1. I know this site is reliable, HyperPhysics, if you know a better/more updated one, please let me know.
it is said:
656.11 nm for hydrogen, treating the nucleus as a fixed center. If you use the reduced mass, you get 656.47 nm for hydrogen and 656.29 nm for deuterium. The difference between the hydrogen and deuterium lines is about 0.2 nm and the splitting of each of them is about 0.016 nm, corresponding to an energy difference of about 0.000045 eV. This corresponds to an internal magnetic field on the electron of about 0.4 Tesla.
I have two preliminary questions
to what is referring "This", to the use of reduced mass or the the difference between H and D?
The value of 0.4 T is referred to B due to the orbit or the internal B of the electron due to the spin?

Is the usual formula to find the magnetic field B of a loop valid to calc B of H1? With Bohr's data I get 0.125 T, where do I go wrong?

Thanks

2.

3. Originally Posted by whizkid
I know this site is reliable, HyperPhysics, if you know a better/more updated one, please let me know.
it is said:
656.11 nm for hydrogen, treating the nucleus as a fixed center. If you use the reduced mass, you get 656.47 nm for hydrogen and 656.29 nm for deuterium. The difference between the hydrogen and deuterium lines is about 0.2 nm and the splitting of each of them is about 0.016 nm, corresponding to an energy difference of about 0.000045 eV. This corresponds to an internal magnetic field on the electron of about 0.4 Tesla.
I have two preliminary questions
to what is referring "This", to the use of reduced mass or the the difference between H and D?
It refers to the common energy difference of 0.000045eV (or, equivalently, the common 0.016nm splitting of lines).

The value of 0.4 T is referred to B due to the orbit or the internal B of the electron due to the spin?

Is the usual formula to find the magnetic field B of a loop valid to calc B of H1? With Bohr's data I get 0.125 T, where do I go wrong?

Thanks
What "usual formula" did you use? You should be carrying out a Zeeman-effect calculation. Is that what you did? If you carry out the most common version of a simplified "freshman physics" calculation of the spin-orbit interaction, you will get about 0.3T. I can't know where you went wrong unless you show your work.

4. Originally Posted by tk421
you will get about 0.3T. I can't know where you went wrong unless you show your work.
The electron falls from level 3s to 2 (is it s or p ?), what is the value of B we must calculate, the one at 3s, r = 9 (Br) or the one at 4 (Br)?
is .4 T the exact value, then?

5. Originally Posted by whizkid
Originally Posted by tk421
you will get about 0.3T. I can't know where you went wrong unless you show your work.
The electron falls from level 3s to 2 (is it s or p ?), what is the value of B we must calculate, the one at 3s, r = 9 (Br) or the one at 4 (Br)?
is .4 T the exact value, then?
You seem to be a bit confused -- there's no "falling" of electrons. The calculation is of the energy difference between two states, and the magnetic field that produces that energy difference.

To perform the "undergraduate textbook" simplified calculation (which gets you to about 80% of the value obtained from more detailed quantum-mechanical calculations), use a quasi-classical approach and compute the magnetic field of the electron as if it arises entirely from a circular orbital motion about the nucleus. Calculate the angular momentum of a 2p electron at 4x the Bohr radius. Plug all that into the equation for magnetic field, and you should get something like 0.3T or thereabouts. Now, mind you, that's all from memory, so I could be off on the numbers a bit here and there, but I'm fairly sure I got the big picture right.

6. See next post

7. Originally Posted by whizkid
Originally Posted by tk421
Plug all that into the equation for magnetic field, and you should get something like 0.3T .
Yes, I'm confused:
Yes, I see. It seems that you are attempting to compute a quantity whose meaning you do not understand. I'm not sure how to help you in that case. Perhaps you should simply start with re-visiting Bohr's calculations of the main spectral lines, and defer an examination of fine structure. But I'll soldier on.

