Question on electron orbitals and wavelengths?

[Austin0]

Question on electron orbitals and wavelengths??

Hi I have a couple of dumb laymans questions:

(1) I have seen various diagrams of electron shells where some of them appear as ,roughly , rotating figure eights. Is there any physical conceptual interpretation of this?
It would appear to mean that the electrons in those shells would interpenetrate inner shells and approach closely to the nucleus ,,but this doesn't seem likely.

(2) Is there any consistant relationship in wavelength between electrons in the same shell??

Is there any correspondence between electron wavelength and the distance from the nucleus, of the shell they occupy?

Thanks Cheers

 

[olgranpappy]

Hi I have a couple of dumb laymans questions:

(1) I have seen various diagrams of electron shells where some of them appear as ,roughly , rotating figure eights. Is there any physical conceptual interpretation of this?

w/out seeing the pictures myself I'm not quite sure, but it sounds to me like maybe someone was trying to draw p-orbitals; the wavefunction of a p-orbital is zero at the origin. there are three orthogonal p orbitals, roughly "pointing in different directions", so maybe that is the "rotating" part.

http://en.wikipedia.org/wiki/Atomic_orbital

It would appear to mean that the electrons in those shells would interpenetrate inner shells and approach closely to the nucleus ,,but this doesn't seem likely.

(2) Is there any consistant relationship in wavelength between electrons in the same shell??

electrons in the same "shell" are supposed to have the same energy.

Is there any correspondence between electron wavelength and the distance from the nucleus, of the shell they occupy?

In a hydrogen atom (or hydrogenlike atom) the expected value of (1/r) is directly proportional to the expected value of the energy. So, higher energies roughly means smaller r.

 

[Austin0]

w/out seeing the pictures myself I'm not quite sure, but it sounds to me like maybe someone was trying to draw p-orbitals; the wavefunction of a p-orbital is zero at the origin. there are three orthogonal p orbitals, roughly "pointing in different directions", so maybe that is the "rotating" part.
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Thanks olgranpappy ---yes those are similar to the diagrams I have seen.
And yes I was referring to some p-orbitals and also some d-orbs.
When you say wavefunction 0 at the origin I would take that to mean zero at the center of the nucleus. But the dumb-bell shaped orbitals seem to approach the nucleus very closely or is that just me trying to apply an inappropriate interpretation to the representation?



In a hydrogen atom (or hydrogenlike atom) the expected value of (1/r) is directly proportional to the expected value of the energy. So, higher energies roughly means smaller
r.

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As I understand it the wavelength in relation to the path of the orbitals is a fundamental harmonic integral of the path length = [roughly circumfrence] I wondered if the same kind of harmonic relationship held for the radius or the distance to the nucleus?

Thanks for your help to a beginner

 

[jtbell]

As I understand it the wavelength in relation to the path of the orbitals is a fundamental harmonic integral of the path length = [roughly circumfrence]

You're thinking of de Broglie's derivation of the radius of the circular orbits in the Bohr model by assuming the circumference is a multiple of the wavelength. This picture has little to do with modern quantum mechanics, which does not contain the notion of a well-defined classical path or orbit, circular or otherwise. All we have is a probability density "cloud" which has the unfortunately-chosen name of "orbital."

You should view the Bohr model and de Broglie's derivation of it in terms of waves along a path, only as a historical stepping-stone towards QM.

 

[Austin0]

You're thinking of de Broglie's derivation of the radius of the circular orbits in the Bohr model by assuming the circumference is a multiple of the wavelength. This picture has little to do with modern quantum mechanics, which does not contain the notion of a well-defined classical path or orbit, circular or otherwise. All we have is a probability density "cloud" which has the unfortunately-chosen name of "orbital."

You should view the Bohr model and de Broglie's derivation of it in terms of waves along a path, only as a historical stepping-stone towards QM.

___________________________________________________________________-
Thanks for the response
One final question
Does the evolved paradigm mean that the concept of harmonics is not valid at all or that it is just not relevant in the context of probability waves?
I hope this question makes some kind of sense.

 

[DeShark]

You should view the Bohr model and de Broglie's derivation of it in terms of waves along a path, only as a historical stepping-stone towards QM.

