PracticeExam3solution-Q1Q1

PracticeExam3solution-Q1Q1 - Solution Key for PSE#3 [1] and...

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Unformatted text preview: Solution Key for PSE#3 [1] and [2] March 15, 2007 1a] Wavenumbers are defined as = 1 , where is in cm . Wavenumbers are related to frequency by = c 100 cm 1 m (1) where c is the speed of light in m s 1 . The eigenvalues for the harmonic oscillator are E n = planckover2pi1 o ( n + 1 2 ) (2) where o = radicalBig k is the angular frequency of the oscillator, where k is the spring constant, and is the reduced mass of the molecule. To find out how o is related to the energies, we can calculate E . For any transition between n and n + 1 , we can easily see that E n n +1 = planckover2pi1 o (3) The energy gap is related to the frequency of a photon by | E | = h p , where p is the frequency of the photon. From trig, we know that = 2 , and so we see simply that o = p (4) remarkably, the molecule oscillates at the exact same frequency required to excite to a higher energy level! As a result, the fundamental frequency of the molecule can be seen to be o = 6 . 51 10 13 s 1 (5) where the period is just T = 1 = 1 . 54 10 14 s = 15 . 4 fs (6) The zero-point energy can be found by plugging n = 0 into equation 2 ZPE = 2 . 16 10 20 J = 0 . 13 eV (7) 1b] For rotational spectroscopy, we know that the eigenvalues for a system are given (in the textbook) as E J = planckover2pi1 2 2 I J ( J + 1) = hcBJ ( J + 1) (8) where clearly B = h 8 2 cI and is in units of cm 1 . Plugging in that B = 0 . 1141619 cm 1 , we can determine I = r 2 , and consequently, determine the radius to be r = 1 . 742035 10 10 m 1 1c] We can define the rotational constant in a different fashion such that E J = planckover2pi1 2 2 I J ( J + 1) = BJ ( J + 1) (9) such that B = planckover2pi1 2 2 r 2 . The energy between the J and the J+1 states is given as E J J +1 = 2 B ( J + 1) (10) more simply, it can be seen that since | E | = h, and B = planckover2pi1 2 2 r 2 , then the radius for a given J J + 1 transition must be given by r = radicalBigg h 4 2 ( J + 1) (11) where we are told that this is a 1 2 transition, so J = 1 , and so, as a result, for the v = 0 and v = 1 states we have r v =0 = 2 . 3648 10 10 m = 2 . 365 r v =1 = 2 . 3736 10 10 m = 2 . 374 (12) with r ave = 2 . 3692 10 10 m = 2 . 369 (13) 1d] The relative populations between two states as weighted by the Boltzmann distribution is given by...
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This note was uploaded on 03/06/2008 for the course CHEM 113A taught by Professor Lin during the Winter '07 term at UCLA.

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PracticeExam3solution-Q1Q1 - Solution Key for PSE#3 [1] and...

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