In the above link, further down, there is the 'right formula' : mu0 * qv/4 pi r^2 or mu0 * eL/ 4 pi mr^3,
and it gives 0.3 T, how does this relate to the value in the quote 0.4 T?
As I said earlier, 0.3T is the result of a simplified calculation, and 0.4T is the result of a more comprehensive numerical one.

To get 0.4T requires abandonment of the Bohr model. Are you aware that the Bohr model is a "toy" model -- it represented a great leap over what preceded it, but it is very crude.

What confuses me is that neither the regular value 656.11 nor the reduced mass 656.47 appears in the picture
an there is 656.3, which, I suppose, becomes 656.316 and 656.284 nm
This is where we have a fundamental problem -- you don't appear to understand what it is that you are trying to calculate, so you are puzzled by things that aren't relevant to the calculation at all. So I find myself in the rather odd position of having to explain to you what the point of your endeavour is.

The simplest Bohr model doesn't predict the existence of the closely spaced lines under examination here. However, a small "upgrade" of the model does. That upgrade is to accommodate the orbital magnetic field. For a 2p electron (n=2), with corresponding radius equal to n^2 times the Bohr radius, you get 4 times the Bohr radius as an effective orbit radius. We'll then idealize the situation as that of a point charge orbiting circularly about a stationary nucleus at precisely that radius (notice all the approximations inherent in that description -- that's part of the reason the calculation will be inexact). A non-relativistic calculation of the magnetic field produced by that orbiting electron yields 0.3T. Notice that the only inputs to that calculation are the Bohr radius, electron rest mass and charge. For the purposes of the approximate calculation, it doesn't matter if you are using just the electron rest mass or the reduced mass -- as you yourself observed, the two mass values hardly differ (by only about 0.5%), thanks to the lopsided mass ratio of proton to electron. That's why we idealize the nucleus as stationary, and the electron orbit as circular.

The calculation then yields the approximate magnetic field generated by the orbital motion of the electron.

You then reconcile that value with the splitting of the lines (not the lines themselves). That splitting is, I thought, the whole point of the exercise (at least, that's what I gathered, based on the very title of your thread).

To calculate (or, rather, estimate) the splitting, you use the calculated magnetic field value. One energy level is augmented by an amount mu*B (where mu is the Bohr magneton, about 58u eV/T), and the other is diminished by that amount. The splitting you get is thus the difference between the two energy levels, or simply 2*mu*B. With our 0.3T estimate, that works out to an expected line splitting of about 0.000035eV, which is close to, but different from, the 0.000045eV actual value.

If you start instead with the experimentally observed 0.000045eV and deduce the magnetic field for our simplified model, you get 0.4T.

Hope that helps!

8. Originally Posted by tk421
....is 656.3, which, I suppose, becomes 656.316 and 656.284 nm
Thanks for your great post, tk.
I know the classical Bohr model, like in the previous thread, I am trying to compare it to the new QM model .

I wondered where 656.3 came from, it comes from the QM formula you say. I suppose that formula is too complicated, that's why you don't mention it.

As to my problem, I know that spin angular and magnetic momentum are both 1/2 h/2pi , that is
mu = h/4pi = 0.07958 h (that should be corrected by g-factor/2 = 1,00165) = 5.79*10^-5 eV/T, and that
mu*B is the delta energy E = 0.0000455eV = 1.1*10^10h =7.29 *10^-24 J
(I gave the value of the wavelengths : 656 .284/316 (delta 0.032nm), energy E= 1.899 177/085 eV (0.000092)

I knew I can find the value of B = E*4pi = 1.38254*10^11 h= 0.0000455*4pi B = 0.0005718 eV

- but, how do you get 0.4T ? I get 0.786 T : 1 T = 0.000727 eV ( 0.0005718 / 0.000727= 0786)
1.7588*10^11 h (1.38254/ 1.7588 = 0.786)

Another thing I do not understand in the link:
the first scheme in the link refers to transition 2p => 1s and the delta energy is the same : 0.000045 eV

as we know that the value of mu does not change in the various orbits but the value of B varies ( I suppose by a factor of n^5) the energy of the split should be 32 (2^5) times greater/ smaller.
- why the value stays the same both in transition 3s => 2p, and 2p => 1s?

