User:Paul V Klimov/Notebook/JuniorLab307L/2008/09/29

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Balmer Series

SJK 00:05, 19 October 2008 (EDT)
00:05, 19 October 2008 (EDT)This is an excellent lab notebook and analysis!  I really enjoyed reading it and looking at your data!  I have a lot of comments below, pushing you further on the analysis, but truly you did a great job.
00:05, 19 October 2008 (EDT)
This is an excellent lab notebook and analysis! I really enjoyed reading it and looking at your data! I have a lot of comments below, pushing you further on the analysis, but truly you did a great job.

Notes by: Paul Klimov and Garrett McMath In this lab, we will be measuring the Rydberg constant by observing the spectrum of several lamps. We will first calibrate with a mercury lamp.

Lab Summary

Theory

Emission Spectra and the Balmer Series:

The emission spectrum of hydrogen was first correctly modeled by a high school teacher, Rydberg, although without theoretical support. He postulated that the emission lines in visible spectrum of Hydrogen could be found from the following relationship:

 \frac{1}{\lambda}=R(\frac{1}{2^{2}}-\frac{1}{m^{2}})

 m \in \mathbb{Z} \geq 3

With the advent of 'classical' quantum mechanics, largely attributed to the workings of Bohr, a theoretical framework for the emission lines was developed. This framework suggested that the emission lines are due to transitions of electrons between excited states and lower energy states.

The Balmer series results from electron transitions between any energy level greater than the second, and the second. Not surprisingly, these were discovered first, because the emissions are in the visible spectrum. Later it was discovered that Hydrogen has many other emissions that lie in the non-visible spectrum, which result from transitions between other states. These series include the Lyman series, the Paschen series, the Pfund Series, the Brackett series, and some others. With slight modification the formula could account for those emissions as well, in the Hydrogen atom. This is the new formula:

 \frac{1}{\lambda}=R(\frac{1}{n^{2}}-\frac{1}{m^{2}})

 m,n \in \mathbb{Z}, m > n

The integers m and n correspond to the energy levels between which the electronic transitions occur. So, for the Balmer series, we would set n=2, and vary m above 2.

The now accepted value of the Rydberg constant is:

R=1.0967758(7)\cdot 10^{7} m^{-1}

Spectroscope:

In addition to exploiting quantum phenomena, this lab makes use of several optics-related concepts. The spectroscope is built around the prism which does essentially all of the work. When light enters the prism, it is immediately refracted. Because each wavelength has a unique index of refraction in the prism, each wavelength splits apart from the others, producing the line spectrum. This light then reflects off of an another edge. I am assuming that the reflection there is 'total internal' in order to maximize the reflected intensity (we could check using Snell's law, if we wanted to). The light then emerges from the prim and heads to your eye where you can observe the spectrum.

Equipment

  • Adam Hilger London Spectrometer. Serial Number 12610
  • Spectrum II Power Supply. Model Sp200 5000V 10mA electro technic products.
  • Helium Tube: Cenco Scientific Company.
  • Hydrogen Tybe: Cenco Scientific Company.
  • Paul V Klimov 00:33, 13 October 2008 (EDT): Deuterium Tube: Cenco Scientific Company.

Calibration

We didn't find the hydrogen vapor lamp, so we will calibrate the device with Helium:

  • 438.793 w
  • 443.755 w
  • 447.148 s
  • 471.314 m
  • 492.193 m
  • 501.567 s
  • 504.774 w
  • 587.562 s
  • 667.815 m

Taken from Hyper physics.

  • We first turned on the helium lamp, and adjusted the aparture on the spectrometer which is facing the bulb.
  • We notice that rotating measuring dial the lines are moved in the spectrometer.
  • To calibrate we use the measuring dial and and turn the prism to line up the bright yellow line at 587nm
  • Then, we went through and checked to make sure the other lines match up with the correct wavelength.
  • this process was pretty painless, as it looked like all of the lines matched up quite well.

