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1121_PartUniversity Physics Solution

# 1121_PartUniversity Physics Solution - 42-2 Chapter 42(b v1...

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42-2 Chapter 42 (b) 9 12 1 1 1 (0.0644 10 m)(1.03 10 rad/s) 66.3 m/s (carbon) v r ω = = × × = 9 12 2 2 2 (0.0484 10 m)(1.03 10 rad/s) 49.8 m/s (oxygen) v r ω = = × × = (c) 12 2 / 6.10 10 s T π ω = = × E VALUATE : From the information in Example 42.3 we can calculate the vibrational period to be 14 r 2 / 2 / 1.5 10 s. T m k π ω π = = = × The rotational motion is over an order of magnitude slower than the vibrational motion. 42.8. r hc E k m , λ Δ = = = and solving for , k 2 r 2 205 N m. π c k m λ ′ = = 42.9. I DENTIFY and S ET U P : The energy of a rotational level with quantum number l is 2 ( 1) / 2 l E l l I = + = (Eq.(42.3)). 2 r , I m r = with the reduced mass r m given by Eq.(42.4). Calculate I and E Δ and then use / E hc λ Δ = to find . λ E XECUTE : (a) 26 27 27 1 2 Li H r 26 27 1 2 Li H (1.17 10 kg)(1.67 10 kg) 1.461 10 kg 1.17 10 kg 1.67 10 kg m m m m m m m m m × × = = = = × + + × + × 2 27 9 2 47 2 r (1.461 10 kg)(0.159 10 m) 3.694 10 kg m I m r = = × × = × 2 2 3: 3(4) 6 2 l E I I = = = = = 2 2 4: 4(5) 10 2 l E I I = = = = = 2 34 2 21 3 4 3 47 2 (1.055 10 J s) 4 4 1.20 10 J 7.49 10 eV 3.694 10 kg m E E E I × Δ = = = = × = × × = (b) / E hc λ Δ = so 15 8 3 (4.136 10 eV)(2.998 10 m/s) 166 m 7.49 10 eV hc E λ μ × × = = = Δ × E VALUATE : LiH has a smaller reduced mass than CO and λ is somewhat smaller here than the λ calculated for CO in Example 42.2 42.10. I DENTIFY : The vibrational energy of the molecule is related to its force constant and reduced mass, while the rotational energy depends on its moment of inertia, which in turn depends on the reduced mass. S ET U P : The vibrational energy is r 1 1 2 2 n k E n n m ω = + = + = = and the rotational energy is 2 ( 1) 2 l E l l I = + = . E XECUTE : For a vibrational transition, we have v r k E m Δ = = , so we first need to find m r . The energy for a rotational transition is [ ] 2 2 R 2 2(2 1) 1(1 1) 2 E I I Δ = + + = = = . Solving for I and using the fact that I = m r r 0 2 , we have 2 2 r 0 R 2 m r E = Δ = , which gives ( )( ) ( ) ( ) 34 16 2 r 2 2 9 4 0 R 2 1.055 10 J s 6.583 10 eV s 2 0.8860 10 m 8.841 10 eV m r E × × = = Δ × × = = 2.0014 × 10 –28 kg Now look at the vibrational transition to find the force constant. ( ) ( ) ( ) 2 28 2 r v v 2 2 16 r 2.0014 10 kg (0.2560 eV) 6.583 10 eV s m E k E k m × Δ Δ = = = × = = = 30.27 N/m E VALUATE : This would be a rather weak spring in the laboratory. 42.11. (a) 2 2 2 2 2 2 1 ( 1) ( 1) , ( ) 2 2 2 l l l l l l l E E E l l l l I I I I + = = Δ = + + = = = = = (b) 2 2 E E l f . h π π I = = = = =

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Molecules and Condensed Matter 42-3 42.12. I DENTIFY : Find E Δ for the transition and compute λ from / . E hc λ Δ = S ET U P : From Example 42.2, 2 ( 1) , 2 l E l l I = + = with 2 3 0.2395 10 eV. 2 I = × = From Example 42.3, 0.2690 eV E Δ = is the spacing between vibrational levels. Thus 1 2 ( ) , n E n ω = + = with 0.2690 eV. ω = = By Eq.(42.9), 2 1 2 ( ) ( 1) .
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