PC Chapter 11

PC Chapter 11 - 11 Molecular Structure 11-1 (a) We add the...

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
169 11 Molecular Structure 11-1 (a) We add the reactions +- + →+ K 4.34 eV K e and -- + →+ I e I 3.06 eV to obtain (29 + + K I K I 4.34 3.06 eV . The activation energy is 1.28 eV . (b) ( 29 σσ σ  = -+   1 37 4 1 26 dU d r rr At = 0 we have = 0 dU dr . Here  =   1 00 1 2 , σ - = 16 0 2 r , - = == 2 0.305 nm 0.272 nm . Then also = ∈ - + - + =-∈+ ∈= - = + =∈ 1 0 0 22 11 44 42 1.28 eV 3.37 eV= 4.65 eV . a aa a U r E EE E Ur (c) σ = - =- 1 4 1 dU Fr d r To find the maximum force we calculate σ = -+= 1 48 2 4 156 42 0 dF d r when σ = rupture 42 156 r - - = - × 136 76 max 19 9 4 4.65 eV 42 42 12 6 41.0 eV nm 0.272 nm 156 156 1.6 10 Nm 41.0 6.55 nN 10 m F Therefore the applied force required to rupture the molecule is + 6.55 nN away from the center.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
170 CHAPTER 11 MOLECULAR STRUCTURE 11-2 πμ  =   12 1 2 K f and using Table 11.2 one obtains ×= 13 1 530 1 5.63 10 2 giving μ - 26 1.22 10 kg . From Equation 11.3, (29 μ - = = + 26 14 16 7.47 u 1.24 10 kg 14 16 . Percent difference - - = × =- × 26 26 1.22 1.24 10 kg 100% 1.63% 1.22 10 kg . Both calculations give μ of the same order of magnitude and a small percent difference. 11-3 For the = 1 l to = 2 l transition, ( 29 [ ] + -+ == h 2 221 111 2 E hf I or = h 2 2 hf I . Solving for I gives ( ( ππ - - × = = = = ×⋅ × h 2 34 4 62 2 2 11 2 6.626 10 J s 1.46 10 kg m 2 2 2.30 10 Hz h I hf f ; μ - = + 26 1.14 10 kg mm , μ 0 0.113 nm I R , same as Example 11.1. 11-4 (a) HI: ( μ - = = + 27 1 127 u 0.992 u 1.65 10 kg 1 127 , ϖ μ = K ( ( ϖ μμ - - - - = × = h h 34 2 2 22 27 11 1 1.055 10 J s 1.45 10 m 320 N m 1.65 10 kg 1.20 10 m 0.012 0 nm A K A (b) HF: ( μ - = = + 27 1 19 u 0.950 u 1.58 10 kg 1 19 ( ( - - - - = × = 34 2 2 32 27 12 1.055 10 J s 8.53 10 m 970 N m 1.58 10 kg 9.23 10 m 0.009 23 nm A A With its smaller value for K , HI has the weaker bond. 11-5 (a) The separation between two adjacent rotationally levels is given by ∆= h 2 El I , where l is the quantum number of the higher level. Therefore λλ λ = × = 65 10 10 65 10 10 10 6 6 6 1.35 cm 8.10 cm 3.00 10 cm s 3.70 GHz 8.10 cm E E c f
Background image of page 2
MODERN PHYSICS 171 (b) == h 2 10 10 E hf I ; (29 ( 29 π π - - ×⋅ × = h 34 9 10 4 52 1.055 10 J s 2 2 3.70 10 Hz 4.53 10 kg m I f I 11-6 HF molecule: = 0 0.092 nm R (a) ( 29 μ - = = = ++ 27 12 1 19 0.95 u 1.58 10 kg 1 19 mm (b) r U ( r ) 0.092 nm 11-7 HCl molecule in the = 1 l rotational energy level: = 0 1.275 R Å, ( 29  =+   h 2 rot 1 2 E ll I . For = 1 l , ϖ h 22 rot 2 I E I , ϖ hh 2 2 2 2 I I ( ( -- -  = = = × × ××   = 2 2 2 27 10 00 4 72 1 u 35 u 0.972 2 u 1.66 10 kg u 1.275 10 m 1 u 35 u 2.62 10 kg m I RR Therefore, ϖ - - = = h 34 12 4 1.055 10 J s 2 2 5.69 10 rad s 2.62 10 kg m I . 11-8 (a) πμ ∆ == 01 2 hK E hf or π μ = 2 K f or = 4 fK where ( ( ( μ μ μ = + = × × + = × × + 27 27 27 26 NO 1 u 127 u 1.66 10 kg u 1.65 10 kg 1 u 127 u 14 u 16 u 1.66 10 kg u 1.24 10 kg 14 u 16 u
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
172 CHAPTER 11 MOLECULAR STRUCTURE so ( 29 ( ( ( π π - - ×= 2 2 13 27 HI 2 2 13 26 NO 4 6.69 10 Hz 1.65 10 kg 292 N m 4 5.63 10 Hz 1.24 10 kg 1 550 N m K K (b) μ  =   h 12 2 22 K AK or μ = h 2 1 A K ( 29( 29 - - - ×⋅ = × = 34 2 2 27 HI 1.06 10 J s 1.519 10 m 292 N m 1.65 10 kg 0.012 3 nm A A Likewise, = NO 0.004 92 nm A . (c) N and O are “cemented” by more electrons. 11-9 ( (29 ( 29 μ - = = = ++ 27 1 u 35 u 35 u 1.62 10 kg 1 u 35 u 36 mm (a) ( 29( 29 μ - -- = = × × = 2 2 27 10 47 2 0 1.62 10 kg 1.28 10 m 2.65 10 kg m IR ( ( 29 ( 29 ( - - - - - - =+ = = × × = ×+ == = = = h h rot 2 34 2 2 23 4 72 3 rot rot 3 rot 3 rot 2 rot 1 2 1.054 10 J s 2.1 10 J 1.31 10 eV 2 2 2.65 10 kg m 1.31 10 eV 1 00 1 2.62 10 eV 2 7.86 10 eV 3 1.57 10 eV E ll I I E lE (b) = 2 2 Kx U , = 0.15 eV U when = 0.01 nm x ( π μπ - - - =  = =   × 2 11 19 13 27 10 m 0.15 eV 1.6 10 J eV 2 480 N m 1 1 480 8.66 10 Hz 1.62 10 K K K f
Background image of page 4
MODERN PHYSICS 173 (c) ( 29 =+ vib 1 2 E v hf ( 29 ( ( -- - - - - - = × × = ×= = = = == × = = 34 13 20 20 0 2 2 20 0
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 18

PC Chapter 11 - 11 Molecular Structure 11-1 (a) We add the...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online