Spectrum of Hydrogen

Paul V Klimov 00:04, 12 October 2008 (EDT): How Measurements Were Made: Measurements were made by lining the spectral line with the cross-hair. To ensure that we got the best measurement possible, we adjusted the aperture on the spectroscope to the point where each line was as thin as possible. In addition, we avoided moving the measuring wheel in the vicinity of any emission as we were promised that this would result in bad data.

Week1

NOTE: The uncertainty was decided on based on the width of the aperture and our discretion. In our first two attempts our uncertainties were greater than in our following attempts. This is because we were getting used to the device.

We noticed that the gears slip quite a bit. For this reason, one person took all measurements from one side, and the other person started from that side and went backwards.

First Attempt. To get the measurement, we lined up the middle of the line with the crosshair

  • 412.1±.3nm
  • 435.4±.1nm
  • 486.3±.1nm
  • 659.5±.1nm

Second Attempt.

  • 653.0±01nm
  • 484.8±.02nm
  • 436.0±.05nm
  • 413.7±.15nm

The spectrometer was re-calibrated at this point, before taking our next round of measurements.

Third Attempt.

  • 656.0±.1nm
  • 485.5±.05nm
  • 433.9±.05nm
  • 411.4±.15nm

Fourth Attempt.

  • 411.0±.05nm
  • 435.6±.05nm
  • 486.0±.02nm
  • 657.4±.02nm

Week2

DATA TAKEN FROM GARRETT's LAB NOTEBOOK!!!

After speaking with Dr. Koch we decided to take another set of data for Hydrogen again calibrating with helium. To ensure the accuracy of our data. In this set of data we calibrated going from the right

First New Attempt(from left, Paul)
*411.0±.1nm
*434.6±.1nm
*486.9±.1nm
*660.4±.3nm
First New Attempt(from right,Paul)
*657.0±.3nm
*486.1±.1nm
*434.3±.1nm
*410.3±.3nm
Second New Attempt(from left,Garrett)
*410.5±.5nm
*434.7±.1nm
*486.7±.1nm
*660.5±.4nm
Second New Attempt(from right,Garrett)
*657.5±.3nm
*486.3±.1nm
*434.2±.1nm
*410.7±.5nm
Third New Attempt(from left,Paul)
*410.7±.3nm
*434.7±.1nm
*486.8±.1nm
*661.0±.6nm
Third New Attempt(from right,Paul)
*657.1±.5nm
*486.0±.1nm
*434.2±.1nm
*410.2±.5nm
Fourth New Attempt(from left,Garrett)
*410.5±.3nm
*434.6±.1nm
*487.0±.1nm
*660.9±.6nm
Fourth New Attempt(from right,Garrett)
*657.9±.5nm
*486.0±.1nm
*434.3±.1nm
*410.6±.5nm

Spectrum of Deuterium

First Attempt

  • 657.4±.03nm
  • 485.0±.04nm
  • 433.3±.04nm
  • 409.8±.05nm

Second Attempt

  • 410.1±.05nm
  • 433.9±.02nm
  • 485.95±.05nm
  • 657.7±.1nm

Possible Sources of Error

  • Gear Backlash: When moving the gear in direct vicinity of the spectral line, it is possible to get incorrect measurements.
  • 2 Possible Parent Distributions: As noted by Dr.Koch, it is possible that we have two different parent distributions, due problems with the gears: One for our measurements starting from the left, and another for our measurements starting from the right. Looking carefully at our data, it is clear that there are, in fact, multiple parent distributions. When we read the spectrum going in the direction opposite of the direction in which the spectrometer was calibrated, we consistently got large errors for some of the wavelengths. For this reason, a large chunk of data will have to be thrown out. This will be discussed and justified later in the report.
SJK 22:49, 18 October 2008 (EDT)
22:49, 18 October 2008 (EDT)This is a really interesting calculation!  More comments later if I can think of anything to add!
22:49, 18 October 2008 (EDT)
This is a really interesting calculation! More comments later if I can think of anything to add!
* Natural Line Width: Due to the the fact that electronic excitations last for short time periods, we expect some spread in the energy, and thus wavelength, of the emitted photon. This is a direct consequence of the Heisenberg uncertainty principle. To determine whether or not this will have any effect on our measurements, we should determine this spread and see if it is within our measuring capabilities (i am guessing it will not be significant enough). One thing I will have to find somewhere (because I don't see how I could calculate it knowing the physics that I know), is the excitation time for the various excitations in the hydrogen atom.

