago, studiot said:Well caught swansont.
John Ye.
This is a clear violation the the definition of an electric dipole, even by your proposals.
An electric dipole is a pair of coupled charges separted by a distance R.
It is defined as the product of the charge and the distance.
In what way does your gorund state H not comply with this definition?
static electron model does not mean that electron is totally stationary. Because of temperature is higher than zero K, the electron has thermal movement, which includes drifting in balanced sphere, and vibrations
Even for the bond electrons in a molecules. They are having thermal movement.
17 minutes 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.
static electron model does not mean that electron is totally stationary. Because of temperature is higher than zero K, the electron has thermal movement, which includes drifting in balanced sphere, and vibrations
Even for the bond electrons in a molecules. They are having thermal movement.
to totally stop electron, reduce the temperature. Some matter will become superconductor.
On 2017/9/8 at 5:49 PM, swansont said:I see no mention of angular momentum, or other details, like the Lamb shift. We already know the Bohr model is incorrect; predicting the energy levels is not sufficient to replace the quantum-mechanical model, which encompasses much more.
This paper does not mention atom's fine structure. The angular momentum is for the quantum theory to explain the fine structure.
the quantum theory has not defined a circle track motion of electron, the momentum is not applicable.
32 minutes 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.
static electron model does not mean that electron is totally stationary. Because of temperature is higher than zero K, the electron has thermal movement, which includes drifting in balanced sphere, and vibrations
Even for the bond electrons in a molecules. They are having thermal movement.
to totally stop electron, reduce the temperature. Some matter will become superconductor.
This paper does not mention atom's fine structure. The angular momentum is for the quantum theory to explain the fine structure.
the quantum theory has not defined a circle track motion of electron, the momentum is not applicable.
I am hoping that this can explain energy levels of exciting hydrogen gas
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22 minutes ago, studiot said:
So you wish to pursue your hypothesis without regard to the rest of Science?
How?
Vibrating or randomly moving electric charge emits energy. This is so called thermal radiation. Any matter does so in universe.
46 minutes ago, Strange said:What creates this “balanced point”? What opposes the electric force between the electrons and the nucleus?
You didn't read my script. Formula 1 tells that F=k(1-R/r)qq/(rr).
F is the force exerted on an electron by a proton. F has a zero point when r =R
This model must be the real physical structure in atom.
using this model, I can easily calculate the spectrum of H and half spectrum of helium.
Quantum model can't.
This is NOT a coincidence.
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10 hours ago, swansont said:
Then your model is wrong, or at least incomplete.
A stationary electron means there would be an electric dipole moment. Do you have any evidence that neutral, ground state atoms possess these dipole moments?
Then the energy will be variable, depending on the temperature. Any evidence that this is true?
Why doesn't the electron, which would be accelerating from these vibrations, radiate?
In terms of electrical dipole, yes, that is true for the atom with fixed electrons.
For atom with electron free to move within balanced sphere, for example, ground state H, the dipole doesn't exist.
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4 hours ago, swansont said:
Then your model is wrong, or at least incomplete.
A stationary electron means there would be an electric dipole moment. Do you have any evidence that neutral, ground state atoms possess these dipole moments?
Then the energy will be variable, depending on the temperature. Any evidence that this is true?
Why doesn't the electron, which would be accelerating from these vibrations, radiate?
Electrons have the lowest energy when they are in the balanced point.
The vibrations are small thermal movement. Which are temperature dependent. The lower the temperature, the less the vibrations.
3 hours ago, Strange said:So why isn’t angular momentum proportional to temperature?
There might be for some electrons which are not fixed. The fixed electrons are those shared by other atoms, e.g. the chemical bond electrons. Fixed ones can not. And it will happen when some external force can make them to drift in balanced sphere.
6 hours ago, Strange said:So why doesn’t this happen in your model?
In my model, electron doesn't need to move for it not to crash into proton. It stays in balanced point, never crash.
If it moves, it's small thermal drifting and vibrations.