Thanks for your kindness and patience

Edit: I was told that here http://physics.nist.gov/PhysRefData/ASD/lines_form.html one can get the exact results, are you able to find the values of the fine structure?

9. Originally Posted by tk421
As I said earlier, 0.3T is the result of a simplified calculation, and 0.4T is the result of a more comprehensive numerical one.
the formula I quoted
gives about 0.4 for 2p and 12.5 for 1s

10. Originally Posted by whizkid
I wondered where 656.3 came from, it comes from the QM formula you say. I suppose that formula is too complicated, that's why you don't mention it.
The real reason is that, as far as I understood what your question was (based mainly on the title you gave it), it was irrelevant.

As to my problem, I know that spin angular and magnetic momentum are both 1/2 h/2pi , that is...
It would help a great deal if you were more specific in stating what your question actually is, because most of what you posted above has almost nothing at all to do with what I thought your question was. So let me be specific about what I was assuming in my answer:

1) I assumed from the thread title that you wanted to understand the origin of the fine structure of hydrogen's spectral lines. That, in turn, led me to assume that you at least understood the primary energy transition that gives rise to the principal line that is being split in the first place. I assumed that you had read the Hyperphysics entry on the fine structure, since you quoted from it.
.
2) I assumed from the phrasing in your first post that you were confused about the calculation in the Hyperphysics pages that alluded to both 0.3T and 0.4T as the magnitude of the internal magnetic field that causes the fine splitting.

3) I assumed that you understand the basic notion of what an approximation is, and why it makes no sense to worry about a 0.1% error in a quantity when you are performing a 30%-accurate calculation.

As I see that many of my assumptions were incorrect, I will ask you to be explicitly clear in re-stating your question, along with a summary of what background assumptions and knowledge you are starting from. Then I'll be better able to tailor my answers.

Pending such clarification, I'll take a stab at adding some commentary to what I've posted already.

The Bohr model was the first to provide a quantitative prediction of the spectral lines emitted by excited hydrogen gas. Spectroscopists had long ago catalogued these lines, and even fitted them to a nice mathematical formula, but it was the Bohr model that provided a deeper explanation of their origin; a particular triumph was Bohr's ability to explain where a constant the spectroscopists used in their fitting (the Rydberg constant) came from. The famous red line, which arises from the transition of a 3s electron to the 2p orbit, is predicted to be about 656.11nm if you use the simple Bohr model and the rest mass of the electron. If you use the reduced mass of the electron, you predict a slightly different value, 656.47nm. What is observed is about 656.3nm. That was deemed close enough agreement to validate Bohr's basic model, and the quantum theory of the atom was off to the races.

As close as the agreement is, the existence of a discrepancy is always a problem for physicists. And there is more trouble in store: There isn't a single line in that transition, there are two very closely spaced ones. That's the fine splitting that I've been addressing, not the principal (averaged) 656.x nm one. And that's why I ignored it, not because it was "too complex," but because it was irrelevant to the question implicit in the thread title.

The splitting of the lines is due to the splitting of the 2p energy levels which, in turn, is due mainly to the orbital magnetic field. I've already carried out that calculation, so I won't repeat it. Within the framework of the assumptions, you get a 0.3T orbital magnetic field. That, in turn, implies an energy splitting of 35u eV. The observed value is 45u eV. The discrepancy is from the neglect of certain factors in the simplified calculation. The simplified calculation is merely to explain why there is splitting, and to give a rough estimate of the expected amount of splitting.

As I said earlier, the 0.4T number comes from running the calculation backward. You start with the measured line splitting and calculate what magnetic flux density would give rise to that observed splitting (again, if that were the whole story). That's where the 0.4T number comes from.

