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University of Washington Department of Chemistry Chemistry 453 Winter Quarter 2014

Lecture 16 2/21/14 EDR 23.5

A. Optional Background in Rotational Mechanics (see also Lecture 9) • Two masses m1 and m2 connected by a weightless rigid “bond” of length r rotates with angular ω=rv where v is the linear velocity. As with vibrations we can render this rotation motion in simpler form by fixing one end of the bond to the origin terminating the other end with reduced mass µ, and allowing the reduced mass to rotate freely • The of the particle is now p22222µvrµω I ω K == = = (16.1) 22µ 2 2 • The term I=µr2 is called the . We will discuss it in detail later. • We can also express the energy K in terms of the . The angular momentum is defined in terms of the cross product between the position vector r and the linear momentum vector p: Lrp=× (16.2) ===prprvrsinθ µ where the angle θ between p and r is ninety degrees for circular motion. • As a cross product, the angular momentum vector L is perpendicular to the plane defined by the r and p vectors:

• We can use equation 16.2 to obtain an expression for the kinetic energy in terms of the angular momentum L: p222LL K == = (16.3) 22µµrI2 2

• The moment of inertia is derived as follows.For two masses m1 and m2: 22 I =+mr11 mr 2 2 (16.4) where r1 and r2 are the distances of m1 and m2 from the center of mass, respectively where,

m2,1 rr1,2 = (16.5) mm12+ mm • Putting 16.5 into 16.4 we obtain I = µr 2 where µ = 12 mm12+

B. Quantum Planar Rigid Rotor: “Particle-on-a-Ring” • Suppose a diatomic rotates in such a way that the vibration of the bond is unaffected by the rotation. Molecular rotation is not naturally treated in Cartersian coordinates so we change to spherical coordinates xr===cosϕ sinθϕθθ ; yr sin sin ; zr cos (16.6) See the figure below for the graphical relationship between the two coordinate systems:

• For simplicity, assume the rigid rotor is confined to a plane. This is not a terribly realistic model for molecular rotations, but it does illustrate an application of Schroedinger’s equation that can be worked out fairly easily. • For a planar rigid rotor we set the angle θ at a constant value of 90 degrees. We imagine that the origin is the center of mass of the rigid rotor and length of the rotor is r and at the point P is located a reduced mass µ. Rotation occurs in the x-y plane around the z axis. As rotation occurs the angle ϕ changes. The notation Lz refers to the fact that because the rotation is around the z axis the angular momentum vector is parallel to the z axis. Schroedinger’s equation is: p2 L2 ψ ()ϕψϕψϕ==z ()E () (16.7) 22µ I ∂ ∂ d 2 • We define Li==− so that L22=− and Schroedinger’s z i ∂ϕ ∂ϕ z dϕ 2 equation becomes: 2 d 2ψϕ( ) −=Eψ ()ϕ (16.8) 2Idϕ 2 • The solution is: ikϕϕ− ik ψ ()ϕψϕψϕ=+Ae+− Ae = +( ) + −( ) (16.9) 2IE where k 2 = 2 • We require that because the rotor is indistinguishable when it is oriented at ϕ versus ϕ+2π: ψ ()ϕψϕπ=+( 2 ) (16.10) ikϕϕ− ik ik()ϕ +−+22πϕπ ik () or… Ae+−+= Ae Ae + + Ae − • For this to occur we require that eik 2π =1, which will only occur if k =±±±0, 1, 2, 3…Because k is positive and negative we need only a single function to describe the , which can now be normalized: 222πππ 12====∫∫∫ψ *2()()ϕψ ϕdAeedAd ϕ−ikϕϕ ik ϕ 22 ϕ π A 000 (16.11) 1 ∴ A = 2π 1 • We substitute ψ ()ϕ = eikϕ into Schroedinger’s equation 16.8 and 2π obtain the energy quantization condition: 22k E = (16.12) k 2I where k =±±±0, 1, 2, 3… L2 • Because E = z equation 16.12 implies that the angular momentum k 2I around the z axis is quantized according to

C. Rigid Rotor – Quantized Angular Momentum • A more realistic model has a rigid linear molecule with moment of inertia I rotating through two angular dimensions, designated by θ and ϕ, which are the two angles used to locate a vector of length r in three dimensional space. See Figure at right which shows θ as the angle between the vector and the z axis: zr= cosθ . The angle ϕ is the angle between the x axis and the projection of the vector onto the x-y plane. Accordingly: xr= sinθ cosϕ and yr= sinθ sinϕ . • Assuming that r is constant, Schroedinger’s equation becomes 1 ⎛⎞p2 L2 pE2 +Ψ=Ψ=Ψϕ θ,,,ϕθϕθϕ (16.13) ⎜⎟θ 2 () () () 2sin2II⎝⎠θ 22 2222 ∂ ⎛⎞∂∂ where it can be shown that pθϕ=−⎜⎟sinθ and p = Lz =− 2 sinθ ∂θθ⎝⎠∂∂ ϕ • Therefore L2 is the total angular momentum: 222 2 ∂ ⎛⎞∂∂ L =−⎜⎟sinθ − 22 (16.14) sinθ ∂θθθϕ⎝⎠∂∂ sin • Note because the rotor is rigid r is a constant: r=R and so all derivatives with respect to r vanish. The problem becomes two dimensional and the wave function is obtained by solving (16.13) with L2 given by (16.14). The wave function has the form… Ψ=ΘΦ(θ ,ϕθϕ) ( ) ( ) (16.15) • One dimensional problems like the particle in the one dimensional box have a single n. The particle in a three dimensional box has three quantum numbers nx, ny, nz. Accordingly the rigid rotor has two quantum numbers l and m. The dependence of the wave functions on l and m is Ψ=ΘΦθ,ϕθϕ (16.16) Jm, ( ) Jm, ( ) m( ) • Schroedinger’s equation can also be solved to get the energy and the wave functions. L22 • The energy is a function of l only: EJJ= =+()1 where J = 0,1,2,3… J 22II