Bohr electrons will crash if circular movement is stopped. So do the electrons in quantum model.
6 hours ago, studiot said:Please stop answering questions I haven't asked.
It wastes your time and mine.
You have more than one misconception and I'm trying to take them one at a time.
Before you come back please look up the following, if you are not familiar with them.
Earnshaw's theorem.
Lennard Jones Potential (This is a theory very similar to yours)
Madelung constants.
Biot- Savart Law
They are all very relevant theory.
thanks. I will if I have time.
5 hours ago, swansont said:Then your model is wrong, or at least incomplete.
A stationary electron means there would be an electric dipole moment. Do you have any evidence that neutral, ground state atoms possess these dipole moments?
Then the energy will be variable, depending on the temperature. Any evidence that this is true?
Why doesn't the electron, which would be accelerating from these vibrations, radiate?
It does. Electrons' thermal movement make them radiate all the time. At absolute zero K temperature, all electrons stop thermal vibration, radiation stops.
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Sorry, it's too late for me, I will go to bed.
Talk to you tomorrow.
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29 minutes ago, John Ye said:
There is no concept of angular momentum in my model. The model assume that electron is static, no circular motion, so angular momentum does not exist.
As for how to explain the fine structure, something new must be introduced
Yes, you're right. The balanced points constitute a balanced sphere. The electron shifts in it, with vibrations in the direction of radius, in normal temperature.
When temperature is extremely low, the electron will be totally static. Without any moving.
In absolute zero k, there will be no any electron movement. But the atoms and molecules are OK.
So,in this model, the motion of electron is NOT a must. It is a temperature dependent feature. In Bohr model and in quantum model, the motion of electron is a must. If electron stopped moving, the atoms would be vanished because electrons crash into proton.
10 minutes ago, studiot said:Doesn't the The Bohr Model only apply to the hydrogen atom?
In any event I specified r = R, whatever value you put on it, so the term in those brackets is still exactly zero.
Now tell me,
If there is another positive charge of any description beyond R, for any reason, will it exert a greater or lesser force on that electron than its own nucleus, since we have just calculated that force to be exactly zero?
Yes. The term must be zero.
Bohr model can only be used with H.
So does quantum model.
If other proton approaches an H atom, it will be exert an attractive force to the electron, the balanced point will be changed. It will no longer be R. In this case, a hydrogen molecule will be formed.
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Just now, swansont said:
I asked about your claim about angular momentum, not energy.
There is no concept of angular momentum in my model. The model assume that electron is static, no circular motion, so angular momentum does not exist.
As for how to explain the fine structure, something new must be introduced
7 minutes ago, swansont said:I asked about your claim about angular momentum, not energy.
How does it get there? If it's moving, there will be no force, so it won't stay there. It will oscillate about an equilibrium point, at best.
Yes, you're right. The balanced points constitute a balanced sphere. The electron shifts in it, with vibrations in the direction of radius, in normal temperature.
When temperature is extremely low, the electron will be totally static. Without any moving.
9 minutes ago, John Ye said:There is no concept of angular momentum in my model. The model assume that electron is static, no circular motion, so angular momentum does not exist.
As for how to explain the fine structure, something new must be introduced
Yes, you're right. The balanced points constitute a balanced sphere. The electron shifts in it, with vibrations in the direction of radius, in normal temperature.
When temperature is extremely low, the electron will be totally static. Without any moving.
In absolute zero k, there will be no any electron movement. But the atoms and molecules are OK.
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4 minutes ago, studiot said:
Would you further agree that your equation 2 contains the same zero when ne = np since their ratio is then one?
I was specifically starting with the simplest possible situation ie one electron and one proton.
Yes. Equation 2 is an extended version of equation 1, for the case when more than one electrons are attached within an atom.
1 minute ago, John Ye said:Yes. Equation 2 is an extended version of equation 1, for the case when more than one electrons are attached within an atom.
