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Micro wave spectroscopy Theory In the experiment entitled Molecular Spectroscopy we learn that to a gross approximation the total energy of a gaseous molecule can be expressed as a sum of electronic vibrational, rotational and translational parts. If the molecule possesses magnetic or electric moments and if there are external fields, further energy terms arise. And of course, interactions among the various motions, for example vibrational-electronic (vibronic), spin-rotation, and rotational-vibrational (rovibrational) must be added as (usually small) correction terms to the total energy of the molecule. The rotational energy of a linear molecule, approximated as a rigid rotor is BJ (J+1), where J = 0,1,2,3… is the rotation quantum number, and where B=h/8π2cI is the rotational constant in reciprocal centimeters (cm-1). For molecules other than hydrides, the moment of inertia I is such that B ≤ 1 cm-1, and the rotational levels are so close together that only “optical” spectrographs or spectrometers of the highest resolving power are capable of detecting the rotational structure on vibrational or vibronic transitions; and even these instruments fail when B << 1 cm-1 as is the case for most molecules containing four or more heavy atoms. Multiplication of B(cm-1) by the velocity of light c shows that in frequency units B is of the order or less than 30 GHz (1 GHz = 1 gigahertz = 109 cycles/sec). This places transitions between the pure rotational energy states of most molecules in the microwave region of the spectrum, where resolution of rotational structure is much better than that in the optical region. Just as in the classical problem where an oscillatory electrical field can cause an electric dipole to rotate with the frequency of the impressed field, the oscillatory electric vector of electromagnetic radiation can similarly “drive” a quantum mechanical rotor possessing a permanent electric dipole moment. Similarly, the magnetic vector of the electromagnetic wave can interact with a rotor possessing a permanent magnetic moment. Thus molecules such as OCS and BrCN, having electric moments, and O2, having a magnetic moment, have rotational spectra in the microwave region, while a molecule such as carbon dioxide does not. For a quantum mechanical rotor, the energy of the electromagnetic field is absorbed only when the frequency of the field is near that corresponding to the energy difference between two discrete energy states of the rotor. The resulting absorption lines for the quantum oscillator constitute the microwave spectrum that one observes. Working backward from the spectrum then, one can ascertain the rotational energy states of the molecule, the moments of inertia of the molecule, and therefore some information about the dimensions of the molecule. In certain cases, one can obtain sufficient information to obtain all the bond distances in the molecule, or for a nonlinear molecule, all the bond distances, bond angles, mass of isotopes(according to reduced mass) and moment of inertia. This is one of the primary goals of microwave spectroscopy. Intensities of Transitions and Selection Rules As mentioned in the Introduction, charges or magnetic moments can interact with the E and H vectors associated with electromagnetic radiation. The selection rules for pure rotational transitions in a linear rotor. These are ΔJ=±1, ΔM=±1 x,y-polarized ΔJ=±1, ΔM=0 z-polarized The rotational energy levels have constant distance between each two rotation levels that equal to 2B, this come out by differences of two levels through energy value as following: âđē = đēJ+1 - đēJ = B J+1(J+1+1) - B J(J+1) âđē=2B(J+1) In this stage there are constant in rotational levels 2B,4B,6B, and so on…. Can be make a table of energy measurements for several values of rotational quantum numbers:J đēJ J-…..J 0 1 2 3 4 0 2B 6B 12B 20B ------0---1 1-----2 2------3 3------4 Absorption spectral line Cm-1 --------First line at 2B Second line at 4B Third line at 6B Fourth line at 8B Stark Effect An atom or molecule possessing electrical charges can interact with a static external electrical field. The result of such interaction is called the Stark Effect. The energy of a stationary dipole in a field is given by, W=−μo.E =−μo.Ecosθ, where θ is the angle between the dipole and the field E. It is convenient to define the field direction in space as the z-axis, since then θ, as before, is just the angle between the space-fixed z and the molecule-fixed z' axes. A rotating dipole, with angular momentum vector perpendicular to μo represents the linear rotor in microwave spectroscopy. If the dipole is rotating very rapidly, the effect of the field largely averages out. However, there is a tendency for the field to change the angular momentum, and therefore the energy of the rotating dipole, a tendency that naturally becomes more pronounced as the angular velocity of the rotor decreases. The quantum mechanical description for this change of angular momentum is one based upon the “mixing” of J-states for the linear rotor. The angular momentum states of the molecule in the field are therefore no longer pure J-states of the free molecule, i.e. J is no longer a “good quantum number”. Because of the form of the Hamiltonian, H'=−μo.Ecosθ, the mixing of states in this case follows the same “selection rules” as do dipole transitions; namely, only states differing in J by ± can mix. The energy level diagram given in Fig. 1 is useful in visualizing the effects of the Stark field. The vertical arrows indicate the