CHAPTER 6: DIFFUSION IN SOLIDS Diffusion- Steady and Non-Steady State

Gear from case-hardened steel (C diffusion) Diffusion - Mass transport by atomic motion ISSUES TO ADDRESS...

• How does diffusion occur? Mechanisms • Gases & Liquids – random (Brownian) motion • Why is it an important part of processing? • Solids – vacancy diffusion or interstitial diffusion • How can the rate of diffusion be predicted for some simple cases? • How does diffusion depend on structure and temperature?

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Simple Diffusion Inter-diffusion • Interdiffusion: In alloys, atoms tend to migrate • Glass tube filled with water. from regions of large concentration. • At time t = 0, add some drops of ink to one end This is a diffusion couple. of the tube. Initially After some time • Measure the diffusion distance, x, over some time. • Compare the results with theory. Adapted from Figs. 6.1 - 2, Callister 6e.

100%

0 Concentration Profiles

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1 Self-diffusion Substitution-diffusion:vacancies and interstitials • applies to substitutional impurities • Self-diffusion: In an elemental solid, atoms also migrate. • atoms exchange with vacancies Number (or concentration*) • rate depends on (1) number of vacancies; of Vacancies at T (2) activation energy to exchange. − ΔE n k T Label some atoms After some time i B ci = =e N C • kBT gives eV A * see web handout for derivation. D

B ΔE is an activation energy for a particular process (in J/mol, cal/mol, eV/atom).

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Substitution-diffusion Inter-diffusion across Interfaces Vacancy Diffusion: • applies to substitutional impurities • atoms exchange with vacancies • rate depends on (1) number of vacancies; (2) activation energy to exchange. • Rate of substitutional diffusion depends on: - vacancy concentration - frequency of jumping.

(Courtesy P.M. Anderson)

increasing elapsed time Why should interstitial diffusion be faster than by vacancy mode of diffusion?

MatSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08 MatSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08

2 Diffusion Mechanisms Processing using Diffusion

• Interstitial diffusion – smaller atoms diffuse between atoms. • Case Hardening: Fig. 6.0, Callister 6e. -Diffuse carbon atoms into the (courtesy of Surface host iron atoms at the surface. Div., Midland-Ross.) -Example of interstitial diffusion is a case hardened gear.

• Result: The "Case" is -hard to deform: C atoms "lock" planes from shearing. -hard to crack: C atoms put the surface in compression. More rapid than vacancy diffusion

MatSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08 MatSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08

Processing using Diffusion Modeling rate of diffusion: flux • Doping Silicon with P for n-type semiconductors: • Flux: • Process: 1. Deposit P rich layers on surface. • Directional Quantity

silicon • Flux can be measured for: Fig. 18.0, - vacancies 2. Heat it. Callister 6e. - host (A) atoms 3. Result: Doped - impurity (B) atoms semiconductor A = Area of flow regions. • Empirically determined: – Make thin membrane of known surface area diffused – Impose concentration gradient mass J ∝ slope silicon – Measure how fast atoms or molecules diffuse M through the membrane time

3 Steady-state Diffusion: J ~ gradient of c Steady-State Diffusion

• Concentration Profile, C(x): [kg/m3] • Steady State: concentration profile not changing with time.

Cu flux Ni flux Adapted from Fig. 6.2(c)

Concentration Concentration of Cu [kg/m3] of Ni [kg/m3]

Position, x

• Fick's First Law: D is a constant! dC • Apply Fick's First Law: Jx = −D dx dC dC • If Jx)left = Jx)right , then = dx dx left right • The steeper the concentration profile, the greater the flux! • Result: the slope, dC/dx, must be constant (i.e., slope doesn't vary with position)!

Steady-State Diffusion Example: Chemical Protection Clothing dC Rate of diffusion independent of time J ~ • Methylene chloride is a common ingredient of paint removers. dx Besides being an irritant, it also may be absorbed through skin. When using, protective gloves should be worn.

• If butyl rubber gloves (0.04 cm thick) are used, what is the C1 C1 Fick’s first law of diffusion diffusive flux of methylene chloride through the glove? • Data: dC -8 2 J = −D – D in butyl rubber: D = 110 x10 cm /s C C2 C = 0.44 g/cm3 C = 0.02 g/cm3 2 dx – surface concentrations: 1 2 – Diffusion distance: x2 – x1 = 0.04 cm x1 x2 D ≡ diffusion coefficient glove x C1 2 t = l b 6D dC ΔC C −C paint skin C - C g 2 1 J = -D 2 1 = 1.16 x 10-5 if linear ≅ = remover x 2 dx x x x x2 - x1 cm s Δ 2 − 1 C2 x1 x2