• For a given value of J, m runs from –J to +lJor m=0, ±1, ±2,… ±J…a total of 2J+1 values. • Because the energy is dependent only on J, there will be 2J+1 wavefunctions corresponding to different values of m, that will have this energy. Each rotational level will be 2J+1 degenerate. • Let’s compare the energy level diagram for a plane rigid rotor to a rigid rotor free to rotate in three dimensional space. For a plane rigid rotor : k 22 Ek==±±;0,1,2… so for k ≠ 0 the energy levels are doubly degenerate. K 2I 2 Also ∆=EE − E =()21 k + so the energy levels are spread farther apart as kk+1 2I the energy increases. See energy level scheme at the right.

• For a rigid rotor in three dimensions each energy level for J ≠ 0 is 21J + degenerate. For ______kn=± J =1 the energy is three-fold degenerate. For J = 2 , it is five-fold degenerate, etc. Note ______k =± 3 ∆=EEJJ+1 − E 2 =++−+⎡⎤()()()JJ12 JJ 1 ______k =± 2 2I ⎣⎦ ______k =± 1 2 =+()22J ______k = 0 2I so the energy level spacing also increases with increasing energy…

______Jm= 2;=±± 0, 1, 2

______Jm==± 1; 0, 1 ______Jm== 0; 0

D. Rigid Rotor- Quantization of Total and Z Angular momentum L22 • Because EJJ==() +1 , the total angular momentum is L quantized J 22II according to LJJ22=+ ()1 or LJJ= ()+1 • The quantum number J has values J=0,1,2,3,4,… 2 • Because Lz also appears in the Schroedinger equation Lz is quantized exactly as 222 in the planar rigid rotor: Lmz = or LmZ = .

• The total angular momentum L is related to its x, y, and z components by: 22 22 LL=++XYZ LL. Therefore in classical mechanics a freely rotating rigid, linear molecule like CO has its angular momentum vector trace out a sphere. • In , because L is quantized according to LJJ=+ ()1 and

Lz is also quantized by LmZ = where mJ= 0,±± 1, 2,… ± . This means that instead of tracing out a sphere, for every value of there are 21J + orientations of the angular momentum vector L. For m>0 the L vector point up relative to the x-y plane, for m<0 the L vector points down, and for m=0 the L vector is perpendicular to the z axis.

• Because only L and Lz are quantized, Lx and Ly are arbitrary. So every quantized state of L and Lz corresponding to a ( ,m) pair, results in a cone pattern in Lx, Ly, Lz space. An example is shown below for = 2 …

• For each cone the length of L is LJJ=+ ()1

=+=22() 1 6

• Each cone has a difference value of Lz

corresponding to LmZ =⇒−−( 2, ,0,,2)

E. Rotational Partition Functions for Planar Rotor and Rigid Rotor: k 22 • For the planar rigid rotor we have Ek==±±;0,1,2… and the partition K 2I function is +∞ +∞ 1/2 1122 22 1⎛⎞2π Ik T 2π IkT qe=≈−−kIkT/2BB edk kIkT /2 ==B B (16.17) rot ∑ ∫ ⎜⎟2 σσk=−∞ −∞ σ⎝⎠ σ • The factor of σ corrects for rotational state redundancies. For heteronuclear diatomics like CO, HCl etc σ=1. For homonuclear diatomics like H2, N2, etc. σ=2. • In equation 16.17 we replaced the summation with an integral assuming 2 kTBkkk∆= E E+1 − E =(21 k +) /2 I. • As a result of equation 16.17 we get for the rotational internal energy: Nk T∂U Nk U===BB and C rot (16.18) rot22 V ∂T • The probability of being in the kth energy state for k>0 is 22 2e− kIkT/2 B Pk = (16.19) qrot where qrot is given by equation 16.17 and the 2 is included because each energy level above the ground state is doubly degenerate. Note the ground state probability (i.e. k=0) is just P0=1/qrot. • But in general when the rigid rotor t is NOT confined to a plane equations 16.17 - 16.19 do NOT apply. To deal with the far more common case where the molecule is not confined to a plane we use the rigid rotor energy expression 2 EJJ=−() +1 (J=0,1,2,3...) in the expression for the rotational partition J 2I function: ∞ 2 1 −+ J ()JIkT1/2 B qJerot =+∑()21 (16.20) σ J =0 2 • In the high temperature limit where kTBJJJ∆= E E+1 − E =21/2() J + I, the summation can be approximated with an integral and we get ∞ 2 1 2 28IkTπ IkT qJedJ=+21−+ JJ()1/2 IkTB ==BB (16.21) rot ∫ () 22 σσσ0 h • This is identical to the classical result as expected because of the replacement of the summation with the integral. • A result of equation 16.21 the internal energy and heat capacity associated with the rotation of a are

UNkTCNkrot= Band V= B (16.22) • Note because each state is 2J+1 fold degenerate the probability of being in the Jth energy state is: 2 e−+ J ()JIkT1/2 B PJJ =+()21 (16.23) qrot where now qrot is given by equation 16.21.