All of the above is focused on the principal energy transition. Now, if you want to go higher up in transition orders, that's fine, but why jump around haphazardly? Get the simplest cases down first, and then move on as your knowledge improves.

ETA: I should say that performing more accurate calculations to close the gap with experiment quickly gets very, very involved. The reason that the Hyperphysics article only goes as far as it does is simply that each digit of refinement will cost you an order-of-magnitude increase in complexity.

11. Originally Posted by tk421
Pending such clarification, I'll take a stab at adding some commentary to what I've posted already
As I said, I studied the Bohr model thoroughly, I only want to check what are the differences with the QM model.

To avoid further misunderstanding ,I'll'ask specific questions so you (or exchemist or anyone) can give me simple answers, and I'll learn what I am looking for:

1- can you show me how you get B (2p) = 0.4 T working backward from the spectral lines i.e. 656.3 +/- 0.016 nm, 1.88913 +/- 0.0000455 eV?

2- can you show me what value you get for B (1s) in the transition 2p=> 1s, lyman series,(in the same link-article at hyperphysics at the top in the same frame) with energy 10.2 eV +/- 0.000045 eV

3- the spinning electron has angular momentum L(e) =h/4pi , magnetic moment mu(e) = h/4pi, right?, has it got a magnetic field B(e)? If not, how come?

Thanks a lot for your patience.

12. Originally Posted by whizkid
Originally Posted by tk421
Pending such clarification, I'll take a stab at adding some commentary to what I've posted already
As I said, I studied the Bohr model thoroughly, I only want to check what are the differences with the QM model.

To avoid further misunderstanding ,I'll'ask specific questions so you (or exchemist or anyone) can give me simple answers, and I'll learn what I am looking for:

1- can you show me how you get B (2p) = 0.4 T working backward from the spectral lines i.e. 656.3 +/- 0.016 nm, 1.88913 +/- 0.0000455 eV?
The easiest is simply to take the ratio 45ueV/35ueV and multiply by 0.3T. That gets you to 0.4T after you round to a single sig fig. As I said, you can't get 0.4T from any simple calculation, other than scaling up the 0.3T value you compute from the quasiclassical model.

2- can you show me what value you get for B (1s) in the transition 2p=> 1s, lyman series,(in the same link-article at hyperphysics at the top in the same frame) with energy 10.2 eV +/- 0.000045 eV
Following the same approach as before, the simplified model says that the B-field depends inversely on the fifth power of n. Based on that simple view, the field should be about 10T (!). I reserve the right to amend this, as I've just scribbled the derivation on a dinner napkin.

3- the spinning electron has angular momentum L(e) =h/4pi , magnetic moment mu(e) = h/4pi, right?, has it got a magnetic field B(e)? If not, how come?
Absolutely it generates a magnetic field. In fact, permanent magnets owe their properties overwhelmingly to spin; the orbital magnetic field component contributes negligibly.

Thanks a lot for your patience.
It would help a great deal if you were to say a bit about what you're after. Is it just practice with physics, or do you want to learn how to calculate the spectral lines to high precision? The answer to your question will determine how best to answer you. For example, I worry that you are too hung up on approximation error, which shouldn't be a big factor if you're starting out in physics. If your goal is, in fact, quantitatively accurate computation, then we need to take a different tack altogether -- band-aiding the Bohr model just isn't the right way to go about it. We'll have to use Dirac's formulation, and getting up to speed there will not happen in a forum.

13. Originally Posted by tk421
I worry that you are too hung up on approximation error, which shouldn't
I wonder where you got that impression from, tk, I am just using the Bohr formulas I know.
Again, I studied the Bohr model in depth, there are some things I do not understand, yet. I'd like to understand them, and then compare (just roughly) with QM. Direct link

Now, I never came across the B-field generated by the spin (Be), can you please tell me more about it and how it interacts with the widely-known B field generated by the orbiting (Bo) we are discussing?