4 Example: C Diffusion in steel plate Example: Diffusion of radioactive atoms • Steel plate at 7000C with geometry shown: • Surface of Ni plate at 10000C contains 50% Ni63 (radioactive) and 50% Ni (non-radioactive). Adapted from Fig. • 4 microns below surface Ni63 /Ni = 48:52 700 C 5.4, Callister 6e. • Lattice constant of Ni at 1000 C is 0.360 nm. Knowns: • Experiment shows that self-diffusion of Ni is 1.6 x 10-9 cm2/sec 3 C1= 1.2 kg/m at 5mm (5 x 10–3 m) below surface. What is the flux of Ni63 atoms through a plane 2 µm below surface? 3 63 63 C2 = 0.8 kg/m at 10mm (4Ni /cell)(0.5Ni /Ni) (4Ni /cell)(0.48Ni /Ni) –2 C = C2 = (1 x 10 m) below surface. 1 9 3 (0.36x10−9m)3 /cell (0.36x10− m) /cell -11 2 27 63 3 41.15x1027Ni63 /m3 D = 3 x10 m /s at 700 C. = 42.87x10 Ni /m =

27 63 3 13 2 (41.15 − 42.87)x10 Ni /m = −(1.6x10− m /sec) • Q: In steady-state, how much carbon transfers € 6 € (4 − 0)x10− m from the rich to the deficient side? 20 63 2 = +0.69x10 Ni /m •s

2 63 How many Ni63 atoms/second through cell? J •(0.36nm) = 9 Ni /s €

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Where can we use Fick’s Law? Non-Steady-State Diffusion

Fick's law is commonly used to model transport processes in • Concentration profile, • foods, C(x), changes w/ time. • clothing, • biopolymers, • pharmaceuticals, • porous soils, • semiconductor doping process, etc. • To conserve matter: • Fick's First Law:

Example The total membrane surface area in the lungs (alveoli) may be on the order of 100 square meters and have a thickness of less than a millionth of a meter, so it is a very effective gas-exchange interface.

2 2 CO2 in air has D~16 mm /s, and, in water, D~ 0.0016 mm /s • Governing Eqn.:

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5 Non-Steady-State Diffusion: another look Non-Steady-State Diffusion: C = c(x,t) • Concentration profile, concentration of diffusing species is a function of both time and position 2 ∂c ∂c C(x), changes w/ time. Fick's Second "Law" D ≈ 2 ∂t ∂x

• Rate of accumulation C(x) • Copper diffuses into a bar of aluminum.

∂C ∂C ∂Jx ∂Jx dx = Jx − Jx dx → dx = Jx − (Jx + dx) = − dx ∂t + ∂t ∂x ∂x Cs C J C B.C. at t = 0, C = Co for 0 ≤ x ≤ ∞ • Using Fick’s Law: ∂ ∂ x ∂ ∂ Fick’s = − = − −D at t > 0, C = C for x = 0 (fixed surface conc.) ∂t ∂x ∂x ∂x 2nd Law S C = Co for x = ∞

2 ∂c ∂ ∂c ∂c Fick's Second "Law" = D ≈ D Adapted from Fig. 6.5, • If D is constant: 2 Callister & Rethwisch 3e. ∂t ∂x ∂x ∂x

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Non-Steady-State Diffusion Example: Non-Steady-State Diffusion FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated • Cu diffuses into a bar of Al. CS temperature and in an atmosphere that gives a surface C content at 1.0 wt%.

If after 49.5 h the concentration of carbon is 0.35 wt% at a position 4.0 mm below the surface, what temperature was treatment done? C(x,t) Solution 2 ∂c ∂c C(x,t) −Co 0.35 − 0.20 x Fick's Second "Law": ≈ D C = = 1− erf = 1− erf(z) ∴ erf(z) = 0.8125 2 o C −C 1.0 − 0.20 2 Dt ∂t ∂x s o Using Table 6.1 find z where erf(z) = 0.8125. Use interpolation. z erf(z) • Solution: 0.90 0.7970 z − 0.90 0.8125 − 0.7970 = So, z = 0.93 z 0.8125 0.95 − 0.90 0.8209 − 0.7970 0.95 0.8209 "error function” Values calibrated in Table 6.1 x x2 Now solve for D z = D = z 2 Dt 2 2 −y 2 4z t erf (z) = e dy x2 (4 x 10−3 m)2 1 h ∫ 0 ∴D = = = 2.6 x 10−11 m2 /s π 4z2t (4)(0.93)2(49.5 h) 3600 s 15 MatSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08 MatSE 280: Introduction to Engineering Materials ©D.D. Johnson 2004, 2006-08

