YF_ISM_42 - MOLECULES AND CONDENSED MATTER 42 42.1 3 2 K...

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42-1 M OLECULES AND C ONDENSED M ATTER 42.1. (a) 4 19 23 3 2 2(7.9 10 eV)(1.60 10 J eV) 6.1 K 2 3 3(1.38 10 J K) K K kT T k × × = = = = × (b) 19 23 2(4.48 eV) (1.60 10 J eV) 34,600 K. 3(1.38 10 J K) T × = = × (c) The thermal energy associated with room temperature (300 K) is much greater than the bond energy of 2 He (calculated in part (a)), so the typical collision at room temperature will be more than enough to break up 2 He . However, the thermal energy at 300 K is much less than the bond energy of 2 H , so we would expect it to remain intact at room temperature. 42.2. (a) 2 0 1 5.0 eV. 4 e U πε r = − = − (b) 5.0 eV (4.3 eV 3.5 eV) 4.2 eV. + = − 42.3. I DENTIFY : The energy given to the photon comes from a transition between rotational states. S ET U P : The rotational energy of a molecule is 2 ( 1) 2 E l l I = + = and the energy of the photon is E = hc/ λ . E XECUTE : Use the energy formula, the energy difference between the l = 3 and l = 1 rotational levels of the molecule is [ ] 2 2 5 3(3 1) 1(1 1) 2 E I I Δ = + + = = = . Since Δ E = hc/ λ , we get hc/ λ = 5 2 = / I . Solving for I gives ( ) ( ) 34 52 2 8 5 1.055 10 J s (1.780 nm) 5 4.981 10 kg m 2 2 3.00 10 m/s I c λ π π × = = = × × = . Using I = m r r 0 2 , we can solve for r 0 : ( ) ( )( ) ( )( ) 52 2 26 27 N H 0 26 27 N H 4.981 10 kg m 2.33 10 kg 1.67 10 kg 2.33 10 kg 1.67 10 kg I m m r m m × × + × + = = × × 0 r = 5.65 × 10 –13 m E VALUATE : This separation is much smaller than the diameter of a typical atom and is not very realistic. But we are treating a hypothetical NH molecule. 42.4. The energy of the emitted photon is 5 1.01 10 eV, × and so its frequency and wavelength are 5 19 34 (1.01 10 eV)(1.60 10 J eV) 2.44 GHz (6.63 10 J s) E f h × × = = = × and 8 9 (3.00 10 m s) 0.123 m. (2.44 10 Hz) c f λ × = = = × This frequency corresponds to that given for a microwave oven. 42.5. Let 1 refer to C and 2 to O. 26 26 1 2 0 1.993 10 kg, 2.656 10 kg, 0.1128 nm m m r = × = × = . 2 1 0 1 2 0.0644 nm (carbon) m r r m m = = + ; 1 2 0 1 2 0.0484 nm (oxygen) m r r m m = = + (b) 2 2 46 2 1 1 2 2 1.45 10 kg m ; I m r m r = + = × yes, this agrees with Example 42.2. 42.6. Each atom has a mass m and is at a distance 2 L from the center, so the moment of inertia is 2 2 44 2 2( )( 2) 2 2.21 10 kg m . m L mL = = × 42.7. I DENTIFY and S ET U P : Set 1 K E = from Example 42.2. Use 2 1 2 K I ω = to solve for ω and v r ω = to solve for v . E XECUTE : (a) From Example 42.2, 23 1 0.479 meV 7.674 10 J E = = × and 46 2 1.449 10 kg m I = × 2 1 2 K I ω = and K E = gives 12 1 2 / 1.03 10 rad/s E I ω = = × 42
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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.
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