rhertz Posted May 1, 2019 Report Posted May 1, 2019 (edited) This formula applies for energy of a photon emitted or absorpted by an atomwith atomic number Z, between discrete energy levels n1 and n2. Is this derivation from Bohr's atom model still being used? Eph = -13.6 Z2 (n2-2 - n1-2) eV Edited June 19, 2019 by rhertz Quote
exchemist Posted May 1, 2019 Report Posted May 1, 2019 (edited) This formula applies for energy of a photon emitted or absorpted by an atomwith atomic number Z, between discrete energy levels n1 and n2. Is this derivation from Bohr's atom model still being used? Eph = -13.6 Z2 (n2-2 - n1-2) eV Note 1: I'm re-studying subjects that I'd learned long ago, but then I forget. Note 2: I wrote "absortion" instead of absortion at the title, but I can't change it. Anyone can do it, please?For teaching purposes, yes, I think so. It works quite well for hydrogen and hydrogen-like ions, i.e. ions with only a single electron, even though its physical picture of the atom was abandoned almost a century ago. For multi-electron atoms and ions, the formula does not work (though can, I seem to recall, be fudged to some degree for alkali spectra with "Zeff" etc), but the reasons why it doesn't work can be illuminating when learning about the elements. Edited May 1, 2019 by exchemist Quote
exchemist Posted May 1, 2019 Report Posted May 1, 2019 It was still being taught 40 years ago, a blast from the past from wiki https://en.wikipedia.org/wiki/Bohr_model Complete with derivations and limitations.Yes, that's what I am saying. The Bohr model is a teaching aid, used in the lower 6th Form or thereabouts, i.e. before the concept of wave-particle duality and orbitals is introduced. You need to cover it so that you can point out why it doesn't work, as that was an important part of the genesis of QM. For the hydrogen atom, the formula for the spectral lines is pretty good. There is no shielding to worry about and spin-orbit coupling is very small (though I see Michelson and Morley seem to have detected a doublet in the Hα line back in 1887, according to my Atkins). Quote
exchemist Posted May 1, 2019 Report Posted May 1, 2019 (edited) I've found interesting theories about the time involved in the transitionsof electrons between states. It seems to vary in a wide range, and involve more than one transitionin complex atoms (with a photon emitted at each transition), but thereis consensus, somehow, that the transition time is aroun 1 nanosecondand involves several wavelengths of the photon. It seems that a problem exists when talking about photons: their propertiescan't be completely separated from electromagnetic waves. So, if a photon as a frequency, it implies oscillations. Where are oscillations, something is vibrating, changing its amplitude witha period (or a variable period in damped oscillations), So, my question is: What oscillates in the photon as it travels at "c" speed? And, as oscillations occur in time axis, in the amplitude axis something changes. So, what changes in amplitude with the oscillating photon? This bring closer and closer the idea of a photon as a packet-wave, isn't it? But the problem with packet-waves is that the simple formula for energy E=hf can'tbe applied to it, as its spectral decomposition has a dispersion of components aroundthe central frequency f. Then, isn`t it a dead end? I'd appreciate some comments about these thoughts.It's not a dead end: it's how the Uncertainty Principle arises. A wave-packet has some uncertainty around both its location and its frequency This is one reason why lines in atomic spectra have finite width. (Though normally one deals with the alternative lifetime/energy expression of the HUP for that.) What oscillates are the electric and magnetic fields. Welcome to wave-particle duality! Edited May 1, 2019 by exchemist Quote
OceanBreeze Posted May 1, 2019 Report Posted May 1, 2019 This formula applies for energy of a photon emitted or absorpted by an atomwith atomic number Z, between discrete energy levels n1 and n2. Is this derivation from Bohr's atom model still being used? Eph = -13.6 Z2 (n2-2 - n1-2) eV Note 1: I'm re-studying subjects that I'd learned long ago, but then I forget. Note 2: I wrote "absortion" instead of absortion at the title, but I can't change it. Anyone can do it, please? Note 2: I wrote "absortion" instead of absortion at the title, but I can't change it. Anyone can do it, please? I'd be happy to change it. Which one would you like; "absortion" or "absortion"? Just kidding, I have already changed it. Quote
exchemist Posted May 1, 2019 Report Posted May 1, 2019 (edited) I'm talking about this example: here you have a wave-packet, which form is obtained by shaping itswaveform with a gaussian envelope. This shaping is neccesary, because if I use a rectangular pulse to shape the sinusoidal wave, theninfinite components appears at its spectral transform. This image is from this link, which has the mathematical details: http://cvarin.github.io/CSci-Survival-Guide/fft.html Unless you are finding the spectral composition of an infinite wave (it gives a Dirac impulseat its frequency), any other waveform that is limited in time generates dispersion at its spectrum. In this simpla case of a wave-packet, you can see how the energy distributes symmetrically aroundthe central frequency f0. This fact inhibits to apply Planck's formula E=hf for the energy of the wave-packet. So, a wave-packet (IMHO) is