From e6304c360c982431f8d3e25c5064b4f28a06b0b2 Mon Sep 17 00:00:00 2001 From: Deepak Mallubhotla Date: Thu, 14 Apr 2022 20:03:08 -0500 Subject: [PATCH] feat: adds 7.12 --- tex/7.12.tex | 78 ++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 78 insertions(+) create mode 100644 tex/7.12.tex diff --git a/tex/7.12.tex b/tex/7.12.tex new file mode 100644 index 0000000..69d20b5 --- /dev/null +++ b/tex/7.12.tex @@ -0,0 +1,78 @@ +\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}