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S18 Ch. 5 Rigid Rotator Chem 4502 Quantum Mechanics & Spectroscopy (Jason Goodpaster) Quantum quote of the day, submitted by Chapter 5 (continued): Rigid Rotator 3502 student David: Trying to capture the physicists precise Lecture 17: mathematical description of the quantum world Chap. 5, Sect. 8: Rigid Rotator - model for molecular rotation with our crude words and mental images is like Chap. 4, Sect. 2: Angular momentum properties MathChap. C, pp. 111-2 Angular momentum vectors playing Chopin with a boxing glove on one hand MathChap. D, pp. 147-8 Spherical coordinates and a catchers mitt on the other. George Johnson, New York Times, 1996 Lecture 18: Chap. 5, Section 9: Rigid Rotator, continued energy levels microwave spectroscopy 1 2 Assuming that H79Br has a vibrational frequency of Classical Rotational Motion ∼2560 cm-1, predict its gas phase IR absorption spectrum. Consider a mass, m, rotating about a point at radius r: Moment of inertia (I) r I = m r2 I corresponds to m in linear motion. Consider 2 masses separated by r, rotating about their center of mass: Figure 13.2 p. 500 I = µ r2 There is a gap centered at ∼2560 cm-1. -1 On either side, absorption lines are spaced by ∼17 cm . µ = reduced mass = m1 m2 m + m These are due to rotational transitions (plus ∆v = 1). 1 2 3 (for 2 masses) Fig. 5.9 p. 1744 Classical Rotational Motion Classical Rotational Motion X Center of mass (CM) : CM Angular momentum (magnitude) defined by X1 X2 m1 m2 L = I ω XCM (m1+m2) = m1X1+m2X2 m > m 1 2 L = angular momentum XCM position of the center of mass I = moment of inertia (µr2) mass-weighted average position of all of the masses ω (omega) = rotational speed or angular velocity radians/sec If m1 = m2, then XCM = ½ (X1+X2) right in the middle In the absence of an external torque (force causing rotation), and µ = ½ (m1) = ½ (m2) angular momentum is conserved. Example: figure skater brings arms in 5 (reducing I) → L remains constant so ω increases (spins faster). 6 Classical Rotational Motion Classical Rotational Motion Angular momentum (direction) Relate our two expressions for angular momentum, L: → → → L = r x p right hand rule → → → L = r x p → → → p = mv = linear momentum Since r and →p are perpendicular for circular motion, magnitude of L is: L = r p = r mv L = m r v We also had (for one rotating mass): L = I ω = m r2 ω ω (rad/s) = v (m/s) 2 π (rad/cycle) 1 / (2 π r) (cycle/m) = v/r Substituting, L = m r2 (v / r) = m r v Same as above MathChapter C pp. 111-112 7 8 Classical Rotational Motion Classical Rotational Motion p. 120 L Rotational kinetic energy z 2 2 r Erot = ½ I ω Analogous to: E = ½ mv 2 2 2 = ½ (I ω) / I = L / 2I = p / 2m 2 (I = mr ) rotation in x-y plane (v) (ω=v/r) ; L= r p = r mv Is skaters Erot conserved when brings arms in? 2 2 mv No: Erot = L / 2I L is the same, but I is reduced so Erot is increased To bring arms in, skater has to do work against the QM Rotation: centrifugal force pulling arms outward. Complementary observables (uncertainty relation) This compensates for the increased rotational kinetic energy. x, px Φ, Lz where Φ is (azimuthal) angle in xy plane 9 10 Quantum Mechanical Rotational Motion Quantum Mechanical Rotational Motion Extend to 3-D. Applications: Schrödinger equation for the rigid rotator: Particle constrained to move on surface of a sphere Ĥ Ψ = E Ψ All the energy is kinetic Gas phase molecule rotating freely about its center of mass Model: Rigid (fixed r) rotator 2-D (planar) motion (in an inherently 3-D system) forbidden ∇2 LaPlacian Operator Uncertainty Principle: if molecule were to rotate only in Cartesian coordinates in the xy plane, both z and pz would be zero Convenient to convert to spherical coordinates (since r is constant) 11 12 Quantum Mechanical Rotational Motion Quantum Mechanical Rotational Motion Spherical Coordinates (MathChapter D p. 147) 2 Using the chain rule, convert ∇ from Cartesian to spherical coordinates: Φ (phi) azimuthal angle (in xy plane) ∇2 = ∂2/∂x2 + ∂2/∂y2 + ∂2/∂z2 Cartesian θ theta angle from z axis r distance from origin Eqn. 5.49 Chap. 5 #30-32, not assigned 0 ≤ Φ ≤ 2 π Since r is constant, can omit first term: 0 ≤ θ ≤ π 0 ≤ r < ∞ generally but here r is fixed 13 14 Quantum Mechanical Rotational Motion Lecture 18: We had: ∇2 = Chap. 5, Sect. 9: Rigid Rotator, continued energy