User:Paul V Klimov/Notebook/JuniorLab307L/2008/09/29
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Balmer SeriesSJK 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. TheoryEmission 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: [math]\displaystyle{ \frac{1}{\lambda}=R(\frac{1}{2^{2}}-\frac{1}{m^{2}}) }[/math] [math]\displaystyle{ m \in \mathbb{Z} \geq 3 }[/math] 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: [math]\displaystyle{ \frac{1}{\lambda}=R(\frac{1}{n^{2}}-\frac{1}{m^{2}}) }[/math] [math]\displaystyle{ m,n \in \mathbb{Z}, m \gt n }[/math] 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: [math]\displaystyle{ R=1.0967758(7)\cdot 10^{7} m^{-1} }[/math] 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
CalibrationWe didn't find the hydrogen vapor lamp, so we will calibrate the device with Helium:
Taken from Hyper physics.
Spectrum of HydrogenPaul 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. Week1NOTE: 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
Second Attempt.
The spectrometer was re-calibrated at this point, before taking our next round of measurements. Third Attempt.
Fourth Attempt.
Week2DATA 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 DeuteriumFirst Attempt
Second Attempt
Possible Sources of Error
This is a really interesting calculation! More comments later if I can think of anything to add! [math]\displaystyle{ \Delta E \Delta \tau \sim \hbar }[/math] [math]\displaystyle{ \Delta E \sim \frac{\hbar}{\Delta\tau} }[/math] Then, relating this to the wavelength: [math]\displaystyle{ E = \frac{hc}{\lambda} }[/math] [math]\displaystyle{ dE = -\frac{hc}{\lambda}\frac{d\lambda}{\lambda} }[/math] Therefore, the uncertainty in wavelength will be given by: [math]\displaystyle{ \Delta \lambda = \frac{\Delta E \lambda}{E} }[/math] 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!
Post Experimental Data AnalysisLooking 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" DistributionSJK 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! 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  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)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 [math]\displaystyle{ ErrorBar_{n}=|\bar{dR_{n}}|=|\frac{\bar{dR_{n}}}{\bar{d\lambda_{n}}} \bar{d\lambda_{n}|} }[/math] [math]\displaystyle{ |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}}} }[/math] This, in turn, allows me to write the Rydberg constant like: [math]\displaystyle{ \bar{R_{n}} \pm ErrorBar_{n} =\bar{R_{n}} \pm \bar{R_{n}}\frac{\bar{d\lambda_{n}}}{\bar{\lambda_{n}}} }[/math] 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)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) 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 The below data is illustrated in Figure 3 Accepted Rydberg Constant: [math]\displaystyle{ R_{act}=1.09677\cdot 10^{7} m^{-1} }[/math] Calculated Rydberg constants, for each transition, including each standard error of the mean: [math]\displaystyle{ R_{3 \rightarrow 2}=1.09526(34) \cdot 10^{7} m^{-1} }[/math] [math]\displaystyle{ % error = 0.138 }[/math] [math]\displaystyle{ R_{4 \rightarrow 2}=1.09716(16) \cdot 10^{7} m^{-1} }[/math] [math]\displaystyle{ % error = 0.036 }[/math] [math]\displaystyle{ R_{5 \rightarrow 2}=1.09658(07) \cdot 10^{7} m^{-1} }[/math] [math]\displaystyle{ % error = 0.018 }[/math] [math]\displaystyle{ R_{6 \rightarrow 2}=1.09635(32) \cdot 10^{7} m^{-1} }[/math] [math]\displaystyle{ % error = 0.038 }[/math] 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)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. [math]\displaystyle{ R_{av}= 1.09634(21) \cdot 10^{7} m^{-1} }[/math] [math]\displaystyle{ % error = 0.039 }[/math] This is the value that will be reported in my summary because I believe it best represents the data. Analysis and DiscussionSJK 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.
[math]\displaystyle{ SEM_{3 \rightarrow 2}=.2056nm. }[/math] [math]\displaystyle{ Deuterium Shift = .180nm }[/math] [math]\displaystyle{ SEM_{4 \rightarrow 2}=.0707nm }[/math] [math]\displaystyle{ Deuterium Shift = .133nm }[/math] [math]\displaystyle{ SEM_{5 \rightarrow 2}=.0289nm }[/math] [math]\displaystyle{ Deuterium Shift =.119nm }[/math] [math]\displaystyle{ SEM_{6 \rightarrow 2}=.1190nm }[/math] [math]\displaystyle{ Deuterium Shift =.112nm }[/math] 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.
Lab QuestionsThe 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: [math]\displaystyle{ R\equiv \mu \frac{e^{4}}{8c\epsilon^{2}_{o}h^{3}} }[/math] 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): [math]\displaystyle{ R_{D}=\frac{R_{H} \mu }{m_{e}} =1.09707\cdot 10^{7} m^{-1} }[/math] [math]\displaystyle{ \mu \equiv \frac{m_{e} (m_{p}+m_{n})}{m_{e}+(m_{p}+m_{n})} = 9.1075\cdot 10^{-31} kg }[/math] 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: [math]\displaystyle{ \lambda_{3} =656.649nm (vs 656.469nm for Hydrogen) }[/math] [math]\displaystyle{ \lambda_{4} =486.407nm (vs 486.273nm for Hydrogen) }[/math] [math]\displaystyle{ \lambda_{5} =434.292nm (vs 434.173nm for Hydrogen) }[/math] [math]\displaystyle{ \lambda_{6} =410.406nm (vs 410.293nm for Hydrogen) }[/math] (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
Algorithms used in CalculationsMATLAB code used for all calculations Algorhithm used for mean: [math]\displaystyle{ \bar{x}_{ij} = \frac{1}{N} \sum_{i=1}^N{x_{ij}} }[/math] Algorithm used for standard deviation: [math]\displaystyle{ \sigma_{x} =\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_{i}-\bar{x})^{2}} }[/math] Algorithm used for standard error of the mean (SEM): [math]\displaystyle{ SEM=\frac{\sigma_{x}}{\sqrt{N}} }[/math] Algorithm used for Percent Error: [math]\displaystyle{ Percent Error= 100\frac{|Actual-Measured|}{Actual} }[/math] References1. 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. | |