And in equation 2, the R is no longer a constant. It depends the number of electrons and protons.
3 minutes ago, John Ye said:Yes. Equation 2 is an extended version of equation 1, for the case when more than one electrons are attached within an atom.
And in equation 2, the R is no longer a constant. It depends the number of electrons and protons.
In equation 1, one electron with one proton, R is constant, its value is Bohr radius. 52.9pm
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55 minutes ago, studiot said:
Once again, you are telling me things I didn't ask for and in this case specifically excluded.
Why do you expect me to read your script if you persistently ignore mine?
I have taken pains to be fair and take notice of what you are proposing specifically I asked if a certain term in one of your equations was zero under certain conditions.
(1−R／r)Would you agree that this is zero when R = r or is your equation wrong?
The answer is yes. It is zero when r=R. Obviously yes.
1 minute ago, John Ye said:
(1−R／r)Would you agree that this is zero when R = r or is your equation wrong?
The answer is yes. It is zero when r=R. Obviously yes.
When r=R, the force = 0, so electron can stay there without any motion.
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On 2017/9/8 at 5:49 PM, swansont said:
I see no mention of angular momentum, or other details, like the Lamb shift. We already know the Bohr model is incorrect; predicting the energy levels is not sufficient to replace the quantum-mechanical model, which encompasses much more.
This paper does not mention atom's fine structure. The angular momentum is for the quantum theory to explain the fine structure.
the quantum theory has not defined a circle track motion of electron, the momentum is not applicable.
my model is "static electron model", so I don't mention the angular momentum.
To explain the spectrum's fine structure in magnetic field, other work and test is needed, which is beyond my capability.
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We don't have any means to detect the light produced by one hydrogen atom.
The light we detected is emitted by the H gas in glass bottle.
We incorrectly thought it was emitted from one single atom.
This leads to all later tragedies in atom theory.
21 minutes ago, studiot said:In your modified coulomb equation you have the term
(1−Rr)Would you agree that this is zero when R = r or is your equation wrong?
Yes, when r = R, the force is 0.
but this formula is only for ONE electron with ONE proton, the case of normal H atom.
the value R is called balanced point in my paper.
if electrons or protons are more than one, the formula is no longer applicable.
to deal with this case, I have another extended formula, formula (2)
20 minutes ago, John Ye said:We don't have any means to detect the light produced by one hydrogen atom.
The light we detected is emitted by the H gas in glass bottle.
We incorrectly thought it was emitted from one single atom.
This leads to all later tragedies in atom theory.
Yes, when r = R, the force is 0.
but this formula is only for ONE electron with ONE proton, the case of normal H atom.
the value R is called balanced point in my paper.
if electrons or protons are more than one, the formula is no longer applicable.
to deal with this case, I have another extended formula, formula (2)
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from my paper, you can see that big atoms ( other than H, He ) are unable to calculate.
The extended Coulumb's law is only applicable to the simplest case which I called "Equal opportunity electron configuration:.
This is why so far this is no atom theory which can give an accurate math other than H atom.
my paper is the same. It gives accurate math for H atom, and for haft He atom. The other half spectrum of He atom is not calculable.
We don't have any means to detect the light produced by one single hydrogen atom.
What we detected is is the light emitted by the H gas in glass bottle.
We incorrectly thought it was emitted from one single H atom.
This leads to all later tragedies in various atom theory.
If, only if, we can hold one electron and one proton in a glass bottle, give energy from outside, detect the emitted light.
What we can detect?
we can see only 92nm wavelength light.
no other light at all.
One single H atom has no quantum effect.
Quantum effect is produced by mass atoms in the H gas bottle.
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56 minutes ago, swansont said:
How does that lead to quantization? i.e. where is your model?
And do you have a citation for observation of H7-?
First, read my paper, I explained how the spectrum quantization happens. I tried my best to explain it.
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A new atom model （static electron configuration model )
in Speculations
Posted
Yes, my work doesn't deal with fine structure things. It only works on basic spectrum.