Zeeman splitting of Balmer alpha lines under 1.5T

[ic1454]

Homework Statement

Suppose that you want measure the Zeeman splitting of the Balmer a-line on the transition $ 2^{2}S_{1/2} \to 3^{2}P_{1/2}$ in a magnetic field of 8 1.5 T. (a) what should be the minimum spectral resolution of the spectrograph so that you can resolve all the spectral features. (b) What is the minimum magnetic field B needed to resolve the Zeeman components with a Fabry-Perot interferometer (plate separation d1 cm, reflectivity of each plate is 99%)?

Relevant Equations

For the second problem $F=\frac{\pi\sqrt{R}}{1-R}$ for fineness

I am not understanding whether it is anomalous Zeeman or paschen back effect.please help.

 

[Charles Link]

See https://courses.physics.ucsd.edu/2016/Spring/physics4e/zeeman.pdf

What is the magnetic field strength? What you wrote looks like a misprint. 1.5 T looks right, not 81.5 T.

 

[techsingularity2042]

You first need to find spin orbit coupling energy and Zeeman splitting energy.
Anomalous Zeeman effect occurs when ΔE_Z ≪ ΔE_SO
and Paschen back effect when ΔE_Z ≥ ΔE_SO.

 

[Charles Link]

I want to add something that might help with the Fabry-Perot part. I think what they do with the resolution part is to consider the graph of $T$ vs. $\lambda$ and observe the peaks and their width, and assume this correlates similarly to a graph of $T$ vs. $\theta$ for a given wavelength. I need to study this further myself, but I think the approximation gets made that $\Delta \lambda \approx C \Delta \theta$ for some $C$.

This might help in the calculation where you try to determine if the interferometer can resolve the peaks between two different wavelengths with their ring patterns. This part IMO is missing from many of the explanations on this topic. It seems they think you are supposed to recognize it, but sometimes it helps to state the not-so-obvious, lest it become the oblivious.

Edit: additional comment or two: For the Fabry-Perot interferometer, it isn't exactly a linear mapping of $\lambda$ vs. $\theta$ but over short ranges it can be treated as reasonably linear, i.e. to first order. and I need to check this one item, but I think it is the case=the maxima occur for $\theta$ for a given $\lambda$ such that $m \lambda =2nd \cos{\theta}$, where $\theta$ is the angle in the material of the refracted ray. If the slab is made with glass on two sides with air in between, then things are simple with $\theta$ the same in the air gap as outside. You then also have $n=1$.

One other item worth mentioning is a very narrow laser line will show up as appearing to have a spread $\Delta \theta$ or $\Delta \lambda$ corresponding approximately to the spread $\Delta \lambda$ of one of the peaks of the transmission vs. wavelength curve. Note that we have a transmission function $T=T(\lambda, \theta)$. If we have a given $\lambda$, there will be peaks in $T$ vs. $\theta$ that have a spread of $\Delta \theta$ for that $\lambda$. If we alternatively look at a given angle, e.g. even normal incidence, there will be peaks in $T$ vs. $\lambda$ with a spread $\Delta \lambda$ for that $\theta$. This is more detail to what I mentioned above that there is basically a relation $\Delta \lambda \approx C \Delta \theta$. This transmission function also has a ridge where we can write $\lambda=\lambda(\theta)$. Give the ridge a constant value of $T$ and we can take a differential to find $\Delta \lambda /\Delta \theta$. I'm just trying to give a little mathematical rigor to something that is normally explained by handwaving. Draw a ridge for $T(\lambda, \theta )$ on a 3-D contour graph of $\lambda$ vs. $\theta$ and it should be apparent how the $\Delta \lambda$ and $\Delta \theta$ are related. There is a ridge in $T(\lambda, \theta)$ for each value of $m$. The ridge is in fact follows the curve $m \lambda=2nd \cos{\theta}$.

Edit: See also https://en.wikipedia.org/wiki/Fabry–Pérot_interferometer

 

[Charles Link]

In my college days, almost 50 years ago, we did look at the sodium doublet with a Michelson interferometer, which works on similar principles as the Fabry-Perot interferometer. Perhaps the biggest difference is the higher reflectivity of the Fabry-Perot, where the Michelson has close to 50% reflectivity, and thereby the bright rings from a spectral line are broad, rather than very narrow as with the high reflectivity $R$ of the Fabry-Perot.

The formula that is used for $T=T(\lambda, \theta)$ with the Fabry-Perot ( see the wiki link above) reminds me very much of the diffraction grating type spectrometer where we use a formula for $I=I(\lambda, \theta )$ in order to compute the resolution by determining the observed linewidth of a laser type spectral line. With the diffraction grating spectrometer we have $I=I_o \frac{\sin^2(N \phi/2)}{\sin^2(\phi/2) }$ where $\phi=(2 \pi/ \lambda) d \sin{\theta}$. It's a very similar calculation like that of above to show resolving power $\lambda/\Delta \lambda=N m$ for the diffraction grating spectrometer.

Relevant to the OP's problem of observing the Zeeman effect in hydrogen, I did observe the Zeeman splitting of the green line in mercury ( Hg) in a magnetic field into 9 components if I remember correctly, with a diffraction grating spectrometer in an upper level undergraduate laboratory spectroscopy course for which I was a teaching assistant.

I should mention in all cases, whether it is hydrogen, sodium, or mercury, it is a gas discharge lamp that is used to observe the spectral lines. For the Zeeman effect, the magnetic field is applied normally with a ring shaped permanent magnet with a gap about one inch across between the north and south poles. The arc lamp is positioned with the part that is being observed in between the pole faces of the magnet.