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chem5314_hwk5_soln

# chem5314_hwk5_soln - Chem 5314 homework#5 out of 33 marks...

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Chem 5314 homework #5 – out of 33 marks Problem 1 – 10 marks Relative to the n = 0 vibrational level of the ground (X) electron state of 79 Br 2 , the vibrational levels (n) of the excited (B) state have energies (in cm - 1 ) as follows: (the table is taken from J. Mol. Spectroscopy 51 , 428 (1974), which is in the library; the symbols X and B are spectroscopic symbols we have not discussed in class) n E (cm - 1 ) 0 15823.47 1 15987.78 2 16148.73 3 16306.26 4 16460.33 5 16610.87 6 16757.83 7 16901.15 8 17040.78 9 17176.65 10 17308.65 11 17436.89 12 17561.14 13 17681.50 14 17797.88 15 17910.18 16 18018.37 17 18122.45 18 18222.46 19 18318.33 20 18410.07 21 18497.66 The dissociation limit (again relative to vibrational level n=0 of the ground (X) electronic state) is 19579.76 cm - 1 for the B state. 1

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The harmonic oscillator has energy levels E n = ( n + 1 2 ) ~ ω The Morse potential has energy levels E n = ( ( n + 1 2 ) - x e ( n + 1 2 ) 2 ) ~ ω where x e is the anharmonicity parameter, which is related to the depth of the well (for a Morse oscillator the well is not infinitely deep because the molecule can dissociate into separate atoms). How well does the data for the B state fit a harmonic approximation? How well does it fit the Morse approximation? You must do some kind of rigorous statistical analysis to quantify your answer. Solution: If we plot Δ E vs (n+1), the harmonic oscillator model gives a horizontal line (zero slope) with a y-intercept of ~ ω and the Morse oscillator model gives a straight line with a y-intercept of ~ ω and a slope of - 2 ~ ωx e . We can see this by doing the math: for the harmonic oscillator, E n = ( n + 1 / 2) ~ ω so that Δ E = ~ ω (independent of n ). For the Morse oscillator, E n = [( n +1 / 2) - x e ( n +1 / 2) 2 ] ~ ω so that Δ E = ~ ω - 2 ~ ωx e (
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