1221_Physics ProblemsTechnical Physics

1221_Physics ProblemsTechnical Physics - 562 Molecules and...

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562 Molecules and Solids (b) The equilibrium distance is the same for both molecules. Ir r I I 14 14 2 14 12 12 2 14 12 12 14 46 46 747 1 46 10 1 59 10 == F H G I K J = F H G I K J = F H G I K J ×⋅ = −− µ . .. u 6.86 u kg m kg m 22 ej (c) The molecule can move to the vJ 19 , bg state or to the 11 1 , state. The energy it can absorb is either E hc hf I hf I + F H G I K J ++ L N M O Q P F H G I K J L N M O Q P λ 1 1 2 99 1 2 0 1 2 10 10 1 2 14 2 14 14 2 14 a f a f , or E hc hf I hf I + F H G I K J L N M O Q P −+ F H G I K J L N M O Q P 1 1 2 11 11 1 2 0 1 2 10 10 1 2 14 2 14 14 2 14 a f a f . The wavelengths it can absorb are then π = c fI 14 14 10 2 = or = + c 14 14 11 2 = . These are: = × ×− × × = 300 10 615 10 101055 10 2 159 10 496 8 13 34 46 . . . ms Hz J s kg m m 2 e j and = × ×+ × × = 6 1 51 0 1 1 1 0 5 0 21 5 91 0 479 8 13 34 46 . . . Hz J s kg m m 2 e j . P43.50 For the N 2 molecule, k = 2297 Nm, m 232 10 26 . k g , r 120 10 10 m , = m 2 ω ==× k 445 10 14 r a d s , × 6 1 0 2 46 1 16 10 1 20 10 1 67 10 . kg m kg m 2 e j . For a rotational state sufficient to allow a transition to the first exited vibrational state, = = 2 2 1 I JJ += a f so I = ×× × = 1 2 2 1 67 10 4 45 10 1055 10 1410 46 14 34
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This note was uploaded on 12/15/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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