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[lectures/latex.git] / solid_state_physics / tutorial / 2n_01s.tex
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30 \begin{document}
31
32 % header
33 \begin{center}
34  {\LARGE {\bf Materials Physics II}\\}
35  \vspace{8pt}
36  Prof. B. Stritzker\\
37  SS 2011\\
38  \vspace{8pt}
39  {\Large\bf Tutorial 1 - proposed solutions}
40 \end{center}
41
42 \section{Indirect band gap of silicon}
43
44 \begin{enumerate}
45  \item \begin{itemize}
46         \item Photon wavelength:\\
47               $E_g=\hbar\omega=\hbar\frac{2\pi}{T}=\hbar 2\pi v
48                   \stackrel{c=v\lambda}{=}\hbar 2\pi\frac{c}{\lambda}$
49               $\Rightarrow \lambda=\frac{\hbar 2\pi c}{E_g}
50                                   =\frac{hc}{E_g}=\ldots=1.11 \, \mu m$
51         \item Photon momentum:\\
52               $p=\hbar k=\hbar\frac{2\pi}{\lambda}=\frac{h}{\lambda}
53                 =\ldots=5.97 \cdot 10^{-28} \, kg\frac{m}{s}$
54        \end{itemize}
55  \item Phonon momentum necessary for transition:\\
56        $\Delta p=\hbar \cdot \Delta k=\hbar \cdot 0.85 \, \frac{2\pi}{a}
57          =\frac{0.85 \, h}{a}=\ldots=1.04 \cdot 10^{-24} \, kg\frac{m}{s}$\\
58        $\rightarrow$ Phonon momentum 3 orders of magnitude below
59                      the momentum necessary for transition!
60  \item \begin{itemize}
61         \item Photon momentum insufficient.
62               Momentum contribution of phonon (lattice vibration) required.\\
63               $\Rightarrow$ Probability of transition very small.
64         \item Recombination energy of electron-hole pairs most probably
65               released as vibrational energy of the lattice.\\
66               $\Rightarrow$ Only direct band gap semiconductors suitable for
67                             effective photon generation.
68        \end{itemize}
69 \end{enumerate}
70
71 \section{Charge carrier density of semiconductors}
72
73 \begin{itemize}
74  \item Calculation of $n$:\\
75 $\forall \epsilon$ of states within conduction band:
76 $\epsilon-\mu_{\text{F}} >> k_{\text{B}}T$
77 $\Rightarrow$
78 $f(\epsilon,T)=
79  \frac{1}{\exp(\frac{\epsilon-\mu_{\text{F}}}{k_{\text{B}}T})+1}\approx
80  \exp(-\frac{\epsilon-\mu_{\text{F}}}{k_{\text{B}}T})$\\
81 Parabolic approximation:
82 $D_c(\epsilon)=\frac{1}{2\pi^2}(\frac{2m_n}{\hbar^2})^{3/2}(\epsilon-E_c)^{1/2}$
83 $\Rightarrow$\\
84 $n=\int_{E_{\text{c}}}^{\infty}D_c(\epsilon)f(\epsilon,T)d\epsilon\approx
85  \frac{1}{2\pi^2}(\frac{2m_n}{\hbar^2})^{3/2}
86  \exp(\frac{\mu_{\text{F}}}{k_{\text{B}}T})
87  \int_{E_{\text{c}}}^{\infty}(\epsilon-E_c)^{1/2}
88  \exp(-\frac{\epsilon}{k_{\text{B}}T})d\epsilon$\\
89 Use: $x=(\epsilon - E_{\text{c}})/k_{\text{B}}T$
90      $\Rightarrow\epsilon=xk_{\text{B}}T+E_{\text{c}}$ and
91      $d\epsilon=k_{\text{B}}Tdx$\\
92 $\Rightarrow$
93 $n=\frac{1}{2\pi^2}(\frac{2m_nk_{\text{B}}T}{\hbar^2})^{3/2}
94  \exp(-\frac{E_{\text{c}}-\mu_{\text{F}}}{k_{\text{B}}T})
95  \underbrace{\int_0^{\infty}x^{1/2}e^{-x}dx}_{=\frac{\sqrt{\pi}}{2}}=
96  \underbrace{2(\frac{m_nk_{\text{B}}T}{2\pi\hbar^2})^{3/2}}_{=N_{\text{c}}}
97  \exp(-\frac{E_{\text{c}}-\mu_{\text{F}}}{k_{\text{B}}T})=
98  N_{\text{c}}\exp(-\frac{E_{\text{c}}-\mu_{\text{F}}}{k_{\text{B}}T})$
99  \item In the same way, calculate $p$:\\
100 $\forall \epsilon$ of states within conduction band:
101 $\mu_{\text{F}}-\epsilon >> k_{\text{B}}T$
102 $\Rightarrow$
103 $1-f(\epsilon,T)=
104  1-\frac{1}{\exp(\frac{\epsilon-\mu_{\text{F}}}{k_{\text{B}}T})+1}\approx
105  \exp(\frac{\epsilon-\mu_{\text{F}}}{k_{\text{B}}T})$\\
106 Parabolic approximation:
107 $D_v(\epsilon)=\frac{1}{2\pi^2}(\frac{2m_p}{\hbar^2})^{3/2}(E_v-\epsilon)^{1/2}$
108 $\Rightarrow$\\
109 $p=\int_{-\infty}^{E_{\text{v}}}D_v(\epsilon)(1-f(\epsilon,T))d\epsilon\approx
110  \frac{1}{2\pi^2}(\frac{2m_p}{\hbar^2})^{3/2}
111  \exp(-\frac{\mu_{\text{F}}}{k_{\text{B}}T})
112  \int_{-\infty}^{E_{\text{v}}}(E_v-\epsilon)^{1/2}
113  \exp(\frac{\epsilon}{k_{\text{B}}T})d\epsilon$\\
114 Use: $x=(E_{\text{v}}-\epsilon)/k_{\text{B}}T$
115      $\Rightarrow\epsilon=E_{\text{v}}-xk_{\text{B}}T$ and
116      $d\epsilon={\color{red}-}k_{\text{B}}Tdx$\\
117 $\Rightarrow$
118 $p=\frac{1}{2\pi^2}(\frac{2m_pk_{\text{B}}T}{\hbar^2})^{3/2}
119  \exp(\frac{E_{\text{v}}-\mu_{\text{F}}}{k_{\text{B}}T})
120  \underbrace{\int_{{\color{red}0}}^{{\color{red}\infty}}x^{1/2}e^{-x}dx}_{=\frac{\sqrt{\pi}}{2}}=
121  \underbrace{2(\frac{m_pk_{\text{B}}T}{2\pi\hbar^2})^{3/2}}_{=N_{\text{v}}}
122  \exp(\frac{E_{\text{v}}-\mu_{\text{F}}}{k_{\text{B}}T})=
123  N_{\text{v}}\exp(\frac{E_{\text{v}}-\mu_{\text{F}}}{k_{\text{B}}T})$
124 \end{itemize}
125
126 \end{document}