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+\pdfoutput=0
+\documentclass[a4paper,11pt]{article}
+\usepackage[activate]{pdfcprot}
+\usepackage{verbatim}
+\usepackage{a4}
+\usepackage{a4wide}
+\usepackage[german]{babel}
+\usepackage[latin1]{inputenc}
+\usepackage[T1]{fontenc}
+\usepackage{amsmath}
+\usepackage{ae}
+\usepackage{aecompl}
+\usepackage[dvips]{graphicx}
+\graphicspath{{./img/}}
+\usepackage{color}
+\usepackage{pstricks}
+\usepackage{pst-node}
+\usepackage{rotating}
+
+\setlength{\headheight}{0mm} \setlength{\headsep}{0mm}
+\setlength{\topskip}{-10mm} \setlength{\textwidth}{17cm}
+\setlength{\oddsidemargin}{-10mm}
+\setlength{\evensidemargin}{-10mm} \setlength{\topmargin}{-1cm}
+\setlength{\textheight}{26cm} \setlength{\headsep}{0cm}
+
+\renewcommand{\labelenumi}{(\alph{enumi})}
+
+\begin{document}
+
+% header
+\begin{center}
+ {\LARGE {\bf Materials Physics I}\\}
+ \vspace{8pt}
+ Prof. B. Stritzker\\
+ WS 2007/08\\
+ \vspace{8pt}
+ {\Large\bf Tutorial 5}
+\end{center}
+
+\section{Charge carrier density of intrinsic semiconductors}
+
+\begin{enumerate}
+ \item Recall the free electron in a box.
+ Write down an expression for the density of states $D(E)$
+ of the free electron gas.
+ {\bf Hint:} The density of states is a function of internal energy $E$
+ such that $D(E)dE$ is the number of states
+ (allowed $k$-values) with energies
+ between $E$ and $E+dE$.
+ For large values of $L$ (side length of the box)
+ the states are quasi-continuous and
+ sums in $k$-space can be replaced by integrals.
+ First calculate the amount of states $dZ'$
+ in-between $k$ and $k+dk$.
+ Therefor calculate the volume of the sperical shell
+ and the volume of a single allowed $k$-point.
+ Neglect terms of the order $(dk^2)$.
+ After that express $dk$ and $k$ by $dE$ and $E$
+ and insert these expressions into $dZ'$.
+ By definition $D(E)=dZ/dE$,
+ where $dZ$ is $dZ'$ devided by the box volume
+ (states per crystal volume).
+ Adjust the expression taking into account
+ the spin of an electron.
+ \item The conduction and valence band in a semiconductor can be approximated
+ by the same parabolic functions of $k$ close to the bandedges.
+ The mass of the electron is replaced by an effective mass
+ of the electron in the conduction band or the hole in the valence band.
+ Show the relation of the effective mass and the curvature of the band.
+ {\bf Hint:} The curvature of a function $f(x)$ is given by the second
+ derivative of this function with respect to $x$.
+ \item Sketch the density of states, the Fermi function and the resulting
+ density of charge carriers (electrons: $m_n$, holes: $m_p$)
+ for $m_n=m_p$ and for $m_n\ne m_p$ for non-zero temperatures.
+ {\bf Hint:} The density of states is given by
+ $D_c(E)=\frac{1}{2\pi^2}(\frac{2m_n}{\hbar^2})^{3/2}
+ (E-E_c)^{1/2}$ for electrons in the conduction band and
+ $D_v(E)=\frac{1}{2\pi^2}(\frac{2m_p}{\hbar^2})^{3/2}
+ (E_v-E)^{1/2}$ for holes in the valence band.
+ $E_c$ is the lowest energy level of the conduction and
+ $E_v$ the highest energy level of the valence band.
+ Thus the bandgap energy is given by $E_g=E_c-E_v$.
+ The density of charge carriers is the product of $D(E)$ and
+ the Fermi function $f(E)$.
+ The Fermi energy $E_F$ adjusts itself in such a way that
+ the amount of electrons and holes equals.
+ Keep in mind that the distribution valid for the holes is
+ $1-f(E)$.
+\end{enumerate}
+
+\section{'Density of state mass' of holes in silicon}
+
+The valence band of silicon is composed by three subbands.
+Two of them contact at the $\Gamma$-point ($k=0$),
+the one for heavy holes ($m_{ph}$) and the one for light holes ($m_{pl}$).
+An additional split-off hole band ($m_{pso}$) is located
+shortly below the first two bands (see Figure).
+
+\begin{enumerate}
+ \item Assume parabolic bands near $k=0$.
+ Write down the total density of states
+ near the maximum of the valence band.
+ Only consider heavy and light holes.
+ \item Write the above result in terms of a density of states expression
+ of a parabolic band with a single uniform effective mass $m_p$.
+ Determine this 'density of state mass' $m_p$.
+ Calculate $m_p$ using the values $m_{ph}=0.49 \, m_e$ and
+ $m_{pl}=0.16 \, m_e$ in which $m_e$ is the electron rest mass.
+\end{enumerate}
+
+\vspace{0.5cm}
+
+\begin{picture}(0,0)(0,140)
+ \includegraphics[width=5.0cm]{silicon_bs.eps}
+\end{picture}
+
+\begin{flushright}
+\begin{minipage}{5cm}
+\end{minipage}
+\begin{minipage}{3cm}
+ \includegraphics[height=3cm]{weihnachtsbaum.eps}
+\end{minipage}
+\begin{minipage}{5cm}
+\begin{center}
+{\Large\bf
+ Merry Christmas\\
+ \&\\
+ Happy New Year!}
+\end{center}
+\end{minipage}
+\end{flushright}
+
+\end{document}
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