Electronic partition function for molecule with degeneracies


by wiveykid
Tags: degeneracies, electronic, function, molecule, partition
wiveykid
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#1
Mar12-12, 02:46 PM
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
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turin
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#2
Mar12-12, 10:44 PM
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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.)
wiveykid
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#3
Mar13-12, 12:12 PM
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

Redbelly98
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#4
Mar13-12, 08:16 PM
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Electronic partition function for molecule with degeneracies


Quote Quote by wiveykid View Post
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 [itex]\frac{\Delta E}{k_B T}[/itex] for the two excited states.

Quote Quote by wiveykid View Post
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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