 \Delta E \Delta \tau \sim \hbar

 \Delta E \sim \frac{\hbar}{\Delta\tau}

Then, relating this to the wavelength:

 E = \frac{hc}{\lambda}

 dE = -\frac{hc}{\lambda}\frac{d\lambda}{\lambda}

Therefore, the uncertainty in wavelength will be given by:

 \Delta \lambda = \frac{\Delta E \lambda}{E}

where absolute values must be taken.

Performing this calculation with a decay time of 10^-8s returns a wavelength uncertainty on the order of 10^-14m. This is not something that we could resolve. Therefore, the natural line width will not be a problem here!

  • Prism Shape: In doing this experiment, we are really trusting that the prism is cut exactly as necessary to match the 'measuring wheel'. A unit of arclength on the wheel must be some specific function of the shape of the prism, as is obvious because the ticks on the measuring wheel expand for certain wavelengths. In altering the shape of the prism, it could be impossible to calibrate the device well for any gas sample. This wasn't a problem for us, as we confirmed during our calibrations.

Post Experimental Data Analysis

Looking at the data, it is easy to see that there are really 2 parent distributions. Therefore, in doing this analysis, I will have to justify which data I will choose to use. All data analysis will be done in MATLAB, and any important algorithms that I use will be mentioned, of course.

Choosing the "Correct" Distribution

SJK 23:42, 18 October 2008 (EDT)
23:42, 18 October 2008 (EDT)This is a really solid discussion and I agree with your decision to throw out the half of the data.  It's very good that you are basing the decision on knowledge about the system and how the calibration and backlash work.  As you say, it's not surprising this gives better results, but it's an important distinction that you're not just throwing the data out because it gives a "wrong" answer.  Very good thinking!
23:42, 18 October 2008 (EDT)
This is a really solid discussion and I agree with your decision to throw out the half of the data. It's very good that you are basing the decision on knowledge about the system and how the calibration and backlash work. As you say, it's not surprising this gives better results, but it's an important distinction that you're not just throwing the data out because it gives a "wrong" answer. Very good thinking!
At first we thought that the direction of calibration was not going to matter, because we thought that the device was only going to return bad data if we moved the measuring wheel back and forth in the immediate vicinity of an emission line. Luckily we took a more extensive set of data the second week of lab which showed us, in contrast to our initial conjecture, that the direction of calibration does indeed matter.

As I mentioned above, it appears that there are two distributions within our data, representing two different parent distributions. One of these distributions corresponds to taking data from the left and the other distribution corresponds to taking data from the right (i.e. 'scrolling' the spectrum to the left from the right; from lower wavelengths to higher). This discrepancy is visible when inspecting our data for the H-alpha emission, which consistently differs by several nanometers between 'right' and 'left' measurements. Due to this discrepancy, I chose to throw out half of our data. I decided to use the data coming only from the right because this is the way in which device was calibrated. Not surprisingly, this set of data matches the real values better.

