# Is The Bohr's Model Of Atoms Still Used To Analyze Emission And Absorption Of Em Energy?

36 replies to this topic

### #1 rhertz

rhertz

Questioning

• Members
• 180 posts

Posted 30 April 2019 - 11:13 PM

This formula applies for energy of a photon emitted or absorpted by an atom

with 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?

Edited by rhertz, 30 April 2019 - 11:15 PM.

### #2 exchemist

exchemist

Creating

• Members
• 2488 posts

Posted 01 May 2019 - 02:16 AM

This formula applies for energy of a photon emitted or absorpted by an atom

with 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 by exchemist, 01 May 2019 - 03:04 AM.

### #3 Flummoxed

Flummoxed

Explaining

• Members
• 607 posts

Posted 01 May 2019 - 03:28 AM

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.

It was still being taught 40 years ago, a blast from the past from wiki https://en.wikipedia...wiki/Bohr_model Complete with derivations and limitations.

### #4 exchemist

exchemist

Creating

• Members
• 2488 posts

Posted 01 May 2019 - 05:32 AM

It was still being taught 40 years ago, a blast from the past from wiki https://en.wikipedia...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).

### #5 rhertz

rhertz

Questioning

• Members
• 180 posts

Posted 01 May 2019 - 09:35 AM

I've found interesting theories about the time involved in the transitions

of electrons between states.

It seems to vary in a wide range, and involve more than one transition

in complex atoms (with a photon emitted at each transition), but there

is consensus, somehow, that the transition time is aroun 1 nanosecond

and involves several wavelengths of the photon.

It seems that a problem exists when talking about photons: their properties

can'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 with

a 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't

be applied to it, as its spectral decomposition has a dispersion of components around

the central frequency f.

Then, isnt it a dead end?

### #6 exchemist

exchemist

Creating

• Members
• 2488 posts

Posted 01 May 2019 - 10:28 AM

I've found interesting theories about the time involved in the transitions

of electrons between states.

It seems to vary in a wide range, and involve more than one transition

in complex atoms (with a photon emitted at each transition), but there

is consensus, somehow, that the transition time is aroun 1 nanosecond

and involves several wavelengths of the photon.

It seems that a problem exists when talking about photons: their properties

can'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 with

a 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't

be applied to it, as its spectral decomposition has a dispersion of components around

the central frequency f.

Then, isnt it a dead end?

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 by exchemist, 01 May 2019 - 10:46 AM.

### #7 rhertz

rhertz

Questioning

• Members
• 180 posts

Posted 01 May 2019 - 10:48 AM

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.

What oscillates are the electric and magnetic fields.

Welcome to wave-particle duality!

I'm talking about this example: here you have a wave-packet, which form is obtained by shaping its

waveform with a gaussian envelope.

This shaping is neccesary, because if I use a rectangular pulse to shape the sinusoidal wave, then

infinite components appears at its spectral transform.

This image is from this link, which has the mathematical details:

http://cvarin.github...-Guide/fft.html

Unless you are finding the spectral composition of an infinite wave (it gives a Dirac impulse

at 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 around

the 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".

### #8 OceanBreeze

OceanBreeze

Creating

• Moderators
• 1032 posts

Posted 01 May 2019 - 11:57 AM

This formula applies for energy of a photon emitted or absorpted by an atom

with 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.

### #9 Flummoxed

Flummoxed

Explaining

• Members
• 607 posts

Posted 01 May 2019 - 02:22 PM

In this simpla case of a wave-packet, you can see how the energy distributes symmetrically around

the 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".

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)

### #10 exchemist

exchemist

Creating

• Members
• 2488 posts

Posted 01 May 2019 - 03:18 PM

I'm talking about this example: here you have a wave-packet, which form is obtained by shaping its

waveform with a gaussian envelope.

This shaping is neccesary, because if I use a rectangular pulse to shape the sinusoidal wave, then

infinite components appears at its spectral transform.

This image is from this link, which has the mathematical details:

http://cvarin.github...-Guide/fft.html

Unless you are finding the spectral composition of an infinite wave (it gives a Dirac impulse

at 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 around

the 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 by exchemist, 01 May 2019 - 03:27 PM.

### #11 exchemist

exchemist

Creating

• Members
• 2488 posts

Posted 01 May 2019 - 03:21 PM

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.

### #12 Flummoxed

Flummoxed

Explaining

• Members
• 607 posts

Posted 02 May 2019 - 10:52 AM

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.

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=pc

Edited by Flummoxed, 02 May 2019 - 10:54 AM.

### #13 exchemist

exchemist

Creating

• Members
• 2488 posts

Posted 02 May 2019 - 11:32 AM

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=pc

But 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 by exchemist, 02 May 2019 - 11:48 AM.

### #14 Flummoxed

Flummoxed

Explaining

• Members
• 607 posts

Posted 03 May 2019 - 12:07 PM

But 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".

E=hf, frequency in these equations is like a gain factor, E= (a constant with appropriate units) x k . k can be anything as long as the units balance, and your constant is selected to give the right results. hf appears an unnecessary complication.

de Broglie wrote some other interesting papers with Bohm, ref pilot waves explaining the double slit experiment results. Plausible perhaps, and do away with the need for wave particle duality.

I think I mentioned somewhere on a previous thread. spin and polarization of photons, I just neglected to post it here.

But when focusing on a photon, what shape of field can be inferred from its known characteristics. It is polarized and has spin, that suggests a disc shaped field. It has spin 1, can the spin rate/frequency increase ?

If a photon is viewed as a half wave blip in a field travelling at c, then you could ascribe it that frequency? rhertz was complaining about the harmonic picture he had, with oscillations which clearly it is not.

I dont think assuming a photon has no oscillating frequency, is a problem, and gives one a clearer picture of a photon. I might be wrong, I often am.

Can anyone recommend a good book on QFT covering in particular QED. My old text books are a bit out of date.

### #15 marcospolo

marcospolo

Understanding

• Members
• 453 posts

Posted 03 May 2019 - 03:56 PM

I've found interesting theories about the time involved in the transitions

of electrons between states.

It seems to vary in a wide range, and involve more than one transition

in complex atoms (with a photon emitted at each transition), but there

is consensus, somehow, that the transition time is aroun 1 nanosecond

and involves several wavelengths of the photon.

It seems that a problem exists when talking about photons: their properties

can'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 with

a 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't

be applied to it, as its spectral decomposition has a dispersion of components around

the central frequency f.

Then, isn`t it a dead end?

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.

### #16 Flummoxed

Flummoxed

Explaining

• Members
• 607 posts

Posted 03 May 2019 - 04:08 PM

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.

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.

### #17 marcospolo

marcospolo

Understanding

• Members
• 453 posts

Posted 03 May 2019 - 04:46 PM

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 by marcospolo, 03 May 2019 - 04:47 PM.