How to Find the "Net Change Ring Area Ratio" for the Zeeman Effect

[Athenian]

Homework Statement

Solve for $\frac{\delta}{\Delta}$ for the respective "orders" and "rings". Definitions would be shortly explained below.

Relevant Equations

$\delta$ = net change in area between two neighboring rings.
$\Delta$ = net change in area between two neighboring orders.

For the image below, each order has 2 rings (with an "invisible" central ring as well). Thus, there are 3 orders and 6 rings for the below image (the center smudged circle does not count).

To find $\delta$ for the 1st order, all I need to do is to square the diameter of the 2nd ring and subtract it to the square of the diameter of the first ring.
$$\delta_{1st \; order} = {d^2}_{2nd \; ring} - {d^2}_{1st \; ring}$$

To find $\Delta$, I can use the below equation.
$$\Delta_{Between \; orders \; 1 \; and \; 2} = {d^2}_{central \; ring \; in \; order \; 2} - {d^2}_{central \; ring \; in \; order \; 2}$$

Note that the diameter of the central ring can be found by taking the average diameter between ring 1 and ring 2 for any given order (e.g. the 1st order).

To my best understanding, as I have 6 rings in total, I am supposed to find 6 "versions" of $\frac{\delta}{\Delta}$. But, I having a hard time approaching this correctly and acquiring all 6 numerical values. Any ideas on how I should approach the problem to acquire all 6 values? What "sets" of rings/orders should go together for the $\delta/\Delta$ ratio?

While I am not certain would this help, the aforementioned ratio is called the "areal ratio".

Thank you for reading through this question!

 

[Charles Link]

Please supply some additional info. : Is this from a Michelson or Fabry-Perot interferometer? I'm puzzled by the size of the central spot=perhaps there is a reason why it has a finite diameter, but I haven't figured that out yet. Edit: It looks like some kind of crystal scattering experiment, perhaps a powder, where the source contains two closely spaced (Zeeman split) spectral lines.

In addition was the path difference $2d$ for the center rays set to zero, or is it some finite but unknown number?

Presumably it is a Zeeman splitting of a single line. What is the element that was used in the experiment?

 

[Athenian]

Thank you, @Charles Link, for your response. Below are my answers to your questions. Hopefully, it helps.

Is this from a Michelson or Fabry-Perot interferometer?

This is from the Fabry-Perot interferometer.

I'm puzzled by the size of the central spot=perhaps there is a reason why it has a finite diameter

Before the current to the coil of the magnet was provided (i.e. I = 0), the rings did not split and the central "spot" did not "exist" or was not visible either. However, as the current increased, the splitting began to occur for the rings for this longitudinal (unpolarized) Zeeman effect. Of course, the central spot became gradually visible too. To view the splitting effect as the current increases, refer to this Google drive link here to preview the images.

where the source contains two closely spaced (Zeeman split) spectral lines.

Usually, there ought to be three lines per order. After rotating the optical axis (I believe), I got (as shown in the image) the longitudinal Zeeman effect (which contains 2 rings per order).

In addition was the path difference 2d for the center rays set to zero, or is it some finite but unknown number?

To find the areal ratio ($\delta/\Delta$), I did not bother to find the diameter of the central "spot" as it just keeps getting larger when the current increases. As I am not entirely sure what you mean by the "path difference 2d from the center rays set to zero", I am also not certain did I properly answer your question here either.

Presumably it is a Zeeman splitting of a single line. What is the element that was used in the experiment?

The above image is for a red cadmium line where $\lambda \approx 644$ nm.

If you need more information, please do let me know. Thank you!

 

[Charles Link]

See https://people.stfx.ca/cadams/physics475/labs/zeeman/QueenMaryUnivLondon_Zeeman effect.pdf
Suggest you google a couple of articles on the topic, and then let us know if there are parts that we can help you with.

