John Ye

Yes, my work doesn't deal with fine structure things. It only works on basic spectrum.

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

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

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

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.

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.

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.

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.

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. 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 )

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.

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. 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!!! 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.

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.

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.

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

Vibrating or randomly moving electric charge emits energy. This is so called thermal radiation. Any matter does so in universe. 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.

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.

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. 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. 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. thanks. I will if I have time. It does. Electrons' thermal movement make them radiate all the time. At absolute zero K temperature, all electrons stop thermal vibration, radiation stops.

Sorry, it's too late for me, I will go to bed. Talk to you tomorrow.

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. 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.

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.

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

(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.

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.

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) ------------------ 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.

First, read my paper, I explained how the spectrum quantization happens. I tried my best to explain it.