6 Example: Processing Solution (cont.): • Copper diffuses into a bar of aluminum. • To solve for the temperature at which D Qd • 10 hours processed at 600 C gives desired C(x). has the above value, we use a rearranged T = • How many hours needed to get the same C(x) at 500 C? form of Equation (6.9a); R(lnDo − lnD) D=D0 exp(-Qd/RT) Key point 1: C(x,t500C) = C(x,t600C). Key point 2: Both cases have the same Co and Cs. • From Table 6.2, for diffusion of C in FCC Fe

-5 2 −11 2 • Result: Dt should be held constant. Do = 2.3 x 10 m /s Qd = 148,000 J/mol D = 2.6 x 10 m /s C(x,t) −C x o = 1− erf Cs −Co 2 Dt ∴ 148,000 J/mol T = (8.314 J/mol-K)(ln 2.3x10−5 m2 /s ln 2.6x10−11 m2 /s) − Note D(T) are T dependent! Values of D are provided. T = 1300 K = 1027°C • Answer:

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Diffusion Analysis Data from Diffusion Analysis

• The experiment: we recorded combinations of t and x that kept C constant.

C(x ,t ) − C x i i o = 1− erf i = (constant here) C C s − o 2 Dti

• Diffusion depth given by: • Experimental result: x ~ t0.58 • Theory predicts x ~ t0.50 from • Close agreement.

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7 Diffusion and Temperature Diffusion and Temperature

• Diffusivity increases with T exponentially (so does Vacancy conc.). • Experimental Data:

Adapted from Fig. 6.7, T(°C) 600 300 1000 Qd 1500 = − 10-8 D Do exp C RT in 2 γ D (m /s) -Fe C in α-Fe 2 Fe Dinterstitial >> Dsubstitutional D = diffusion coefficient [m /s] Fe -14 in C in α-Fe Cu in Cu 2 10 in α Al D = pre-exponential [m /s] (see Table 6.2) -Fe Al in Al o γ C in γ-Fe -Fe in Fe in α-Fe Q = activation energy [J/mol or eV/atom] Al Fe in γ-Fe d Zn in Cu R = gas constant [8.314 J/mol-K] 10-20 T = absolute temperature [K] 0.5 1.0 1.5 1000 K/T

Q Q Adapted from Fig. 6.7, Callister & Rethwisch 3e. d 1 1 Note: ln D = ln D − log D = log D − d (Data for Fig. 6.7 from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals 0 R T 0 2.3R T Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)

Example: Comparing Diffuse in Fe Connecting Holes, Diffusion, and Stress Is C in fcc Fe diffusing faster than C in bcc Fe? z-axis

600 300 T( C)

1500 1000 ° (Table 6.2) 10-8 • fcc-Fe: D =2.3x10–5(m2/2) C 0 in 2 C Q=1.53 eV/atom D (m /s) γ-Fe in –12 2 α-Fe T = 900 C D=5.9x10 (m /2) Fe Fe -14 in 10 in Al –7 2 α • bcc-Fe: D =6.2x10 (m /2) γ -Fe in 0 -Fe Q=0.83 eV/atom Al T = 900 C D=1.7x10–10(m2/2) 10-20 0.5 1.0 1.5 1000 K/T Red octahedral fcc Green + Red octahedral bcc • FCC Fe has both higher activation energy Q and D0 (holes larger in FCC). • BCC and FCC phase exist over limited range of T (at varying %C). • Hence, at same T, BCC diffuses faster due to lower Q. FCC represented as a BCT cell, with relevant holes. • Cannot always have the phase that you want at the %C and T you want! which is why this is all important.

8 Other Types of Diffusion (Beside Atomic) Inter-diffusion (diffusion couples) interface • Flux is general concept: e.g. charges, phonons,… Cu flux Ni flux • Charge Flux – 1 dq 1 d(Ne) e = electric chg. C1 Cu flux Ni flux j = = N = net # e- cross A q A dt A dt Concentration Concentration 3 of CuConcentration [kg/m ] C0 Concentrationof Ni [kg/m3] Defining conductivity σ dV Solution: Fick’s 2nd Law j = −σ of Cu [kg/m3] of Ni [kg/m3] (a material property) dx V −V x C x s = erf 2 I 1 V V Position, x Ohm’s Law j = = = σ V0 −Vs 2 σt q Position, x A A R l

C x − C x 1 dH 1 d(N ) 1 0 • Heat Flux – ε N = # of phonons with Assuming DA = DB: = erf C = (C +C /2 Q = = C − C 2 Dt 0 1 2) (by phonons) A dt A dt avg. energy ε 1 0 dT Defining thermal conductivity κ x Q = −κ Solution: Fick’s 2nd Law Curve is symmetric about C , (a material property) dx For C2 0 erf (−z) = −erf (z) T − T x κ x s = erf Or w/ Thermal Diffusivity: h Q c T = = V ρ T − T 2 ht cV ρ 0 s

