Fourier transform spectroscopy

Transcription

1 Atomic and Molecular Spectroscopy 017 Purpose Fourier transform spectroscopy The purpose with this laboratory exercise is to experimentally study molecular spectra in emission and absorption. Using a diffractive grating spectrometer with CCD detector you will acquire a low resolution spectrum from a gas flame in the visible wavelength interval. Here the characteristic band structure in molecules, i.e. transitions between vibrational levels in two different electronic states will be studied. In this case the finer structure due to the molecular rotation is not resolved. Utilizing a Fourier transform spectrometer, a high resolution absorption spectrum of CO, in the IR, will be acquired. Analyzing this spectrum will show the R- and P-branches in the rotational-vibrational spectrum for transitions within an electronic ground state. An application of Fourier transform spectroscopy is analyzing gas emission from cars. This will also be studied during the laboratory exercise. Preparation prior to the lab Before the laboratory exercise the following sections in Atomic and Molecular Spectroscopy, S. Svanberg should be studied: Electronic-, rotational- and vibrational energy structure 4.. Rotational- and vibrational transitions 6.. Grating spectrometer 6..4 Fourier transform spectrometer Also read the instructions carefully and do the following preparatory exercises. 1. Why is a molecular spectrum built up by more lines than an atomic spectrum?. In what wavelength intervals do the following transitions occur: a. pure rotational transitions in molecules? b. transitions between vibrational levels within a given electronic state? c. transitions between different electronic levels? 3. What values can the vibration quantum number hold? What is the lowest vibrational energy? What is the energy difference between two consecutive levels? 4a. State the rotational energy in a diatomic molecule utilizing the rotational constant (B). 4b. How does B depend on the moment of inertia (I) and the equilibrium distance between the atomic nucleus? 5. If one can measure with sufficient resolution you should see that the 1

2 rotational constant decreases when the rotational quantum number increases. Why? 6. The spectrum for the vibrational transition 0 in CO is shown in the figure below. The wavenumbers for the first three transitions in the P- and R-branch respectively are given in the table below. The rotational energy can be written E rot = B(+1)= hcb σ (+1) where B is the rotational constant. Traditionally the constant is called B in the ground state and B in the excited vibrational state. We are going to determine the energy difference between the vibrational levels (σ 00 ) and the rotational constant(s) from the data utilizing three approximations with increasing precision. The final analysis will be done during the laboratory exercise. Note that this manual and the book use different notations for the vibrational and rotational quantum numbers (rot: J=l, vib: v=n). Wavenumber /cm a) Determine σ 00 and B σ = B σ = B σ (unit: cm -1 ) by assuming that the rotational constants are equal in the two vibrational levels. Hence only data for two transitions should be utilized. b) Refine the analysis and determine σ 00 and the rotational constants B σ and B σ by forming and solving a set of equations. Only three experimental values should be used. c) Show, by using the selection rule 1, that the wavenumber (σ) for all possible transitions can be written

3 Atomic and Molecular Spectroscopy 017 ( B ' B ") m ( B ' B " m 00 ) where m = +1 = 1,, 3,... in the R-branch and m = - = -1, -, -3,... in the P-branch. d) Determine σ 00 and the rotational constants B σ and B σ by fitting a second order polynomial to all the experimental data given in the table. (For example, use Matlab and the function polyfit.) e) Determine the equilibrium distance between the atoms in both of the vibrational states. 3

4 Energy structure in molecules The energy of a molecule can be separated into three parts E= E elec + E vib + E rot. The largest contribution comes from the electronic energy, E elec, which originates from the electron movement in the potential due to the nuclei. This part is analogous to the quantized energy structure in an atom and its magnitude is in the region of 1-10 ev. The potential energy due to the nuclei is dependent on the distance between the atomic nuclei. Instead of defining an unambiguous binding energy it results in a potential relation depicted in Figure 1. Traditionally, the electronic ground state in a molecule is denoted X, while the excited states are characterized by A, B, C etc. In this laboration we will focus on the vibrationaland rotational structure of the molecule. Figure 1 The potential energy in a diatomic molecule as a function of distance between the atomic nuclei. E S () represents the total energy for two completely separated atoms. R 0 represents binding distance. Even though the atoms are bound together at the distance R 0, where the potential energy is minimized, the nuclei will vibrate with the center of vibration at this equilibrium state. The vibrational energy (E vib ) results in a first order fine structure in the order of 0.1 ev. The simplest model to describe the vibrational motion is a harmonic oscillator, i.e. the nuclei are connected by a spring. The model implies that the real (and complicated) potential relation is approximated by a parabola. A seen in Figure this approximation holds when vibration is small and close to R 0. 4

