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

[Sebas4]

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)?

 

[DrClaude]

In the absence of external fields, there is no energy associated with the spin state, so $E(s) = 0$ for all spin states. Therefore,
$$
Z_\mathrm{int} = \sum_s e^{-E(s) / k_B T} = \sum_s 1 = 2
$$