HW#2_solutions

HW#2_solutions - ECE594I, Prof. Brown, Fall Quarter 2009...

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ECE594I, Prof. Brown, Fall Quarter 2009 1 Homework #2 Solutions: THz Propagation Phenomenology 1. Vapor State: Molecular Rotational Transitions. . (a) A quick Wikipedia search shows that N 2 O is the linear molecule (i.e., “rotor”) with bond lengths shown below. The moment of inertia is then given by 3 2 1 2 23 3 2 2 13 3 1 2 12 2 1 m m m d m m d m m d m m I + + + + = where m 1 , m 2 and m 3 are the masses of atoms 1, 2, and 3, respectively and can be designated arbitrarily provided d ij is the corresponding distance between atom i and atom j. For the two nitrogen atoms m ~ 14 m p and for the oxygen m = 16 m p , where m p is the proton rest mass. For the ground state rotational transition, we find is ) / ( 2 I h U h = = ν or ν = 25.2 GHz, and any harmonic of this is also an allowed rotational transition So between 100 GHz and 1.0 THz, the lowest frequency transition will be the 4 th harmonic (at 100.7 GHz) and the highest will be the 39 th harmonic (at 981.5 GHz), for a total of 36 rotational transitions. (b) Although greatly simplified by being planar, phosgene is more complicated than a simple rotor since we need to calculate all the interatomic distances. Designating m 1 as the central carbon, m 2 as the oxygen, and m 3 and m 4 as the chlorines, we have by inspection d 12 = 1.18 A, d 13 = 1.74 A, d 14 = 1.74 A. By trigonometry (law of cosines) we have for d 23 (oxygen-chlorine distance): (d 23 ) 2 = (d 12 ) 2 + (d 13 ) 2 -2 d 12 d 13 cos α , where α is the included angle = (360 – 111.8)/2 = 124.1 o . For d 34 (chlorine-chlorine distance) we have: (d 34 ) 2 = (d 13 ) 2 + (d 14 ) 2 -2 d 13 d 14 cos β where β = 111.8 ο . These yield d 23 = 2.59 A and d 34 = 2.88 A. We then substitute into the following expression noting there are six unique pair-wise terms for four atoms: 4 3 2 1 2 34 4 3 2 24 4 2 2 23 3 2 2 14 4 1 2 13 3 1 2 12 2 1 m m m m d m m d m m d m m d m m d m m d m m I + + + + + + + + = =3.53x10 -45 [MKS] So that the ground state rotational frequency (about the axis perpendicular to the plane) is ) / ( 2 I h U h = = = 4.76 GHz. This is in the microwave “C band”, and is so low in frequency because chlorine is relatively heavy. 2. Liquid State: Debye Model.
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This note was uploaded on 12/02/2009 for the course ECE 000 taught by Professor O during the Spring '09 term at UCSB.

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HW#2_solutions - ECE594I, Prof. Brown, Fall Quarter 2009...

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