levels; microwave spectroscopy Hamiltonian operator in Cartesian coordinates: Lecture 19: Chapter 6: Hydrogen Atom Factoring out 1/r2 and using I = µ r2 , obtain Hamiltonian operator in spherical coordinates (fixed r) 15 16 Quantum Mechanical Rotational Motion Quantum Mechanical Rotational Motion Schrödinger equation for the rigid rotator: We had: Ĥ Y(θ,Φ) = E Y(θ,Φ) call wave functions Y(θ, Φ) spherical harmonics Are the spherical harmonics also eigenfunctions of any other operator? Recall for kinetic energy (here, call it K): K = ½ I ω2 = L2 / 2 I for rotational motion 2 2 Ĥ E = rotational Analogous to: K = ½ mv = p / 2m for linear motion kinetic energy Write the operator corresponding to L2 : So, a rotating molecule in a state characterized by one of the Y(θ,Φ) eigenfunctions will have a well-defined rot. kinetic energy. L2 = 2 I K So, Lˆ 2 = 2 I Kˆ Kˆ = Ĥ since have no potential energy Will any of its other properties also be well-defined? ˆL 2 = 2 I Ĥ 17 18 Quantum Mechanical Rotational Motion Rigid Rotator Ĥ Y(θ, ) = E Y(θ, ) Schrödinger equation We had: Φ Φ Use separation of variables. Apply boundary conditions Also, since ˆL 2 = 2 I Ĥ Y(θ, Φ) = Y(θ, Φ+2π) Solution yields 2 quantum numbers: ˆL 2 = -2 J = 0, 1, 2, ... unitless (we will call this ℓ for H atom) Natural unit of angular momentum is (i.e., J•s). in rigid rotator, J determines total (= kinetic) energy The spherical harmonics, Y(θ,Φ), will also be eigenfunctions of ˆL 2 with eigenvalues L2 = 2I E. also determines magnitude of angular momentum m = 0, ±1, ±2, ... ±J (so, 2J+1 degenerate states for a given J) Ĥ Y(θ,Φ) = E Y(θ,Φ) ˆL 2 Y(θ,Φ) = 2I E Y(θ,Φ) 19 determines direction of angular momentum vector20 Rigid Rotator Rigid Rotator Look at the lowest energy solution: The energy levels turn out to be: ½ 2 J=0, m=0 Y(θ,Φ) = 1 / (4π) E=0 EJ = ћ J (J+1) J = 0, 1, 2, ... 2I J=0 no rotation so rotational kinetic energy = 0 where I = µr2 moment of inertia angular momentum = 0 (also linear momentum = 0) Rotational energy levels are quantized. Does this violate the uncertainty principle? In contrast, classically the rotational kinetic energy No. can have a continuous range of values: 2 Y(θ,Φ) is a constant; spherically symmetric Erot = ½ I ω ω (omega) = rotational speed (rad/sec) The position (angular orientation) is completely unknown. 21 22 Rigid Rotator Model for rotating Rigid Rotator diatomic molecule We had: E = ћ2 J (J+1) We had: E = ћ2 J (J+1) J J 2I 2I -1 2 In units of wave numbers (cm ), where I = µr µ = m1 m2 ћ = h/2π m1 + m2 ∼ 2 EJ = EJ / hc = ћ J (J+1) = h J(J+1) 2I hc 8 π2 I c ∼ So, by measuring the rotational energy levels EJ , ∼ ∼ Rotational -1 B -1 EJ (cm ) = B J (J+1) constant (cm ) we can determine r, the bond length of the molecule. For polyatomics, can also measure bond angles. In units of E/h, B = h / ( 8 π2 I ) 23 24 ∼ Rigid Rotator J E (cm-1) g ∼ Rigid Rotator 20B 9 ∼ ∼ -1 Selection rule EJ (cm ) = B J (J+1) to absorb or emit a photon and change rotational level, ∼ 7 Selection rule for 12B the molecule must have a permanent dipole moment. absorption or emission of a photon: ∼ Homonuclear diatomics (O2, N2) 6B 5 or symmetric molecules (e.g., benzene, methane) ∆J = ± 1 ∼ 2B 3 do not have a microwave (pure rotational) spectrum 0 1 Rotational spectrum (but they do have discrete rotational levels). shows absorption lines ∼ separated by 2B Degeneracy g = 2J+1 since m = 0, ±1, ..., ±J Fig. 5.10 p. 178 25 26 Rigid Rotator Rigid Rotator 12 16 The linked image cannot be displayed. The file may have been moved, renamed, or deleted. Verify that the link points to the correct file and location. Calculate the J=0 to J=1 rotational energy spacing of C O, Selection rule and its rotational constant, in cm-1. The equilibrium bond (classical picture): length is 1.13 Å. ∼ 1.66 x 10-27 kg/amu a molecule with a B = h / 8 π2 I c h = 6.63 x 10-34 Js permanent dipole ∼ ∼ 2 -1 I = µr moment can absorb EJ (cm ) = B J (J+1) c = 3.00 x 108 m/s rotational energy from µ = m1 m2 / (m1+m2) the oscillating electric field of the (microwave) radiation. Answer: -26 µ = m1 m2 / (m1+m2) = [12•16/28] amu = 6.9 amu = 1.14 x 10 kg I = µr2 = 1.14 x 10-26 kg (1.13 x 10-10 m)2 = 1.46 x 10-46 kg m2 ∼ B = h / 8 π2 I c = 6.63 x 10-34 Js / ((8 π2 1.46x10-46 kg m2)( 3.00 x 1010 cm/s)) = 1.9 cm-1 = rotational constant Engel p.
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