5.76 Lecture #21 2/28/94 Page 1. Lecture #21: Rotation of Polyatomic Molecules I

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1 5.76 Lecture # /8/94 Page Lecture #: Rotaton of Polatomc Molecules I A datomc molecule s ver lmted n how t can rotate and vbrate. * R s to nternuclear as * onl one knd of vbraton A polatomc molecule can have R orented along an bod fed drecton smmetrc and asmmetrc tops A polatomc molecule can stretch an bond or bend an bond par Normal modes of vbraton A lot of ver complcated classcal mechancs. TODAY: Derve R R R = + + I I I and evaluate matr elements n KJM bass set, where I, I, and I are called prncpal components (.e. egenvalues) of the 3 3 moment of nerta tensor, I, and are analogous to n an AB datomc. µr AB. Center of mass.. rgd bod rotaton T n terms of ω (angular veloct), m, (,, ) (postons of atom n center of mass bod frame) / ω Ιω I a 0 0 T IT prncpal aes 0 I b I c 3. and matr elements n KJM bass. 4. Smmetrc tops - prolate and oblate energ level formulas. Consder a rgd N-bod sstem. Each atom has mass m and bod-fed coordnate q (defned relatve to an arbtrar bod-fed orgn). Our frst task s to locate the center of mass, because we epect to separate the 3N degrees of freedom nto 3 center of mass translatons, 3 rotatons about the center of mass, and 3N 6 vbratons. Center of Mass: 3 Cartesan component equatons. 0 3N = m q q = atoms ( ) CM In fact, the defnton of bod fed as sstem s not even obvous for vbratng molecule.

2 5.76 Lecture # /8/94 Page Eample: 0 N 3 (projected onto plane) Take advantage of smmetr whenever possble! pck C 3 (3-fold rotaton as) as as as locate orgn at N atom locate at φ = 0 (.e. n plane) q = ( r,, φ ) = ( R,,0 ) ( ) ( ),, = Rsn,0, Rcos C 3 as π 3 q = R,, ( ),, = Rsn, Rsn, Rcos 3 4π 3 q =,, R R ( ) 3 3, 3, 3 = Rsn, Rsn, Rcos 3 qn = (0,0,0) N Now solve for center of mass. CM = CM = 0 are trval equaton ( ) ( ) ( 0 ) 0= m = 3m Rcos + m CM CM N CM CM 3m = Rcos 3m + mn = M So we have coordnates of all atoms relatve to new orgn now at center of mass, epressed n terms of unknown bond coordnates, R and. Rcos 3m q = Rsn,0,Rcos m M Rcos=Rcos M 3m M N M q = Rsn, 3 Rsn,Rcos m N M q 3 = Rsn, 3 Rsn,Rcos m N M q N = 0,0, 3m N M Rcos

3 5.76 Lecture # /8/94 Page 3 Net we need to wrte out and put t nto a convenent form. T = + V T free rotor, thus V = 0 = mv Want to re-epress all v s n terms of q and ω where ω s the drecton and magntude of the angular veloct of the rgd bod rotatons. (All atoms eperence the same ω.) ω m q q ω q ω cos = = ω ω center of mass q q q need veloctes for T v = q ω (rght hand rule for mnus sgn) v = q ω sn q, ω known. Must solve for sn. sn = q q = q q q ω ω q / so v = q ω sn = q ω q ω ( ) = m qω q ω ( ) Go to Cartesan coordnates (alwas safe for settng up quantum mechancal amltonan operator).

4 5.76 Lecture # /8/94 Page 4 = m ( + + )( ω +ω +ω) ( ω + ω + ω) a bt of algebra = m ( + ) ω + ( + ) ω + ( + ) ω ωω ωω ω ω Reformulate as matr dagonalaton problem! Defne I m + I = m ( ) etc. perpendcular dstance squared from as ( ) etc. = ω I ω Ths s a compact form for mess equaton above! ω ω ω ω I I I I I I I ω = ω ω ω sm. I I I s a real, smmetrc matr. It can be dagonaled (b a coordnate transformaton, a rotaton about center of mass) to gve I 0 0 a TIT= 0 Ib I a I b I c b defnton and are called the prncpal moments of nerta. = ω T T IT T ω T T = ( )( )( ) I c

5 5.76 Lecture # /8/94 Page 5 T T ω ω ω = ω a ω b ω c a = b c We fnd a specal bod fed coordnate sstem wth orgn at the center of mass whch causes I to be dagonal. Usuall possble to fnd prncpal aes b nspecton.. One as s as of hghest order rotatonal smmetr, called b conventon.. Another as s to C n and to a σ v plane. E.g. f σ() ests, then m = m = 0 (,, ). 3. 3rd s to frst aes. because there s alwas an dentcal nucleus at (, +, ) and at So when I s dagonal the nuclear rotatonal angular momentum s defned as J = I ω = ( I ω + I ω + I ω ) a a b b c c should actuall use notaton of R or N Ja Jb J c = J I J = + + I I I a b c lke p m J I (The recprocal or nverse of a dagonal matr s trval.)

6 5.76 Lecture # /8/94 Page 6 We can now defne three rotatonal constants A = h cm (E/hc) c 8π I a B = h cm Note that we wll sample rotatonal constants wth c 8π I b I averaged over specfc vbratonal state, not at C = h the equlbrum geometr. Want equlbrum cm geometr, get strange average. c 8π I c A B C (agan, b defnton) One obtans A, B, C b pckng bond lengths and angles, specfng atomc masses, and dagonalng I. For each change n masses (sotopc substtuton) or teratve change n geometr, I must be redagonaled. Eample: Prncpal Moments for N 3 C as must be one prncpal as 3 estence of reflecton plane 3 3 so I = R sn m + m m = 3mRsn (the dstance of each atom from as specfed) 3mm 3m σ ( ) mples I = R cos R sn v + M prncpal component to that plane. You show that I = I N 3 (for an smmetrc top). [General rule, ever molecule wth C n rotaton as wth n 3 has two equal moments of nerta!] Specal case of D d S as: ccloctatetrene and allene 4 allene C 3 4

7 5.76 Lecture # /8/94 Page 7 for Smmetrc top. B conventon, I = I, I s unque. J J J = + + I I I I = I J J = J + J J ( ) J = J + J + = J J + I I I I = J + J I I I Use JKM smmetrc top bass functons whch are just lke JΩM functons for a datomc molecule. So tpes of smmetrc top: E = J( J + ) + K I I I lke a datomc molecule projecton of J onto unque (.e. smmetr) as of bod (lke Ω). I I a s unque, I b = I c > I a, prolate top, lke a cgar. Coeffcent of K s > 0 because A > B b defnton. E prolate = BJ(J +) + (A B)K hc. I I c s unque. I a = I b < I c, oblate top, lke a Frsbee. Coeffcent of K s < 0. E oblate hc = BJ(J +) (B C)K J = 0,,, K = 0, ±, ±J denote as (J,K) or J K possble levels

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