5 Atomic and Molecular Spectroscopy 017 Figure Quantized energy levels due to the vibrating atoms. The dotted line shows how the potential can be approximated with a second order relation (harmonic oscillator) close to the equilibrium distance. For a given electronic configuration and a given vibrational level (n value) the nuclei can finally also rotate around the common center of gravity. This yields a further energy contribution (E rot ) in the range of magnitude ev. Figure 3 summarizes the energy structure in a diatomic molecule. Figure 3 Energy structure in a diatomic molecule A quantum mechanical treatment where the molecular vibration is approximated by a harmonic oscillator and where the rotation is described by a rigid rotator 5

6 results in the following expression for the total energy: E E elec E vib E rot E elec 1 0 ( n ) ( 1). R The vibration is described by the quantum number n = 0, 1,, 3... and the resonance frequency 0. The rotational motion is described by the quantum number = 0, 1,, 3...; = m 1 m /(m 1 + m ) is the reduced mass. Table 1 states the magnitude of the different contributions in energy, wavelength and temperature. In a gas in thermal equilibrium the average kinetic energy of a molecule is: E 3 k kt. The last row in Table 1 thus states at what temperature a collision between two molecules might excite the different energy levels. This, in turn, explains why the heat capacity (C v ) for gases is different within different temperature ranges. Table 1 Energy contributions in a diatomic molecule. Electron Vibration Rotation E / cm E / ev 5 0, 0,001 / m 0, T / K (Note that it happens to be that the energies stated in cm -1 and as temperature approximately have the same numerical value, i.e. 1/(100hc) = /(3k) = 4.810!) 0 Elastic rotator To describe the rotational energy in a molecule it is convenient to introduce the rotational constant B: B where R I 0 I R 0 is the moment of inertia of the molecule. In Figure we see that, in the harmonic-oscillator approximation, B will not depend on n since R 0 is constant in the symmetric parabolic potential. In the real, asymmetric potential, on the other hand, B should decrease for increasing n since R 0 increases. Figure 3 shows that B decreases in excited states due to the same reason. Even within a given vibrational state the distance of equilibrium, R 0, between the nuclei should increase due to the centrifugal force when the molecule rotates 6

7 Atomic and Molecular Spectroscopy 017 faster (higher ) so that E rot decreases. Due to this the upper rotational levels are successively shifted downwards. This improvement, of the model rigid rotator, is called elastic rotator. One can show that the rotational energy now instead can be described by E rot = B(+1) - D (+1) where D is the so called centrifugal distortion constant. 3 4B 0 D where 0 is the vibration frequency of the molecule. Since D << B the elastic rotator will differ noticeable from the rigid rotator not until the -values get reasonably high. This can also be expressed as it is required very high experimental precision to see the effect. Figur 4 Rotational levels for rigid and elastic rotator. 7

