YF_ISM_42

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) 41 9 23 3 2 2(7.9 10 eV)(1.60 10 J eV) 6.1K 2 3 3(1.38 10 J K) K Kk T 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. IDENTIFY: The energy given to the photon comes from a transition between rotational states. SET UP: The rotational energy of a molecule is 2 (1 ) 2 El l I =+ = and the energy of the photon is E = hc/ λ . EXECUTE: Use the energy formula, the energy difference between the l = 3 and l = 1 rotational levels of the molecule is [] 22 5 3(3 1) 1(1 1) 2 E II Δ= + − + = . Since Δ E = hc/ , we get hc/ = 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 23 . 0 01 0 m / s I c π ×⋅ = × × = . Using I = m r r 0 2 , we can solve for r 0 : ( ) 52 2 26 27 NH 0 26 27 4.981 10 kg m 2.33 10 kg 1.67 10 kg 2.33 10 kg 1.67 10 kg Im m r mm ×+ × + 0 r = 5.65 × 10 –13 m EVALUATE: 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 51 9 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 × = × This frequency corresponds to that given for a microwave oven. 42.5. Let 1 refer to C and 2 to O. 26 26 120 1.993 10 kg, 2.656 10 kg, 0.1128 nm r = × = . 2 10 12 0.0644 nm (carbon) m rr ⎛⎞ ⎜⎟ + ⎝⎠ ; 1 20 0.0484 nm (oxygen) m + (b) 4 6 2 11 2 2 1.45 10 kg 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 4 4 2 2( )( 2) 2 2.21 10 kg m . mL m L × 42.7. IDENTIFY and SET UP: Set 1 KE = from Example 42.2. Use 2 1 2 KI ω = to solve for and vr = to solve for v . EXECUTE: (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 = and = gives 12 1 2 / 1.03 10 rad/s EI × 42

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42-2 Chapter 42 (b) 91 2 11 1 (0.0644 10 m)(1.03 10 rad/s) 66.3 m/s (carbon) vr ω == × × = 2 22 2 (0.0484 10 m)(1.03 10 rad/s) 49.8 m/s (oxygen) × × = (c) 12 2 / 6.10 10 s T πω × EVALUATE: From the information in Example 42.3 we can calculate the vibrational period to be 14 r 2/ 2 / 1 . 51 0 s . Tm k π = × The rotational motion is over an order of magnitude slower than the vibrational motion. 42.8. r hc Ek m , λ Δ= = = and solving for , k 2 r 2 205 N m. π c km ⎛⎞ ⎜⎟ ⎝⎠ 42.9. IDENTIFY and SET UP: The energy of a rotational level with quantum number l is 2 (1 )/ 2 l El l I =+ = (Eq.(42.3)). 2 r , Im r = with the reduced mass r m given by Eq.(42.4). Calculate I and E Δ and then use / Eh c to find .
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This note was uploaded on 03/14/2012 for the course MAE 162D taught by Professor Shaefer during the Spring '11 term at UCLA.

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

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