Calculations and Uncertainty

Figure 1: Rydberg Constant vs. Quantum Number. Error bars are our uncertainties as decided on by judgement (Steve Koch:This is good that you describe the error bars...however, you can provide more information.  As a reader, I don't know whether the bards represent your full "reasonable range" (something like a 95% confidence), or an estimate of a 68% confidence interval, or something else)
Figure 1: Rydberg Constant vs. Quantum Number. Error bars are our uncertainties as decided on by judgement (Steve Koch:This is good that you describe the error bars...however, you can provide more information. As a reader, I don't know whether the bards represent your full "reasonable range" (something like a 95% confidence), or an estimate of a 68% confidence interval, or something else)
Figure 2: Rydberg Constant vs. Quantum Number. Error bars are the calculate standard error of the mean for each point (Steve Koch: In contrast to figure 1, in this case it is enough information, because SEM implies 1 sigma and normal distribution.)
Figure 2: Rydberg Constant vs. Quantum Number. Error bars are the calculate standard error of the mean for each point (Steve Koch: In contrast to figure 1, in this case it is enough information, because SEM implies 1 sigma and normal distribution.)

As suggested by Dr. Koch, the Rydberg constant should be calculated and averaged for each quantum number separately, at first. Then, only if the constants seem to be distributed randomly, versus increasing quantum number, can one average them. However, if there is a clear trend then one cannot average the values. It is clear why this is so, because a trend in such a distribution would imply that there is some systematic error that would cause us to get progressively worse results consistently. In our case, a trend is expected because for at higher quantum numbers, the emission lines are significantly harder to resolve and thus measure accurately.

Very interestingly, however, this did not seem to be the case! (see Figure 1) The constants, as calculated for each quantum number separately did not have an obvious trend and seemed to be distributed fairly randomly. I must mention that there seemed to be one outlier (relatively speaking since its percent error was still incredibly low!) for the 3->2 transition. This is not very surprising because we had trouble measuring this emission, as it seemed to be thicker than any of the others, even at the highest resolution that we could obtain. At first we thought that we could be observing the 'fine structure' that would results in a splitting of emission bands. However, after looking into this we found that it would be very unlikely given that the wavelength split due to this phenomenon would be roughly 1/100 of a nanometer. This is a resolution that we could not achieve due to technical limitations and perhaps human error.

I also included error bars that were made from my best attempts at doing error propagation, using our measured uncertainties. NOTE:The uncertainty was estimated based on how well we could resolve the line, and our personal judgement. The uncertainty for each measurement was completely independent of all other measurements. (This was discussed with Dr. Koch in class both lab days). I used the following expression to estimate the magnitude of error bars on the calculated value of R for its respective quantum number. Notice that I have defined a new quantity eta as a constant for each quantum number, n, to make the calculations cleaner. Also notice that I will use the average wavelength for each quantum number and the average uncertainty in measuring that wavelength. Averages are denoted by bars in the below calculations.

SJK 23:49, 18 October 2008 (EDT)
23:49, 18 October 2008 (EDT)Yes, you are doing this correctly, good job!  I don't know if you noticed, but what you arrive at is that the fractional uncertainty in R (dR/R) is equal to the fractional uncertainty in lambda.  Your intuition about framing the result in terms of R was a good one.  Lots of the formulas for error propagation simplify a lot in terms of fractional uncertainties. See wikipedia
23:49, 18 October 2008 (EDT)
Yes, you are doing this correctly, good job! I don't know if you noticed, but what you arrive at is that the fractional uncertainty in R (dR/R) is equal to the fractional uncertainty in lambda. Your intuition about framing the result in terms of R was a good one. Lots of the formulas for error propagation simplify a lot in terms of fractional uncertainties. See wikipedia

 ErrorBar_{n}=|\bar{dR_{n}}|=|\frac{\bar{dR_{n}}}{\bar{d\lambda_{n}}} \bar{d\lambda_{n}|}

 |dR_{n}|=\eta_{n}\frac{|-1|}{\bar{\lambda^{2}_{n}}}\bar{d\lambda_{n}} = \bar{R_{n}}\frac{\bar{d\lambda_{n}}}{\bar{\lambda_{n}}}

This, in turn, allows me to write the Rydberg constant like:

 \bar{R_{n}} \pm ErrorBar_{n} =\bar{R_{n}} \pm \bar{R_{n}}\frac{\bar{d\lambda_{n}}}{\bar{\lambda_{n}}}

