Chapter 4 Rigid Models and Angular Momentum Eigenstates Homework Solutions

Transcription

1 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)( ) ()(4) ()( ) ()() ()(4) ( 1)(). Ae sin i (a) L /I L 1 1 sin( ) I I sin( ) sin ( ) Aˆ Bˆ I Calculation of A sin sin i i i Ae sin sin Ae sin cos Ae sin cos 1 1 i i 1 sin Ae sin cos Ae sin cos sin sin sin i 1 Ae sin ( sin ) cos sin cos sin i i Ae sin cos Ae sin (1 sin Ae sin i

2 Calculation of B 1 1 i 1 i Ae sin i Ae sin sin sin sin 4Ae i Calculation of L /I L Aˆ Bˆ Ae sin 4Ae I I I 6 i 6 Ae sin I I i i Terefore, te Energy is: 6 ll ( 1) E I I (b) L We've actually already accomplised tis. Te energy is: L 6 E I I 6 L I( E) I 6 l( l1) I Note: Te above result illustrates tat is a wavefunction wit l = (c) L z Lˆz m i ˆ i i i Lz Ae sin Asin ie Ae sin i i Tus we see tat, for tis wavefunction, mz = -

3 . Te energy levels for a D Rigid Rotor (see PP) are given by: m Em m0, 1,, etc. I x10 19amu mf m 1 F m F amu 1.58x10 m m F F 6 I r 1.58x10 1.4x10 m.18x10 m fin minit E Efin E init x10 1 J 4 m x10 J s I I I.1810 x m 46 c E c 1 E 1.04x10 J 5.cm 4 10 c (6.6x10 J s)(.00x10 cm / s) 1 4. Rotational Raman: Transition J'' J' = J''+ E cb ( J')( J' 1) cbj ''( J'' 1) E BJ ( ')( J ' 1) BJ ''( J '' 1) c 1 1 First line: J'' = 0 J' = B()() B(0)(1) 6B50.cm B 8.7 cm Second line: J'' = 1 J' = B()(4) B(1)() 10B8.7 cm B 8.7 cm x10 m s s B I.4x10 m Ic 8 Bc 8(.14) (8.7 cm )(.00x10 cm / s) 47 7 mhmbr (1.00 amu)(79.0 amu) 1.66x amu 1.64x10 m m 1.00amu 79.0amu 1amu H Br I.4x10 m x10 I r r x m x m nm A

4 5. Any molecule wit a dipole moment will ave a ational microwave spectrum. DO ave a ational microwave spectrum: HO, H-CC-Cl, cis-1,-dicloroetylene, NH DO NOT ave a ational microwave spectrum: H-CC-H, benzene E 6. BJ ( ')( J ' 1) BJ ''( J '' 1) c (a) First line: J'' = 0 J' = 1 1 B(1)() B(0)(1) B.84 cm B 1.9cm 4 6.6x10 m s s B I 1.46x10 m Ic Bc cm x cm s (.14) (1.9 )( / ) 7 mm C O (1.0 amu)(16.0 amu) 1.66x amu 1.14x10 m m 1.0amu 16.0amu 1amu C O I 1.46x10 m x10 I r r x10 m 1.1x10 m0.11nm1.1 A 1 46 (b) 7t. Line: J'' = 6 J' = 7 B(7)(8) B(6)(7) 14B 14(1.9 cm ) 6.9cm (c) 5t. line: J" = 4 J' = 5 nd. line: J" = 1 J' = 1 1 cbj ( J 1) EJ kt NJ (J 1) e (J 1) e Te transition intensity is proportional to NJ were J is te initial state of te absorption. cb x J s x cm s cm kt x J K K ( )( / )(1.9 ) ( / )(98 ) 9.9x10 (4)(5) cb (4 5) N4 (4 1) e 9 (1)() cb (1 ) 1 (11) I 18 cb 18(9.9x10 ) e e.54 I N e