 

[Athenian]

Thank you for the response, @Charles Link. Ironically, I have read the same article you sent here. A good article that relatively encompasses the scope of my question can be found here too: https://www.nikhef.nl/~h73/kn1c/praktikum/phywe/LEP/Experim/5_1_10.pdf.

To put it shortly, I am supposed to find a $\delta/\Delta$ vs. $B$ graph. While I am able to find the magnetic field for a given current, I am trying to figure out how many areal ratios I ought to have for the figure.

For example, in a PDF provided by PHYWE (which I will attach here as it is made available for free online), they found the following ring diameters and then got corresponding ($\delta/\Delta$) as shown below:

Graph for the above:

What I find odd is how does the paper find numerous $B$'s for 1 image on the Zeeman effect. In addition, how does the author find 6 $\delta/\Delta$ values for the 6 corresponding rings?

 

[Charles Link]

They did the experiment 6 separate times, each with a different coil current on the electromagnet to have 6 separate magnetic field $B$ values.
I still haven't figured out how to analyze the center spot, but it gets a number between two integers $n_o$, and then the first maximum has order $n_1$ as the first integer less than that.
You will get an average $\delta$ and $\Delta$ for each $B$. These you use to make the line graph of $\delta / \Delta$ vs. $B$.

I think the analysis of the central spot in detail, (perhaps more than you need), for the Fabry-Perot etalon is semi-complicated because of the multiple reflections. For a Michelson interferometer the intensity pattern is more readily computed. Here I think all you need is to figure out the maxima starting with $n_1$.

I believe there is a typo on the bottom of page 5 on the left side of the page: It should read $\delta_{a,b}^p=r_{p,a}^2-r_{p,b}^2$.

The $\delta$ and $\Delta$ are readily computed for a set of rings for a given $B$.

 

[Athenian]

They did the experiment 6 separate times, each with a different coil current on the electromagnet to have 6 separate magnetic field B values.

For my case, I did the experiment with 9 different coil currents going from I = 0 A to I = 8 A (increasing at an increment of 1 A). Therefore, I ought to have 9 data points for my line graph, correct?

I still haven't figured out how to analyze the center spot, but it gets a number between two integers $n_0$, and then the first maximum has order $n_1$ as the first integer less than that.

From what I gather, my instructor only requests that I calculate the "delta ratio" values for the three orders (which doesn't include the center spot). Therefore, perhaps analyzing the center spot is unnecessary unless it is for some reason needed to find the "delta ratio" of the three orders.

(Edit):

I think the analysis of the central spot in detail, (perhaps more than you need), for the Fabry-Perot etalon is semi-complicated because of the multiple reflections. For a Michelson interferometer the intensity pattern is more readily computed. Here I think all you need is to figure out the maxima starting with $n_1$.

I just noticed this portion of the edited post. Looking over at the lab activity content, I do agree with your viewpoint. Thank you for the clarification on this front. That said, the below questions still remain.

You will get an average δ and Δ for each B. These you use to make the line graph of δ/Δ vs. B.

Allow me to first assume an arbitrary value for B (e.g. B = 800 mT). Thus, to attain the average $\delta$ and $\Delta$ for each $B$, I need to first find the individual components to the "delta ratios" before calculating the average number. Below is my approach.

*Note that the equational definitions for $\delta$ and $\Delta$ are defined in the relevant equation section.

Does the above thought process look fine? Particularly, I am concern about the last $\delta/\Delta$ I wrote down as finding the net change in area between orders 3 and 1 just seems really off. Usually, it ought to be finding the net change in area between orders 4 and 3. However, as order 4 does not exist, this is proving to be a bit problematic.

Note that the above equation applies to the below image.

How is my thought process coming together? Are there any errors?

Thank you for your help!

 

[Charles Link]

For the last $\Delta$, use the average of the previous two $\Delta$'s. (The $\Delta$ 's for a given $B$ should all be nearly the same). Otherwise, it looks very good.