In addition to preparing error bars by the above method, I also made another plot (see Figure 2) where the error bars are the standard error of the mean for that quantum number. I was actually surprised to see the agreement between the two methods by which I made error bars. Although the magnitudes are off by a bit, their relative 'intensities' are similar. Hopefully this means I am doing something right here.SJK 23:52, 18 October 2008 (EDT)
23:52, 18 October 2008 (EDT)Yes, I think you are doing this correctly!  I think the reason the figure 1 error bars are bigger is because you and Garrett had a more conservative estimate of the uncertainty.  I think your estimate is probably more like the standard deviation of the population--as opposed to the SEM, which is smaller than that by 1/sqrt(N)
23:52, 18 October 2008 (EDT)
Yes, I think you are doing this correctly! I think the reason the figure 1 error bars are bigger is because you and Garrett had a more conservative estimate of the uncertainty. I think your estimate is probably more like the standard deviation of the population--as opposed to the SEM, which is smaller than that by 1/sqrt(N)
I am also pleased to see that some of the error bars include the actual value of the Rydberg constant.SJK 23:56, 18 October 2008 (EDT)
23:56, 18 October 2008 (EDT)First of all, I agree that your data are really good!  Since you have good error analysis, I'm going to push you on the statistical interpretations.  If your SEM version represents 68% confidence interval, how many would you expect to overlap?  Also, What are the chances that your first two quantum numbers are consistent with each other?  Or what is the chance that the accepted value is consistent with your distribution for the red line?  How many sigma is it away?
23:56, 18 October 2008 (EDT)
First of all, I agree that your data are really good! Since you have good error analysis, I'm going to push you on the statistical interpretations. If your SEM version represents 68% confidence interval, how many would you expect to overlap? Also, What are the chances that your first two quantum numbers are consistent with each other? Or what is the chance that the accepted value is consistent with your distribution for the red line? How many sigma is it away?


Results

Figure 3: Rydberg Constant vs. Quantum Number. Error bars are the calculated standard error of the mean for each point. The magenta line is the mean of all the values and the green lines are the standard errors about that mean (Steve Koch:Cool graph!)
Figure 3: Rydberg Constant vs. Quantum Number. Error bars are the calculated standard error of the mean for each point. The magenta line is the mean of all the values and the green lines are the standard errors about that mean (Steve Koch:Cool graph!)

The below data is illustrated in Figure 3

Accepted Rydberg Constant:

R_{act}=1.09677\cdot 10^{7} m^{-1}

Calculated Rydberg constants, for each transition, including each standard error of the mean:

R_{3 \rightarrow 2}=1.09526(34) \cdot 10^{7} m^{-1}

%error = 0.138

R_{4 \rightarrow 2}=1.09716(16) \cdot 10^{7} m^{-1}

%error = 0.036

R_{5 \rightarrow 2}=1.09658(07) \cdot 10^{7}  m^{-1}

%error = 0.018

R_{6 \rightarrow 2}=1.09635(32) \cdot 10^{7} m^{-1}

%error = 0.038

Since there did not appear to be a clear trend between calculated Rydberg constant and quantum numbers, the above constants were averaged. This produced the following result and standard error of the mean (for the last two digits), marked off by parentheses. ***Please let me know If I denoted this correctly with the parentheses***SJK 00:01, 19 October 2008 (EDT)
00:01, 19 October 2008 (EDT)Yes, I think you did this correctly, and it is a clear presentation.  You did not specify that you did a weighted mean, so I'm guessing you didn't or you would have mentioned it.  I think this would make your result even better (it gives more weight to numbers with less uncertainty). However, Given my question above (about how far apart the first two data points are from each other, relative to their uncertainties, I'd question the validity of averaging these.  Your precision is so good, that it appears you've surpassed what looks to me as some systematic error--small, but larger than your random error.
00:01, 19 October 2008 (EDT)
Yes, I think you did this correctly, and it is a clear presentation. You did not specify that you did a weighted mean, so I'm guessing you didn't or you would have mentioned it. I think this would make your result even better (it gives more weight to numbers with less uncertainty).

