# Electronic partition function for molecule with degeneracies

by wiveykid
Tags: degeneracies, electronic, function, molecule, partition
 P: 3 1. The problem statement, all variables and given/known data A atom had a threefold degenerate ground level, a non degenerate electronically excited level at 3500 cm^-1(setting the energy orgin as the ground electronic state energy of the atom ) and a threefold degenerate level at 4700 cm^-1 . Calculate the electronic partition function of this atom at 2000K 2. Relevant equations qel= sumnation{i=0inf}[gel*exp[-(Ei-Ei-1)/(kbT)]] 3. The attempt at a solution I see three levels in the problem I believe, given the problem statement, that g0=3, g1=1 @3500cm-1, g2=3@4700cm-1 I came up with the equation qel= 3+ 1*exp[-(E1-E0)/(kbT)] + 3*exp[-(E2-E1)/(kbT)] T= 2000K, kb= boltzmann constant ~1.38e-23 I am having trouble finding the energy values for each excited state. I am not sure if E=hv applies to this problem. Also I am not too confident in the equation I found. If anyone understands degeneracy, electronic partition functions or excited electronic stateI would greatly appreciate any help
 HW Helper P: 2,327 I would assume that you just use E = h f = h c / lambda = hbar c k , no? (I'm assuming that the inverse lengths that the problem specifies are wavenumbers of photons that would be emitted from the corresponding energy differences.)
 P: 3 yes I guess I can just multiply those numbers by c to get the excited state energies. Doing this it seems that all the exponentials go to zero because kbT is so small and c is so large. I guess the electronic partition function ends up being equal to 3
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## Electronic partition function for molecule with degeneracies

 Quote by wiveykid yes I guess I can just multiply those numbers by c to get the excited state energies. Doing this it seems that all the exponentials go to zero because kbT is so small and c is so large. I guess the electronic partition function ends up being equal to 3
There is more to the calculation than just kBT and c. Try calculating the value of $\frac{\Delta E}{k_B T}$ for the two excited states.

 Quote by wiveykid I came up with the equation qel= 3+ 1*exp[-(E1-E0)/(kbT)] + 3*exp[-(E2-E1)/(kbT)]
That's pretty much correct, except that it is E2-E0 in the final term.

 I am having trouble finding the energy values for each excited state. I am not sure if E=hv applies to this problem.
Yes, it does.

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