 

[Charles Link]

Be sure and see the complete post 6 above, with the typo on page 5 of the "link".
Meanwhile, I don't know that $I=0$ will give a data point, so you might have 8 data points.

 

[Athenian]

For the last $\Delta$, use the average of the previous two $\Delta$'s. (The $\Delta$ 's for a given $B$ should all be nearly the same). Otherwise, it looks very good.

Thank you for the clarification! So, I just need to evaluate the three equations I had earlier (with the last equation having the $\Delta$ change you mentioned) then and find the average number. The calculation process should be relatively straightforward. I'll give it a go at it.

Be sure and see the complete post 6 above, with the typo on page 5 of the "link".
Meanwhile, I don't know that $I=0$ will give a data point, so you might have 8 data points.

Thank you for pointing this out! I'll be sure to keep this in mind.

 

[Athenian]

@Charles Link, here's the update on the ongoing process and I have run into some (hopefully) simple questions that are eluding me. But, first, some background data.

Below are some of the data I acquired:

Order 1:	Order 2:	Order 3:
Ring 1 = 40 units	Ring 1 = 70 units	Ring 1 = 90 units
Ring 2 = 50 units	Ring 2 = 76 units	Ring 2 = 96 units
Central Ring = 45 units	Central Ring = 73 units	Central Ring = 93 units

*Central ring is obtained by "(Ring 1 + Ring 2)/2" (i.e. taking the average).With the above data, I get the below values:

$\delta_{Ring \, 1, \; Ring \, 2}^{Order \, 1} = {50}^2 - {40}^2 = 900$	$\delta_{Ring \, 2, \; Ring \, 2}^{Order \, 1} ={76}^2 - {73}^2 = 876$	$\delta_{Ring \, 3, \; Ring \, 2}^{Order \, 1} = 1116$
$\Delta_{Central \; Ring}^{Order \, 2, \; Order \, 1} = {73}^2 - {45}^2 = 3304$	$\Delta_{Central \; Ring}^{Order \, 3, \; Order \, 2} = {93}^2 - {73}^2 = 3320$	$(\Delta_{Central \; Ring}^{Order \, 2, \; Order \, 1} + \Delta_{Central \; Ring}^{Order \, 3, \; Order \, 2})/2 = (3304+3320)/2 = 3312$
$\frac{\delta}{\Delta} = 0.272$	$\frac{\delta}{\Delta} = 0.264$	$\frac{\delta}{\Delta} = 0.337$

As seen with the last order, the numerical value of $\delta/\Delta$ is much (or, at least, comparatively) higher than the former two. Is this fine? Sometimes, with other data sets, I get a 0.1 difference (e.g. 0.2 to 0.3). Interestingly enough, though, I did find that rather than finding the average of the two $\Delta$'s in the third order, dividing the sum of two $\Delta$'s by $1.5$ made my numerical value much closer to the former two ratio values. Then again, I do not believe this means anything "special".

In short, I was wondering should I be concerned about the blown-up value for the last "delta ratio" due to taking the average of $\Delta$'s for the 3rd order? If I should be concerned about the result, how should I go about dealing with it?

Once again, thank you so much for all your help thus far.

 

[Charles Link]

One typo in middle: 76^2-70^2=876. It should be a 70, and not 73.
This looks reasonably good. The sig figs are minimal in the measurement of the radii. Perhaps you could try to get another sig fig. If that was a 90.5 and a 95.5, it would give you approximately 900 for the 3rd $\delta$. (The problem I think is the $\delta$, and not the $\Delta$).

 

[Charles Link]

One suggestion: Since $r_a^2-r_b^2=(r_a-r_b)(r_a+r_b)$, you really need an accurate measurement of $r_a-r_b$. Suggest you measure this separately as the width of the tracks, rather than trying to reference $r_a$ and $r_b$ all the way back to the origin. (e.g. to measure something one inch wide, just measure it, and don't use a reference point that is 100 feet away).