However, Given my question above (about how far apart the first two data points are from each other, relative to their uncertainties, I'd question the validity of averaging these. Your precision is so good, that it appears you've surpassed what looks to me as some systematic error--small, but larger than your random error.

 R_{av}= 1.09634(21) \cdot 10^{7} m^{-1}

%error = 0.039

This is the value that will be reported in my summary because I believe it best represents the data.

Analysis and Discussion

SJK 00:03, 19 October 2008 (EDT)
00:03, 19 October 2008 (EDT)I like the way you are approaching this, and I agree with your analysis that you don't have enough data to draw a good conclusion, but that you have some hope you could do it.
00:03, 19 October 2008 (EDT)
I like the way you are approaching this, and I agree with your analysis that you don't have enough data to draw a good conclusion, but that you have some hope you could do it.
  • Deuterium:When we looked at the spectrum of deuterium, we were supposed to find that some wavelengths had been shifted due to the atoms heavier nucleus. Unfortunately I will not be able to make any strong conclusions because of our limited data size. However, it is certainly worth it to examine whether or not we could have measured the differences in wavelength, if we had a larger set of data. To do so, I will first calculate the SEM of the measured wavelengths for each quantum number and see if they exceed the wavelength shifts in deuterium, for each respective quantum number.

SEM_{3 \rightarrow 2}=.2056nm.

DeuteriumShift = .180nm

SEM_{4 \rightarrow 2}=.0707nm

DeuteriumShift = .133nm

SEM_{5 \rightarrow 2}=.0289nm

DeuteriumShift = .119nm

SEM_{6 \rightarrow 2}=.1190nm

DeuteriumShift = .112nm

Immediately we see that 2 out of 4 wavelength shifts in Deuterium lie within our SEM. However, two of them lie outside of it suggesting that they could have been measured by us. However, given that our SEMs vary quite a bit, and also our Deuterium measurements, I am skeptical as to whether or not this would actually be possible.

The Deuterium emissions that we measured seem to jump below and above the wavelength emissions that we measured for Hydrogen. This suggests to me that it would not be possible to tell the difference between the spectra of the isotopes, because we know that the Deuterium spectrum should consistently have higher wavelength emissions. However, as I already stated, no real conclusions can be made because we have too little data (as we spent all of our time the second day taking better data for the hydrogen emissions), which leaves far too much room for interpretation. If I decide to come back to this lab in the future for the lab write up, this is definitely something that I want to clear up by taking lots of data for Deuterium.

  • Fine Structure: At one point we thought that we could resolve two red emissions that were right next to each other. At first we thought that we were observing the energy splitting due to spin-orbit coupling; an effect caused by the interaction of the intrinsic spin of the electron and the electrons orbit about the nucleus. However, we later learned that such an split would be much more subtle than even the wavelength shift in deuterium -- on the order of a hundredth of a nanometer. Therefore, there is absolutely no way we could have resolved this.
  • Red Doublet: Although the doublet described above could not be accounted for by the spin-orbit coupling, it could definitely be caused by some impurities in the gas.
  • Sodium Doublet: Although there was no sodium lamp for us to use, I can surmise what we could have seen based on our measuring capabilities. As given in the 307 Lab Manual, the well researched Sodium doublet appears at 589.0nm and 589.6nm. Given that our worst SEM was .2056nm, I think that the doublet could have been resolved and measured.

Lab Questions

The spectrum of hydrogen should be only slightly different from that of deuterium. The reason there should be any difference at all is because of the greater mass of the nucleus in deuterium. This causes the reduced mass of the system to change, which in turn changes the Rydberg Constant. Below I will calculate this strictly from the reduced mass. However, I believe that the derivation could be done from even more basic principles, which don't necessarily invoke Bohr's theory at all. The way to do that is to simply consider energy and momentum conservation. We know that the energy transition has to account for the photon energy and the recoil energy of the nucleus. This, together with momentum conservation gives you two relations with two unknown variables, which can be solved without problems.

