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Chapter 7. a Quantum Mechanical Model for the Vibration and Rotation of Molecules

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Chapter 7. a Quantum Mechanical Model for the Vibration and Rotation of Molecules Chapter 7. A Quantum Mechanical Model for the Vibration and Rotation of Molecules Harmonic oscillator: Hooke’s law: F kx, x is displacement Harmonic potential: 1 2 V (x) Fdx kx 2 k is force constant: d 2V k (curvature of V at equilibrium) 2 dx x0 Newton’s equation: d 2x m F kx (diff. eqn) dt 2 Solutions: x = Asint, (position) 1/ 2 (k/m) (vibrational frequency) 1 Verify: d 2 Asin t d lhs m mA cost mA2 sin t dt 2 dt k m Asin t kx rhs m Momentum: dx p m mAcost dt Energy: p2 E V 2m (mAcost)2 1 k(Asin t)2 2m 2 m 2 A2 (cost)2 (sin t)2 2 m 2 A2 kA2 2 2 When t = 0, x = 0 and p = p = mA max When t = , p = 0 and x = x = A max Energy conservation is maintained by oscillation between kinetic and potential energies. 2 Schrödinger equation for harmonic oscillator: 2 2 d 1 2 2 kx E 2m dx 2 Energy is quantized: 1 E 0, 1, ... 2 where the vibrational frequency 1/ 2 k m Restriction of motion leads to uncertainty in x and p, and quantization of energy. Wavefunctions: y2 / 2 (x) N H (y)e where 1/ 4 mk y x, 2 Normalization factor 3 1/ 2 N 2! H (y) is the Hermite polynomial H0 (y) 1 H1(y) 2y 2 H2 (y) 4y 2 ... H1 2yH 2H1 (recursion relation) Let’s verify for the ground state 2 x2 / 2 0 (x) N0e 2 2 d 1 2 2 x 2 / 2 lhs N0 2 kx e 2m dx 2 2 d 2 x 2 / 2 2 1 2 2 x 2 / 2 N0 e x kx e 2m dx 2 2 2 x 2 / 2 2 2 2 x 2 / 2 2 1 2 2 x 2 / 2 N0 e x e kx e 2m 2 2 4 2 2 k 2 2 2 x / 2 N0 x e 2 2m 2m Note that 4 1/ 4 mk 2 It is not difficult to prove that the first term is zero, and the second term as / 2. 1 2 2 rhs N e x / 2 2 0 So, rhs lhs, it is a solution. One can also show normalization: 2 2 2 x2 0 dx N0 2e dx 2 2 1 0 4 where we have taken advantage of the even symmetry of the Gaussian 2 2 eax eax and used the integral formula ax2 e dx 0 4a For higher states, the wavefunctions are essentially the product of a Gaussian and a polynomial: 5 The Gaussian is always an even function, but the polynomial can either be even f (x) f (x) or odd f (x) f (x). As a result, the wavefunctions are either even or odd, depending on . Properties of the oscillator i. Zero point energy ( 0) 1 E (quantum effect) 0 2 Even at 0 K, there is still vibration. ii. Orthonormality: * dx iii. Average displacement * (x)x (x)dx 0 6 2 No matter if (x) is even or odd, (x) is even, and thus the integrant is odd. Integration of an odd function in , is always zero. iv. Average momentum * d (x) (x)dx 0 i dx Because the derivative of an even function is an odd function, and vice versa. v. Nodes The state has 1 nodes (due to Hermite polynomial) vi. Energy separation E E1 E 11/ 2 ( 1/ 2) (equal separation). vii. Tunneling Wavefunction penetrates into classically forbidden regions. A quantum phenomenon. Example: A harmonic oscillator has a force constant of 475 N/m and a mass of 1.61×10-27 kg. Calculate the ZPE and the photon frequency needed to bring the oscillator from ground state to the first excited state. 