With this change, I think you might find your data set to be much more accurate.

I processed the data from your image above, and I get somewhat different ratios of numbers:
For the $r_a-r_b$ I get 5, 4, and 3.5, while for the $r$ to the center of the order I get 43,64, and 83. This really doesn't seem to offer much improvement. I have to wonder if they made some approximations in their write-up. I need further study of this.

Edit: I reviewed their calculations, and I did some of my own. They did use the small angle approximation in a number of places, but that should work reasonably well. One other comment is their derivation ignored refraction, which has no effect on their results, of the ratio $\delta/\Delta$, but to be more exact, the constructive interference condition is $m \lambda=2nd \cos{\theta_m}$, where $\lambda$ is the wavelength in air, $n$ is the index of refraction, and $\theta_m$ is the angle in the material. In any case, the results you get with their method should be reasonably accurate.

 

[Athenian]

I just got on Physics Forums and noticed your new post. Thank you so much for the informative explanation here. It is incredibly helpful. Inputting all the necessary data onto an excel sheet, the average $\delta/\Delta$ is looking great and within expectations.

On that note, I do have a quick question. Unlike the previous Zeeman effect image I showed last time, I have an image of a Zeeman effect that possesses only 2 orders (the 3rd order is too blurred and ultimately unreliable). Below is the image.

When there were 3 orders, I could find the last $\Delta$ by taking the average of the former two. However, with 2 orders, how would I be able to find the latter (i.e. 2nd) $\Delta$ for this given case?

Once again, thank you for all your help thus far. I sincerely appreciate it.

 

[Charles Link]

The $\delta$ is supposed to be constant for the entire data set, and the $\Delta$ is also supposed to be constant for the entire data set, (i.e. for a given $B$). If you just have one $\Delta$, use it in both places for that data set.

 

[Athenian]

The $\delta$ is supposed to be constant for the entire data set, and the $\Delta$ is also supposed to be constant for the entire data set, (i.e. for a given $B$). If you just have one $\Delta$, use it in both places for that data set.

Thank you for the fast response. If you do not mind me clarifying, for a given $B$, I should be making the below calculations, correct?

*Note the the red coloring is just there to indicate they are the same quantities.

 

[Charles Link]

Just to summarize results in terms of wavelength, when you have the spectral line split in two, the result is $\frac{\Delta \lambda}{\lambda}=(\frac{\delta}{\Delta })(\frac{\lambda}{2 \mu t} )$. (Edit: Made a correction here. The factor that I added, $\frac{\lambda}{2 \mu t} \approx \frac{1}{5000}$ ). The $\Delta \lambda$, and thereby $\frac{\Delta \lambda}{\lambda}=(\frac{\delta}{\Delta})(\frac{\lambda}{2 \mu t} )$ is proportional to the magnetic field strength $B$.

 

[Athenian]

Great, this helps a lot in reinforcing my understanding on the calculations I have been doing thus far. Once again, you have been an amazing help and thank you so much for taking the time to assist me through the physics content. On that note, I'll get back to calculating and see what results I acquire!

 

[Charles Link]

One more thing worth noting: The Zeeman effect is the splitting of the observed atomic spectral lines of some atoms (with an arc lamp of that element) when a magnetic field is applied. The splitting can be observed with a Fabry-Perot etalon, but more commonly, this splitting is observed in the spectrum of intensity vs. wavelength that is measured with a diffraction grating spectrometer. On the graph, it is called a spectral line, because that's what it looks like=a spike in intensity at one specific wavelength. (also from photographs taken in the focal plane of the spectrometer, it shows up as a line on the photo, in the shape of the entrance slit of the spectrometer). When a magnetic field is applied, the spike in the graph becomes two or more closely spaced spikes or lines. If I remember correctly, the green line of mercury splits into 9 components when a magnetic field is applied.

