# Statistical mechanics - Helmholtz free energy and Z

1. 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.

3. 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?

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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 $$F=-k\cdot T\cdot \ln(Z)$$?