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Unformatted text preview: 24Jan2011 Chemistry 21b Spectroscopy Lecture # 9 Rotation of Polyatomic Molecules The rotational spectra of molecules can be classified according to their principal moments of inertia. Assume that the molecule rotates as a rigid body, that is, the relative nuclear positions are fixed. The moment of inertia I of any molecule about any axis through the centerofmass is then given by I = summationdisplay i M i R 2 i (9 . 1) where M i and R i are the mass and perpendicular distance of atom i from the axis. Recall that the centerofmass position for a given axis may be found from the equation: R ( C.M. ) = summationdisplay i M i R i / summationdisplay i M i . In general, for any rigid three dimensional body, there exists a 3 3 moment of inertia tensor (that is, matrix). If we use any three axes defined by ( xyz ), the moment of inertia tensor has diagonal elements defined as I xx = i M i ( y 2 i + z 2 i ) (using the xaxis, for example), and offdiagonal elements given by I xy = i M i x i y i , etc. (Note the minus sign in the offdiagonal terms). Clearly, the momentofinertia tensor is Hermetian, and so can be diagonalized to yield three (potentially) distinct eigenvalues. These eigenvalues are called the principal moments of inertia, and the eigenvectors corresponding to the diagonalized coordinate system are referred to as the principal axes of the molecule. Mathematically, the eigenvectors of the matrix in any coordinate system form the direction cosines that relate the arbitrary coordinate system to the principal axis coordinate system, that latter of which must rotate with the molecule. One can always find one axis, called the caxis, about which the moment of inertia has its maximum value, and another axis, labeled the aaxis, about which I has its minimum value. It can be shown that the a and c axes must be mutually perpendicular. Thus, according to convention, the principal axes are ordered: I c I b I a . (9 . 2) Why do we care? If we use these principal axes, then the components of the rotational angular momentum P along these axes can be shown to be P a = I a a , P b = I b b , P c = I c c , (9 . 3) and the kinetic energy operator for a rigidrotor simply becomes T rot = P 2 a 2 I a + P 2 b 2 I b + P 2 c 2 I c 71 B F F F (c) (a) (b) r BF 0.13 nm 60 Figure 9.1: The structure and principal axes of boron trifluoride (BF 3 ). In all other coordinate systems things are much more complicated! As a simple example, let us calculate the moments of inertia for the molecule BF 3 . This molecule is planar, with the three F atoms arranged around a central B atom at angles of 120 and a BF distance of 1.30 A. (See Figure 9.1). The center of mass is clearly at the B atom. One principal axis is perpendicular to the plane of the molecule, and since this axis will possess the largest moment of inertia, it is the caxis. A second axis will pass through one of the BF bonds, and the third will be mutually perpendicular to the first...
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This note was uploaded on 01/03/2012 for the course CH 21b taught by Professor List during the Fall '10 term at Caltech.
 Fall '10
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 Atom, Mole

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