 

[Charles Link]

Please see the correction in post 17. One other item on interest is that if the etalon is about one millimeter in thickness, the result is that we are working with interference orders $m$ of approximately 5000 for the first ring, 4999 for the second ring, 4998 for the 3rd ring, etc.

In computing the Bohr magneton, I get $\mu_B=(\frac{hc}{\lambda})(\frac{\lambda}{2 \mu t})(\frac{\delta}{\Delta})(\frac{1}{2B})$. Further computation needs the index of refraction $\mu$ of the etalon (perhaps it is air in the gap), and the thickness of the etalon. In any case, this appears to be in the right ballpark. The write-up in the "link" appears to have an algebra error with a factor of 2, but I could be mistaken. For (13) in their write-up, I think the 2 belongs in the denominator. It appears to be a typo that they did a couple of times, because when I did the arithmetic using their numbers, and putting the 2 in the denominator, the Bohr magneton computed correctly.

 

[Athenian]

Thank you for the additional clarification as well as noting some of the algebra errors in the given link! I'll look over the content again. Also, for the Bohr magneton equation you acquired, I am assuming that is for the anomalous Zeeman effect, correct?

 

[Charles Link]

Please see any additions I may have added to post 20. Cadmium has the normal Zeeman effect, where the effect is due to the orbital angular momentum, rather than the spin. The paper has that calculation computed correctly, other than what I think is simply a typo. (see post 20).(When computing the anomalous case, $\Delta E=g_s \mu_B m_s B$, with $g_s=2$. For the normal case, they (correctly) used $g_L= 1$ and $(\Delta) m_L=1$,so that $\Delta E=\mu_B B$, for (12) of their paper).

 

[Charles Link]

Per post 20, to compute the Bohr magneton $\mu_B$ from the data, the values of the thickness $t$ and the index of refraction $\mu$ of the etalon substrate (perhaps air) are needed. Did they give you values for those? I look forward to seeing how your data comes together. :)

 

[Athenian]

the values of the thickness $t$ and the index of refraction $μ$ of the etalon substrate (perhaps air) are needed.

Yes, you are correct. The value of thickness $t$ is $3$ mm and the index of refraction $\mu$ would be $1.456$ for a normal Zeeman effect with a red cadmium line (i.e. $\lambda = 643.85$ nm). However, for an anomalous Zeeman effect, the index of refraction changes to $1.452$ for the case of a green cadmium line (i.e. $\lambda = 508.59$ nm). This is what I was able to gather via a linked PDF down below.

With that noted, for the case of a normal Zeeman effect where
$$\mu_B = \frac{hc}{2\mu t B} \bigg(\frac{\delta}{\Delta} \bigg)$$
I can use the value $1.456$ for $\mu$.

For the case of the anomalous Zeeman effect where

$\mu_B=(\frac{hc}{\lambda})(\frac{\lambda}{2 \mu t})(\frac{\delta}{\Delta})(\frac{1}{2B})$

which simplifies to the below equation
$$\mu_B = \frac{hc}{4\mu t B} \bigg(\frac{\delta}{\Delta} \bigg)$$
I can use the refractive index of $1.452$ instead to find $\mu_B$.

Does this all sound correct?

Note that to find the above values, I used the below (expanded?) PDF that I shared last time around.
Link: https://web.phys.ntu.edu.tw/asc/FunPhysExp/ModernPhys/exp/ZeemanEffect.pdf

 

[Athenian]

On second thought, as mentioned once, I was creating a $\delta/\Delta$ versus $B$ graph. In other words, for slope, I get:
$$m = \frac{y}{x} = \frac{\delta}{\Delta B}$$

Therefore, to find the value of $\mu_B$ from a graph of a normal Zeeman effect, I can get the below equation.
$$\mu_B = \frac{hc}{2\mu t B} \bigg(\frac{\delta}{\Delta}\bigg) = \frac{hc}{2\mu t } \cdot m$$

The above is the "graphical" approach to finding $\mu_B$ that I just figured out.

