Oxidation Rate Example

$$\begin{array}{rl}B=B_0\exp \left(-E_a/k_{B}T\right)& B/A=B_0/A_0\exp \left(-E_a/k_{B}T\right)\\ B& =\left(772\mu m^2/hr\right)e^{-(1.23\mathrm{eV})/k_{B}T}\\ B/A& =\left(3.71\times 10^6\mu m/hr\right)e^{-(2.00\mathrm{eV})/k_{B}T}\\ k_B& =1.38\times 10^{-23}\mathrm{J}/\mathrm{K}\\ T& =\left(273^{\circ }\mathrm{C}+1100^{\circ }\mathrm{C}\right)=1373\mathrm{K}\\ k_{B}T& =\left(1.38\times 10^{-23}\mathrm{J}/\mathrm{K}\right)(1373\mathrm{K})=1895\times 10^{-23}\mathrm{J}=1.9\times 10^{-20}\mathrm{J}\\ 1\mathrm{eV}& =1.6\times 10^{-19}\mathrm{J}\\ \frac{1.9\times 10^{-20}\mathrm{J}}{1.6\times 10^{-19}\mathrm{J}/\mathrm{eV}}\approx 0.12\mathrm{eV}& \end{array}$$ $$\begin{array}{rl}& B=772e^{-1.23/0.12}\simeq 0.03\mu m^2/hr\\ & \frac{B}{A}=3.71\times 10^6{e}^{-2/0.12}\simeq 0.21\mu m/hr\end{array}$$ $$\underset{r}{\to }\{\begin{array}{l}A=0.14\mu m\\ B=0.03\mu m^2/hr\end{array}$$

we can check if we are in the diffusion limited regime For $\frac{4B}{A^2}t\gg 1\Rightarrow x=\sqrt{Bt}$

$$\frac{4B}{A^2}t=\frac{4(0.03)}{(0.14{)}^2}3=18⟶x=\sqrt{B}\sqrt{t}\simeq 0.3\mu m$$

Otherwise

$$\begin{array}{c}{c}_1=4\left[(0.025{)}^2+(0.14)(0.025)\right]\simeq 0.017\mu m^2\\ x=\frac{1}{2}\left[\sqrt{(0.14{)}^2+4(0.03)(3)+0.017}-(0.14)\right]\\ x\simeq 0.24\mu m\end{array}$$