1211_Physics ProblemsTechnical Physics

# 1211_Physics ProblemsTechnical Physics - 552 P43.12...

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552 Molecules and Solids P43.12 (a) Minimum amplitude of vibration of HI is 1 2 1 2 2 kA hf = : A hf k == ×⋅× = 6626 10 669 10 118 10 00118 34 13 11 .. Js s 320 N m m n m ej e j . (b) For HF, A = = 872 10 970 772 10 000772 34 13 12 s Nm n m e j . Since HI has the smaller k , it is more weakly bound. P43.13 µ = + × = × −− mm 12 27 27 35 36 1 66 10 1 61 10 kg kg E k vib J e V × × = = 1055 10 480 161 10 574 10 0358 34 27 20 . . P43.14 (a) The reduced mass of the O 2 is = + × = × 16 16 16 16 8 8 1 66 10 1 33 10 27 26 u u u u kg kg af a f a f . The moment of inertia is then Ir == × × = × 22 6 1 0 2 46 1 33 10 1 20 10 1 91 10 . kg m kg m 2 e j . The rotational energies are E I JJ rot 2 kg m =+ = ×⋅ + = 2 34 2 46 2 1 2 1 91 10 1 a f a f . . . Thus EJ J rot J + 291 10 1 23 . a f . And for J = 012 ,,, E rot eV, eV × 0 3 64 10
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