8 Molecular spectra For light to be emitted or absorbed the total angular momentum of the molecule has to change since the photon that is emitted/absorbed at a transition carries one momentum quanta. Since the vibrational states are not associated with any angular momentum, this implies that the rotational state and/or the electronic state must change. To be able to state selection rules we therefore have to separate certain cases. Pure rotational transitions, i.e. within a vibrational level in a given electronic state. Wavelengths in the microwave range. a. Homonuclear molecules (e.g. O, N ), that lack permanent dipole moment, do not have rotational spectra since their rotation cannot interact with electromagnetic radiation. b. = 1 Rotational-vibrational transitions, i.e. within a given electronic state. Wavelengths in IR. a. Homonuclear molecules, that lack a permanent dipole moment, do not have rotational-vibrational spectra since their vibration cannot interact with the light. b. n = 1 (harmonic-oscillator approximation),, 3,... c. = 1 (+1 yield R-branch, -1 yield P-branch) Transitions between electronic states. Wavelengths in visible or UV. a. Analogous with the situation in atoms, selection rules can be stated for electronic transitions in molecules, utilizing the total eletron angular momentum and spin. b. n = 0, 1,, 3,... c. = 0, 1 = 0 is forbidden in a pure rotational-vibrational transition, i.e. within an electronic state but permitted in conjunction with electronic transitions since the total angular momentum of the molecule might be different in the lower and upper electronic state respectively. = 0 transitions give rise to another branch, the Q-branch. If B is equal in the two electronic states the Q-branch is reduced to one spectral line. Electronic spectra Transitions where the electronic state in the molecule is changed appear in the visible or the ultraviolet wavelength interval, as is the case for neutral atoms. If the absorption or emission spectra is studied with moderate resolution the fine structure due to the rotational levels cannot be seen, but instead a number of so called bands. These bands represent different combinations of the vibration 8

9 Atomic and Molecular Spectroscopy 017 quantum numbers (n 1, n ) in the electronic states (1, ); see Figure 3. If we neglect the rotational energy contribution to the total energy we can state: E E i elec i ( n 0i i 1/ ) where 0i is the resonance frequency in state i. The frequency, f, for a transition is then: f (E f 0 elec 1 ( n 1 ( n1 01 1/ ) E 01 1 elec n 0 ) ( n In the harmonic-oscillator approximation n = 0 or 1. We now can separate the cases where the resonance frequency is equal or different in the two levels. 0 1/ )) / h f f n f f0 f, f0 f0 01, f In this (unrealistic) case we only get three spectral lines in the band structure. Let now instead the resonance frequencies be different and write n = n 1 +n. This will result in the characteristic band structure: 1 f f0 ( n1 0 n 0) Now, instead of the three lines n = -1, 0 or 1, we get several sets of equidistant transitions: 1-0, -1, 3-, , 0-0, 1-1, -, , and 0-1, 1-, -3, Rotational-vibrational spectrum In the infrared wavelength region rotational-vibrational transitions occur, that is transitions between different levels within a given electronic state. During the laboration you will study the transitions 0-1 and 0- in the ground state of CO. With the Fourier transform spectrometer you will be able to measure the wave numbers with high precision giving the possibility to analyze the energy structure with decreasing level of approximation. 9

10 Figure 5 Energy level diagram of the lowest rotational-vibrational levels in a diatomic molecule with absorption transitions from n = 0 to n = 1. The spectrum has two branches. The R-branch with = +1 and the P-branch with = -1. In the lower part of the figure a spectrum is shown with the corresponding spectral lines. The rotational constants are set to equal (B σ "=B σ '=B σ ) which leads to equidistant spacing between the lines, except the void where 0 should have occurred. Approximation 1 If the distortion term, D, is neglected and if the B-constants are assumed to be identical in the two vibrational states the wave number () for a transition between two vibrational states n and n (n > n ) is given by: where B '( ' 1) B ''( '' 1) (1) ( n' n") 10

11 Atomic and Molecular Spectroscopy 017 Utilizing the selection rule = 1 results in = +1, = 00 + B σ ( +1) = 0,1, (a) = -1, = 00 - B σ = 1, (b) If ( +1) and - is replaced by m, the equations (a) and (b) can be combined to: = 00 + B σ m, m = 1, (3) Lines with positive m-values ( = +1) form the R-branch, while those with negative m- values ( = -1) form the P-branch; see Figure 5. Approximation According to previous discussions we know that the distance of equilibrium (R 0 ) should increase slightly with increasing vibration energy which leads to that B decreases for increasing n. If the rotational constants are denoted B and B in the upper and lower state respectively, instead of equation 3 we get: 00 ( B ' B ") m ( B ' B ") m (4) m = +1 =1,, 3,... m = - = -1, -, -3,... R-branch P-branch Note that if B and B are equal the quadratic term vanish and equation 3 still holds. Since B B the distance between the spectral lines will decrease successively in the R-branch while the distances will increase in the P-branch. For sufficiently high a conversion of the R-branch occurs resulting in the formation of a so called band head. Equation 4 has a parabolic shape and is called a Fortrat parabola; see Figure 6. Figure 6 Fortrat parabola. 11

