Electronic partition function for molecule with degeneracies

[wiveykid]

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

 

[turin]

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]

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]

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 k_BT and c. Try calculating the value of $\frac{\Delta E}{k_B T}$ for the two excited states.

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.

 

[paranormal]

!

Hello,

Thank you for your question. I can provide some guidance on how to approach this problem.

Firstly, degeneracy refers to the number of energy levels that have the same energy value. In this case, the threefold degenerate ground level means that there are three energy levels with the same energy value. Similarly, the threefold degenerate level at 4700 cm^-1 means that there are three energy levels with the same energy value of 4700 cm^-1.

Now, for the electronic partition function, we need to consider the energy levels and their degeneracies. The formula you have provided is correct, but we need to know the energy values for each level in order to calculate the partition function.

To find the energy values, we can use the equation E = hv, where E is the energy, h is the Planck's constant, and v is the frequency. We can also use the relation between frequency and wavenumber, which is v = c/λ, where c is the speed of light and λ is the wavelength.

Using these equations, we can find the energy values for the excited levels. For example, the non-degenerate excited level at 3500 cm^-1 would have an energy value of E = (3500 cm^-1)(c/λ), where c = 3.00 x 10^8 m/s and λ = 3.00 x 10^10 cm. Similarly, for the threefold degenerate level at 4700 cm^-1, we would have E = (4700 cm^-1)(c/λ).

Once we have the energy values for each level, we can plug them into the partition function equation and solve for qel. Your equation is correct, but we need to use the correct energy values for each level.

I hope this helps. If you have any further questions, please let me know. Good luck with your homework!