Chapter37

# Chapter37 - MOLECULES AND SOLIDS 37 2 2 EXERCISES Section...

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37.1 MOLECULES AND SOLIDS EXERCISES Section 37.2 Molecular Energy Levels 16. The energies of rotational states (above the j = 0 state) are given by Equation 37.2, where for the HCl molecule, h 2 263 /. I = meV (from Example 37.1). Thus, El l Il l rot meV. =+ = + () / . 12 1 2 6 3 2 1 2 h For l = 012 ,, , and 3, E rot meV meV and = 0 263 789 1578 ,. , . . 17. INTERPRET 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. DEVELOP Using Equation 37.2, the difference in energy between the l = 1and l = 0 states is E II rot = hh 22 2 11 1 0 The photon wavelength corresponding to this transition is λ = hc E rot EVALUATE Substituting the value of I given in the problem, we get ππ == = = ×× hc E hc I cI 2 84 3 1 0 / (/ m s)(1.95 10 6 2 34 1 055 10 348 kg m Js mm ×⋅ = ) . . ASSESS Transition between adjacent rotational energy levels requires absorbing a photon in the microwave region (frequency f , 10 11 Hz). 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 = × × hc E cI l // ( / . 2 2 3 10 1 75 10 7 h ms)( kg × = mJ s 2 )/ ( . ) 5 1 055 10 34 62 5 .m . µ 19. INTERPRET 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. DEVELOP Using Equation 37.2, the difference in energy between the l = 1and l = 0 states is E 10 2 0 = The energy of the absorbed photon equals the difference in energy: Eh f h c rot Thus, h 2 5 1240 168 738 10 I E hc = eV nm cm e . .VJ 118 10 23 . We can find I readily from the above equation. EVALUATE The rotational inertia is I E × = h 2 34 2 23 1 055 10 9 (. . . ) J 41 10 46 kg m 2 ASSESS The value of I is reasonable for a molecule. We can estimate the bond length of the molecule using Im R = 2 . With m , 10 26 kg, we get R , 03 .nm , which is also a reasonable value. 20. The spacing between adjacent vibrational energy levels is h ω × × = hf ) 4 136 10 1 32 10 15 14 eV s Hz) 0 546 .e V . 37

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37.2 Chapter 37 21. INTERPRET This problem is about the vibrational motion of the N molecule. 2 We’re given the energy difference between the adjacent levels, and asked about its corresponding classical vibrational frequency. DEVELOP The quantized vibrational energy levels are given by Equation 37.3: En vib =+ (/ ) 12 h ω Therefore, the energy difference between the adjacent levels is Eh f vib == h . EVALUATE The classical vibrational frequency is f E h ×⋅ vib eV eV s 0 293 4 136 10 708 10 15 13 . . .H z ASSESS The frequency associated with vibrational motion is f vib Hz, , 10 13 which is higher than that associated with rotation, f rot Hz. , 10 11 This means that a more energetic photon must be absorbed by the molecule in order to excite the vibrational modes.
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Chapter37 - MOLECULES AND SOLIDS 37 2 2 EXERCISES Section...

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