While I do agree with this point, isn't it true that this approach is pretty much what is used in the path integral formulation? i.e. you consider classical trajectories, work out their contribution to the amplitude (position amplitude in this case) and then use that to do your calculations. I haven't solved the hydrogen atom or any atoms in this manner, but I have an inkling feeling that the Bohr trajectories will almost certainly be the paths which contribute most to the real amplitude. Since I have never used the path integral formulation and have never taken the time to sit down and figure out the details, I honestly don't know. If anyone could point out that I'm right or wrong, it would help enormously.

As for the original problem of wavelengths being correlated (the same?) for electrons in the same orbital, I would say no. While the energy of the electron is a fixed constant $$E_{nj} = \frac{-13.6 \ \mathrm{eV}}{n^2} \left(1 + \frac{\alpha^2}{n^2}\left(\frac{n}{j+\frac{1}{2}} - \frac{3}{4} \right) \right) \,$$ depending on "n" and "j", and since energy = frequency, the frequency of both electrons will be the same. However, I do think that the wavelength will vary. As has been stated, the electron can be found in a spread out "cloud". If a measurement on the particles position is performed then at this point, its energy will be given by its kinetic energy (momentum squared over 2 times the mass) + the potential. If it is found in an area with higher potential (further from the nucleus), then the electron must give up some of its kinetic energy. Then, according to DeBroglie, $$\lambda=hp$$ and so its wavelength will be lower, despite it being in a given energy level. If you follow this through for another electron in the same level, then the two electrons will have the same frequency, but not necessarily the same wavelength. For a classical analogy, one could think about the velocity of the electron reducing as it goes into an area of higher potential, but using this as any more than an analogy will cause trouble.

 

[Manchot]

The problem with talking about a "wavelength" is that it isn't really well-defined for a bound state. The wavefunction will be spread out over a spectral range.

 

[DeShark]

The wavefunction will be spread out over a spectral range.

I don't quite understand what you mean. Could you elaborate please?

 

[jtbell]

Consider the hydrogen wave functions listed here:

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html

(scroll down to "Normalized Hydrogen Wavefunctions.")

Consider the one for n = 2, l = 1, m_l = +/- 1. What is its wavelength?

 

[Austin0]

As for the original problem of wavelengths being correlated (the same?) for electrons in the same orbital, I would say no. While the energy of the electron is a fixed constant $$E_{nj} = \frac{-13.6 \ \mathrm{eV}}{n^2} \left(1 + \frac{\alpha^2}{n^2}\left(\frac{n}{j+\frac{1}{2}} - \frac{3}{4} \right) \right) \,$$ depending on "n" and "j", and since energy = frequency, the frequency of both electrons will be the same. However, I do think that the wavelength will vary. As has been stated, the electron can be found in a spread out "cloud". If a measurement on the particles position is performed then at this point, its energy will be given by its kinetic energy (momentum squared over 2 times the mass) + the potential. If it is found in an area with higher potential (further from the nucleus), then the electron must give up some of its kinetic energy. Then, according to DeBroglie, $$\lambda=hp$$ and so its wavelength will be lower, despite it being in a given energy level. If you follow this through for another electron in the same level, then the two electrons will have the same frequency, but not necessarily the same wavelength. For a classical analogy, one could think about the velocity of the electron reducing as it goes into an area of higher potential, but using this as any more than an analogy will cause trouble.[/QUOTE]

Thanks for a very clear and helpful explantion. The one area I am still not clear on is whether or not in either case the term wavelength has any connotation of an integral harmonic relationship to its state or location or if that is completely a concept I should just scrap.

 

[DeShark]

Thanks for a very clear and helpful explantion. The one area I am still not clear on is whether or not in either case the term wavelength has any connotation of an integral harmonic relationship to its state or location or if that is completely a concept I should just scrap.

Honestly, I don't think the concept of wavelength really means all that much. It's merely the rate of change of the phase factor with respect to space. Frequency is the rate of change of the phase with respect to time. Phase is what causes interference which is one of, if not the most important, factors of quantum mechanics. The fact that material things can interfere is utterly bizarre to me. Energy and frequency are the same except for a conversion factor (planck's constant).

I'd scrap the concept which relates integral wavelengths coinciding with "orbits". It's interesting that the concept can be applied to the hydrogen atom to get the right spectrum, but it's not really good for anything else. Learn proper quantum mechanics, with all its state vectors and wave functions. This is the only formalism which has stood up to the test of time.

