Chapter 5 Molecular Vibrations and Time-Independent Perturbation Theory Homework Solutions

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1 Chapter 5 Molecular Vibrations and ime-independent Perturbation heory Homework Solutions. 4 y Form of solution: n 3 4 y anx a axax a3x a4x... n Specific Recursion Relations dy 3 aax3a3x 4 a4x... a 6a3x a4x... 4 y a 6a3x a4x... 4 a axax... 4a 4ax 4 ax... Equate powers of x to get: a 4a a a 6a 4a a a /3 3 3 a 4 a a a / General Recursion Relations: FYI only (not requested in the problem) y dy m m a x m m m ma x m m ( m) ma x m m Let's conert summation index to start at by deing: n = m- m = n+ ( n)( n) a x n n n 4 y ( )( ) n 4 n n n a 4 n n x ax n ax n n n n Equate coefficients of equal exponent terms: 4 ( n)( n) a 4 n a a n n a ( n)( n) n Note: You will d that this general recursion formula reduces to the specific relations aboe if you use it with n =, or

2 . x / k Axe (a) Normalization x / x x Axe A x e A x e A 4 / /4 / A A a (b) <x> x / 3 x because integrand is odd x x x Axe A x e (c) <x > x / 4 x 4 x x x x Axe A x e A x e / / / A 8 8 (d) d d A xe x / A x x / x / x / x / e e A e x x / d Axe x / x / x / p pˆ Axe A xe [ e ( x )] i i x 3 x 3 x A [ e ( xx )] A xe x e A i i i = : Both integrals anish because the integrands are odd functions of x (e) d d Ae x x Ae x x ( x/) e x / 3 A e ( x 3 x) x / x / x /

3 d x / x / 3 p Axe A e ( x 3 x) x 4 4 x x ( ) 3 ( ) 3 A e x x A x e x e A 3 A A (f) (g) p p 4 4 kx k k 3 3 k 3 k 3 3 V x kx 3. x k y 9kx kx y 3 3 x E Ex Ey nx x ny y nx x ny 3x 3 nx x 3ny x nx 3ny x n x n y E g ħx 3 ħx 4 ħx 3 5 ħx 6 ħx 4 7 ħx ħx 3 6

4 4. (a) Force Constant 7 mm C O ( amu)(6 amu).66x amu.4x m m amu 6amu amu C O k c (3.4)(3. / )(7.4 6 k c x cm s cm x 9 / s 9 m / s / m 9 N / m (b) ZPE and Energy Spacing c (3.4)(3.x cm/ s)(7 cm ) 4.9x s 4 ZPE: E x Js x s 3 kj.5x J x6.x mol x.9 kj / mol 3 J E: E x Js x s kj x J x x mol x kj mol J / 3 5. k 3 N / m3( m/ s ) / m 3 / s 7 mcl mcl mcl 35amu.66x 6 7.5amu.9x m m amu Cl Cl k 3 / s 556cm 6 c (3.4)(3.x cm / s).9x cm, 334 cm, 667cm 3 4 E(,,,) 3 4 hc 59cm (a) 349cm 334cm 667 cm 667 cm E hc cm x J s x cm s cm 34 (,,, ) 59 (6.63 )(3. / )(59 ) kj x J x x mol x kj mol J / 3

5 E(,,,) hc cm (b) 349cm 334cm 667 cm 667 cm E hc cm x J s x cm s cm 34 (,,, ) 5844 (6.63 )(3. / )(5844 ) kj x J x x mol x kj mol J / 3 E E(,,,) E(,,,) 5844cm 59cm 3335cm hc hc hc E hc cm x Js x cm s cm 34 (3335 ) (6.63 )(3. / )(3335 ) kj x J x x mol x kj mol J / I : 5cm (a) 7 mm I I mi 7 amu.66x.5x m m amu k c I I (3.4)(3. / )(5.5 5 k c x cm s cm x 7 / s 7 m / s / m 7 N / m (b) Intensity Ratio: = 3 o C = 573 K 5 hc x J s x cm s cm k x J K K 34 (6.63 )(3. / )(5 ) 3 (.38 / )(573 ).54 I N e e I N e e E / k (3/) hc / k hc / k e E / k (/) hc / k.54 e.58

6 8. he Spectroscopic Dissociation Energy (De) is the energy gap from the ery bottom of the potential well to the separated atoms. he hermodynamic Dissociation Energy (D) is from the lowest rational energy leel to the separated atoms. he two Dissociation Energies are related by: De D E( ) D hc 34 E( ) hc 6.63x Js(3.x cm/ s)(99 cm ) 3 kj.97x J x6.x mol x 7.9 kj / mol 3 J De D E( ) D hc 48 kj / mol7.9 kj / mol 446 kj / mol x x x x E * H Asin B Asin Asin B Asin a a a/ a a a/ a E A B sin xa B sin x a/ a a/ a () () () 9. a/ AB x sin x AB x sin x 4 4 a/ a a a AB sin a AB sin a sin a Note: sin( ) sin a a a sin and sin( a) sin sin a a a a a E A B A B A Ba A Ba Ba Ba Ba A B 4 4 a 4 a 4 () h 3 E E E B 8ma a

