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Unformatted text preview: 11Feb2011 Chemistry 21b – Spectroscopy Lectures # 17 – The Vibrational Modes of Periodic Solids In the infrared spectra of functional groups we covered last time, we saw that for so-called molecular solids (or liquids) – in which the interactions between molecules are weaker than the intra-molecular forces – the vibrational frequencies of various parts of the molecule were little changed depending on the details of the molecular structure. This is both the strength and the weakness of infrared spectroscopy. What about a material like diamond, or silver, in which a single atomic nucleus is repeated on an essentially infinite lattice; or more complex solids such as NaCl or GaAs? We will think about the electronic properties of such materials (that is, what makes diamond an insulator, silver a metal, gallium arsenide a semiconductor, etc.) later in the class after we have covered the electronic spectroscopy of polyatomics. Consider a cube of material of length L , say 1.5 cm on a side, with atoms spaced at 1.5 ˚ A intervals. This yields N =10 24 atoms in the sample. Even if we restrict ourselves to a purely harmonic force field among the atoms, our normal mode analysis tells us that there will be 3 N- 6=O(10 24 ) vibrational modes of the sample! There are a number of good texts that consider this problem, one of the most readable is from our own David L. Goodstein, States of Matter (Dover Books, NY), from which these notes are unabashedly modeled. Mathematically, and in the classical limit, we can write ¨ U i =- 3 N summationdisplay j =1 λ ij U j , (17 . 1) where U j are the positions (or displacements) of the atoms in space, and ¨ U i are the corresponding accelerations. The λ ij represent the forces between the atoms, and so will involve things like the spring constants for the bonds (the second derivative of the potential) and the masses of the oscillators. Based on our earlier work, we could try oscillatory solutions of the form U i = U i e iωt , which would lead to 3 N equations as before an a 3 N × 3 N determinant of the form Det | ( λ ij- ω 2 δ ij ) | = 0 (17 . 2) that could be solved to yield the 3 N eigenvalues (remember three would be translational and three rotational) provided the λ ij are known or can be calculated. With the eigenvalues in hand, the eigenfunctions (that is, the normal modes) can be obtained. This worked well for small molecules, but is clear not a sensible way to proceed in the present case! As we will see, the key for regular solids is to use the periodicity in the lattice to advantage, and so solve not for the normal modes themselves per se; but to look for classes or types of solutions that can be distinguished from each other (by symmetry, typically)....
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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