Diffusion — Units of D and an MgO Diffusion-Barrier Lifetime: 04-BS-11, May 2013, Q3

Question 3: Diffusion — Units of D and an MgO Diffusion-Barrier Lifetime (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. MgO barrier thickness $L=400\ \mu\text{m}=0.0400$ cm; nickel layer to be consumed $x=1\ \mu\text{m}=1\times10^{-4}$ cm; diffusion coefficient of Ni in MgO at 1400°C, $D=9\times10^{-12}\ \text{cm}^2/\text{s}$; Ni lattice constant $a_{Ni}=3.6\times10^{-8}$ cm.

Find. (a) Units of $D$. (b) Time for a 1 μm layer of Ni to be removed by diffusion through the barrier.

Fig. Q3 — Ni | MgO (400 μm) | Ta diffusion couple with the assumed linear, steady-state Ni concentration profile across the barrier.

Approach

Part (a) is a dimensional check on Fick's first law. Part (b) treats the barrier as a steady-state, linear-concentration-profile diffusion problem (Fick's first law) and converts the resulting flux into a time by a mass balance on the nickel being stripped from the source layer.

(a) Units of D. Fick's first law is $J=-D\,\dfrac{dC}{dx}$, where flux $J$ has units mol/(cm$^2\cdot$s) and concentration gradient $dC/dx$ has units (mol/cm$^3$)/cm $=$ mol/cm$^4$. Solving for $D$: $$D=\frac{J}{dC/dx}\ \Rightarrow\ [D]=\frac{\text{mol}/(\text{cm}^2\cdot\text{s})}{\text{mol}/\text{cm}^4}=\boxed{\text{cm}^2/\text{s}}.$$
(b) Steady-state flux through the barrier. With a linear concentration profile from the Ni/MgO interface (concentration equal to nickel's own atomic density, $C_{Ni}$) to the Ta/MgO interface (sink, $C\approx0$), Fick's first law gives $$J=D\,\frac{C_{Ni}}{L}\quad\text{(atoms per unit area per unit time)}.$$
Mass balance on the stripped layer. Removing a layer of thickness $x$ of pure Ni (same atomic density $C_{Ni}$) removes $C_{Ni}\,x$ atoms per unit area, so $$t=\frac{C_{Ni}\,x}{J}=\frac{C_{Ni}\,x}{D\,C_{Ni}/L}=\frac{xL}{D}.$$ The nickel atomic density $C_{Ni}$ (which the given lattice constant would fix, via 4 atoms/unit cell for FCC nickel) cancels — it is needed only if the actual atomic flux (atoms/s) were requested, not the time.
Evaluate. $$t=\frac{xL}{D}=\frac{(1\times10^{-4})(0.0400)}{9\times10^{-12}}=\boxed{4.44\times10^5\ \text{s}}=123.5\ \text{hr}=5.14\ \text{days}.$$

Quantity	Result
Units of $D$	cm²/s
Time to strip 1 μm of Ni	4.44×10⁵ s = 123.5 hr = 5.14 days

Concept & Theory

Topic. Steady-state diffusion (Fick's first law) through a ceramic diffusion barrier.

Key relations. Fick's first law $J=-D\,dC/dx$; steady-state linear profile $J=D\Delta C/L$; mass balance $t=(\text{amount removed})/J$.

Why this works. A thin diffusion barrier reaches a steady (time-independent) linear concentration gradient quickly compared with the long service life being asked about, so Fick's first law (rather than the time-dependent second law) applies; because both the source concentration and the flux scale with the same nickel atomic density, that density cancels out of the time-to-deplete calculation.

Common pitfall. Assuming the given nickel lattice constant must be used somewhere — extra data are sometimes supplied that cancel out of the specific quantity asked for; always check whether a quantity is actually needed rather than forcing it into the calculation.

Source. Callister & Rethwisch, Materials Science and Engineering, 9th ed., Ch. 5 (diffusion).

Question 3 of 8: Diffusion — Units of D and an MgO Diffusion-Barrier Lifetime

Question 3: Diffusion — Units of D and an MgO Diffusion-Barrier Lifetime (20 marks)

Approach