chapter 8 to 9 - Chem140B Physical Chemistry II Lecture...

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Chem140B – Physical Chemistry II Lecture Homework #6 March 10, 2009 due March 17, 2009 Q8.4) If the rotational levels of a diatomic molecule were equally spaced and the selection rule remained unchanged, how would the appearance of the rotational-vibrational spectrum in Figure 8.16 change? Only the two peaks corresponding to the 0-1 and 1-0 transitions would be observed. P8.15) The rotational constant for 2 D 19 F determined from microwave spectroscopy is 11.007 cm –1 . The atomic masses of 19 F and 2 D are 18.9984032 and 2.0141018 amu, respectively. Calculate the bond length in 2 D 19 F to the maximum number of significant figures consistent with this information. () 0 22 2 0 34 0 2 27 1 1 10 1 11 0 ; 88 6.6260755 10 J s 2.0141018 18.9984032amu 8 1.6605402 10 kgamu 11.007cm 2.99792458 10 cms 2.0141018 18.9984032 9.1707 10 m hh Br rB r r πμ π −− == × = × ×× × × × + × P8.16) Calculate the moment of inertia, the magnitude of the rotational angular momentum, and the energy in the J = 1 rotational state for 1 H 2 in which the bond length of 1 H 2 is 74.6 pm. The atomic mass of 1 H is 1.007825 amu. ( ) ( ) 2 2 7 1 0 48 2 34 34 34 2 2 34 2 48 2 1.007825 amu 1.6605402 10 74.6 10 m 2 1.007825 4.657 10 kg m 1 2 1.0554 10 J s = 1.4914 10 J s = 1.49 10 kg m s 2 1.0554 10 J s 1 2 2 4.657 10 kg m J Ir JJ J E I μ × × × × × =+ = × × × × + = = = 21 2.39 10 J × 1 2
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P8.19) Because the intensity of a transition to first order is proportional to the population of the originating state, the J value for which the maximum intensity is observed in a rotational- vibrational spectrum is not generally J = 0. Treat J in the equation n n g g eJ e JJ kT Ik T J 00 12 0 2 21 == + −+ εε bg = () as a continuous variable. a. Show that 22 2 2 ( 1 )2 0 2 2 J J JI k T J k n d n J ee dJ IkT ⎛⎞ ⎜⎟ + ⎝⎠ =− = T 2 (1 ) 2 ) 2 0 2 ) 2 ) 2 2 2 ) 2 1 1 2 2 2 J T T T T T T n d J I k T n e dJ dJ J e J =+ + + + + = = = b) Show that setting 0 0 J n d n = gives the equation 2 2 max 20 2 J + = = 2 2 2 ) 2 ) 2 0 2 2 ) 2 2 cancelling on both sides of the equation, 2 0 2 J T T T n d nJ dJ I kT J e + = + = = = = c) Show that the solution of this quadratic equation is max 2 14 1 2 J = = In Problems P19.20 and P19.21 we assume that the intensity of the individual peaks is solely
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This note was uploaded on 06/14/2009 for the course CHEM CHEM 2080 taught by Professor Pathivan during the Spring '05 term at Indian Institute of Technology, Chennai.

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chapter 8 to 9 - Chem140B Physical Chemistry II Lecture...

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