91Ó°ÊÓ

Formula for Gibb's Factor is given as:

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

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

where, $\mathrm{μ}$is chemical potential, $\mathrm{Î\mu }$is energy occupied, $k$is Boltzmann's constant, $T$is temperature.

Substitute $\mathrm{Î\mu }=0and\mathrm{μ}=0$for unoccupied state.

$$\begin{gathered} G\left(s\right)=e^{\frac{-(0-0)}{kT}} \\ =e^0 \\ =1 \end{gathered}$$

Substitute $\mathrm{Î\mu }=-1$for occupied state.

role="math" localid="1647153963926" $$\begin{gathered} \tilde{G}\left(s\right)=e^{\frac{-(-1-\mathrm{μ})}{kT}} \\ =e^{\frac{(1+\mathrm{μ})}{kT}} \end{gathered}$$

The ratio of probability of unoccupied state to occupied state is:

role="math" localid="1647154152958" $\frac{G\left(s\right)}{\tilde{G}\left(s\right)}=\frac{1}{e^{\frac{(1+\mathrm{μ})}{kT}}}$

Formula for chemical potential is given as:

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

Substitute $\frac{V}{N}=\frac{kT}{P_e}$where $P_e$is partial pressure of electron gas.

role="math" localid="1647154444368" $\mathrm{μ}=-kT×\ln \left(\frac{kT{Z}_{in}}{P_e{\mathrm{ν}}_Q}\right)$

For mono atomic gas, substitute $Z_{in}=1$

Ratio of probability can be written as,

$$\begin{gathered} \frac{G\left(s\right)}{\tilde{G}\left(s\right)}=\frac{1}{e^{\frac{(1+\mathrm{μ})}{kT}}} \\ =e^{\frac{-1}{kT}}×e^{\frac{-\mathrm{μ}}{kT}} \end{gathered}$$

Substitute $e^{\frac{-\mathrm{μ}}{kT}}=\frac{kT}{P_e{\mathrm{ν}}_Q}$in above equation,

role="math" localid="1647155185719" $$\begin{gathered} \frac{G\left(s\right)}{\tilde{G}\left(s\right)}=e^{\frac{-1}{kT}}×\frac{kT}{P_e{\mathrm{ν}}_Q} \\ =\left(\frac{kT}{P_e{\mathrm{ν}}_Q}\right)e^{\frac{-1}{kT}} \end{gathered}$$

Hence, the ratio of probabilities of unoccupied state to occupied state is $\left(\frac{kT}{P_e{\mathrm{ν}}_Q}\right)e^{\frac{-1}{kT}}$ .