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\begin{document}
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\heading{E4213: Chemical equilibrium $A \rightleftarrows A^+ + e^-$}
\auname{Yaron de Leeuw}
{\bf The problem:}
\Dn
$N$ atoms of type $A$ were put into a box of volume $V$.
The system has reached a state of thermochemical equilibrium
\[
A \rightleftarrows A^+ + e^-
\]
The solution should be expressed by the masses $m_e$,$m_A$, the ionization energy $\varepsilon$, the temperature $T$ and the gas density $n=\frac{N}{V}$.
\Dn
(1) Find the percent of ionized atoms, if the electrons and $A$ atoms form a classical (Boltzmann) gas. ($T \gg T_1 $).
(2) Find the percent of ionized atoms, if the electrons form a low temperature Fermi gas, while the $A$ atoms continue to form a classical gas. ($T_1 \gg T \gg T_0 $).
(3) Define $T_0$ and $T_1$.
\Dn\Dn
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{\bf The solution:}
\Dn
The free energy of the system after $N_i$ atoms have been ionized is:
\[ F = F_{A} (N-N_i) + F_{A^+}(N_i) + F_{e^-}(N_i) \]
So the probability for $N_i$ ionized atoms should behave like:
\[ p(N_i) \propto e^{-\beta F} \]
The most probable value for $N_i$ will be the minimum of the free energy:
\begin{align*}
\frac{dF}{dN_i} = 0 \\
- \mu_{A} + \mu_{A^+} + \mu_{e^-} =0 \label{eq:chem_eq}
\end{align*}
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\Dn
(1) The electrons form a classical Boltzmann gas. Because the spin of the $A$ atoms is unknown, we choose to consider the atoms and the electrons to be "spinless". The generalization of the degeneracy for particles with non zero spin is trivial . The single particle partition function for an atom $A$ is:
\[Z_A = \frac{V}{{\lambda_T}_A^3} \]
Where ${\lambda_T}_A$ is defined by
\[ {\lambda_T}_A = \sqrt{\frac{2\pi}{m_A T}} \]
For the ionized $A^+$ atoms, the partition function will be:
\[Z_{A^+} = \frac{V}{{\lambda_T}_{A^+}^3}e^{-\beta\varepsilon} \]
The multiparticle partition function for $N$ atoms of type $A$ is:
\[ Z_{AN} = \frac{Z_A^N}{N!} \]
So the chemical potential $\mu$ is:
\[ \mu_A = -\frac{d}{dN}T\ln Z_{AN} = T\ln \frac{N}{Z_A}\]
The same calculation will yield similar results for $A^+$ and $e^-$. Putting these results into the chemical equilibrium equation, we obtain:
\begin{align*}
&-T\ln \frac{N-N_i}{Z_A} + T\ln \frac{N_i}{Z_{A^+}} + T\ln \frac{N_i}{Z_{e^-}} =0 \\
&\frac{N-N_i}{N_i^2} = \frac{Z_A}{Z_{A^+}Z_{e^-}} \\
&\frac{N_i^2}{N^2} + \frac{1}{N}\frac{Z_{A^+}Z_{e^-}}{Z_A}\frac{N_i}{N}-\frac{1}{N}\frac{Z_{A^+}Z_{e^-}}{Z_A} = 0 \\
\frac{N_i}{N} &= \frac{1}{2N}\frac{Z_{A^+}Z_{e^-}}{Z_A}\left( -1 \pm \sqrt{ 1 + \frac{4N Z_A}{Z_{A^+}Z_{e^-}} } \right) \\
\frac{N_i}{N} &= \frac{1}{2}\frac{V}{N}\left(\frac{Tm_e(m_e+m_A)}{2\pi m_A}\right)^{\frac{3}{2}}e^{-\beta\varepsilon}\left( -1 + \sqrt{ 1 + \frac{4N}{V}\left(\frac{2\pi m_A}{Tm_e(m_e+m_A)}\right)^{\frac{3}{2}}e^{\beta\varepsilon} } \right)
\end{align*}
In the last step we have chosen the positive solution, as the number of ionized atoms cannot be less than $0$.
%%%%%%%%%%%%%%%
\Dn
(2) For a Fermi gas we have the following expression:
\[ \frac{N}{V} = \frac{1}{\lambda_T^3} F_{3/2}(e^{\beta\mu}) \]
Putting in the chemical potential equation we obtain:
\begin{align*}
\frac{N_i}{V} &= \frac{1}{{\lambda_T}_e^3} F_{3/2}(e^{\beta(\mu_A-\mu_A^+)}) = \frac{1}{{\lambda_T}_e^3} F_{3/2}\left(\frac{\frac{N-N_i}{Z_A}}{\frac{N_i}{Z_{A^+}}}\right) = \frac{1}{{\lambda_T}_e^3} F_{3/2}\left(\frac{N-N_i}{N_i}\frac{Z_{A^+}}{Z_A}\right) \\
&= \frac{1}{{\lambda_T}_e^3} F_{3/2}\left(\frac{N-N_i}{N_i}\left(\frac{m_{A^+}}{m_A}\right)^{\frac{3}{2}}e^{-\beta \varepsilon}\right) \approx \frac{1}{{\lambda_T}_e^3} F_{3/2}\left(\frac{N-N_i}{N_i}e^{-\beta \varepsilon}\right)
\end{align*}
The first order approximation of $F_{3/2}$ gives exactly the same solution as before.
%%%%%%%%%%%%%%%
\Dn
(3) The necessary condition for Boltzmann's approximation is
$n\lambda_T^3 \ll 1$.
Putting in $\lambda_T$, this gives
\[
n\left(\frac{2\pi}{m T}\right)^{\frac{3}{2}} \ll 1 \quad \Rightarrow \quad
n\left(\frac{2\pi}{m}\right)^{\frac{3}{2}} \ll T^{\frac{3}{2}} \quad \Rightarrow \quad
T \gg n^{\frac{2}{3}}\frac{2\pi}{m}
\]
For the electrons, this temperature will be:
\[ T_1 = n^{\frac{2}{3}}\frac{2\pi}{m_e} \]
And for the $A$ atoms this will be:
\[ T_0 = n^{\frac{2}{3}}\frac{2\pi}{m_A} \]
Because the mass of the atoms is much larger than the mass of the electrons, $T_1 \gg T_0$, there exists a wide range of temperatures which satisfy the conditions for section (2).
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\end{document}