7 . d d H x kx x kx H H 4 4 () () () 4 hus the perturbation is: H x kx () () ( ) / 4 h x x / x 4 E * H d Ae x kx Ae A e x kx 4 x k x A x e A x e II 4 x 4 x 3 I A x e A x e A 8 / / k k x k x I A x e A x e A k 4 / / k k k k k E II w 4 k 4 E E E 4 k k () 3 3 Note: he simplification I'e shown (putting I in terms of ħ) isn't really necessary.

8 . Experimental HCl Frequencies: Fundamental: 885cm First Oertone: 5664cm En n n xe hc () E E cm xe xe xe hc hc 4 4 () E E cm xe xe 6xe hc hc 4 4 Eliminate xe ~ : 3() - () 3(885) (5664) 99cm Use () x e 53cm x e 53cm xe.8 99cm. Vibrational Contributions to hermodynamic Properties Preliminary Calculations: Partition Function and Deriatie = 8 o C = 73 K q e e 34 hc (6.63x J s)(3.x cm / s)(7 cm ) 389 K 3 k.38x J / K 389 K K ln Q ln q Nln( q ) Nln Nln e e N e N

9 ln Q dln Q N N e d Vn, e e N e Enthalpy: H ln Q ln Q ln Q H k k k U lnv V, N, N V, N ln Q e e H k k N k nna Vn, e e e e nr nr nr H ZPE H therm e e Alternate form for H (to simplify taking next deriatie) e e H nr nr nr nr H H e e e 389 K H ( mol)(8.3 J / mol) ( mol)(8.3 J / mol)(389 K).363 e 33 J 477 J 8 J 8. kj ZPE therm Note that the ZPE contribution to H (and to U ) is independent of temperature, the thermal contribution aries with temperature. For example, at = o C = 73 K, the thermal contribution to H = 34,6 J Constant Pressure Heat Capacity, C P C p H dh C d Pn, V

10 d d CP nr nr nr e d d e d V nre e nr e e d e nr e.363 e ( mol)(8.3 J / mol K) J / K.363 e Note: Equipartition of Energy predicts that the contribution should be nr = (8.3) = 6.6 J/K. We are fairly close to the limit S kln Q (a) the low frequency and (b) the high temperature. U because of Entropy: S Note: U = H = 8 J = 8. kj kn k ln Q kn ln e knna knn Aln e nr nr ln e.363 ( mol) 8.3 J / mol K.363 ( mol)(8.3 J / mol K) ln e 3.74 J / K U 8 J S kln Q 3.74 J / K J / K 73K Gibbs Energy: G ln Q G kln Q k kln Q klnq A lnv, N G kln Q (73 K)( 3.74 J / K) 474 J 4.7 kj

11 3. () = a () = b (3) = b (a) Actiity of Fundamental modes We will use the "shortcut" shown in the lecture (and in many other courses) IR Actiity (a) is NO IR actie because none of x, y, z belong to the a representation (b) IS IR actie because x belongs to the b representation 3(b) IS IR actie because y belongs to the b representation Raman Actiity All three rations ARE Raman actie because eery representation has one or more of quadratic functions (x, y, z, xy, xz, yz). (b) Actiity of Combination modes (b) + 3,,,,,, a xa xb B,, axaxa A IR Actiity x M e x BxBxA A i y i z i M e y B xb xa A A M e z B xa xa B A his mode IS IR actie because Mi x transforms as A Raman Actiity x BxAxAB A Same for y and z integrals xy B xa xa B A xz B xb xa A yz BxB xa A A his mode IS Raman Actie because the xz integral transforms as A (b) - 3,,,,,, axbxa B,, axaxb B IR Actiity x M e x BxBxB B A i

12 M e y B xb xb B A y i z i M e z B xa xb A A his mode is NO IR actie because none of the three transition moments transform as A Raman Actiity x BxAxB A A Same for y and z integrals xy B xa xb A xz B xb xb B A yz Bx B x B B A his mode IS Raman Actie because the xy integral transforms as A

Chapter 4 Rigid Models and Angular Momentum Eigenstates Homework Solutions

Chapter 4 Rigid Models and Angular Momentum Eigenstates Homework Solutions Capter 4 Rigid Models and Angular Momentum Eigenstates Homework Solutions 1. A i j k B i 4j k AB AB AB AB()() ( 1)(4) ()( ) 4 x x y y z z i j k i j k AxB A A A 1 x y z Bx By Bz 4 i j k 5i 1 j 14k ( 1)(