Question 7 of 7: Point Defects and Diffusion — Vacancy Concentration and Dopant Diffusion Rate

Question 7: Point Defects and Diffusion — Vacancy Concentration and Dopant Diffusion Rate (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.

Part (a) — elastic vs. plastic deformation. An elastic deformation is a small, reversible distortion of the crystal in which every atom's displacement from its lattice site is proportional to the applied stress (Hooke's law) and the atomic bonds themselves are only stretched or bent, never broken and reformed; removing the stress restores the original lattice exactly, with the stored strain energy fully recovered. A plastic deformation is a large, permanent (irreversible) distortion produced once the stress exceeds the crystal's yield point: whole planes of atoms slip past one another via the motion of dislocations, breaking and reforming interatomic bonds along the slip plane, so the crystal does not return to its original shape when the stress is removed — the mechanism (dislocation glide) is fundamentally different from, and only possible because of, the line defects absent from an ideally elastic (defect-free) crystal.

Given (b). Vacancy-formation energy $E_v=1.0\ \text{eV}$ for Na; target concentration $n/N=1/100000=10^{-5}$; Boltzmann relation, eq. (18): $\dfrac{n}{N-n}=\exp(-E_v/k_BT)$.

Find (b). the temperature $T$ giving this vacancy fraction.

Given (c). Diffusion of As in Si, Table T7: $D_0=0.32\ \text{cm}^2/\text{s}$, $E=3.56\ \text{eV}$; $T=1100\ \text{K}$; Arrhenius relation, eq. (19): $D=D_0\exp(-E/k_BT)$.

Find (c). the diffusion coefficient $D$ at 1100 K.

Approach. Both parts are direct evaluations of the exam's own Boltzmann-type activation laws (eqs. 18 and 19); part (b) is solved by inverting for $T$, part (c) by substituting for $D$.

1. Part (b) — vacancy temperature. With $n/(N-n)=10^{-5}$ exactly, the reciprocal ratio $(N-n)/n=10^{5}$ is exact regardless of the $n\ll N$ approximation, so eq. (18) inverts cleanly: $$\frac{N-n}{n}=\exp(E_v/k_BT)\ \Longrightarrow\ T=\frac{E_v}{k_B\ln(10^5)}$$ With $E_v=1.0(1.60219\times10^{-19}\ \text{J})=1.60219\times10^{-19}\ \text{J}$, $k_B=1.38062\times10^{-23}\ \text{J/K}$, $\ln(10^5)=5\ln10=11.513$: $$T=\frac{1.60219\times10^{-19}}{(1.38062\times10^{-23})(11.513)}$$ $$\boxed{T\approx1008\ \text{K}}$$ — far above sodium's melting point (371 K), so with a 1.0 eV formation energy this vacancy fraction cannot actually be reached in solid Na; the Boltzmann-statistics answer to the question as set is nevertheless 1008 K (see the note below).
2. Part (c) — As diffusion rate in Si at 1100 K. $k_BT$ at 1100 K in eV: $k_BT=8.6173\times10^{-5}(1100)=0.09479\ \text{eV}$, so $E/k_BT=3.56/0.09479=37.56$. Substituting into eq. (19): $$D=D_0e^{-E/k_BT}=(0.32\ \text{cm}^2/\text{s})\,e^{-37.56}$$ $$\boxed{D\approx1.56\times10^{-17}\ \text{cm}^2/\text{s}}$$ — an extremely slow diffusion rate, as expected for the large activation energy ($3.56\ \text{eV}$) of a substitutional dopant in the strongly-bonded Si lattice; it is why practical As-doping of Si is normally carried out at higher temperatures ($>1400\ \text{K}$) or via ion implantation rather than simple thermal diffusion at 1100 K.