37_InstSolManual_PC

# 37_InstSolManual_PC - MOLECULES AND SOLIDS E XERCISES...

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Unformatted text preview: MOLECULES AND SOLIDS E XERCISES Section 37.2 Molecular Energy Levels 16. The energies of rotational states (above the j = state) are given by Equation 37.2, where for the HCl molecule, 2 / 2.63 meV I = h (from Example 37.1). Thus, 2 1 rot 2 ( 1) /2 ( 1)2.63 meV. E l l I l l = + = + h For 0,1, 2, l = and 3, rot 0, 2.63 meV, 7.89 meV, and 15.78 meV. E = 17. I NTERPRET The oxygen molecule must absorb a photon in order to make a transition to the excited rotational energy state. We are interested in the wavelength of the photon. D EVELOP Using Equation 37.2, the difference in energy between the 1 l = and l = states is 2 2 rot 1(1 1) 2 E I I ∆ = +- = h h The photon wavelength corresponding to this transition is rot / . hc E λ = ∆ E VALUATE Substituting the value of I given in the problem, we get 8 46 2 2 34 2 2 (3 10 m/s)(1.95 10 kg m ) 3.48 mm / 1.055 10 J s hc hc cI E I π π λ-- × × ⋅ = = = = = ∆ × ⋅ h h A SSESS Transition between adjacent rotational energy levels requires absorbing a photon in the microwave region (frequency 11 10 Hz). f , 18. The energy difference between adjacent rotational levels is proportional to the upper j-value (see Example 37.1 and the solution to Problem 30), so 8 47 2 34 / 2 / 2 (3 10 m/s)(1.75 10 kg m )/5(1.055 10 J s) hc E cI l λ π π-- = ∆ = = × × ⋅ × ⋅ = h 62.5 m. μ 19. I NTERPRET The gas molecules must absorb a photon in order to make a transition to the excited rotational state. We are given the wavelength of the photon, and asked to find the rotational inertia of the molecule. D EVELOP Using Equation 37.2, the difference in energy between the 1 l = and l = states is 2 2 1 1(1 1) 2 E I I → ∆ = +- = h h The energy of the absorbed photon equals the difference in energy: rot / . E hf hc λ ∆ = = Thus, 2 5 23 1 1240 eV nm 7.38 10 eV 1.18 10 J 1.68 cm hc E I λ-- → ⋅ = ∆ = = = × = × h We can find I readily from the above equation. E VALUATE The rotational inertia is 2 34 2 46 2 23 1 (1.055 10 J s) 9.41 10 kg m 1.18 10 J I E--- → × ⋅ = = = × ⋅ ∆ × h A SSESS The value of I is reasonable for a molecule. We can estimate the bond length of the molecule using 2 . I mR = With 26 10 kg, m- , we get 0.3 nm, R , which is also a reasonable value. 20. The spacing between adjacent vibrational energy levels is 15 14 (4.136 10 eV s)(1.32 10 Hz) hf ϖ- = = × ⋅ × = h 0.546 eV. 37.1 37 37.2 Chapter 37 21. I NTERPRET This problem is about the vibrational motion of the 2 N molecule. We’re given the energy difference between the adjacent levels, and asked about its corresponding classical vibrational frequency. D EVELOP The quantized vibrational energy levels are given by Equation 37.3: vib ( 1/2) E n ϖ = + h Therefore, the energy difference between the adjacent levels is vib . E h f ϖ ∆ = = h E VALUATE The classical vibrational frequency is 13 vib 15 0.293 eV 7.08 10 Hz 4.136 10 eV s E f h- ∆ = = = × × ⋅ A SSESS The frequency associated with vibrational motion is 13 vib 10 Hz,...
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## This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.

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37_InstSolManual_PC - MOLECULES AND SOLIDS E XERCISES...

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