5 (d) Te maximum intensity transition is from te value of J wit te igest value of NJ We can determine is from basic calculus. We will differentiate NJ wit respect to J and set te derivative equal to 0 to determine Jmax ( ) / ( dn J J cb kt J J) cb/ kt J d ( J J) cb/ kt d d 0 (J 1) e (J 1) e e (J 1) dj dj dj dj ( J J ) cb/ ( kt cb J J ) cb/ kt 0 (J 1) e (J 1) e kt ( ) / 0 ( 1) cb J e J cb kt J kt (Jmax 1) 15 cb 9.9x10 kt J Jmax = max 7. H-CC-H rcc = 1.1 Å, rch = 1.05 Å Terefore, rc = 0.50(1.1) = Å rh = 0.50(1.1) = Å Calculation of I and B i i C C H H x10 1x10 m 46 I m r m r m r (1.0 amu) A (1.00 amu) A i 14.6amu A.7x10 m 1amu 1A 4 6.6x10 m s s B 1.18cm Ic 8(.14) (.7x10 m )(.00x10 cm / s) 1 (a) Line 1: J''=0 J'=: B( J')( J' 1) B( J'')( J'' 1) B()() B(0)(1) 6B 6(1.18 cm ) 7.08cm 1 1 Line : J''=1 J'=: B( J ')( J ' 1) B( J '')( J '' 1) B()(4) B(1)() 10B 10(1.18 cm ) 11.8cm 1 1 (b) Line 1: J''=0 J'=: B( J ')( J ' 1) B( J '')( J '' 1) B()(4) B(0)(1) 1B 1(1.18 cm ) 14.16cm 1 1 Line : J''=1 J'=4: B( J ')( J ' 1) B( J '')( J '' 1) B(4)(5) B(1)() 18B 18(1.18 cm ) 1.4cm 1 1

6 (c) T = 100 o C = 7 K Line 0: J''=19 J'=1 Line 5: J''=4 J'=6 cb x J s x cm s cm kt (1.8x10 J / K)(7 K) 4 ( )( / )( ) 4.56x10 (19)(0) cb (19 1) 19 ( 19 1) 9 (4)(5) cb I(4 6) N4 (4 1) e 9 I N e e 4.e 4.(0.194) cb 60(4.56x10 ) 8. NO, T = 150 o C = 4 K, = Ia =.07x m, Ib = 6.0x m, Ic = 6.50x m 4 6.6x10 J s 4 6.6x10 J s 4 6.6x10 J s a 1.14 K Ik x m x J K 8 a / b 0.651K Ik x m x J K 8 b / c 0.61K Ik x m x J K 8 c / Preliminary Calculations: Partition Function and Derivative q 1/ 1/ 1/ 1/ T.14 (4 K) 45 a b c (1.14 K)(0.651 K)(0.61 K) 1/ N T ln( Q ) ln q nnaln( q ) nnaln a b c nn 1/ / Aln( T ) ln ln a b c V, N / dln T nn A A [ln( Q )] nn T dt T 1/ 1/ [ln( Q )] lnv T, N 0 i.e. te ational partition function is independent of volume.

7 Internal Energy: U ln Q nn A U kt kt n( NAk) T nrt T T Vn, mol 8.1 J / mol K 4 K J 10.6 kj Note: Tis result illustrates equipartition of Internal Energy. Eac of te tree ations contributes (1/)RT to U. Entalpy: H ln Q ln Q H kt kt U 0 nrt 10.6kJ T lnv V, N T, N Constant Pressure Heat Capacity H d CP nrt nr mol (8.1 J / mol K ) 4.9 J / mol K T dt Vn, U d Note: CV nrt nr 4.9 J / mol K T dt Vn, Notice tat te ational contribution to bot te constant volume and constant pressure eat capacities are te same. It is only te translational contribution to te two eat capacities wic is different. Entropy: H nrt U U S kln Q knna ln( q ) nrln( q ) T T T mol8.1 J / mol K ln 45mol8.1 J / mol K 14.9 J / K 4.9 J / K J / K 160 J / K Helmoltz Energy: A A ktln Q kt nna ln( q ) nrtln( q ) ( mol)(8.1 J / mol K) 4K ln(45) J 57.1kJ Gibbs Energy: G ln( Q ) G kt ln Q kt kt ln( Q ) 0 V Tn, A 57.1kJ

8 9. T T T I T r q CT R 8 Ik 8 k 8 k C is a constant wic cancels out wen taking ratios. (a) mhmh mh 1.0amu H 0.5amu mh m H mm D D md.0amu D 1.0amu md m D Note: Because we only need te ratio of reduced masses, it is not necessary to convert to SI units. q ( D ) CD T 1.0 D q ( H ) C T 0.5 H H q (D)= x q (H) = (1.70) =.40 (b) T1 = 5 o C = 98 K, T = 000 o C = 7 K q ( T) C T T 7K q ( T) C T T 98 K q (T)=11.0 x q (T1) = (11.0)(1.70) = 18.7