PracticeExam3solution-Q1Q1 - Solution Key for PSE#3[1 and[2...

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Solution Key for PSE#3 [1] and [2] March 15, 2007 1a] Wavenumbers are defined as ˜ ν = 1 λ , where λ is in cm . Wavenumbers are related to frequency by ν = c ˜ ν · 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 ω o ( n + 1 2 ) (2) where ω o = radicalBig k μ is the angular frequency of the oscillator, where k is the spring constant, and μ is the reduced mass of the molecule. To find out how ω o is related to the energies, we can calculate Δ E . For any transition between n and n + 1 , we can easily see that Δ E n n +1 = planckover2pi1 ω o (3) The energy gap is related to the frequency of a photon by | Δ E | = p , where ν p is the frequency of the photon. From trig, we know that ω = 2 πν , and so we see simply that ν o = ν 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 ν o = 6 . 51 10 13 s 1 (5) where the period is just T = 1 ν = 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 π 2 cI and is in units of cm 1 . Plugging in that B = 0 . 1141619 cm 1 , we can determine I = μr 2 0 , and consequently, determine the radius to be r 0 = 1 . 742035 10 10 m 1
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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 μr 2 0 . The energy between the J and the J+1 states is given as Δ E J J +1 = 2 B ( J + 1) (10) more simply, it can be seen that since | Δ E | = hν, and B = planckover2pi1 2 2 μr 2 0 , then the radius for a given J J + 1 transition must be given by r 0 = radicalBigg h 4 π 2 μν ( 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 Å r v =1 = 2 . 3736 10 10 m = 2 . 374 Å (12) with r ave = 2 . 3692 10 10 m = 2 . 369 Å (13) 1d] The relative populations between two states as weighted by the Boltzmann distribution is given by n i n j = g i g j e ( ǫ i ǫ j ) / (
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