How Do You Calculate the Partition Function for Positronium at T=20000K?

[wilsonaj4]

First off, I'm glad I'm finally a member on this board. It has helped me TREMENDOUSLY over the past few years with various problems I've had. You guys/gals are awesome and hopefully I can make some contributions to this site.

1. Homework Statement
A. Write down the partition function for positronium, assuming that only the levels illustrated in the diagram exist. Evaluate the partition function for T=20000K. Remember to include the degeneracies for each level. (I couldn't figure out how to copy and paste the diagram I was given, so I googled the one below. It's the same diagram, but the diagram I was given stops at the N=3 energy level)

B. Write down an expression for the probability that the atom will exist in the state given by N=3, l=1, and determine that probability for T=20000K

C. Find an expression for the mean energy and evaluate that expression for T=20000K

Homework Equations

1. E_n = 6.803eV (1 - 1/n²)
2. Z= ∑e^-E(s)/KT
3. P(s)= 1/z * e^-E_n/KT
4. ∑ E(s)* P(s)

The Attempt at a Solution

A. So, we haven't done anything even close to this in class, so I'm a little coonfused, but to start, I substituted eq1 into the partition function in eq2 to get Z= ∑e^{-6.803eV (1 - 1/n2)/KT}. After this, I'm completely stuck because we didn't really cover degeneracy in class very well. I know I'm supposed to multiply the expression by the degeneracy, but I'm not exactly sure how to do that.

B. Once A is found, this should be straightforward
C. Once A is found, this should be straightforwardMy work is due by 11/25 at 5pm (EST), any help would be greatly appreciated!

 

[mfb]

You can treat the degenerate states like different states with a very similar energy (they just happen to have a difference of exactly zero here), I think. The real states are not degenerate either.

5 pm where? In my time zone, you posted the thread after 5 pm ;).

 

[wilsonaj4]

You can treat the degenerate states like different states with a very similar energy (they just happen to have a difference of exactly zero here), I think. The real states are not degenerate either.

5 pm where? In my time zone, you posted the thread after 5 pm ;).

So for my n in my energy equation, I can just use the degeneracy?

and I've edited my post to include time zone, thanks!

 

[wilsonaj4]

So, I've rewritten the partition function using what I think is correct regarding the degeneracy.

Z= e^{-6.803(1-1/9)/KT} + e^{-6.803(1-1/4)/KT} + e^{-6.803(1-1/1)/KT}
= e^-6.047/KT + e^-5.102/KT + e⁰
= (4)(1+ e^-6.047/KT + e^-5.102/KT)

I multiplied the partition function by 4 since it's four-fold degenerate.

am I at least on the right track?

 

[mfb]

I don't see where you used the degeneracy now.

 

[wilsonaj4]

I don't see where you used the degeneracy now.

Sorry, I should've multiplied the partition function by 4 since it's four-fold degenerate. Is that correct?

 

[mfb]

4? One specific part of it, yes.