2.3: Self-Assessment- Electronic Materials + Answer
- Page ID
- 408598
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Problem 1
Indium phosphide (InP) is a semiconductor with a band gap, \(\mathrm{E}_g\), of \(1.27 \mathrm{eV}\). Calculate the value of the absorption edge of this material. Express your answer in meters.
- Answer
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For absorption of incoming radiation, the following must be true:
\[\mathrm{E}_{\text {radiation }}=\mathrm{E}_{\mathrm{g}} \nonumber \]
Using the Planck relationship gives the wavelength of the absorption edge:
\begin{aligned}
&\mathrm{E}_{\text {radiation }}=\frac{\mathrm{hc}}{\lambda} \\
&\lambda=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{g}}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.27 \times 1.6 \times 10^{-19}}=9.74 \times 10^{-7} \mathrm{~m}
\end{aligned}
Problem 2
Chemical analysis of a germanium (\(\mathrm{Ge}\)) crystal reveals antimony ( \(\mathrm{Sb}\) ) at a level of \(0.0002\) atomic percent.
a. Assuming that the concentration of thermally excited charge carriers from the Ge matrix is negligible, calculate the density of free charge carriers (carriers \(/ \mathrm{cm}^3\) ) in this \(\mathrm{Ge}\) crystal.
- Answer
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Each \(\mathrm{Sb}\) atom will donate an electron to the conduction band; we have only to determine the number of \(\mathrm{Sb}\) atoms \(/ \mathrm{cm}^3\) of \(\mathrm{Ge}\). The atomic volume of the host crystal (\(\mathrm{Ge}\)) is given on your PT as \(13.57 \mathrm{~cm}^3 / \mathrm{mole}\).
\(\# \mathrm{Ge}\) atoms \(/ \mathrm{cm}^3=\frac{6.02 \times 10^{23} \text { atoms }}{1 \text { mole }} \times \frac{1 \text { mole }}{13.57 \mathrm{~cm}^3}=4.44 \times 10^{22}\) atoms \(/ \mathrm{cm}^3\)
\(\# \mathrm{Sb}\) atoms \(/ \mathrm{cm}^3=4.44 \times 10^{22} \times 0.0002 \times 10^{-2}=8.87 \times 10^{16} \mathrm{Sb} / \mathrm{cm}^3\)
Thus, the number of free charge carriers is \(8.87 \times 10^{16} / \mathrm{cm}^3\); they are created by the donation of one electron by each \(\mathrm{Sb}\) atom to the conduction band of the host \(\mathrm{Ge}\) crystal.
b. Draw a schematic energy band diagram for this material and label the valence band, conduction band, band gap, and the energy level associated with the \(\mathrm{Sb}\) impurity.
- Answer
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