12 Approximation 3 If the distortion term cannot be neglected, instead of equation 4 we get: ( B ' B ") m ( B ' B " D ' D ") m ( D ' D ") m ( D ' D " m (5) 3 00 ) Equation 5 show that is given by a fourth order polynomial in m. Relative intensities temperature measurements At thermodynamic equilibrium the population of the different rotational levels, within a given vibrational level, is given by the Boltzmann distribution. B( 1) / kt N ( 1) e We have here regarded that each rotational state has +1 magnetic sublevels and assigned the population of the lowest rotational level ( = 0) to 1. The relative intensity of a spectral line is generally given by: I () A N where () is the relative detection efficiency and A is the transition probability. If the A value is assumed to be the same for all rotational transitions and is constant it is seen that the relative intensity is a direct measure of the relative population of different rotational levels, which in turn depends on the temperature. The observed relative intensities in the R- and P-branches can therefore be utilized for non-contact temperature measurements. The highest populated rotational level in the lower vibrational state at a given temperature is approximately given by: max R 0 k T 1 In fact the probability for absorption depends on both the -value and. One can show that a better model for the relative intensity is: I ( " ' 1) e B"( " 1)/ kt where " is the rotational quantum number in the lower state and ' is the rotational quantum number in the upper state. This results in different relative intensities in the R- and P-branch. B"( " 1) / kt IR (" ) e, " 0,1,,... ' " 1 B"( " 1) / kt IP " e, " 1,,3,... ' " 1 4 1

13 Atomic and Molecular Spectroscopy 017 Figure 7. Relative intensities in the R- and P-branch in CO. Note that the R- branch is slightly more intense. Car exhaust Air pollution contributes to deteriorated health for many. The traffic is believed to cause between 300 and 000 cancer incidences in Sweden each year. The most common pollutions in car exhaust are: Carbon monoxide (CO) Carbon monoxide is a gas without smell or colour that is formed when the combustion in an engine cylinder is incomplete. When engine temperature is low the content of CO in the exhaust is especially high. The health risks due to CO are that it decreases the oxygen transporting ability of the blood. CO also affects the heart and artery system. The problem with too high concentrations of CO has been reduced with improved emission control. Hydrocarbons (C x H y ) During combustion of petrol and diesel oils the exhaust contains hydrocarbons due to incomplete combustion. Other harmful substances, such as ozone, are created when the hydrocarbons spread in the air. Hydrocarbons are also deposited in crop fields and through the food accumulated in the human digestive system. Hydrocarbons can yield irritation in eyes and throat. Certain types are directly carcinogenic while others are suspected to increase the risk for developing cancer. The ozone decreases the physical performance something which is serious for especially asthmatic people. Nitrous oxides (NO x ) Nitrous oxides are formed during combustion, especially in engines that are subject to high load. The car traffic contributes largely to the total emission of nitrous oxides in urban areas. Nitrous oxides forms, together with hydrocarbons, ozone. Nitrous oxides can result in decreased breathing capability, especially difficult for asthmatic people. 13

