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Unformatted text preview: Ch21b Winter 2011: Solutions for Final Exam Prepared by Jason Crowley, Scarlett Dong & Geoff Blake Problem 1 Use group theoretical arguments to answer the following. You can be brief! a). Imagine a perfect surface of a transparent dielectric solid in the (x,y) plane, with the space fixed zaxis being normal to the surface. Suppose we trap water molecules on this surface so that they are not free to rotate, and further that they are oriented perpendicular to it. If you try to take the vibrational spectrum of these adsorbed molecules, what polarizations of the applied E&M field are needed to detect the three fundamental vibrations and their overtones and combination bands? Figure 14.3 will be helpful here! The fundamental vibrations are A 1 and B 2 , and thus any combination bands or overtones must also be A 1 or B 2 . Since the overall symmetry must be A 1 and the (x,y,z) operators are B 1 , B 2 , and A 1 , all of the A 1 modes are permitted via the zcomponent of the field and the B 2 mode(s) via the ycomponent. Thus, with linearly polarized light along the zaxis only the symmetric stretch and bend fundamentals would be observed, with no absorption observed at all for xpolarized light (and only the asymmetric stretch for the ycomponent, of the fundamental vibrations). b). Assess the polarization of the 1 A 2 1 A 1 transition in H 2 CO and the 1 B 2u 1 A g transition in ethylene (CH 2 =CH 2 ), including vibrational coupling if necessary. Is the latter transition electric dipole allowed? Both transitions involve singlet to singlet spin states, we we are good on that front. For formaldehyde, we use the C 2v group. We need the overall symmetry of the matrix element(s) to be A 1 and the (x,y,z) operators are B 1 , B 2 , and A 1 . Thus, the transition is not allowed by strict electric dipole selection rules, and we need to consider vibronic coupling. From the class notes, we have that the vibrations are A 1 , B 1 , and B 2 , which yields composite symmetry products of A 2 (A 1 ,B 1 ,B 2 ) = (A 2 , B 2 , B 1 ). The latter two can couple with the y and xcomponents of the field, respectively, to produce transitions. The group for ethylene, D 2h , is more complex but still nondegenerate; so each symmetry type is again its own inverse. All the position operators are u , and so the geradeungerade rules are fulfilled for this transition. The (x,y,z) operators are (B 3u , B 2u , B 1u ), and so the origin band (no vibrations) is permitted through the ycomponent. What about vibrations? These are tabulated in the Chapter 3 problem sets of Harris and Bertolucci, and give for the vibronic symmetry products in the excited state: B 2u (A g, B 1g ,B 2g ,A u ,B 1u ,B 2u ,B 3u ) = B 2u ,B 3u ,A u ,B 2g ,B 3g ,A g ,B 1g (see the character table). The gerade states we can through out immediately, and only B 2u and B 3u states (from the A g and B 1g vibrations) are connected to the ground state through the y and xcomponents, respectively. Problem 2...
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