1212_Physics ProblemsTechnical Physics

1212_Physics ProblemsTechnical Physics - Chapter 43 P43.15...

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Chapter 43 553 P43.15 In Benzene, the carbon atoms are each 0.110 nm from the axis and each hydrogen atom is 0 110 0 100 0 210 .. . += nm nm af from the axis. Thus, Im r = 2 : I × × = × −− −−− 6 1 99 10 0 110 10 6 1 67 10 0 210 10 1 89 10 26 9 2 27 9 2 45 . kg m kg m kg m 2 ej . The allowed rotational energies are then E I JJ EJ J J rot 2 rot Js kg m J e V eV where =+ = ×⋅ × × + = = 2 34 2 45 24 6 2 1 1055 10 2 1 89 10 1 2 95 10 1 18 4 10 1 1 84 1 0123 bg . . ., , , , µ The first five of these allowed energies are: E rot eV, 111 eV, 221 eV, and 369 eV = 03 69 ,. µµµ . *P43.16 We carry extra digits through the solution because part (c) involves the subtraction of two close numbers. The longest wavelength corresponds to the smallest energy difference between the rotational energy levels. It is between J = 0 and J = 1 , namely = 2 I λ π == = hc E hc I Ic h min = 2 2 4 . If is the reduced mass, then Ir × = × ∴= ×× µµ πµ 29 2 20 22 0 8 34 23 0 127 46 10 1 624 605 10 4 1 624 605 10 2 997 925 10 6626075 10 2901830 10 . . m m m s mkg 2 2 e j e j (1) (a) 35 27 1007825 34968853 0 979 593 1 626 653 10 = + × uu u
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