transition J=1←0,ΔM=0with and without the Start field. Note that this transition is shifted to higher energy by an amount proportional to the square of the molecular dipole moment, μ0. The Stark effect is thus useful in determining dipole moments. Figure 1.Energy level diagram for ridged rotator, illustrating the stark effect. Hyperfine Structure In addition to angular momentum due to molecular rotation, some molecules possess nuclear angular momentum because of the nuclear spin of one or more of their nuclei. The proper description of the rotational states must account for both the molecular and nuclear-spin angular momentum contributions to the total angular momentum. If no interaction exists between the orientation of the nucleus and that of the molecule, the energy of a given molecular rotation state designated by J is unaffected by the nuclear spin, I. Such interactions exist however, and lead to hyperfine structure in molecular spectra. In most atoms, hyperfine structure mainly arises because of interaction between magnetic moments of the nuclei and of the electrons. In most molecules the electronic magnetic moment is zero, and angular-dependent electrostatic terms are often the largest in the hyperfine interaction. These terms arise because of the interaction of electric charges in the nuclei with electrons in the molecule. The angular dependence of the interaction naturally vanishes if either charge distribution is spherically symmetric, or this is the case for S-states of atoms or when the nuclei themselves have spin I<1. Expanding the interaction energy in a multipole expansion, the lowest-order non-zero angular-dependent term is that where the electric quadrupole moment of the nucleus interacts with the gradient of the electrostatic potential due to the electrons. The dependence of this interaction upon the orientation of the quadrupolar nuclei gives rise to small energy corrections to the rotational states of the molecule. For a linear molecule, the nuclear quadrupole hyperfine energy terms are given by, F is the total angular momentum quantum number inclusive of nuclear spin I. F takes on all values I+J,I+J−1……I−J. WQ vanishes for I=0 or I=1/2 since in both these cases the nuclear quadrupole moment Q is zero. The quantity ∂2V/∂z'2, usually denoted q, is the second derivative of the electrostatic potential at the quadrupolar nucleus along the molecular axis due to all charges outside a small sphere containing the nucleus. The correction term to take into account electrons that penetrate into the nucleus can be shown to be very small. The selection rules for transitions among the hyperfine levels are as follows :- ΔJ=0,±1 ΔF=0,±1 ΔI=0 A transition J=1←0 with I=32 will thus be split into three lines corresponding to as shown in Figure 2. Note that the ΔJ=+1 and ΔF=0,±1 sign of eqQ is important in determining the appearance of a hyperfine multiplet. A detailed examination of electric quadrupole interactions reveals considerable information relating to the electronic charge distribution in molecules. One of the principal difficulties in interpretation of the data is in the separate determination of the quantities in the product eqQ, known as the quadrupole coupling constant. However, by measuring this product for a series of molecules incorporating a given nucleus, Q thus remaining constant, the relative variation of q with molecular structure can be investigated. Figure 2. Energy level diagram for linear rigid rotator. Rotational Motion The rotational motion of a molecule is determined by the moments of inertia and the angular momenta. 1. Classically, any object has three orthogonal principal moments of inertia (diagonals of inertia tensor) with corresponding simple expressions for the rotational energy and angular momentum. 2. This carries over to quantum mechanics, and it is customary to classify the rotational properties of molecules according to the values of the principle moments of inertia. The principle moments of inertia are designated Ia, Ib, and Ic in order of increasing magnitude Linear Molecules & Symmetric Tops A molecule which is linear or has an axis of rotational symmetry is called a symmetric top Either Ic = Ib > Ia or Ic > Ib = Ia At this time the same equation of two linear molecule has been using for the distance between the spectral lines 2B but with small value of bond distortion constant (D) to give up the real estimation of bond according(non rigid rotate).example// O=C=S,O=N-H,H≡C-Cl • Linear molecules (methyl halids)have a small I about the axis of the molecule so they are of the first type and are called Prolate symmetric tops • Other molecules, e.g., benzene, have the largest moment of inertia about the symmetry axis & are called oblate symmetric tops • Molecules which are spherically symmetric, e.g, methane have three equal moments of inertia and are called spherical tops – Molecules with Ic ≠ Ib ≠ Ia are asymmetric tops Prelaboratory Questions 1) Calculate the exact abundances and masses of the isotopic variants of OCS. 2) Describe the normal mode vibrations of OCS. Look up the vibrational frequencies and indicate any degeneracies, where appropriate. 3) With the available experimental apparatus, only rotational transitions of OCS with are observed. How could transitions with ΔM'J=O ; ΔM'J=±1 be observed? 4) Calculate the stark field (volts cm-1) required to produce a stark splitting of 100 MHz for the most abundant isotropic species (use the literature parameters given in the notes). 