91Ó°ÊÓ

As the system of a single Hydrogen atom/ion, basically has two possible states such as Unoccupied and occupied. Considering the two independent spin states of the electron, now the system has two occupied states one with the electron in each spin configuration.

The Gibbs factor is given as:

$P\left(s\right)=e^{-(\mathrm{Î\mu }-\mathrm{μ})/kT}$

Consider a system made up of a single hydrogen atom that can exist in three states, one of which is unoccupied and the other two of which are occupied (spin up and spin down):

Unoccupied state:

Because the chemical potential is 0 and the state's energy is also zero, the Gibbs factor is:

$P\left(s\right)=e^{-(0-0)/kT}=1$

Occupied state (when the electron is in the ground state):

In this situation, replace $\mathrm{Î\mu }$with ionisation energy -I (notice that the factor is multiplied by two because we have two electrons):

$P^{\text{'}}\left(s\right)=2e^{(I+\mathrm{μ})/kT}$

The total of the two Gibbs factors is the Grand partition function, which is:

$\AE \mu =1+2e^{(I+\mathrm{μ})/kT}$

The Gibbs factor divided by the Grand partition function equals the probability of the first state (the ionised state, where the state is not occupied):

$\mathcal{P}\left(s\right)=\frac{P\left(s\right)}{\AE \mu }=\frac{1}{1+2e^{(I+\mathrm{μ})/kT}}$

The probability of the second state (when the state is occupied) equals the Gibbs factor divided by the Grand partition function:

${\mathcal{P}}^{\text{'}}\left(s\right)=\frac{P^{\text{'}}\left(s\right)}{\AE \mu }=\frac{2e^{(I+\mathrm{μ})/kT}}{1+2e^{(I+\mathrm{μ})/kT}}$

The ratio of probability of the unoccupied to the occupied states is therefore:

$$\begin{gathered} \frac{P\left(s\right)}{P^{\text{'}}\left(s\right)}=\frac{1}{2e^{(I+\mathrm{μ})/kT}} \\ \frac{P\left(s\right)}{P^{\text{'}}\left(s\right)}=\frac{1}{2}{e}^{-(I+\mathrm{μ})/kT} \\ \frac{P\left(s\right)}{P^{\text{'}}\left(s\right)}=\frac{1}{2}{e}^{-I/kT}{e}^{-\mathrm{μ}/kT}\left(1\right) \end{gathered}$$

The chemical equation is:

$\mathrm{μ}=-kT\ln \left(\frac{V{Z}_{int}}{N{v}_Q}\right)$

The volume per particle can be written as (from the ideal gas law):

$\frac{V}{N}=\frac{kT}{P}$

Therefore,

$\mathrm{μ}=-kT\ln \left(\frac{kT{Z}_{int}}{P{v}_Q}\right)$

Because we have two spin states in a monoatomic gas, the internal partition function is 2, hence the chemical potential can be reduced to:

$$\begin{gathered} \mathrm{μ}=-kT\ln \left(\frac{2kT}{P{v}_Q}\right) \\ -\frac{\mathrm{μ}}{kT}=\ln \left(\frac{2kT}{P{v}_Q}\right) \end{gathered}$$

Exponentiation both sides:

$e^{-\mathrm{μ}/kT}=\frac{2kT}{P{v}_Q}$

Substitute into (1):

$$\begin{gathered} \frac{P\left(s\right)}{P^{\text{'}}\left(s\right)}=\frac{1}{2}{e}^{-I/kT}\left(\frac{2kT}{P{v}_Q}\right) \\ \frac{P\left(s\right)}{P^{\text{'}}\left(s\right)}=e^{-I/kT}\left(\frac{kT}{P{v}_Q}\right) \end{gathered}$$

This is the same result of the previous problem, that means there is no effect on the fraction if we include the spin degeneracy.