\documentclass[10pt,a4paper,draft,final,oneside]{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{ amssymb }
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% template that does not use Revtex4
%%% but allows special fonts
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%
%%% Please use this template.
%%% Edit it using e.g. Notepad
%%% Ignore the header (do not change it)
%%% Process the file in the Latex site
%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Page setup
\topmargin -1.5cm
\oddsidemargin -0.04cm
\evensidemargin -0.04cm
\textwidth 16.59cm
\textheight 24cm
\setlength{\parindent}{0cm}
\setlength{\parskip}{0cm}
% Fonts
\usepackage{latexsym}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{bm}
% Math symbols I
\newcommand{\sinc}{\mbox{sinc}}
\newcommand{\const}{\mbox{const}}
\newcommand{\trc}{\mbox{trace}}
\newcommand{\intt}{\int\!\!\!\!\int }
\newcommand{\ointt}{\int\!\!\!\!\int\!\!\!\!\!\circ\ }
\newcommand{\ar}{\mathsf r}
\newcommand{\im}{\mbox{Im}}
\newcommand{\re}{\mbox{Re}}
% Math symbols II
\newcommand{\eexp}{\mbox{e}^}
\newcommand{\bra}{\left\langle}
\newcommand{\ket}{\right\rangle}
% Mass symbol
\newcommand{\mass}{\mathsf{m}}
\newcommand{\Mass}{\mathsf{M}}
% More math commands
\newcommand{\tbox}[1]{\mbox{\tiny #1}}
\newcommand{\bmsf}[1]{\bm{\mathsf{#1}}}
\newcommand{\amatrix}[1]{\begin{matrix} #1 \end{matrix}}
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}
% Other commands
\newcommand{\hide}[1]{}
\newcommand{\drawline}{\begin{picture}(500,1)\line(1,0){500}\end{picture}}
\newcommand{\bitem}{$\bullet$ \ \ \ }
\newcommand{\Cn}[1]{\begin{center} #1 \end{center}}
\newcommand{\mpg}[2][1.0\hsize]{\begin{minipage}[b]{#1}{#2}\end{minipage}}
\newcommand{\Dn}{\vspace*{3mm}}
% Figures
\newcommand{\putgraph}[2][0.30\hsize]{\includegraphics[width=#1]{#2}}
% heading
\newcommand{\exnumber}[1]{\newcommand{\exnum}{#1}}
\newcommand{\heading}[1]{\begin{center} {\Large {\bf Ex\exnum:} #1} \end{center}}
\newcommand{\auname}[1]{\begin{center} {\bf Submitted by:} #1 \end{center}}
\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\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}