5) Major contaminants in the vacuum system are N2, O2, Ar, CO2, and H2O (assuming it is not a smoggy day). Would you expect any of these species to give lines in the 8-12 GHz portion of the microwave spectrum? Laboratory Experiments Procedure The purpose of the experiment is to study a portion of the microwave spectrum of the linear molecule OCS. Various modifications of the molecule are possible because of the presence of isotopes of O, C, and S in natural abundance. Prominent in the microwave spectrum are 16O12C32S, 16O12C34S, 16 13 32 O C S, 16O12C33S. In addition, rotational transitions in thermally populated excited vibrational levels of the most abundant species 16O12C34S can also be detected. The spectra allow the study of: 1) The rotational states of OCS and determination of the bond distances in the molecule; 2) The Stark splitting, the dependence of this splitting on E, and a determination of the dipole moment of OCS; 3) The nuclear quadrupole splitting due to the 33S nucleus in the molecule 16O12C33S, and a determination of the quadrupole coupling constant. Assignment 1) Measure the line positions for the 16O12C32S, 16O12C34S, 16 13 32 O C S, 16O12C33S (in this case refer to the diagram to calculate the line position in the absence of quadrupole splitting), and (if possible), 18O12C32S species and calculate the bond distances. 2) Measure the stark splitting for the calculate the dipole moment. 16 O12C32S species and 3) Measure the line positions for the 16O12C33S multiplet and calculate. You should think about a method to determine the line spacing with reasonable accuracy. 4) Measure the position of the line for the 16O13C32S species and that of a nearby less intense line. Verify that the latter line arises from a vibrationally excited species by cooling the stark cell with dry ice. Notes on the Microwave Spectroscopy Experiment Calculating moments of inertia, I, from the known bond distances is straightforward given one of the formulas discussed below. You should do this to determine where to look for the lines for each isotopic species: 16O12C32S, 16O12C34S, 18O12C32S, 16 13 32 O C S, 16O13C33S. What are the relative abundances and do they correctly correspond to the observed relative intensities? The predicted line positions should help to identify the line due to a vibrationally excited species. This line can be positively identified by cooling the cell to reduce the population of this species and thus its relative intensity. Use dry ice, but be careful not to freeze the o-ring seals at the ends of the cell. What information can you get from this line position? When you have collected your data and converted it to moments of inertia, you should use it to calculate the bond distances. Line positions for any two isotopic species can be used to obtain two bond distances. What assumption is made in doing this calculation? The bond distances can be calculated to all pairwise combinations of the isotopic species for which you have obtained line positions. Which pairs should give the most accurate bond distance values? Why? The moments-of-inertia to bond-distance calculation are not straightforward. These notes outline one way that you might precede. Two isotopic species differing only in their sulfur masses, m3, are considered as an example. A change of subscripts allows these calculations to be applied to isotopic species differing at oxygen. The case of isotopic species differing at carbon is different and is left as an exercise. Moments of Inertia from Bond Distances We'll start with the relation defining the COM m1r1+m2r2+m3r3=0 (1) And the definition of the moment of inertia about the COM I=m1r12+m2r22+m3r32. The bond distances are l23=r3−r2orr3=r2−l23 (2) l12=r2−r1orr1=r2−l12 (4) and let l13=r3−r1=l12+l23 (3) (5) Substituting (3) and (4) into (1) to eliminate r1 and r3 m1r2−m1l12+m2r2+m3r2+m3l23=0 Mr2=m1l12−m3l23 (6) Where M=m1+m2+m3 (7) Substituting (3) and (4) into (2) to eliminate r1 and r3 I=(m1r2−l12)2+(m2r22+m3r2+l23)2 I=Mr22+2(r2m3l23−m1l12)+m1l122+m3l232 (8) Now, we may substitute (6) into (8) to obtain I=M(m1l12−m3l23)2/M2+2[(m1l12−m3l23)(m3l23−m1l12)]/M +m1l122+m3l232 I=m1l122+m3l232−(m1l12+m3l23)2/M (9) This is an expression for the moment of inertia in terms of the bond distances. It may be rewritten: I=1/M.[m12l122+m1m2l122+m1m3l122+m1m3l232+m2m3l232+m32l232− m12l122+2m1(m3l12l23−m32l232)] =1/M.[m1.m2l122+m2m3l232+m1m3.(l122+2l12l23+l232)] I=1/M.[m1m2l122+m2m3l232+m1m3l132] (10) Where (5) has been used in the last step. So far, we have not considered isotopic substitution so the results are perfectly general. Bond Distances from Moments of Inertia Starting with the moments of inertia I and I' for two species differing in mass m3, we wish to obtain the bond distances. One approach starts with equation (10): I=1/M.[m1m2l122+m2m3l232+m1m3l132] (10) For the isotopic species I'=1/M.[m1m2l122+m2m'3l232+m1m'3l132] (11) Multiplying (10) and (11) by m'3M and M3M' respectively m'3MI=m1m2m'3l122+m2m3m'3l232=m1m3m'3l132 (12) 2 2 2 m3M'I'=m1m2m3l12 +m2m3m'3l23 +m1m3m'3l13 (13) Subtracting (13) from (12) m'3MI−m3M'I'−m1m2(m'3−m3).l122 (14) or l122=m'3MI−m3M'I'/(m1m2m'3−m3) (15) This may be substituted into (10) using (5) and the resulting quadratic equation solved for l23. Diagrammatic figure of molecular bonds. Example// Rotation spectrum of 76Br19F has a serious lines in equal distance from each other (0.71433cm-1).calculate the rotation constant, moment of inertia, and length bond of molecule. Answer // 1-Equal distance= 2B =0.71433cm-1 ⇒ The rotation constant=B=0.35717 cm-1 2- Moment of inertia=h/8π2BC ⇒ I=7.83x10-96 kg.m2 3- length bond of molecule= r= (I/đ)0.5=0.1755nm