Q. 7.3 - An Introduction to Thermal Physics

Question

Consider a system consisting of a single hydrogen atom/ion, which has two possible states: unoccupied (i.e., no electron present) and occupied (i.e., one electron present, in the ground state). Calculate the ratio of the probabilities of these two states, to obtain the Saha equation, already derived in Section 5.6 Treat the electrons as a monotonic ideal gas, for the purpose of determining $\mu$. Neglect the fact that an electron has two independent spin states.

Step-by-Step Solution

Answer

The ratio of probabilities of unoccupied state to occupied state is $\left(\frac{kT}{P_e{\nu }_Q}\right)e^{\frac{-1}{kT}}$.

Step 1. Formula Gibb's Factor

Formula for Gibb's Factor is given as:

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

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

where, $\mu$is chemical potential, $\epsilon$ is energy occupied, $k$ is Boltzmann's constant, $T$ is temperature.

Step 2. Ratio of Probabilities

Substitute $\epsilon =0 and \mu = 0$ for unoccupied state.

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

Substitute $\epsilon =-1$ for occupied state.

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

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

$\frac{G\left(s\right)}{\tilde{G}\left(s\right)}=\frac{1}{ e^{\frac{(1+\mu )}{kT}}}$

Step 3. Substitution in chemical potential formula

Formula for chemical potential is given as:

$\mu =-kT\times \ln \left(\frac{V{Z}_{in}}{N{\nu }_Q}\right)$

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

$\mu =-kT\times \ln \left(\frac{kT{Z}_{in}}{P_e{\nu }_Q}\right)$

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

$$\begin{gathered} \frac{-\mu }{kT}=\ln \left(\frac{kT}{P_e{\nu }_Q}\right) \\ {e}^{\frac{-\mu }{kT}}=\frac{kT}{P_e{\nu }_Q} \end{gathered}$$

Step 4. Final calculation

Ratio of probability can be written as,

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

Substitute $e^{\frac{-\mu }{kT}}=\frac{kT}{P_e{\nu }_Q}$ in above equation,

$$\begin{gathered} \frac{G\left(s\right)}{\tilde{G}\left(s\right)}= e^{\frac{-1}{kT}}\times \frac{kT}{P_e{\nu }_Q} \\ =\left(\frac{kT}{P_e{\nu }_Q}\right)e^{\frac{-1}{kT}} \end{gathered}$$

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