# Hydrogen fine structure

1. Mar 17, 2014

### bobie

1. The problem statement, all variables and given/known data
This is not homework, I am trying learn fine structure.
This site is reliable : http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html, can you tell me if the content of "hydrogen fine structure" is still valid or dated?.
Even if it were classical oudated model, I'd like to understand it. Just a few preliminary questions:

- the wavelength of transition 3s => 2p is given as 656.11 or (with reduced mass) 656.47 in the text, but in the picture it becomes 656.3, why that?
- the real values of fine structure are then 656.316 and 656.284?
- B field is indicated (at 2p) 0.4 T while in the third frame it becomes 0.3 T, which is right?

2. Relevant equations
- the right formula to find B is: μ0* qv /4πr2 ?, does it work both in mks and cgs?

Last edited: Mar 17, 2014
2. Mar 17, 2014

### Staff: Mentor

For the absolute values and at this level of precision, you have to account for relativistic effects and the refractive index of air, the difference could come from a different treatment of those.
Spectral line database

The fine-structure is known for several decades now and nothing changed.

The magnetic field is not a constant value everywhere, those calculated numbers are a rough estimate and cannot agree perfectly.
Formulas work independent of the unit system.

3. Mar 18, 2014

### bobie

I was referring to the article "Hydrogen fine structure" in this site http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html.
I would like to know if the data in there are a good approximation of what are considered today the real values.

-In particular if the formula they give in the 3rd and 4th frame to calculate the value of B
$B = \frac{\mu0*qv}{4πr^2}$
which gives 12.5 T for 1s level and 0.39 for 2p.

- Also I do not understand why (in the first frame) the value of the spin-orbit split is the same (i.e. 0.000045 eV) for the 2p => 1s transition ( at the top of the frame) and the 3s => 2p (at the bottom). Since B is 32 times greater and only the value of the spin magnetic moment is always the same (1/2*h/2π) I expected the delta value of the splits of the fine structure to be different.

I'd appreciate any help on these issues. Thanks

Last edited: Mar 18, 2014
4. Mar 18, 2014

### Staff: Mentor

All those Tesla values are not "real" field strengths. They are a classical approach to a quantum mechanical effect. They can lead to some interesting insights, but don't think something could be replaced by a magnetic field of that strength.

The split comes from the shift of 2p in both cases.

5. Mar 19, 2014

### bobie

Thanks for the direct link, mfb, I was going mad trying to post it properly. How did you do it ?
Sorry, but I do not understand the bolded bit, to what is something referring?

Can I get a good approximation of B dividing the energy of the split E (0.000045 eV) by $\mu$ since $\mu*B = E$ ?
B = 0.000045*4pi T ?

Last edited: Mar 19, 2014
6. Mar 19, 2014

### Staff: Mentor

Looking for a single value for a magnetic field strength is as meaningful as looking for a single value for the speed of cars (yes, as general as that). So, how fast are cars?
It just does not make sense to reduce this to a simple number. You can still say "a typical speed is 50 to 100km/h", this is equivalent to the values you see for B at those pages.

7. Mar 20, 2014

### bobie

My question was more modest, I probably should have quoted the article in my first post.
"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.

0.000045 eV = 7.2 *10-24 J
in another thread, here, I just learned that $\mu$ = ≈3.2*10-23 J/T

E =7.2*10-24 J, now E= mu*B, but E/ itex]\mu[/itex] it is not near 0.4
Where is my mistake?

Last edited: Mar 20, 2014
8. Mar 20, 2014

### Staff: Mentor

E=µB is valid if B is constant everywhere, but that is not the case.

9. Mar 20, 2014

### bobie

Thanks, mfb, what baffled me (and that's why I went on posting) is that maybe in QM things are different , but they state that it works all the same in Bohr's model, so they must have worked it out some way
.

Last edited: Mar 20, 2014
10. Mar 20, 2014

### Staff: Mentor

Bohr's model is wrong, even if it gives some right values for hydrogen-like atoms.

11. Mar 20, 2014

### Staff: Mentor

The magnetic moment of the electron is $\mu_\mathrm{e} = -9.284764 \times 10^{-24}\ \mathrm{J/T}$. If you add the additional g-factor, $g \approx 2$, you will get the right answer for the energy splitting.

12. Mar 21, 2014

### bobie

That value is Bohr magneton 9.274 multiplied by g/2: 1.00116
I have been told that :
where L is √3/2 $\hbar$

Last edited: Mar 21, 2014
13. Mar 21, 2014

### Staff: Mentor

Yes, I made a small mistake there with the g-factor. The magnetic moment to use to calculate the splitting is $\mu = g \mu_\mathrm{B} = -2 \mu_\mathrm{e}$.

If there is an L, it means you are considering orbital angular momentum, which is not what you are supposed to do here.

The idea is to find the equivalent uniform B field such that the two spin orientations of a free electron differ in energy exactly the same as the fine structure. Take the field to be aligned along +z. The energy of the electron is given by
$\hat{H} = - \mu \cdot \mathbf{B} = 2 \mu_\mathrm{e} B M_\mathrm{s}$
such that
$\Delta E = -2 \mu_\mathrm{e} B$

14. Mar 21, 2014

### bobie

In the other thread, dauto (in the post I quoted) corrected me when I considered the value of Le on the z-axis, which I supposed the direction perpendicular to the plane of the orbit,
the point here, is to find a justification, if not a derivation of B in the classical model, the way yhey did at hyperphysics. Do you know if it is possible to ask them?
If B = 0.4 then $\mu$ must be 1.8

I hope you could help understand the role of the spin angular momentum Le:
- how do we establish that it exists? is there an experiment in which it is possible to separate it from the magnetic moment of the spin $\mu$, isn't it possible that all shifts are caused by magnetic momentum?
- the angular momentum of the orbit is greater in upper levels Lo (n*$\hbar$) ,can you tell me if the angular or magnetic momenta have always the same value $1.0016*\hbar$? Also the magnetic moment is intrinsic?

Last edited: Mar 21, 2014
15. Mar 27, 2014

### Staff: Mentor

It took me a while to figure this one out! The value of 656.3 nm is what you get when you take the spectrum of hydrogen in air. The value of 656.47 nm is the wavelength in vacuum, and hence the one you get when starting from the energy levels of H and the Rydberg formula.

When taking the spectrum in air.

Neither! As I explained previously, the value of 0.4 T is what you obtain when you take the actual value of the fine structure splitting and try to figure out what field you you need such that a free electron would have such an energy difference between its parallel and anti-parallel states. The value of 0.3 T is what you get with the simple classical model to explain the origin of the fine structure. As mfb said, the actual magnetic field is not uniform.