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Topic: Spectroscopy and the Boltzman Expression  (Read 2349 times)

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Offline asdaf7477

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Spectroscopy and the Boltzman Expression
« on: April 25, 2010, 06:18:29 AM »
1. Constructing the nuclear spin wavefunctions, how do you account for the alternation in the populations of the rotational energy levels of the ground electronic state of the H2 molecule. For the H atom, we know I = ½.

2. Also, Using the Boltzmann expression, and including nuclear spin and orientational degeneracy, how do you calculate the relative populations of the J = 6 and J = 7 rotational states in the v(Nu)= 0 level of the ground electronic state of D2 at 300 K. For the ground electronic state of molecular deuterium Be = 30.429 cm-1, D = 0.011 cm-1 and alpha = 1.049 cm-1.

Answering question 2, I approached it by doing the following:
N(J)/N(J0) = N(7)/N(6) = gNS(7)/gNS(6) = 15/13
 :delta: E = {E(J=7)} - {E(J=6)}



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