P.S I am afraid I still do not understand your reply to first question. We start from the only experimental data we have : an emission of 1.889 eV with a split of DeltaE = 0.000045 eV, we know that DeltaE is the product of mu(e) (=1/4pi) by B(o) , shouldn't we divide DeltaE (0.000045) by mu (0.000045/(h/4pi)?
if :

14. Originally Posted by whizkid
Originally Posted by tk421
I worry that you are too hung up on approximation error, which shouldn't
I wonder where you got that impression from, tk, I am just using the Bohr formulas I know.
Yes, I understand that. The reason for my concern is that you might have expectations that the model simply cannot meet. We'll just have to see if my worries are without foundation (I hope so).

Again, I studied the Bohr model in depth, there are some things i do not understand, yet. I'd like to understand them, and then compare (just riughly) with QM.
Ok.

Now, I never came across the B-field generated by the spin, can you please tell me more about it and how it interacts with the widely-known B field generated by the orbiting?
This property was not known to Bohr or anyone else in 1913, so it doesn't appear in his model. For a reasonably good summary, I recommend the Wikipedia article: Electron magnetic dipole moment - Wikipedia, the free encyclopedia

Qualitatively, we can pretend that the electron is a spinning ball of charge. That constitutes a current, which generates a magnetic field. If you base a computation on a classical model, you get an answer that is off by almost exactly two.

P.S I am afraid I still do not understand your reply to first question. We start from the only experimental data we have : an emission of 1.889 eV with a split of DeltaE = 0.000045 eV, we know that DeltaE is the product of mu(e) =1/4pi by B(orbit) , shouldn't we divide Delta by h/4pi? if
Getting 0.4T from 0.3T is much, much simpler (and much, much less intellectually profound) than you think. The energy difference is proportional to B. We computed a 0.3T B first from a simple model, and predicted a corresponding energy shift. When we actually measure the energy shift, we find that it is larger than computed from our simple model, by a factor very close to 4/3. So, multiply the B we computed by that 4/3 factor.

15. Originally Posted by tk421
I know that article very well, but it refers to the magnetic moment of e mu(e), which interacts with magnetic field of the orbit B(o)

I was asking about a magnetic field B generated by the spin like the one generated by the orbiting around the nucleus we are discussing in this thread:
- The orbiting (around the nucleus) generates an angular momentum L(o) , a magnetic field B(o), and a magnetic moment mu(o), is that right?, now
- The spinning (around itself) generates a magnetic moment mu(e) an angular momentum L(e) , is there a B (e)?

16. Originally Posted by whizkid
Originally Posted by tk421
I know that article very well, but it refers to the magnetic moment of e mu(e), which interacts with magnetic field of the orbit B(o)

I was asking about a magnetic field B generated by the spin like the one generated by the orbiting around the nucleus we are discussing in this thread:
- The orbiting (around the nucleus) generates an angular momentum L(o) , a magnetic field B(o), and a magnetic moment mu(o), is that right?, now
- The spinning (around itself) generates a magnetic moment mu(e) an angular momentum L(e) , is there a B (e)?
This is one of the reasons I have been asking for what your goal and background are. In the absence of that information, I have to assume something. As a default, I'm targeting my explanations at the level of an undergraduate. In this case, that target student would know the relationship between magnetic moment and field, so pointing to information about the moment is all I need to do. Since, again, my assumption is wrong, please help me fix it so that we don't waste each other's time. I think it's commendable that you are willing to learn this material; we can greatly increase the utility of my explanations if I have a better idea of what you're after, and what you already know.

So, let's start here: Magnetic dipole - Wikipedia, the free encyclopedia

Now, once again, you must be very careful about mixing classical and quantum ideas. Nevertheless, we can get rough order-of-magnitude answers without working too hard, and sometimes (as with the simple upgrade of the Bohr model) we can do better than that.

 Bookmarks
##### Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement