Electronic partition function for molecule with degeneracies

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SUMMARY

The electronic partition function for an atom with a threefold degenerate ground state and two excited states can be calculated using the formula qel = Σ(gi * exp[-(Ei - E0)/(kB * T)]). For this atom at 2000K, the ground state has g0 = 3, the first excited state at 3500 cm-1 has g1 = 1, and the second excited state at 4700 cm-1 has g2 = 3. The correct equation for the partition function is qel = 3 + 1 * exp[-(E1 - E0)/(kB * T)] + 3 * exp[-(E2 - E0)/(kB * T)], where E1 and E2 are derived from the wavenumbers using E = h * c / λ. The final result indicates that the partition function approaches 3 due to the dominance of the ground state at this temperature.

PREREQUISITES
  • Understanding of electronic states and degeneracy in quantum mechanics
  • Familiarity with the Boltzmann constant (kB) and its application in statistical mechanics
  • Knowledge of the relationship between energy, frequency, and wavenumbers (E = h * f = h * c / λ)
  • Basic proficiency in thermodynamic concepts related to partition functions
NEXT STEPS
  • Calculate the energy values for excited states using E = h * c / λ for specific wavenumbers
  • Explore the implications of degeneracy on electronic partition functions in different systems
  • Investigate the behavior of partition functions at varying temperatures and their physical significance
  • Learn about the application of partition functions in predicting molecular behavior in statistical mechanics
USEFUL FOR

Students and researchers in physical chemistry, quantum mechanics, and statistical mechanics who are studying molecular energy levels and partition functions. This discussion is particularly beneficial for those working with electronic states and degeneracies in atoms.

wiveykid
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Homework Statement


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


Homework Equations



qel= sumnation{i=0inf}[gel*exp[-(Ei-Ei-1)/(kbT)]]

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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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.)
 
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
 
wiveykid said:
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.

wiveykid said:
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.
 
Last edited:

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