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\exnumber{4213}
\heading{Chemical equilibrium for $A{ \rightleftarrows}A^++e^-$} 
\auname{ Yuval Tsedek}


\textbf{The question:}\\


$N_0$ atoms of type $A$ are placed in an empty box 
of volume $V$, such that their initial density is $n_0=N_0/V$. 
The ionization energy of the atoms is $\varepsilon_0$.
The box is held in temperature $T$, and eventually 
a chemical equilibrium ${A \rightleftarrows A^{+} + e^{-}}$ 
is reached. The fraction of ionized atoms is ${x=N^{+}/N_0}$.
The masses of the particles are $m_e$ for the electron, 
and $m_{A^+} \approx m_A$ for the atoms and the ions. 
 
\Dn

(a) Define temperature $T_0$ such that ${T \gg T_0}$ is a sufficient 
condition for treating the gas of atoms in the Boltzmann approximation. 

\Dn

(b) Assuming the Boltzmann approximation for both the atoms 
and the electrons, write an equation for $x$. 
Write its {\em approximate} solution assuming ${x\ll1}$. 
Write the condition for the validity of the latter assumption.  

\Dn

(c) Assuming that ${x\ll1}$, write a condition on the density $n_0$, 
that above $T_0$ it was legitimate to treat the electrons 
in the Boltzmann approximation. Note: the condition is a simple 
inequality and should be expressed using $(m_e,m_A,\varepsilon_0)$. 


\Dn

Assume that the condition in (3) breaks down.
It follows that there is a regimes ${T_0 \ll T \ll T_1}$ 
where the atoms can be treated in the Boltzmann approximation, 
while the electrons can be treated as a low temperature quantum gas. 

\Dn

(d) Write an equation for $x$ assuming that the electrons 
can be treated approximately as a zero temperature Fermi gas. 
Exotic functions should not appear.
You are not expected to solve this transcendental equation.

\Dn

(e) What would be the equation for $x$ if the electrons 
were Bosons instead of Fermions.
    

 Note:
 
 
$\cdot$ Express all the final answers using $(m_e,m_A,n_0,\varepsilon_0,T)$,  
and {\em elementary} functions. \\ Exotic functions should not appear.
It is allowed to use the notation $\lambda_e(T)=(2\pi/\mass_e T)^{1/2}$.\\
$\cdot$You can assume that the spin of $A,e^-$ is 1/2 and that $A^+$ is spinless.


\Dn


 \textbf{Solution:}\\
 
 (a) The condition for Boltzmann approximation is $|z|{\ll}1$. Because of the monotonicity of
 the Polylogarithm function $Li_\alpha(z)$, in the case of 3D the condition for Boltzmann approximation can by written as  $ \lambda_A^{-3}(T) \gg n_A$  where the thermal wavelength of the A particles is  $ \lambda_A(T)=({2\pi}/{m_AT})^{1/2}$ . We don't know the value of $n_A$ but we do know that $n_A{\leq}n_0$ therefore Boltzmann holds for $T\gg{T_0}$ where 
 
\begin{equation}
T_0=\dfrac{2{\pi}n_0^{2/3}}{m_A}
\end{equation}

\Dn


(b) In chemical equilibrium   $\mu_A=\mu_{A^+}+\mu_{e^-}$. The chemical potential in the Boltzmann approximation is $\mu=-T\ln (Z/N)$, where $Z$ is the partition function of a single particle. In our case we have 3 different partition functions for $A,A^+$ and $e^-$. It follows that in equilibrium:

\begin{equation}
\ln(Z_{A}/N)=\ln({Z_{A^+}}/N_{A^+})+\ln(Z_{e^-}/N_{e^-})\Rightarrow
\end{equation}

\begin{equation}
\dfrac{Z_{A}}{N_A}=\dfrac{Z_{A^+}}{N_{A^+}}\dfrac{Z_{e^-}}{N_{e^-}}
\end{equation}

Using the relations $N_{A^+}=N_{e^-}=xN_0 \text{ and } N_A=(1-x)N_0$ we get: 

\begin{equation}
 \dfrac{x^2}{1-x}=\dfrac{Z_{A^+}Z_{e^-}}{Z_{A}N_0}
\end{equation}

Using $m_{A^+}{\approx}m_A$ the partition function of $A^+$ ions is $Z_{A^+}=V\lambda_{A}^{-3}$. For the electrons $Z_{e^-}=2V{\lambda_{e^-}^{-3}}$ (the 2 factor comes for the spin) and for the A atoms we used $Z_{A}=2V{\lambda_A}^{-3}e^{\beta\varepsilon_0}$, where $\epsilon_0$ is the binding energy required to ionize $A$. Finally we get the equation

\begin{equation}
\dfrac{x^2}{1-x}=\dfrac{1}{n_0\lambda_e^3}e^{-\beta\varepsilon_0}
\end{equation}

For $x\ll1,$ we use first order approximation for x to get:  

\begin{equation}
x^2=\dfrac{1}{n_0\lambda_e^3}e^{-\beta\varepsilon_0}
\end{equation}

\Dn


(c)  When Boltzmann approximation holds for the electrons, the electron density obeys  ${n_e}\ll({\lambda_{e^-}^3})^{-1}$  therefore $x={n_e}/{n_0}\ll({n_0\lambda_{e^-}^3})^{-1}$ and from (6) we get 

\begin{equation}
n_0\lambda_{e^-}^3{\ll}e^{\beta\varepsilon_0}
\end{equation}

\Dn

(d) Again we are in chemical equilibrium   $\mu_A=\mu_{A^+}+\mu_{e^-}$ with the same chemical potential for $A,A^+$. The difference is in the chemical potential of the electrons that goes to Fermi energy as the temperature goes to zero, $\mu_{e^-}\approx\varepsilon_F$ so now the chemical equilibrium equation get the form

\begin{equation}
T\ln(Z_{A}/N_A)=T\ln({Z_{A^+}}/N_{A^+})+\varepsilon_F\Rightarrow
\end{equation}

\begin{equation}
\dfrac{2x}{1-x}=e^{\beta(\varepsilon_F-\varepsilon_0)}
\end{equation}

where $\varepsilon_F=(6\pi^2n_e)/{2m_e}=(3\pi^2xn_0)/m_e$

\Dn


(e) For Bosons $\mu_{e^-}=0$ so that the chemical equilibrium equation is
\begin{equation}
T\ln(Z_{A}/N_A)=T\ln({Z_{A^+}}/N_{A^+})+0
\end{equation}

which leads to 

\begin{equation}
xe^{\beta\varepsilon_0}=1-x\Rightarrow
\end{equation}


\begin{equation}
\dfrac{x}{1-x}=e^{-\beta\varepsilon_0}  
\end{equation}


\end{document}