7 k 475N / m 5.431014 s1 m 1.611027 kg ( N / m kg / s2 ) 1 E 0.51.0541034 Js 5.431014 s1 0 2 2.861020 J 20 E 2E0 5.7210 J E 5.721020 J 8.631013 s1 h 6.6261034 Js 3476nm ( c) Example: The IR spectrum of H35Cl is dominated by a peak at 2886 cm-1. Calculate the force constant of the molecule. Reduced mass 135 0.97amu 1.661027 kg / amu 1.611027 kg 1 35 2 2c~ 23.143.01010 cm/ s 2886cm1 5.431014 s1 2 k 2 1.611027 kg 5.431014 s1 475kg / s2 475N / m 8 What about DCl? 2 35 1.89amu 1.661027 kg / amu 3.141027 kg 2 35 2HCl k 475kg / s2 1 3.891014 s1 3.141027 kg 2 HCl 3.891014 s1 1 ~ 2063cm1 ~ 2c 23.143.01010 cm/ s 2 HCl Isotope effect is important to assign spectral lines. Rigid rotors Angular momentum vector l r p I (r fixed) where the angular velocity and moment of inertia are d 2 , Ir dt The rotational energy (classical): l 2 E 2I 9 Schrödinger eqn. for 2D motion: 2 2 2 E 2 2 2mx y Use polar coordinate system ( x r cos, y rsin , it becomes for 2D rigid rotor: 2 d 2 E 2I d 2 General solution: 1/ 2 1/ 2 1 im 2IE m () e , m 2 2 Cyclic boundary condition: () ( 2) So, ( 2 ) Neim( 2 ) Nei2m eim N(ei )2m eim (1)2m () 2m has to be even integers, i.e. m 0, 1, 2, ... 10 Quantization m22 E (2D) m 2I Angular momentum: lz m (quantized angular momentum) Angular momentum operator ˆ Lz xˆpˆ y yˆpˆ x x y i y x i is a common eigenfunction of both operators: 1/ 2 1/ 2 ˆ 1 im 1 im Lz e im e m i 2 i 2 2 2 1/ 2 2 ˆ 1 im (m) H 2 e 2I 2 2I Uncertainty principle (just like p and x) lz / 2 11 3D rotors: Spherical coordinates: x rsin cos y rsin sin zr cos d dxdydz r2 sindrdd Hamiltonian: 2 Hˆ 2 2m 2 2 2 2 2 2 2 2m x y z 2 1 2 Lˆ2 r 2m r r2 2mr2 The differential operator 2 is called Laplacian. If r is fixed, the 3D rigid rotor Hamiltonian: Lˆ2 Hˆ (rigid rotor) 2I where 12 2 ˆ2 2 1 1 L 2 2 sin sin sin Schrödinger eqn. Lˆ2 E 2I The solution is the spherical harmonics: m m (,) Yl (,) Nlm Pl (cos)m () where (2l 1) (l m)! N lm 2 (l m)! 1/ 2 1 im m () e (2D rotor) 2 m and Pl (cos) is the associated Legendre function. The quantization of two angular coordinates leads to two quantum numbers: l = 0, 1, 2, ... m = 0, ±1, ±2, ..., ±l (2l+1) 13 Rotational energy (3D rotor) l(l 1)2 E l 2I Since E is independent of m, each l level is 2l + 1 degenerate, which can be removed by an external field (Zeeman effect). m When m 0, Pl (cos) becomes the Legendre polynomial: P0 (cos) 1 P1(cos) cos 2 P2 (cos) 3cos 1/ 2 … For m 0 m m m d Pl (cos) sin Pl (cos) d cos m Examples of spherical harmonics: 0 Y0 1 0 Y1 cos 1 Y 1 sin ei 1 2 1 Y 0 3cos2 1 2 2 14 ˆ2 ˆ They are eigenfunctions for both L and Lz : ˆ2 m 2 m L Yl (,) l(l 1) Yl (,) ˆ m m LzYl (,) mYl (,) 0 Verification for Y1 cos 2 ˆ2 2 1 1 L 2 2 sin cos sin sin 2 1 2 0 sin sin 2 2sin cos 22 sin Lˆ2 22 Hˆ 2I 2I Lˆ cos 0 z i The spherical harmonics are orthonormal 2 m * m sind dYl (,) Yl (,) llmm 0 0 Note the volume element and ranges. 15 Spherical harmonics have nodes due to Legendre functions. For , the angular node can be determined by cos 0, 90o Vector model: The angular momentum can be considered as a vector L . In quantum mechanics, it is quantized in i. magnitude: L l(l 1) ii. orientation: Lz m Only certain directions of are allowed. Example (l = 1, ml = 0, ±1): 0 Y1 cos Example: Calc. amplitude of L, Lz and E for a rotor with I = 4.6x10-48 kg m2 and l = 1. 16 34 34 L l(l 1) 2 1.05510 Js 1.4910 Js 34 Lz ml 0, 1.05510 Js (3-fold degen.) The angle between and the z-axis for ml 1: L 1 cos z , 45o L 2 L L2 (1.491034 Js)2 E 2.41021 J 2I 2(4.61048 kgm2 ) Example. A 3D rotor absorbs at 11 cm-1. Calculate its moment of inertia. (2 0)2 E E E 1 0 2I E h hc~ so 2 I hc~ 2c~ 1.051034 Js 2 3.14 3.01010 cm/ s 11cm1 5.061047 kgm2 17 .
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