lecture11_2011

lecture11_2011 - 28Jan2011 Chemistry 21b Spectroscopy...

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Unformatted text preview: 28Jan2011 Chemistry 21b Spectroscopy Lecture # 11 Vibration-Rotation Spectra of Diatomic Molecules What happens to the rotation and vibration spectra of diatomic molecules if more realistic potentials are used to describe the interatomic interaction? So for we have just assumed that R = R e , and that no change in the bond length is possible. Clearly, for a realistic interatomic potential, such as those outlined in Figure 4.1, considerable changes in bond length with occur as one moves up in energy from the ground vibrational state. As a one-dimensional equation, the radial nuclear motion described by Eq. (6.4) can be solved numerically quite simply, and there are in fact computer routines which invert experimental data to yield diatomic potentials V ( R ) = E el ( R ) with spectroscopic precision. Considerable physical insight, however, can be gained by using simple, analytical potentials for V ( R ), and analyzing the resulting energy level expressions. One of the simplest and most useful potentials was developed by Morse in the early part of this century, which well proceed to examine now. 11.a. The Morse Potential Clearly, an appropriate potential must go to small values as R , have a minimum at R = R e , and go to very large values as R 0. A simple potential which does this, the Morse potential, is given by V ( R ) = D e (1 e a ( R R e ) ) 2 , (11 . 1) where the dissociation energy D e , the equilibrium bond length r e , and the curvature of the potential near the minimum a are three adjustable parameters. The major qualitative problem with the Morse potential is that is does not go to as it should as R 0. However, we know from our examination of the harmonic oscillator that the wavefunctions tail away rapidly to zero in the classically forbidden regions, and so as long as the potential is sufficiently large at small R this should not pose a significant problem. If the Morse potential is placed into the radial nuclear motion equation, Eq. (6.4) on page 47 of the notes, we find: d 2 S ( R ) dR 2 + braceleftbigg J ( J + 1) R 2 + 2 h 2 [ E D e D e e 2 a ( R R e ) + 2 D e e a ( R R e ) ] bracerightbigg S ( R ) = 0 , (11 . 2) where the substitution F ( R ) = 1 R S ( R ) has been made. Like all such problems in quantum mechanics, the challenge is in finding the right substitution of variables that allows Eq. (11.2) to be recast in a form that has well-known solutions. Well only briefly outline the procedure here. The first set of substitutions involve the equations y = e a ( R R e ) and A = J ( J + 1) h 2 2 R 2 e . (11 . 3) 86 If these are subsituted into Eq. (11.2), we find d 2 S dy 2 + 1 y dS dy + 2 a 2 h 2 parenleftbigg E D e y 2 + 2 D e y D e AR 2 e y 2 R 2 parenrightbigg = 0 (11 . 4) For A negationslash = 0, the last ( R 2 e /R 2 ) term must be expanded in order to generate an equation containing only y . The first three terms of the Taylor expansion are given by....
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lecture11_2011 - 28Jan2011 Chemistry 21b Spectroscopy...

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