14 Particles The combustion of fuel results in particles, in particular soot. Emission from diesel engines is the major part of the particle emission from vehicles. Increased doses affect breathing. Emission control by catalytic converter Most vehicles on the market today have either a Otto- or a Diesel engine. Most of the Otto engines have a so called three way catalytic converter that reduces the emission of NO x, C x H y and CO. The condition for optimal catalytic performance is that the engine runs with a constant air/fuel-mixture. This condition is not fulfilled in a Diesel engine since this engine runs with an abundance of air. In a three way catalytic converter the three following reactions occur. CO + O CO Oxidation of carbon monoxide C H 4 + 3O CO + H O Oxidation of hydrocarbon NO + CO N + CO Reduction of nitrous oxide In spite of the catalytic emission control present in almost all cars of today the emission of nitrous oxides hardly decreases. There are several reasons to this: 1. Cold engine starts and short trips. A cold catalytic converter does not operate optimally.. The catalytic converter is aging fast which reduces the efficiency. 3. The emission control is not working properly at high engine loads and high rpm:s (revolutions per minute). Laboratory exercises Electronic spectra As an example of electronic molecular spectra you will study the emission from the blue green part of a butane gas burner using a grating spectrometer. 1. Get familiar with the grating spectrometer. You can open the cover and look inside. What type of grating spectrometer is it?. The light from flame needs to be focused onto the slit of the grating spectrometer. Use a large lens between the flame and the slit. As help you can use another bright light source to find the correct positions for the lens and source before you change to the flame. 3. Start to study the spectrum using a grating with only 300 lines/mm, i.e. with low resolution. The light from the flame is a result of 14

15 Atomic and Molecular Spectroscopy 017 chemical reactions where the reactive products are created in excited states, which disintegrates (so called chemo luminescence). In the blue-green range you see light from e.g. the C molecule which originates due to the following reactions: CH * C C H C* C hc/λ The three almost unresolved peaks in C are called the Swan band and arise from transitions where n = 0, 1 in the molecule. Identify the peaks using the tabulated data provided by the lab instructor. 4. Change to the grating with 400 lines/mm and acquire a new spectrum of one of the peaks. Is the spectrum shaded to the red or the violet? Why does shading occur? Which of the states has the largest rotational constant (B or B )? Discuss what state (upper or lower) that has the largest equilibrium distance between the atoms. Rotational-vibrational spectra Now we will study a rotational-vibrational spectrum for CO within the electronic ground state by observing the vibrational transitions n = 0 to n = 1 at the wavelength 4.5 m (00 cm -1 ). Furthermore we will study emission from cars using molecular spectroscopy. 1. Get acquainted with the Fourier transform spectrometer and the software controlling it. Can you identify the most important parts of the spectrometer? See Appendix B for information about the software.. Acquire background spectra (no sample in sample compartment) with different spectral resolutions (0.5, 0.5, 1 cm -1 ). Note how the structure appears when the resolution is increased. Try to figure out what gases in the air that you see in the spectra. 3. Measure the absorption of the CO cell. Acquire a spectrum with high resolution of the transition n = 0 to n = 1. Identify the lines in the R- and P-branch with respect to the rotation quantum number "and '. Assign an m-value to each line; see equation 3 and 4. Determine the wave numbers of the lines ( obs ) by fitting a Gaussian profile in the computer program Gfit, See Appendix B about Gfit. 4. Use MatLab to plot obs as a function of m. 5. Analyze the measured data using MatLab. First assume that the molecule is a rigid rotator. Determine 00, the rotational constants B and B, as well as the equilibrium distances. Study the residual of this fit conclusion? 6. Improve your model! 15

16 Appendix A: Fourier Transform Spectrometer A Fourier transform spectrometer is a Michelson interferometer where one of the mirrors can move with a constant speed. Hence the optical system is simple, the interferometer consists of two mirrors, one detector and a beam splitter. Figure A1 shows how a beam from A hits the beam splitter C. Some part of the light (Ray 1) is reflected by the beam splitter towards the mirror M 1. The ray is reflected from mirror M 1 back through C to the detector. The rest of the light (Ray ) is transmitted through the beam splitter and is reflected on mirror M. Ray is then reflected at C towards the detector. Fixed mirror Moving mirror Ray 1 Light source Ray Beam splitter Detector Figure A1 A collimator is irradiated with monochromatic light yielding a parallel ray of light. The ray is split into two components in the beam splitter. Following reflection in the mirrors another passage through the beam splitter occurs and thereafter the rays are added on the detector. All radiation incident on the interferometer is registered by the detector or is reflected back to the light source. If both mirrors in the interferometer are positioned at equal distances to the detector the two light rays propagate exactly the same path length and arrives at the detector in phase, thus interfere constructively. The two components that interfere at the light source are on the other hand completely out of phase. In this case all incident energy is transferred to the detector. If the movable mirror (M ) is translated a quarter of a wavelength optical retardation of a half of a wavelength is achieved since the light propagates back and forth. This results in complete destructive interference of the detected signal. So when the moveable mirror is translated the detector registers darkness followed by brightness. 16

