feat: adds 7.12

tex/7.12.tex

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+\documentclass{article}
+
+% set up telugu
+\usepackage{fontspec}
+\newfontfamily\telugufont{Potti Sreeramulu}[Script = Telugu]
+\usepackage{polyglossia}
+\setdefaultlanguage{english}
+\setotherlanguage{telugu}
+
+%other packages
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{physics}
+\usepackage[binary-units=true]{siunitx}
+\usepackage{todonotes}
+\usepackage{luacode}
+\usepackage{titling}
+\usepackage{enumerate}
+
+% custom deepak packages
+\usepackage{luatrivially}
+\usepackage{subtitling}
+
+\usepackage{cleveref}
+
+\begin{luacode*}
+	math.randomseed(31415926)
+\end{luacode*}
+
+\newcommand{\kb}{k_{\mathrm{B}}}
+
+\title{Problem 7.12}
+\subtitle{Semiconductors}
+\author{\begin{telugu}హృదయ్ దీపక్ మల్లుభొట్ల\end{telugu}}
+% want empty date
+\predate{}
+\date{}
+\postdate{}
+
+% !TeX spellcheck = en_GB
+\begin{document}
+	\maketitle
+
+	We're looking at an example of a doped superconductor.
+	We have $N - M$ atoms of silicon, and $M$ atoms of phosphorus.
+	Each Si atom contributes one electron, and two states (at $\pm \flatfrac{\Delta}{2}$).
+	We'll take the energy gap $\Delta = \SI{1.16}{\eV}$ for example.
+	
+	The phosphorous atoms contribute \emph{two} electrons and two states, at $-\flatfrac{\Delta}{2}$ and $\flatfrac{\Delta}{2} - \epsilon$.
+	The impurity energy guy $\epsilon = \SI{0.044}{\eV}$, and is much smaller than the energy gap.
+	
+	So the ground state here has $N + M$ electrons, so the $N$ valence bands at $-\flatfrac{\Delta}{2}$ and $M$ impurity band states at $-\flatfrac{\Delta}{2} - \epsilon$ are filled, and the $N - M$ conduction band states at $\flatfrac{\Delta}{2}$ are empty.
+
+	\subsection*{(a)}
+	Derive a formula for the number of electrons as a function of temperature $T$ and chemical potential $\mu$ for the energy levels of our system.
+	
+	\section{Solution} \label{sec:solution}
+	
+	\subsection*{(a)}
+	So we have $N + M$ electrons.
+	Easy.
+	
+	Sethna probably wants us to write the right hand side of this too.
+
+	For each energy $E$, the single electron occupation is our Fermi distribution:
+	\begin{equation}
+		f(T) = \frac{1}{e^{\beta \left(E - \mu \right)} + 1}
+	\end{equation}
+	So if we add this up by the number of electrons and number of available states, we get
+	\begin{equation}
+		N + M = \frac{N}{e^{\beta \left(-\flatfrac{\Delta}{2} - \mu \right)} + 1} + \frac{M}{e^{\beta \left(\flatfrac{\Delta}{2} - \epsilon - \mu \right)} + 1} + \frac{N - M}{e^{\beta \left(\flatfrac{\Delta}{2} - \mu \right)} + 1}.
+	\end{equation}
+	What this really gives us is an implicit relationship for $\mu(T)$, because those are the two undetermined quantities.
+	
+	\newpage
+	\listoftodos
+
+\end{document}