Of course, for the anomalous Zeeman effect, I get the below equation to find to $\mu_B$.
$$\mu_B = \frac{hc}{4\mu t B} \bigg(\frac{\delta}{\Delta}\bigg) = \frac{hc}{4\mu t } \cdot m$$

 

[Charles Link]

Very good. It looks correct. The anomalous case and the exact factor to use in the denominator looks a little complicated=I will need to study that paper further, but I think you might have it right. Very good so far. Keep up the good work. :)

Try plugging in a couple of values if you have them, and see what you get for $\mu_B$. It helps to have the slope of the graph to get the average, but a single value of $\frac{ \delta}{\Delta }$ and $B$ will suffice.

Note: In this case, the graph passes through the origin, so the slope is simply the best straight line fit through the points including the origin=in principle the slope is simply getting an average value of $\frac{\delta}{\Delta}/B$.

 

[Athenian]

Thank you for the words of advice! Because of your kind help, I was able to successfully complete the provided lab exercise and found the process to be quite entertaining as well. I learned a lot from this exchange of dialogue and you were an incredible blessing to help walk me through the challenging subject matter. Once again, thank you so much and I look forward to learning more on this topic in the future! :)

 

[Charles Link]

Very glad to hear. :) I would be interested in seeing a couple more results, like the value(s) you got for the Bohr magneton, $\mu_B$, and even a couple of the values for $\delta$, $\Delta$, and $B$ if you have time to share them. Keep up the good work.

Meanwhile, for a couple of simple derivations:

For the same wavelength at adjacent maxima ,$m \lambda=2nd \cos{\theta_m}$, and $(m-1) \lambda=2nd \cos{\theta_{m-1}}$ gives by subtracting that $\lambda=2nd (\cos{\theta_m}-\cos{\theta_{m-1}}) \approx 2nd ((1-\theta_m^2/2)-(1-\theta_{m-1}^2/2)) \rightarrow \Delta (\theta^2 )=\frac{\lambda}{nd}$. Seeing that $\theta$ is proportional to $r$, (and $\theta^2$ is proportional to $r^2$), this gives the parameter which they call $\Delta=C \, \Delta (\theta)^2=C (\frac{\lambda}{nd})$ .

For two closely spaced wavelengths at the same maximum, $m \lambda_a=2nd \cos{\theta_{ma}}$
and $m \lambda_b=2nd \cos{\theta_{mb}}$, so that $m \Delta \lambda=2nd (\cos{\theta_{ma}}-\cos{\theta_{mb}})=nd \Delta (\theta')^2$ where the small angle approximation is used again on the cosines. This gives $\Delta (\theta')^2=m \frac{\Delta \lambda}{nd}$. With $m \approx \frac{2nd}{\lambda}$, we have $\Delta (\theta')^2=\frac{2 \Delta \lambda}{\lambda}$.
Note that the parameter $\delta=C \, \Delta (\theta')^2=C (\frac{2 \Delta \lambda}{\lambda})$.

Note: This $\delta$ or $\Delta (\theta')^2$ would almost give us what we need, but we calibrate $\Delta (\theta')^2$ using $\Delta (\theta)^2$ from above. This way we don't need to determine the proportionality constant $C$, and we don't need to get an absolute measure of the angle $\theta'$. We do need to know the factor $\frac{\lambda}{nd}$ for this method though. Note also that $d$ is chosen large enough for this experiment so that $\Delta$ is about five times bigger than $\delta$.

Finally, with simple algebra, computing $\frac{\delta}{\Delta}$ from above, $\frac{\Delta \lambda}{\lambda}=(\frac{\delta}{\Delta})(\frac{\lambda}{2nd})$.

See also post 17. I changed notation here, using $n$ for index of refraction, and $d$ for etalon thickness. My formulas are all consistent with the formulas that the "link" derives. I do think my derivations are a little simpler than those of the "link". :)

 

[Athenian]

Wow, thank you for taking the time to provide a step-by-step derivation! I found it both quite help and intriguing as well!