17 Atomic and Molecular Spectroscopy 017 A Fourier transform spectrometer can be used to study both emission spectrum from a light source and absorption spectrum from a gas sample placed in the light path. The latter is the most common modality and this is what we will use. For absorption measurements a light source with continuous spectrum in a broad wavelength range is utilized. The gas sample is placed between the beam splitter and the detector. Since the light path in the interferometer is in air absorption due to air will occur even outside the sample cell. Therefore a measurement is done in two steps: first a background is acquired without the sample cell, secondly a measurement is done with the cell place in the sample compartment. The difference of the measurements will then be a measure of the absorption in the sample under investigation. Specifications for the instrument Beam splitter Wave number / cm -1 Wavelength / m KBr Quartz Detector Wave number / cm -1 Wavelength / m DTGS InGaAs Si (Red) Si (Blue) Fourier analysis For monochromatic light with the wavelength and wave number =1/ the phase difference at the detector is given by x where x is the path length difference between the two light paths. The refractive index is here set to 1. According to the formula for interference between two waves the total intensity I (x) is given below. I ( x) I01 I0 I01I0 cos( ). If I01 I0 I0 we can write: I ( x) I0(1 cos( x)) 17

18 If one of the mirrors is moving with a constant speed (v/) we get, since x vt, a time dependent signal which oscillates around a mean value of I I ( x) I (1 cos v t). Thus we get a signal, the so called interferogram, with an oscillation that is dependent on the wave number of the incident radiation. For a monochromatic light source, e.g. a HeNe-laser, this means that the detector registers an interferogram with cosine-shape (Figure A). Relative intensity Mirror displacement (λ) Figure A Intensity as a function of mirror translation for monochromatic light. A polychromatic light source emits light in many frequencies. Each frequency gives rise to a cosine shaped signal and the resulting interferogram is the sum of all of these. Only where the distance from both of the mirrors to the beam splitter is equal we get constructive interference for all wavelengths. At all other distances some wavelengths interfere constructively while others do not. With Fourier analysis one can convert the intensity as a function of the mirror displacement (interferogram) to the intensity as a function of the wave number (the spectrum). If the light source intensity distribution is I() the mean intensity that reaches the detector is proportional to I()d and a slowly varying function of that takes into account the transmission and reflection coefficient in beam splitter and mirrors. The modified source function is denoted B(). We now instead detect: I I ( x) B( )(1 cos x) d B( ) d B( ) cos(x ) d 0 0 B( )cos(x) d where I is the constant level with no modulation present. The second term gives all information about the spectrum. The lower integration limit can be set to