not a valid representation of a "photon". OK, but then one needs to take into account the uncertainty principle and the probablistic nature of a wave function. The envelope is made up of a range of frequencies, meaning there is a range of frequency results that might be obtained by making a single measurement, i.e. there is an uncertainty about the frequency. The relative contribution of each element of the frequency mix is proportional to the likelihood of that frequency being measured. The expectation value will be the central frequency. It's a probability distribution, in effect, a bit like an electron orbital in an atom, except this is a distribution of frequency (hence momentum and energy) rather than position. P.S. I like the pictures very much. This is actually one of the nicest illustrations of the uncertainty principle I have ever seen. Edited May 1, 2019 by exchemist Quote
exchemist Posted May 1, 2019 Report Posted May 1, 2019 I think what you are saying is a Photon has energy it does not have frequency, or at least can not be represented by a series of harmonics, making up a localized wave packet. Blue shift represents an increase in energy (due to what, doppler effect). A photon acquires energy from increasing its speed(momentum) beyond c, which is (apparently)impossible, or it increases its momentum or frequency due to an increase in inertia whilst still travelling at c. Red shift represents a decrease in momentum or frequency, and energy. A photon loses energy via decreasing its speed, (possible in a gravitational field, Einstein said so!) or it loses momentum/frequency due to a reduction in inertia. For a photon to lose energy implies it is losing one or all of the following, momentum, frequency or speed. Inertia and speed give momentum. Frequency assumes the photon has a frequency, which you apparently don't agree with. You are therefore viewing the photon as a packet of energy according to E=pc not E=hf I tend to agree a photon is not an oscillation passing through space, and I don't think it can be proven to be either. It is a packet of energy equivalent to E=pc. E = hf is equivalent but the f does not represent a true oscillating frequency as in ac wire, it is merely a gain figure. I am not sure that is helpful (or even right, but it is how I see it)I would not agree with this. See my other post. I don't see the difficulty in assigning a range of frequencies to a photon, given the way the uncertainty principle applies to all such cases. Quote
exchemist Posted May 2, 2019 Report Posted May 2, 2019 (edited) Uncertainty in momentum x position is not the same as uncertainty in frequency x position. At best stating a photon has frequency appears an unprovable mathematical assumption/analogue, not a physical one, especially when you include additional harmonics. How would you go about detecting the harmonics and base frequency of a "single" photon, if it had any? I do not think you can. All you can detect is the energy and momentum, and polarisation of the photon, nothing more. To ascribe different colours, different frequencies appears wrong, they can only be regarded as packets of energy. E=pc might mathematically give the same result as E=hf. A photon has no charge or magnetic field, it only has inertial energy, the only way this energy level can fluctuate is if it interacts with perhaps virtual particles, Compton scattering for example, or is absorbed and remitted by some other substance. There is no need to ascribe a photon a frequency, an energy level is sufficient relating to its momentum. To assign frequencies to photons radio waves are used. A radio wave has frequency, which may consist of a wave of photons, some of which are absorbed by receivers. A radio wave is not a single photon, unless the definition of a photon has changed. Virtual particles separated momentarily via a moving magnetic field around the transmitter likely acquire enough energy to produce photons on their decay, which radiate as a radio wave consisting of electromagnetically neutral photons, that only have momentum to transfer on the receiving end. A little bit of speculation is involved here, but I do not think an individual photon can be ascribed any other properties than E=pcBut frequency is proportion to momentum, by de Broglie's relation: p =h/λ. Since c=νλ, this can be rewritten as p=hν/c. Which fits both E=pc and E=hν. So if you accept uncertainty in momentum you de facto accept a proportional uncertainty in frequency, unless you contend that de Broglie, and thus the whole of QM, has got it wrong. Is that your position? To look at this another way, since EM radiation is observed (a) to have a frequency and ( b ) to consist of photons, it is a bit hard to maintain with much credibility that a photon has no frequency. Whether you can measure the frequency of a single photon or not is a matter of practical experimental technique and does not say anything fundamental about the nature of photons. After all, we cannot directly measure the interatomic distance between two individual atoms, but nobody disputes that since you can measure the interatomic distances in an array of them in a crystal (X-ray diffraction), then the same must apply to the individuals making up the array. P.S. There is one more property of photons that you have not included in your list: angular momentum, or "spin". Edited May 2, 2019 by exchemist Quote
