Question 7 of 7: Point Defects — Color Centers, Schottky Defects, Dopant Diffusion

Question 7: Point Defects — Color Centers, Schottky Defects, Dopant Diffusion (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $0.0001\%=1\times10^{-6}$ of Cu lattice sites are Schottky defects at $T=1000\ \text{K}$. Figure P7 is a semi-log plot of $D$ (cm²/s, decades $10^{-21}$ to $10^{-1}$) against $10^3/T$ (K$^{-1}$, printed decreasing from 2.2 on the left to 0.0 on the right, with 500/1000/2000 K marked on the top axis). It shows measured points for As in Si with a fitted solid line, extended as a dashed line to $10^3/T=0$.

Find. (a) meaning of color center + two coloring methods; (b) Schottky formation energy; (c) purpose of As diffusion; (d) the diffusion constant $D_0$ read from the graph, and the activation energy $E$ from its slope.

[Figure not reproduced: Fig. 8 — The printed Figure P7 line, digitized and redrawn with $10^3/T$ increasing to the right. The dashed extrapolation meets $10^3/T=0$ at $D_0\approx10^{-0.48}\approx0.33$ cm²/s, and the slope is $-8.82$ decades per unit of $10^3/T$. See the official exam paper.]

Approach. (a)/(c) recall standard point-defect physics; (b) invert the Schottky-defect Boltzmann factor, eq. (18); (d) with eq. (19), $D=D_0e^{-E/k_BT}$, the diffusion constant $D_0$ is the value of $D$ where $1/T\to0$, so it is read (without calculation) where the dashed extrapolation meets the $10^3/T=0$ axis. $E$ then follows from the slope of $\log_{10}D$ vs. $1/T$.

1. Part (a) — color centers. A color center (e.g. an F-center) is a point defect — typically a halide-ion (anion) vacancy that has trapped an electron in its place — that introduces a discrete, allowed optical-absorption level inside the otherwise-transparent band gap of an alkali halide crystal; the trapped electron's absorption of a particular visible wavelength is what gives the normally colorless crystal (e.g. NaCl, KCl) its color. Two standard methods to create color centers: (1) irradiation of the crystal with X-rays, $\gamma$-rays, or high-energy electrons, which knocks halide ions out of their lattice sites and leaves trapped electrons behind; (2) additive coloration — heating the crystal in the vapor of its own alkali metal (e.g. Na vapor for NaCl), so excess alkali atoms diffuse in, ionize, and leave their electrons trapped at newly-created anion vacancies.
2. Part (b) — Schottky defect formation energy. Eq. (18), $\dfrac{n}{N-n}=\exp(-E_v/k_BT)$, with fractional defect concentration $n/N=1\times10^{-6}\ll1$ so $n/(N-n)\approx n/N$: $$E_v=-k_BT\ln\!\left(\frac nN\right)=k_BT\ln\!\left(\frac{N}{n}\right)=(8.62\times10^{-5}\ \text{eV/K})(1000\ \text{K})\ln(10^6)$$ $$\boxed{E_v\approx1.19\ \text{eV}}$$ a physically reasonable Schottky vacancy-formation energy for copper (literature values run $\approx1.0$–$1.3\ \text{eV}$), which supports both the given defect fraction and the use of eq. (18) without the extra factor of 2 sometimes seen for defect pairs.
3. Part (c) — purpose of As diffusion into Si. Arsenic is a group-V element with one more valence electron than Si; substituting an As atom onto a Si lattice site contributes a nearly-free (shallow donor) electron to the conduction band. Diffusing As into Si is therefore the standard way to create n-type doping — controllably raising the free-electron concentration (and hence the Fermi level, cf. Question 5) to engineer the conductivity and carrier type needed to build semiconductor devices such as diodes and transistors (e.g. forming a p–n junction against an existing p-type region).
4. Part (d) — diffusion constant and activation energy from Figure P7. In eq. (19), $D=D_0\exp(-E/k_BT)$, the exponential goes to 1 as $1/T\to0$. The diffusion constant $D_0$ is therefore simply the value of $D$ where the line meets $10^3/T=0$, which is what the printed dashed extension is for. Reading that intercept on the right-hand ($10^3/T=0.0$) axis, the line arrives about half a decade below $10^{0}$: $$\boxed{D_0\approx10^{-0.48}\approx0.3\ \text{cm}^2/\text{s}}$$ For the activation energy, eq. (19) in base-10 form is $\log_{10}D=\log_{10}D_0-\dfrac{E}{2.303\,k_B}\cdot\dfrac1T$. With $x=10^3/T$, the slope of $\log_{10}D$ vs. $x$ is $-E/(2.303\times10^3\,k_B)$. The printed line falls from $D\approx10^{-4.0}$ at $x=0.40$ to $D\approx10^{-19.0}$ at $x=2.10$, i.e. by 15.0 decades over 1.70 units (slope $-8.82$ decades per unit): $$E=8.82\times2.303\times10^3\times(8.62\times10^{-5}\ \text{eV/K})$$ $$\boxed{E\approx1.75\ \text{eV}}$$ Check: at 1000 K ($x=1.0$) the line gives $D=10^{-0.48-8.82}=10^{-9.3}\approx5\times10^{-10}$ cm²/s, and $D_0e^{-E/k_BT}=0.33\,e^{-1.75/0.0862}=5\times10^{-10}$ cm²/s. The two agree.