The equation for the constant is:

 R\equiv \mu \frac{e^{4}}{8c\epsilon^{2}_{o}h^{3}}

To find the new constant for deuterium,Rd, we can do the following, using the known Rh (which was calculated for an infinitely massive nucleus):

 R_{D}=\frac{R_{H} \mu }{m_{e}} =1.09707\cdot 10^{7} m^{-1}

 \mu \equiv \frac{m_{e} (m_{p}+m_{n})}{m_{e}+(m_{p}+m_{n})} = 9.1075\cdot 10^{-31} kg

where the greek symbol mu is the reduced mass of the system. Clearly the reduced mass of the system will be less with the extra neutron in the nucleus. Given that R is inversely proportional to the wavelength, we know that the wavelength should increase in deuterium. These are my calculated wavelengths for deuterium:

λ3 = 656.649nm(vs656.469nmforHydrogen)

λ4 = 486.407nm(vs486.273nmforHydrogen)

λ5 = 434.292nm(vs434.173nmforHydrogen)

λ6 = 410.406nm(vs410.293nmforHydrogen)

(I compared my calculations to Hyperphysics, where they reported a wavelength shift of .179nm for the H-alpha emission, as compared to my .180nm shift)

The differences in wavelength are all on the order of a tenth of a nanometer. While our SEM's suggest that the Deuterium 4->2 and 5->2 transitions could have been measurably different from those of Hydrogen, our limited data suggested otherwise. However, no conclusions were made because of our small data size for Deuterium.

Things to try in the future

  • One thing I noticed was that the hole where the light entered and exited the prism was quite large. It would be interesting to see by how much the spectra would improve if the gap between the hole and the scope was somehow closed off, so as to exclude ambient lighting. Perhaps this could be accomplished with some aluminum foil.
  • The spectroscope is quite a beast as we saw and can provide for some extremely accurate measurements. In the future it would be interesting to see how great its resolving power is, by trying out various gasses with tightly spaced emissions. And as a goal, I guess that I would like to see if it would be possible to optimize conditions to the point where we could tell the difference between hydrogen and deuterium with confidence. SJK 23:39, 18 October 2008 (EDT)
    23:39, 18 October 2008 (EDT)I like this idea and don't know the answer...looks like you'd be close to being able to do it!
    23:39, 18 October 2008 (EDT)
    I like this idea and don't know the answer...looks like you'd be close to being able to do it!
  • It would be awesome if we could get our hands on some bulbs with hydrogen 'contaminated' with deuterium, or visa versa, so that we could compare the spectra simultaneously. Maybe if someone has a glove box filled with hydrogen (i wouldn't bet on it haha) we could do it ourselves.SJK 23:39, 18 October 2008 (EDT)
    23:39, 18 October 2008 (EDT)
    23:39, 18 October 2008 (EDT)

Algorithms used in Calculations

MATLAB code used for all calculations

Algorhithm used for mean:

 \bar{x}_{ij} = \frac{1}{N} \sum_{i=1}^N{x_{ij}}

Algorithm used for standard deviation:

\sigma_{x} =\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_{i}-\bar{x})^{2}}

Algorithm used for standard error of the mean (SEM):

SEM=\frac{\sigma_{x}}{\sqrt{N}}

Algorithm used for Percent Error:

Percent Error= 100\frac{|Actual-Measured|}{Actual}

References

1. Hyper Physics: Helium Spectrum. Used for calibration.

2. Wikipedia: Spin-Orbit Coupling. Used in discussion of energy level splitting

3. An Introduction to Error Analysis by John R. Taylor. 2nd Edition. Used for error analysis

4. Junior Lab 307 Lab Manual by Dr. Gold.


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