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Harmonic Oscillator Molecular Vibrations (16.9) Diatomic Molecules I

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Harmonic Oscillator Molecular Vibrations (16.9) Diatomic Molecules I Section VIII: Vibrational Spectroscopy Vibrational Spectroscopy Simple Harmonic Oscillator Molecular Vibrations (16.9) Diatomic molecules i. IR Spectra (16.9, 16.10) ii. Raman Spectra (16.13) iii. Anharmonicity (16.11) iv. Vibration-Rotation (Rovibrational) Spectroscopy (16.12) Polyatomic molecules (16.14) i. Vibrational Selection Rules ii. Vibration-Rotation (Rovibrational) Spectroscopy (16.15) iii. Raman Spectra (16.16) iv. Anharmonicity Applications Harmonic Oscillator We can treat the vibrations in a diatomic molecule as an single oscillator of mass µ that obeys Hooke’s Law re m1 m2 re r re x r µ x µ Suppose the bond between the two atoms behaves as a spring obeying Hooke’s law: F = -kx where k is a force constant, with units of N m-1, and x is the vibrational coordinate, a measurement of displacement away from equilibrium, x = r - re. If µ is the reduced mass, we rewrite Hooke’s law as: d 2x µ '&kx dt 2 and rearrange into a second-order differential equation: d 2x k d 2x k % x ' % $2x = 0 , $ ' dt 2 µ dt 2 µ Classical Harmonic Oscillator One may use the series method to solve this equation for differential equations, or just try some simple trial functions. In fact, x(t) = Asin(Tt) is a solution, yielding: dx k 1/2 p(t) ' µ ' µ TAcos(Tt), where T ' dt µ Thus, the position of the particle varies in time harmonically (i.e., sin Tt) with frequency < = T/2B. The particle is also stationary (i.e., p = 0) when the displacement x has its maximum value, A, which is the amplitude of the motion. The potential energy is written as dV '&Fdx '%kxdx 1 1 V ' kx 2 ' kA 2sin2 Tt 2 2 The kinetic energy is written as p 2 (µ TAcosTt)2 1 E ' ' ' µ T2A 2cos2Tt K 2µ 2µ 2 and since T = (k/µ)1/2, it can be rewritten: 1 E ' kA 2cos2Tt K 2 and the total energy is therefore constant! 1 1 1 E ' kA 2sin2Tt % kA 2cos2Tt ' kA 2 2 2 2 Q.M. Harmonic Oscillator Using the Schroedinger equation with the known potential energy for the harmonic oscillator, we write 2 2 £ d 1 2 & % kx R ' ELR 2µ dx 2 2 Solving this Schroedinger equation is a bit messy, as it involves the use of Hermite polynomials. Through some effort, it can be shown that the wavefunctions for this equation are found to be of the form: Tµ R ' N H ( "x)exp[(&"x 2)/2] " ' L L £ NL is a normalization factor 1 " 1/2 1/2 NL ' 2LL! B and HL is a Hermite polynomial L 2 L 2 d exp(y ) HL(y) ' (&1) exp(y ) dy 2 Hermite Polynomials: H0(y) = 1 H1(y) = 2y 2 H2(y) = 4y -2 3 H3(y) = 8y -12y 4 H4(y) = 16y -48y2 +12 5 3 H5(y) = 32y -160y +120y 6 4 2 H6(y) = 64y -480y +720y -120 Q.M. Harmonic Oscillator, 2 The wavefunction for the vibrational ground state is: 2 R0 ' N0H0( "x)exp[(&"x )/2] Substituting in: Tµ " ' H ( "x) ' 1 £ 0 1 " 1/2 1/2 " 1/4 N0 ' ' 200! B B we finally have: " 1/4 "x 2 R ' exp & 0 B 2 which is a gaussian function. A table of harmonic oscillator wavefunctions are shown below, along with a graphical representation: Energies of the Harmonic Oscillator Now, the eigenvalue comes from the Schroedinger equation, with the energy of the state with the vibrational quantum number, L = 0, 1, 2, ..., 4 equal to: £ k k E ' (2L % 1) ' £ (L % ½) L 2 µ µ or if we let k T ' £ e µ we have: EL ' Te (L % ½) or in more traditional frequency units: 1 k E ' h< (L % ½), < ' L 0 0 2B µ # Even if v = 0, energy is not equal to zero (i.e., this is the zero point energy -there is always some vibrational contribution to energy). # Wavefunctions also extend outside of the potential (-ve EK is implied by this!) - this leads to the concept of tunneling Introduction: Vibrational Spectroscopy In this section, we will focus upon the region of the spectrum ranging from 100 - 5000 cm-1 (1.2 - 60 kJ mol-1). The energy of light in this region excites vibrations of molecules which absorb it (i.e., molecule is excited to a state with increased vibrational energy). Since rotational energies of molecules are smaller than vibrational energies, rotational transitions are often simultaneously excited. In infra-red spectroscopy, light of all different frequencies passes through a sample and the intensity of the transmitted light is measured at each frequency. Samples can be gaseous, suspended in a Nujol mull or pressed into a KBr pellet. NaCl or KBr windows are typically used on the