19 Atomic and Molecular Spectroscopy 017 since B( ) 0 for all negative. I(x) is defined as the modulated part of the interferogram which yields I ( x) B( ) cos( x) d and the spectrum is achieved through the inverse Fourier transform of this relation B ( ) I( x) cos( x) dx There are though two fundamental differences from these equations in real applications. First the spectrum is acquired over a finite distance, i.e. x = L. In this instrument L is maximum 8 cm. The finite distance limits the resolution of the instrument so that the smallest structure that can be resolved is 1/L, that is 1/8 = 0.15 cm -1 in this case. Secondly, discrete sampling points, with the interval x, are used to calculate the Fourier transform which results in a Fourier sum instead of a Fourier integral. In this instrument the minimum step size is cm -1. Fourier transform spectroscopy compared to grating spectroscopy It is instructive to compare the Fourier transform technique with a more traditional spectroscopy modality such as one based on grating spectrometers. An interferometer does not separate the light into its frequency components before it is registered. This implies that each point in the interferogram consists of information about all wavelengths in the light source. So if 8000 sampling points are utilized to acquire the interferogram this means that all wavelengths are registered 8000 times. This yields the same data collection efficiency as a spectrograph or a CCD detector. On the other hand, if we compare with a monochromator where the intensity is measured at the exit slit as a function of the grating angle, one achieve a lot with the FT spectrometer. A monochromator that collects 8000 sampling points only measures the wavelength once. This is usually referred as the Fellgett advantage. In a spectrograph the resolution depends on the slit width since the spectral lines are images of the entrance slit. To achieve a higher resolution a thin slit is required which reduces the intensity. Furthermore, a lot of light is lost in the spectral orders that are not detected. In an interferometer, on the other hand, no slits are needed and the only light lost is the portion if the light that is reflected back to the light source (approximately half). This leads to that more energy can be registered than what is possible with a dispersive spectrometer which increases the signal-to-noise ratio in the spectrum. This is referred to as the Jacquinot advantage. The Fellgett and Jacquinot advantages imply that a Fourier transform spectrometer can acquire a high-quality spectrum in the infrared wavelength region in just a fraction of the time required for a dispersive spectrometer to do 19

20 the same thing. Additionally, in a Fourier transform spectrometer the resolution of the detected intensity can be increased by increasing the scanning length of the moving mirror. In a dispersive instrument the wavelength determination exactitude and precision depends on 1. Calibration using external reference wavelengths, i.e. one calibrates the measured spectrum with wavelengths acquired at another time.. The mechanical ability of the instrument to repeat grating movements and aperture translation exactly during and between different measurements. In comparison, the Fourier transform spectrometer has an internal frequency standard, usually a HeNe-laser (Figure A3). The moving mirror and the detector sampling interval are timed through the interference fringes that arise from the monochromatic light from the laser. All wave numbers in the resulting spectrum are then calculated using the known laser frequency. This is called the Conne advantage. A Fourier transform spectrometer can reach a precision in wave number better than 0.01 cm -1. This means that spectra can be compared irrespective of if these are acquired with five minute interval or five year interval. Fixed retro-reflective mirror Moveable retroreflective mirror Figure A3 The ray trace of the HeNe-laser in the FT-spectrometer. In a FT-spectrometer the resolution is the same for all wave numbers while the resolution of the grating spectrometer is wavelength dependent. Finally, a disadvantage of the FT-spectrometer is that the light source has to be very stable since fluctuations in the light source (i.e. if B() is time dependent) affect the inverse transform. 0

21 Appendix B: Gfit and Butane lines How to find the peaks in Gfit? 1. Import your saved spectrum i. Open "Gfit" from the desktop. ii. Click "Use Windows Open File Dialog". iii. Click OK. iv. Find your file.. Zoom to the appropriate area i. Click "zoom". ii. Click just to the left of your region of interest. iii. Click to the right of you region of interest. 3. Find peaks i. Click "Fit Control" ---> "Set parameters auto start values" ii. Click on the "Yes" at the "Set Intensity Limit Graphically" iii. Click at the desired threshold to find peaks (at the lowest intensity you want the program to consider peaks, where you click in the x-direction doesn't matter). iv. Click "Test Fit". v. If the program found peaks where you don't want peaks, click "Clear plot" and try to increase the threshold slightly. vi. When the program only found peaks that you want (perhaps some are still missing), click "OK". vii. It is important that you don't lack any peaks in between other found peaks (if some at the edges of your branches are missing that is alright. So to add missing peaks click "Peak". Left click on all peaks you want to add except for the last peak you want to add; this one you right click on. viii. When you found all the peaks you want, but no peaks that you don't want, continue to the next step. 4. Fit Gaussian profiles to the peaks. i. Click "Fit 1/repeat" 5. To the left you got two tables, the left contains the wavenumbers and the right contains the intensities. Mark these two columns. The data is now automatically copied to the clipboard. You can now paste it into a text file or directly in to MATLAB.

22 Atomic and Molecular Spectroscopy 01 3