I How Does Electron Spin Affect the Partition Function in Saha's Equation?

AI Thread Summary
The discussion centers on the derivation of Saha's equation for ionizing hydrogen atoms, specifically examining the role of electron spin in the partition function. The equation shows the relationship between the pressures of protons and hydrogen atoms, incorporating variables like temperature and ionization energy. It is established that when the electron is bound to the proton, it can occupy two spin states, leading to a partition function value of Z_int = 2. The partition function is defined as the sum of the exponential terms for all states, which simplifies to 2 when considering the spin states of the electron with zero energy in the absence of external fields. This clarification highlights the connection between electron spin and the partition function in the context of Saha's equation.
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Hey, I have a question about proving Saha's equation for ionizing hydrogen atoms.
The formula is
\frac{P_{p}}{P_{H}} = \frac{k_{B} T}{P_{e}} \left(\frac{2\pi m_{e} k_{B}T}{h^2} \right)^{\frac{3}{2}}e^{\frac{-I}{k_{B} T}}
with
P_{p} pressure proton's,
P_{H} pressure hydrogen atoms,
m_{e} mass electron,
T temperature resevoir,
I ionization energy,
k_{B} Boltzmann's constant,
h Planck's constant.

When the electron is bounded, the electron has two spin states.
So the three situations are, the electron is bounded and is in one of the spin states, the electron is bounded and is in the other spin state, and the electron is not bounded by the proton.
The system of interest is the proton.
When the electron is bounded by the proton, the energy of the system of interest is -I.
I have derived that
\frac{P_{p}}{P_{H}} = \frac{1}{2}\frac{k_{B} T}{P_{e}} Z_{int} \left(\frac{2\pi m_{e} k_{B}T}{h^2} \right)^{\frac{3}{2}}e^{\frac{-I}{k_{B} T}}
with Z_{int} the partition function (for rotational/vibrational motion) of the electron is.
If you compare the two equations, you see that Z_{int} = 2.
I don't get why Z_{int} is 2.

The partition function is
Z = \sum_{s}^{} e^{\frac{-E\left(s \right)}{k_{B} T}}
summing over all states (represented by letter s).
The partition function is not all the possible states the electron can have right?

So how does Z = \sum_{s}^{} e^{\frac{-E\left(s \right)}{k_{B} T}} is equal to the number 2 (assuming that two spin states the electron can have, have non zero energy)?
 
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In the absence of external fields, there is no energy associated with the spin state, so ##E(s) = 0## for all spin states. Therefore,
$$
Z_\mathrm{int} = \sum_s e^{-E(s) / k_B T} = \sum_s 1 = 2
$$
 
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