sample cells, since there is no absorbance in the IR region. In Raman spectroscopy, scattered light is observed as opposed to transmitted light. The sample if often held in a capillary tube, and the anti-Stokes and Stokes scattering are detected at right angles to the sample with various types of photon counters. The intensity of the scattered light is about 1/1000 of the incident beam. sample cell ir detector ir sousource reference cell incident light sample tube visible or ir laser scattered light visible photon counter Vibrational Spectra of Diatomics We have already seen that under the simple harmonic oscillator approximation, vibrational energy levels, Ev, of a diatomic molecule are given by 1 k E ' h<0 (L % ½), <0 ' L 2B µ where the reduced mass is µ = m1m2/(m1 + m2) and k is the force constant, which typically has units of aJ Å-2. Force constants increase with increasing bond order. O2, NO, CO and N2 have bond orders of 2, 2.5, 3 and 3, respectively. Molecule k (aJ Å-2) Molecule k (aJ Å-2) Molecule k (aJ Å-2) HCl 5.16 F2 4.45 CO 18.55 HF 9.64 O2 11.41 N2 22.41 Cl2 3.20 NO 15.48 note that 1 aJ = 10-18 J (a signifies atto / 10-18) The strength of the spring representing the bond results from a subtle balance of nuclear repulsions, electron repulsions and nuclear-electron attractions - none of which are affected by isotopic substitution (so neither is k). A common shape for the potential energy is a parabola, V = ½kx2 Infra-Red Vibrational Spectroscopy Similar to rotational spectroscopy, term values are used to describe the spectrum. The vibration term G(v) has dimensions of wavenumber in most cases: E L ' G(L) ' T(L % ½) hc N.B.: another unfortunate & inconsistent convention is that T is the vibration wavenumber in cm-1 (not frequency) A transition between lower and upper states described by the vibrational wavefunctions RLO and RLN is given by ( R ' R) µR)) dx L m L L x represents displacement from equilibrium (i.e., r - re) For a molecule without a permanent electric dipole moment (i.e., homonuclear diatomic), µ = 0, so RL = 0, and IR active vibrational transitions are forbidden. For a heteronuclear diatomic, µ … 0 and varies with x. The variation goes as a Taylor series expansion (e = equilibrium configuration): dµ 1 d 2µ µ ' µ % x % x 2 % ... e dx 2! 2 e dx e (dµ/dr)e µ re r IR Spectroscopy, 2 This means the transition moment integral can be written as ) ( )) dµ ) ( )) R ' µ R R dx % R x R dx % ... L e m L L m L L dx e Since RvO and RvN are eigenfunctions of the same Hamiltonian, they are orthogonal, and when LO … LN ( R) R)) dx ' 0 m L L which means that dµ ) ( )) R ' R x R dx % ... L m L L dx e The first term in this series will be non-zero if )L ' ±1 which is the vibrational selection rule. Once again, the transition is always )L = L(upper) - L(lower) = +1 Vibrational spectra are typically observed as absorption spectra. At normal temperatures, the intensities of the transitions decrease as LO increases, since the population NL of the vth vibrational level is related by N0 by N E L ' exp & L N0 kT IR Spectroscopy, 3 Each vibrational transition gives rise to a band. All bands with LO … 0 are referred to as hot bands, since the populations of the lower levels of such transitions, and therefore transition intensities, increase with temperature. The term line is reserved for transitions between rotational levels associated with the two vibrational levels. The lines are referred to as fine structure of a band, which is commonly observed in vibrational-rotational spectra of molecules in the gas phase. 2 Transition intensities are proportional to *Rv* , and 2 therefore to (dµ/dx)e (dµ/dr)e µ re r # As r 6 0, µ 6 0, since the nuclei are in the same position. # As r 6 4, µ 6 0, since the nuclei dissociate into atoms. # The maximum values of µ occurs at r < re in the figure above, meaning that there is a negative slope at this point, and a negative slope at re. If the maximum of µ occurs at r > re, then dµ/dr is positive, and so is the slope at re. # If the maximum µ is at re, then dµ/dr = 0. In this case, the )L = 1 transitions are still allowed, but will have zero intensity (selection rules tell us which transitions may occur, but nothing about their intensities!!!) Vibrational Raman Spectroscopy Both homonuclear and heteronuclear molecules are Raman active, since the polarizability varies during vibrational motion, giving the vibrational Raman effect.
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