91Ó°ÊÓ

A hypothetical atom that has just two states: a ground state with energy zero and an excited state with energy 2 eV.

The partition function is equal to the total of all Boltzmann factors, i.e.

$Z=\underset{s}{∑}{e}^{-E\left(s\right)/kT}$

Given two states, a ground state with zero energy and an excited state with e =2eV, the partition function is

$$\begin{gathered} Z=e^0+e^{-\mathrm{Ï\mu }/kT} \\ Z=1+e^{-\mathrm{Ï\mu }/kT} \end{gathered}$$

Substituting $\mathrm{Ï\mu }$and Boltzmann's constant $k=8.617×{10}^{-5}\mathrm{eV}/\mathrm{K}$

$$\begin{gathered} Z=1+e^{-2\mathrm{eV}/\left(8.617×{10}^{-5}\mathrm{eV}/\mathrm{K}\right)T} \\ Z=1+e^{-23210\mathrm{K}/T} \end{gathered}$$

Using python to plot the partition function as a function of temperature, the code is:

The graph is:

At temperature of T= 300 K, the value of the partition function is:

$$\begin{gathered} Z_{(300\mathrm{K})}=1+e^{-23210\mathrm{K}/300\mathrm{K}} \\ {Z}_{(300\mathrm{K})}=1+2.512×{10}^{-34} \end{gathered}$$

At temperature of T= 3000 K, the value of the partition function is:

$$\begin{gathered} Z_{(3000\mathrm{K})}=1+e^{-23210\mathrm{K}/3000\mathrm{K}} \\ {Z}_{(3000\mathrm{K})}=1.000436 \end{gathered}$$

At temperature of T= 30000 K, the value of the partition function is:

$$\begin{gathered} Z_{(30000\mathrm{K})}=1+e^{-23210\mathrm{K}/30000\mathrm{K}} \\ {Z}_{(30000\mathrm{K})}=1.461 \end{gathered}$$

At temperature of T= 300000 K, the value of the partition function is:

$$\begin{gathered} Z_{(300000\mathrm{K})}=1+e^{-23210\mathrm{K}/300000\mathrm{K}} \\ {Z}_{(300000\mathrm{K})}=1.925 \end{gathered}$$