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Raman

Raman

In isolated molecules,

In isolated molecules, these vibrations are associated with the interatomic bonds and the molecular geometry. In the simplest case of one mass and one spring, the kinetic T and potential V energies of the system are 1 2 1 2 T= mx , V= kx , (4) 2 2 where k is the spring constant of the bond and x and x are, respectively, the displacement from the equilibrium position and its time derivative. The resonant frequency of this simple system is obtained solving the secular equation 2 k− mω = 0 . (5) Consider now a molecule formed by a set of masses (atoms) linked by springs (bonds), specified by a set of generalized coordinates q j (e.g., bond lengths and angles). For small oscillations around the equilibrium positions, the kinetic and potential energy of the system can be expressed as 1 1 T= ∑m q q , V= ∑k q q . 2 2 j,k jk j k jk j k j,k (6) These equations are formally equivalent to eqns. (4), the only difference being that the mass M = (m ij ) and spring constant K = (k ij ) are now matrixes. The secular equation (5) becomes 2 K− ω M= 0, (7) which has nontrivial solutions only for 2 ( ) Det K − ω M = 0. (8) The solutions ω 2 to eqn. (8) are the eigenvalues that simultaneously diagonalize K and M, and thus correspond to the resonant frequencies of the molecule. Let's consider the CO 2 molecule (O=C=O) as an example. For simplicity, we'll limit the analysis to the linear vibration modes. Consider the C atom as the origin of the coordinates; the generalized coordinates describing the molecule are the positions of the two O atoms (x 1 and x 2 ). The center of mass of the molecule lies at mOx1 + mOx2 x CM = , M= 2mO + m C. (9) M And has a velocity 2

m x O CM = ( x1 + x 2) . (10) M The kinetic and potential energy of the system are 1 1 1 1⎛ ⎞ 1⎛ ⎞ T= m x + m x + Mx = m x m x x x 2 2 2 2 + + + + ⎜ M 2 ⎝ ⎠ ⎟ ⎝⎜ M ⎠ ⎟ M 2 2 2 2 2 2 mO 2 mO 2 m O1 O2 CM ⎜ ⎟ ⎟ O 1 ⎜ O O 2 1 2, 1 2 1 2 V= kx1 + kx 2. 2 2 (11) Thus, ⎛ 2 2 mO m ⎞ ⎜ ⎛ ⎞ O ⎜⎜ ⎜ m O 2 ⎜⎜ + ⎜⎜ M M ⎜⎝ ⎠⎟ ⎛k 0⎞ = ⎜ 2 2 k =⎜ ⎟. m 0 k (12) O m ⎟ ⎛ ⎞ ⎜⎝ ⎠ O 2 m ⎟⎟ O + ⎜ M ⎜ M ⎟⎟ ⎝ ⎝ ⎠ ⎟⎠⎟ ( m ) ⎜ ⎟ ij , ( ij) Solving eqn. (8) yields the resonant frequencies of the molecule 2 k ω 1 = , m O 2m + m ω k. 2 O C 2 = mOmC (13) Mode 1 is the symmetric mode, in which both O atoms move toward the center or away from it at the same time, i.e. both O are oscillating with frequency ω 1 and a 180° phase shift. Mode 2 is the antisymmetric mode, in which both O atoms lie still while the C atom oscillates between the O atoms with frequency ω 2 . Larger molecules have many more vibrational modes but the mode frequencies and displacements can be calculated in the same way. In general, an N‐atom molecule has 3N degrees of freedom of which 3 correspond to solid translation of the whole molecule and 3 (or 2, for linear molecules) correspond to solid rotations. Thus the number of vibrational modes is n= 3( N− 2). (14) 3

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