Statistical mechanics - Helmholtz free energy and Z

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SUMMARY

The discussion focuses on calculating the temperature of a neutral carbon atom with a 9-fold degenerate ground state and a 5-fold degenerate excited state, given that 10% of the atoms are in the excited state. The calculated temperature is 5900 K. The user attempts to verify the Helmholtz free energy (F) using the relation F = -k*T*log(Z) and the probability expression P(s) = exp(-F/kT)/Z, but encounters discrepancies in the expected results. The main issue arises from the application of the Helmholtz free energy formula in relation to the partition function (Z) and the probability of the excited state.

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  • Knowledge of Boltzmann statistics and probability distributions in thermal equilibrium.
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Students and researchers in physics, particularly those studying statistical mechanics, thermodynamics, and quantum mechanics. This discussion is beneficial for anyone looking to deepen their understanding of the Helmholtz free energy and its applications in calculating thermodynamic properties.

Niles
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Homework Statement


The neutnral carbon atom has a 9-fold degeerate ground level and a 5-fold degenerate excited level at an energy 0.82 eV above the ground level. Spectroscopic measurements of a certain star show that 10% of the neutral carbon atoms are in the excited level, and that the population of higher levels is negligible. Assuming thermal equilibrium, find the temperature.

The Attempt at a Solution



Using the fact that P(s) = 0.1 = ..., I have found that T = 5900 K.

Now I want to verify F = -k*T*log(Z) and P(s) = exp(-F / kt)/Z by plugging in the same values, but it doesn't make sense. This is my F:

F = - k \cdot 5900K \cdot \ln \left( Z \right) = - k \cdot 5900K \cdot \ln \left( {9 + 5 \cdot e^{\frac{{ - 0.82}}{{k \cdot 5900K}}} } \right)

I insert this is in P(s) = exp(-F/kT)/Z:

P(s) = \frac{1}{Z} \cdot \exp \left( {\frac{{ - F}}{{kT}}} \right) = \frac{1}{{9 + 5 \cdot e^{\frac{{ - 0.82}}{{k \cdot 5900K}}} }} \cdot \exp \left( {\ln \left( {9 + 5 \cdot e^{\frac{{ - 0.82}}{{k \cdot 5900K}}} } \right)} \right). I have NOT multiplied with 5 since we are using F.

I can see that it gives one, since e takes the logarithm, but still - it doesn't make sense, since it should equal 0.1. Can you see where my error is?
 
Last edited:
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Is the problem that P(s) = \frac{1}{Z} \cdot \exp \left( {\frac{{ - F}}{{kT}}} \right) only works when I write F=E-TS and NOT F=-k\cdot T\cdot \ln(Z)?

What I do know is that there is something fundementally wrong in the way I use this expression. It would be great if you could point out my error.
 
Last edited:

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