Question 3 of 7: Diffusion — Units of $D$ and a Ceramic Barrier Lifetime
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-11, Properties of Materials — May 2017. 3 hours, closed-book examination (approved Casio or Sharp calculator only). Any five of the seven questions constitute a complete paper (only the first five in the answer book are marked); all questions are of equal value. All seven questions are solved below for completeness.
Reference texts: Callister & Rethwisch, Materials Science and Engineering: An Introduction, 9th ed. (dislocations and slip / Schmid’s law, mechanical behaviour, ionic crystal geometry and the radius-ratio rule, X-ray diffraction, diffusion and Fick’s laws, polymer molecular weight, ASTM grain size, strengthening and annealing, steel heat treatment and hardenability, casting defects, ceramics, glasses and composites).
Question 3: Diffusion — Units of $D$ and a Ceramic Barrier Lifetime (20 marks)
Approach. Rearrange Fick’s first law $J=-D\,dc/dx$ for $D$ and substitute the units of flux and concentration gradient.
Units of each term. Flux $J$ is atoms crossing unit area per unit time, $\left[\frac{\text{atoms}}{\text{cm}^2\cdot\text{s}}\right]$; the concentration gradient $dc/dx$ is $\left[\frac{\text{atoms/cm}^3}{\text{cm}}\right]=\left[\frac{\text{atoms}}{\text{cm}^4}\right]$.
Solve for $D$. $$[D]=\frac{[J]}{[dc/dx]}=\frac{\text{atoms}/(\text{cm}^2\cdot\text{s})}{\text{atoms}/\text{cm}^4}=\boxed{\frac{\text{cm}^2}{\text{s}}}.$$ The atom count cancels, leaving length$^2$/time.
Part (b) — time to remove 1 µm of nickel
Given. Barrier thickness $L=300\ \mu\text{m}=0.030$ cm; nickel to be removed $x=1\ \mu\text{m}=1\times10^{-4}$ cm; $D=9\times10^{-12}$ cm$^2$/s; nickel lattice constant $a_{Ni}=3.6\times10^{-8}$ cm (FCC).
Find. The time for a 1 µm thickness of nickel to diffuse away through the MgO barrier.
Fig. Q3(b) — Steady-state concentration of Ni across the MgO barrier: the source side sits at the Ni atomic density $n_{Ni}$ and the tantalum side acts as a sink ($c\approx0$), giving a linear gradient and a constant flux $J=D\,\Delta c/L$.
Approach. Treat the barrier as being at steady state (linear concentration drop from the Ni source density to zero at the tantalum sink), then equate the number of atoms that must cross per unit area to the flux × time.
Nickel atomic density (context). FCC nickel has 4 atoms per cell, so $$n_{Ni}=\frac{4}{a_{Ni}^3}=\frac{4}{(3.6\times10^{-8})^3}=8.57\times10^{22}\ \text{atoms/cm}^3.$$
Steady-state flux through the barrier. With $c=n_{Ni}$ at the source and $c\approx0$ at the sink over thickness $L$, $$J=D\frac{n_{Ni}}{L}.$$
Atoms to remove and the time. Removing a thickness $x$ of nickel over unit area requires $x\,n_{Ni}$ atoms to cross, so $$t=\frac{x\,n_{Ni}}{J}=\frac{x\,n_{Ni}}{D\,n_{Ni}/L}=\frac{xL}{D}.$$ The nickel density — and hence the lattice constant — cancels: the lattice constant is a distractor for the time asked. Substituting, $$t=\frac{(1\times10^{-4})(0.030)}{9\times10^{-12}}=\boxed{3.33\times10^{5}\ \text{s}}\approx92.6\ \text{h}\ (\approx3.9\ \text{days}).$$
Check: the model assumes a fully developed steady-state linear profile across the barrier and a perfect sink at the tantalum interface ($c\to0$); these are the standard assumptions for a thin diffusion-barrier lifetime estimate and make the answer independent of $a_{Ni}$.