My formulas are all consistent with the formulas that the "link" derives. I do think my derivations are a little simpler than those of the "link". :)

I most certainly agree! As you quoted once, "It is mostly the clarity and simplicity that makes for good physics, rather than the complexity and degree of difficulty". Couldn't agree more! Once again, thank you for providing the wonderful derivation here!

Very glad to hear. :) I would be interested in seeing a couple more results, like the value(s) you got for the Bohr magneton, $\mu B$, and even a couple of the values for $δ$, $Δ$, and $B$ if you have time to share them. Keep up the good work.

Sure, most definitely. Here is a rundown version of the data I acquired. After following along your instructions on the methods of finding $\frac{\delta}{\Delta}$, I found that my "deltas ratio" values ranged from $0.03$ to $0.3$. Of course, the larger $B$ is, the larger the ratio becomes. Similarly, the smaller $B$ is, the smaller the ratio becomes.

From there, I plotted a $\delta/\Delta$ vs. $B$ graph as shown below.

The above equation (with uncertainties) for the linear line is as follows:

$$\frac{\delta}{\Delta} = (0.00393 \pm 2.096 \times 10^{-5})B - (0.02721 \pm 0.01058)$$

Thus, with the linear equation available, I know that:
$\frac{\delta}{\Delta B} = m = 0.00393 \pm 2.096 \times 10^{-5}$
y-intercept $(b) = -0.02721 \pm 0.01058$

Finding the Bohr magneton, I can use the below equation previously mentioned on Post #25.
$$\mu_B = \frac{hc}{2 \mu t B} \bigg( \frac{\delta}{\Delta} \bigg) = \frac{mhc}{2 \mu t}$$

Plugging my numbers in, I get the below value for the Bohr magneton:
$$\mu_B = 8.93427 \times 10^{-27} J/T$$

However, the above value is off by three orders of magnitude. The actual value of the Bohr magneton is the one shown below.
$$\mu_B = 9.273 \times 10^{-24} J/T$$

While the slope does not properly reflect the solution I wanted, I did find it quite interesting that the data does.

Case and point, I used a data point (the last one) I had on the graph to find the Bohr magneton. The data point contains a $\delta/\Delta$ value of $0.3$ and a $B$ value of $800$ mT. Plugging into the equation of the Bohr magneton mentioned earlier, I get the value below.
$$\mu_B = 8.52507 \times 10^{-24} J/T$$

This time around, the number acquired is on the same order of magnitude as the actual value of the Bohr magneton. Thus, the data found is considered relatively accurate.

With all the information above, I am still trying to figure out why using the numerical value of the slope in my calculations gave me incorrect values for the Bohr magneton. I think I may have to plug in all the data points and try to identify the issue. Of course, if you have any thoughts on the matter, I would love to hear about it too.

That said, the above is the information/data I acquired. If you would like to know more, please let me know and I will be happy to share!

 

[Charles Link]

$B$ on the graph is in milliTeslas, but $B$ in the formulas is in Tesla. That is your factor of 1000. When you take the slope, your x-axis must be corrected for this. (Include the units on your slope computation, and you will see that everything is consistent). In any case, very good. :) I think your data is quite excellent. :)

additional note: When you draw the best straight line to find the slope, it should pass through the origin if possible. Looks like this method gets you a value for $\mu_B$ that is accurate to about $\pm 10$ % or slightly better, which is quite good.

 

[Charles Link]

Note on the above: If you draw the best fit line that passes through the origin, and use the slope, the answer won't be very much different than the answer you get when you put in the values of $\frac{\delta}{\Delta}=.3$ and $B=.8$.
(If you draw the line through the origin and through the point with $\frac{\delta}{\Delta}=.3$ and $B=.8$, and compute the slope, you get the same answer as by plugging in the values).