marcospolo Posted May 3, 2019 Report Posted May 3, 2019 I've found interesting theories about the time involved in the transitionsof electrons between states. It seems to vary in a wide range, and involve more than one transitionin complex atoms (with a photon emitted at each transition), but thereis consensus, somehow, that the transition time is aroun 1 nanosecondand involves several wavelengths of the photon. It seems that a problem exists when talking about photons: their propertiescan't be completely separated from electromagnetic waves. So, if a photon as a frequency, it implies oscillations. Where are oscillations, something is vibrating, changing its amplitude witha period (or a variable period in damped oscillations), So, my question is: What oscillates in the photon as it travels at "c" speed? And, as oscillations occur in time axis, in the amplitude axis something changes. So, what changes in amplitude with the oscillating photon? This bring closer and closer the idea of a photon as a packet-wave, isn't it? But the problem with packet-waves is that the simple formula for energy E=hf can'tbe applied to it, as its spectral decomposition has a dispersion of components aroundthe central frequency f. Then, isn`t it a dead end? I'd appreciate some comments about these thoughts.Here's a thought. All three proposed models of what light is, do not fit the observations, so according to the scientific Method, all three must be officially pronounced as being wrong. Its that simple guys.You DO NOT NEED to provide a better model of hypothesis to be able to rule out ones that don't work! That is a dodge to allow pseudo science to hang on to its false theories.So light is clearly NOT a particle of any kind, especially a particle that has no size and no mass, which is a flat out contradiction, and a logical impossibility, so scratch that one right now. Einstein was wrong to present the concept of light being a particle. Next, its NOT a wave. A wave is NOT a thing. Its not what light is. It may be indicative of HOW light permeates across distances, but "wave" is NOT wave light IS.Can you can go to the beach and bring home a bucket of waves? So the chance that light is a "wave packet", of pure probability equations, even MORE of a impossibility than the first two. This proposal is something that a crack head might claim, its NOT rational. Which is why it fits into the QM world of make-believe. So, EInstein's nobel prize winning paper on photoelectric effect is just wrong. This is NOT the way light works, and not the way electricity can be generated by a photo cell. There will be a correct explanation, Einstein's is NOT it! Back to the drawing board, this crop of explanations are all wrong. Quote
marcospolo Posted May 3, 2019 Report Posted May 3, 2019 (edited) Would you agree light has energy, momentum, and can be polarized. Light also moves in a straight line, assuming it isn't deflected by something. I would NOT agree at all. Light may be the result of some energy, but not necessarily IS that energy.next, its irrational to say that something that has NO MASS, can possibly possess the property of Momentum. This is patently just wrong.Next, can 'light" be polarized? well something is happening with those filters, but its a trap to say you have change light, you probably have changed the observable effects that light has. Not the light itself. We have NO IDEA what light IS. We only can see the effects. But does light MOVE in a straight line? well, two possibilities (there may be more) we don't really KNOW that "light" itself is doing any moving at all! This idea of light traveling was agreed on by committee, way back in the early days, not by any observed evidence.Next, clearly, the effects of whatever light is able to generate when related to objects, (there is no "light" in space, you MUST have an object, even dust before you can "see" light) and its obvious that the effects of light are blocked by solid objects, unless they are transparent. so it seems then that light "travels" in a straight line. But the effects of light always are traced by straight paths between the emitter and the object. I for one would dearly like to figure out what light really is, because we have no decent hypothesis at this stage. I don't think I have any chance of understanding it really. Edited May 3, 2019 by marcospolo Quote
OceanBreeze Posted May 4, 2019 Report Posted May 4, 2019 You admit you don't have any chance of understanding it (since you are not a physicist) Yet you declare it is "patently just wrong" No, you are patently just wrong. Quote
marcospolo Posted May 4, 2019 Report Posted May 4, 2019 I don't know how light works, and neither do you. But its not hard to figure out the the current theories cant be right. And Ive briefly explained why the three ideas,( particle, wave and wave packet ) are just bad guesses. Once again, I suggest you reply with real reasons why my statement are incorrect, and not bring up some interpretation of an observation that may or may not support your cherished ideas. Tell me where I'm wrong, I discussed claims about what light is, not someone's observations. Why are my criticisms of the three claims wrong? Quote