 

[Austin0]

Honestly, I don't think the concept of wavelength really means all that much. It's merely the rate of change of the phase factor with respect to space. Frequency is the rate of change of the phase with respect to time. Phase is what causes interference which is one of, if not the most important, factors of quantum mechanics. The fact that material things can interfere is utterly bizarre to me. Energy and frequency are the same except for a conversion factor (planck's constant).

I'd scrap the concept which relates integral wavelengths coinciding with "orbits". It's interesting that the concept can be applied to the hydrogen atom to get the right spectrum, but it's not really good for anything else. Learn proper quantum mechanics, with all its state vectors and wave functions. This is the only formalism which has stood up to the test of time.

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Thanks I understand, I think, the concepts of interference and absence of definite locality.. And also, that to gain any kind of deep or quantitative grasp of the dynamics would necessitate digging in and learning the terms and math of QM.
At the same time, I think it is more than merely interesting or coincidental that the harmonic conceptualization seems to conform to reality in the case of hydrogen.
My conceptualization of the earlier more, more localized version, is of a string without fixed ends but still displaying integral harmonics.
The current model , less localized , I see as a drumhead without fixed borders . Having much more complex harmonics but still on some level manifesting them.
My interest in the interference relationship was not just between the electrons but also between the electron and the nucleus. Having looked at the reference above , helpfully provided by jtbell, it appears to me that the electron probability domains display not only symetry, but an apparent symetric integral spacing relative to the nucleus.
This would seem to indicate the possiblity that the electron/proton, interference/phase relationship does have a spatial [wavelength] component, within limits of indefiniteness.
I understand that this may not be quantitatively useful with more complex atoms but it is conceptually very intrigueing
I do plan to extend my knowledge of the maths of QM and am appreciative of your input which has been both relevant and clear Thanks

 

[DeShark]

My conceptualization of the earlier more, more localized version, is of a string without fixed ends but still displaying integral harmonics.

This isn't a bad picture. There are huge similarities in the maths between solving the schroedinger equation and solving wave equations with boundary conditions (in fact the schroedinger equation is a wave equation). The similarities are so similar that advancements in the mathematics of one will almost certainly lead to benefits in the other. I had heard for example that the similarity between the diffusion equation and the schroedinger equation is so good that feynman's path integral approach provided new ways of doing statistical physics, since it mirrored in many ways the random walk. However, the complex coefficient in the schroedinger equation causes it to be different in many aspects, namely that the solutions may (and in some cases must) be complex functions. Thus the analogy between waves and harmonics and quantum mechanics is a useful tool, but not always sufficient to explain some of the weirdness. I can't recommend a better book for getting to grips with QM than feynman's lectures - third volume. It really is as easy as it gets in my opinion.

 

[Austin0]

[ There are huge similarities in the maths between solving the schroedinger equation and solving wave equations with boundary conditions (in fact the schroedinger equation is a wave equation). I can't recommend a better book for getting to grips with QM than feynman's lectures - third volume. It really is as easy as it gets in my opinion.[/QUOTE]
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Thanks for the tip on Feynman. I am sure it is a good one if his QM explication is even half as clear and straight forward as his coverage of SR.

 

[qatadah]

got a question:

A hydrogen atom has 1 electron, and according 2 Lorenz rule, it is supposed to accelerate so fast, having a very high speed, and that would make it leave the atom making H+. The chemistry teacher says its impossible, theoritically it is possible. But practically that doesn't happen.
Need explanation??
it should happen, shouldn't it??

 

[qatadah]

Hydrogen electron Q.

got a question:

A hydrogen atom has 1 electron, and according 2 Lorenz rule, it is supposed to accelerate so fast, having a very high speed, and that would make it leave the atom making H+. The chemistry teacher says its impossible, theoritically it is possible. But practically that doesn't happen.
Need explanation??
it should happen, shouldn't it??

 

[0xDEADBEEF]

Three points

Austin0:
1) The higher shells do have electron probabilities "inside" the lower shells.

2) I am not sure about the resonance condition, but you might want to look into Sommerfeld formalism. It does allow many more orbits for the electrons and still produces correct results. It's a semi classical treatment that is still in use for QM calculations today, but it is more complicated than simple resonance conditions.

qatadah:
3) The electron cannot and will not escape when it is bound to the nucleus. Your teacher is right. You claim that it is theoretically possible, but it is not, so show your math to prove it.