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PC Chapter 11

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

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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 ( 29 σ σ σ ∈  = - + 13 7 4 12 6 dU dr r r At = 0 r r we have = 0 dU dr . Here σ σ = 13 7 0 0 1 2 r r , σ - = 1 6 0 2 r , ( 29 σ σ - = = = 1 6 2 0.305 nm 0.272 nm . Then also ( 29 ( 29 - - = - + = - + = -∈+ ∈= - = + =∈ 12 6 1 6 1 6 0 0 0 0 0 0 2 2 1 1 4 4 4 2 1.28 eV 3.37 eV= 4.65 eV . a a a a r r U r E E E r r E U r (c) ( 29 ( 29 ( 29 σ σ σ ∈  = - = - 13 7 4 12 6 dU F r dr r r To find the maximum force we calculate ( 29 ( 29 σ σ σ ∈  = - + = 14 8 2 4 156 42 0 dF dr r r when ( 29 σ = 1 6 rupture 42 156 r ( 29 ( 29 ( 29 - - = - = - × = - = - 13 6 7 6 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.

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170 CHAPTER 11 MOLECULAR STRUCTURE 11-2 π μ = 1 2 1 2 K f and using Table 11.2 one obtains π μ × = 1 2 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 ( 29 - - - × = × = - × 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 ( 29 [ ] + - + = = h 2 2 2 1 1 1 1 2 E hf I or = h 2 2 hf I . Solving for I gives ( 29( 29 π π - - × = = = = × × h 2 34 46 2 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 2 1 2 1.14 10 kg m m m m , μ = = 1 2 0 0.113 nm I R , same as Example 11.1. 11-4 (a) HI: ( 29( 29 μ - = = = × + 27 1 127 u 0.992 u 1.65 10 kg 1 127 , ϖ μ = 1 2 K ( 29 ( 29 ϖμ μ - - - - × = = = = × × = × = h h 1 2 34 2 22 2 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: ( 29( 29 μ - = = = × + 27 1 19 u 0.950 u 1.58 10 kg 1 19 ( 29 ( 29 - - - - × = = × × = × = 34 2 23 2 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 E l I , where l is the quantum number of the higher level. Therefore ( 29 λ λ 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
MODERN PHYSICS 171 (b) = = h 2 10 10 E hf I ; ( 29 ( 29 π π - - × = = × = × h 34 9 10 45 2 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( 29 ( 29 μ - = = = = × + + 27 1 2 1 2 1 19 0.95 u 1.58 10 kg 1 19 m m m m (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 l l I . For = 1 l , ϖ = = h 2 2 rot 2 I E I , ( 29 ϖ = = h h 1 2 2 2 2 2 I I ( 29( 29 ( 29 - - - = = = × × × × + + = × 2 2 2 27 10 1 2 0 0 1 2 47 2 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 m m I R R m m Therefore, ( 29 ϖ - - × = = = × × h 34 12 47 2 1.055 10 J s 2 2 5.69 10 rad s 2.62 10 kg m I .

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

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