Kirkendall Effect: What is DA > DB? Diffusion in Compounds: Ionic Conductors

• Kirkendall studied Mo markers in Cu-brass (i.e., fcc Cu70Zn30). • Unlike diffusion in metals, diffusion in compounds involves • Symmetry is lost: Zn atoms move more readily in one direction second-neighbor migration. (to the right) than Cu atoms move in the other (to the left). • Since the activation energies are high, the D’s are low unless vacancies are present from non-stoichiometric ratios of atoms. Cu70Zn30 Cu Cu O Ni O Ni O Ni O t = 0 e.g., NiO There are Ni O Ni O Ni2+ O2– Ni Zn Schottky defects 2+ 2– 2– Mo wire markers O Ni O Ni O + O

concentration Ni – Ni2+ O Ni O Ni After diffusion O Ni O Ni O Ni O

• When diffusion is asymmetric, interface moves away markers, • The two vacancies cannot accept neighbors because they have i.e., there is a net flow of atoms to the right past the markers. wrong charge, and ion diffusion needs 2nd neighbors with high

• Analyzing movement of markers determines DZn and DCu. barriers (activation energies). • Kirkendall effect driven by vacancies, effective for T > 0.5 Tmelt.

9 Diffusion in Compounds: Ionic Conductors Ionic Conduction: related to fuel cells • D’s in an ionic compound are seldom comparable because of size, change and/or structural differences. • Molten salts and aqueous electrolytes conduct charge when placed in • Two sources of conduction: ion diffusion and via e- hopping from ions of variable electric field, +q and –q move in opposite directions. valency, e.g., Fe2+ to Fe3+, in applied electric field. • The same occurs in solids although at much slower rate. e.g., ionic • Each ion has charge of Ze (e = 1.6 x 10–19 amp*sec),

• In NaCl at 1000 K, DNa+~ 5DCl– ,whereas at 825 K DNa+ ~ 50DCl–! so ion movement induces ionic conduction • This is primarily due to size rNa+ = 1 A vs rCl–=1.8 A. • Conductivity σ = n µ Z e is related to mobility, µ, which is related to D via the Einstein equations: µ = ZeD / k T e.g., oxides B • Hence 2 2 2 2 4+( 2– 7 nZ e nZ e −Q/ RT • In uranium oxide, U O )2, at 1000 K (extrapolated), DO ~ 10 DU. σ ionic = D = Doe • This is mostly due to charge, i.e. more energy to activate 4+ U ion. kBT kBT • Also, UO is not stoichiometric, having U3+ ions to give UO , so that the 2-x 2 2 2- nZ e Q anion vacancies significantly increase O mobility. log ~ ln D So, electrical conduction can be used 10 σ ionic o − kBT 2.3RT determine diffusion data in ionic solids. e.g., solid-solutions of oxides (leads to defects, e.g., vacancies) 2+ 3+ 2+ • If Fe1-xO (x=2.5-4% at 1500 K, 3Fe -> 2Fe + vac.) is dissolved in MgO e.g., What conductivity results by Ca diffusion in CaO at 2000 K? under reducing conditions, then Mg2+ diffusion increases. • CaO has NaCl structure with a= 4.81 A, with D(2000 K)~10-14m2/s, and Z=2. + 2+ + • If MgF2 is dissolved in LiF (2Li -> Mg + vac.), then Li diffusion increases. 2 2 −5 4 cell 28 3 nZ e 1.3x10 All due to additional vacancies. n 2+ = = 3.59x10 / m Ca cell (4.81x10−10 m)3 σ = D ~ kBT ohm − cm

Example: solid-oxide fuel cell (SOFC) Ceramic Compounds: Al2O3

• SOFC is made up of four layers, three of which are ceramics (hence the name).

• A single cell consisting of these four layers stacked together is only a few mm thick.

• Need to stack many, many together to have larger DC current.

Overall: H2 + 1/2O2 → H2O

Holes for diffusion

+ – H2 → 2H +2e 2– + – O + 2H + 2e → H2O Image: http://www.ip3.unipg.it/FuelCells/en/htfc.asp

S. Haile’s SOFC (2004 best): Thin-film of Sm-doped Ceria electrolyte (CeO2, i.e. SmxCe1-xO2-x/2) and BSCF cathode (Perovskite Ba0.5Sr0.5Co0.8Fe0.2O3-d) show high power densities – over 1 Unit cell defined by Al ions: 2 Al + 3 O 2 W/cm at 600 C – with humidified H2 as the fuel and air at the cathode.

10 Summary: Structure and Diffusion

Diffusion FASTER for... Diffusion SLOWER for...

• open crystal structures • close-packed structures

• lower melting T materials • higher melting T materials

• materials w/secondary • materials w/covalent bonding bonding

• smaller diffusing atoms • larger diffusing atoms

• cations • anions

• lower density materials • higher density materials

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