Dubbelosix Posted May 4, 2019 Report Posted May 4, 2019 (edited) One of the seemingly strong factors supporting a wavelength for photons might be doppler shift. Most galaxies are red shifted and assumed moving away from us and a small number are blue shifted indicating they are moving towards us. Could blue shift be explained by Zwickys tired light, it would appear not. http://www.astronomy.com/magazine/ask-astro/2017/12/blueshifted-galaxies Is there any other way blue shift could be explained? In my model of fluid dynamics in the unified theory linking it to Friedmann cosmology, the imbalance would not ensure all galaxies have to move away from us, its a complicated situation where many galaxies had bombarded in the early universe and either move parallel towards us, showing a blue shift, but they are much fewer than those observed through the redshift implying itself, a massive preference in which may a universe should expand. Recall, the Friedmann equation is a fluid equation. So is Bernoulli's equation, but what we learned from united the two, was a mechanical reason. Before then, we have attributed it to 1. Vacuum pressure and/or various pressure interacting2. A constant of integration from Einsteins equations 3. Vacuum energy resulting in an accelerated phase... But keep in mind, if we want to look for mechanical reasons, the simplest should be sought out first. Very clever men, scientists no less, have provided wonderful and speculative models. Model's should not be too speculative and the main goal of a scientist should not be concerned with pure math but how they manifest theoretical implications in mechanical and simple ways. Einstein was right, ''a theory should be no more complicated than it has to be.'' Edited May 4, 2019 by Dubbelosix Quote
Dubbelosix Posted May 4, 2019 Report Posted May 4, 2019 Something is wrong with this value. According to it, we only have about 6 million years before a full collision. And the effects, being Andromeda 10 times more massive than the Milky Way, would start to affect everyoneof the 100 million stars of the Milky Way much sooner that that, perhaps in 2 or 3 million years, Not too much time, in cosmic scales. I thought it was 7 but hey devils in the details and not all our memories and faculties are always around :P But yeah, you are right, there is a massive redshifted galaxies that dwarfs the blue-shifted types. Quote
OceanBreeze Posted May 4, 2019 Report Posted May 4, 2019 Something is wrong with this value. According to it, we only have about 6 million years before a full collision. And the effects, being Andromeda 10 times more massive than the Milky Way, would start to affect everyoneof the 100 million stars of the Milky Way much sooner that that, perhaps in 2 or 3 million years, Not too much time, in cosmic scales. Yes, that figure is way off. The exact closing speed is difficult to measure so these are just estimating: The two galaxies are closing at over 100 km/sec, less than one thousandth the speed of light. Andromeda is presently more than 2 million light years away, so it will take roughly 4 Billion years to collide with our Milky Way galaxy. The experts figure it will be unlikely for any two stars to collide, but things will definitely get moved around, with some stars being ejected altogether. However, there is nothing to worry about! In about 3 billion years our sun is expected to begin the process of going into red giant phase. When it reaches a 40% increase in brightness, in 3.5 billion years, the oceans will be boiled away and Earth will be reduced to a lifeless dry world. But, by that time, if humans still exist, their technology should be advanced enough for them to find, and move to any planet they like in the new super galaxy of over a trillion stars. (I thought somebody said they wanted science fiction) Quote
Dubbelosix Posted May 4, 2019 Report Posted May 4, 2019 (edited) Yes, that figure is way off. The exact closing speed is difficult to measure so these are just estimating: The two galaxies are closing at over 100 km/sec, less than one thousandth the speed of light. Andromeda is presently more than 2 million light years away, so it will take roughly 4 Billion years to collide with our Milky Way galaxy.editThe experts figure it will be unlikely for any two stars to collide, but things will definitely get moved around, with some stars being ejected altogether. However, there is nothing to worry about! In about 3 billion years our sun is expected to begin the process of going into red giant phase. When it reaches a 40% increase in brightness, in 3.5 billion years, the oceans will be boiled away and Earth will be reduced to a lifeless dry world. But, by that time, if humans still exist, their technology should be advanced enough for them to find, and move to any planet they like in the new super galaxy of over a trillion stars. (I thought somebody said they wanted science fiction) Am I wrong in recollecting that a complete submergence will occur around 10 billion years? It's like throwing snooker balls from a very large snooker table, to find only some will interact... but that could be devastating to our galaxy, depending on how violent systems are thrown out of their orbits by Andromeda, even in respect of it being a slight smaller galaxy? edit: including bodies thrown out of our galaxy and towards more vital parts... I hope it does not distrupt us, or that is the day of Armageddon. Edited May 4, 2019 by Dubbelosix Quote
Dubbelosix Posted May 4, 2019 Report Posted May 4, 2019 Just to add, that's plenty years to become a fully space dwelling race. Can we get there? Save our own asses just in case? I hope so. Quote
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