PY3090 PY3090 –– 66 1

PY3090PY3090Preparation of MaterialsPreparation of MaterialsLecture 6Lecture 6

Colm StephensColm StephensSchool of PhysicsSchool of Physics

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DiffusionDiffusion

DiffusionDiffusionMass transport by atomic motionMass transport by atomic motion

MechanismsMechanismsGases & Liquids Gases & Liquids –– random (Brownian) random (Brownian) motionmotionSolids Solids –– vacancy diffusion or vacancy diffusion or interstitial diffusioninterstitial diffusion

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Interdiffusion: In an alloy, atoms tend to migrate from regions ofhigh concentration to regions of low concentration.

Initially

DiffusionDiffusion

After some time

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• Self-diffusion: In an elemental solid, atomsalso migrate.Label some atoms After some time

DiffusionDiffusion

A

B

C

DA

B

C

D

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Diffusion MechanismsDiffusion MechanismsVacancy Diffusion:

• atoms exchange with vacancies• applies to substitutional impurities atoms • rate depends on:

--number of vacancies--activation energy to exchange.

increasing elapsed time

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Diffusion MechanismsDiffusion Mechanisms

Interstitial diffusion Interstitial diffusion –– smaller atoms can smaller atoms can diffuse between atoms.diffuse between atoms.

More rapid than vacancy diffusion

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• Case Hardening:--Diffuse carbon atoms

into the host iron atomsat the surface.

--Example of interstitialdiffusion is a casehardened gear.

• Result: The presence of C atoms makes iron (steel) harder.

Processing Using DiffusionProcessing Using Diffusion

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Carburising of steelCarburising of steel

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Carburising of steelCarburising of steel

Fe3C

(cem

entit

e)

1600

1400

1200

1000

800

600

4000 1 2 3 4 5 6 6.7

L

γ

(austenite)

γ +L

γ +Fe3C

α +Fe3C

α + γ

L+Fe3C

δ

(Fe) Co, wt% C

1148°C

T(°C)

α 727°C = T eutectoid

1.4%

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• Doping silicon with phosphorus for n-type semiconductors:• Process:

3. Result: Dopedsemiconductorregions.

silicon

Processing Using DiffusionProcessing Using Diffusion

2. Heat it.

1. Deposit P richlayers on surface.

silicon magnified image of a Si chip

light regions: Si atoms

light regions: Al atoms

0.5mm

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DiffusionDiffusionHow do we quantify the amount or rate of How do we quantify the amount or rate of diffusion?diffusion?

Measured empiricallyMeasured empiricallyMake thin film (membrane) of known surface areaMake thin film (membrane) of known surface areaImpose concentration gradientImpose concentration gradientMeasure how fast atoms or molecules diffuse through the Measure how fast atoms or molecules diffuse through the membranemembrane

( )( ) smkgor

scmmol

timearea surfacediffusing mass) (or molesFlux 22=≡≡J

M =mass

diffusedtime

J ∝ slope

dtdM

A1

AtMJ ==

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SteadySteady--State DiffusionState Diffusion

dxdCDJ −=

Fick’s first law of diffusionC1

C2

x

C1

C2

x1 x2

D ≡ diffusion coefficient

Rate of diffusion independent of timeFlux proportional to concentration gradient =

dxdC

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12 linear ifxxCC

xC

dxdC

−−

=ΔΔ

≅

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Example: Chemical Protective Example: Chemical Protective Clothing (CPC)Clothing (CPC)

MethyleneMethylene chloride is a common ingredient of paint chloride is a common ingredient of paint removers. Besides being an irritant, it also may be absorbed removers. Besides being an irritant, it also may be absorbed through skin. When using this paint remover, protective through skin. When using this paint remover, protective gloves should be worn.gloves should be worn.If butyl rubber gloves (0.04 cm thick) are used, what is the If butyl rubber gloves (0.04 cm thick) are used, what is the diffusive flux of diffusive flux of methylenemethylene chloride through the glove?chloride through the glove?Data:Data:

diffusion coefficient in butyl rubber: diffusion coefficient in butyl rubber: DD = 110= 110 x10x10--88 cmcm22/s/s

surface concentrations:surface concentrations:C2 = 0.02 g/cm3

C1 = 0.44 g/cm3

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scmg 10 x 16.1

cm) 04.0()g/cm 44.0g/cm 02.0(/s)cm 10 x 110( 2

5-33

28- =−

−=J

Example (cont).Example (cont).

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12- xxCCD

dxdCDJ

−−

−≅=D

tb 6

2=

gloveC1

C2

skinpaintremover

x1 x2

• Solution – assuming linear conc. gradient

D = 110x10-8 cm2/s

C2 = 0.02 g/cm3

C1 = 0.44 g/cm3

x2 – x1 = 0.04 cm

Data:

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Diffusion and TemperatureDiffusion and Temperature

• Diffusion coefficient increases with increasing T.

D = Do exp⎛

⎝⎜

⎞

⎠⎟−

Qd

RT

= pre-exponential [m2/s]= diffusion coefficient [m2/s]

= activation energy [J/mol or eV/atom] = gas constant [8.314 J/mol-K]= absolute temperature [K]

DDo

Qd

RT

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Diffusion and TemperatureDiffusion and Temperature

D has exponential dependence on T

Dinterstitial >> DsubstitutionalC in α-FeC in γ-Fe

Al in AlFe in α-FeFe in γ-Fe

1000K/T

D (m2/s) C in α-Fe

C in γ-Fe

Al in Al

Fe in α-Fe

Fe in γ-Fe

0.5 1.0 1.510-20

10-14

10-8T(°C)15

00

1000

600

300

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Example: At 300ºC the diffusion coefficient and activation energy for Cu in Si are

D(300ºC) = 7.8 x 10-11 m2/sQd = 41.5 kJ/mol

What is the diffusion coefficient at 350ºC?

⎟⎟⎠

⎞⎜⎜⎝

⎛−−==−∴

121

212

11lnlnln TTR

QDDDD d

T ⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

101

202

1lnlnand1lnlnTR

QDDRQDD dd

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Example (cont.)Example (cont.)

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ −

−= −

K 5731

K 6231

K-J/mol 314.8J/mol 500,41exp /s)m 10 x 8.7( 211

2D

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

1212

11exp TTR

QDD d

D2 = 15.7 x 10-11 m2/s

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2

2

xCD

tC

∂∂

=∂∂

NonNon--steady State Diffusionsteady State Diffusion

The concentration of diffusing The concentration of diffusing species is a function of both time species is a function of both time and position and position C C = = CC((xx,,tt))In this case In this case FickFick’’ss Second LawSecond Law is is usedused

Fick’s Second Law

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NonNon--steady State Diffusionsteady State Diffusion

B.C. at t = 0, C = Co for 0 ≤ x ≤ ∞

at t > 0, C = CS for x = 0 (const. surf. conc.)

C = Co for x = ∞

• Copper diffuses into a bar of aluminum.

pre-existing concentration, Co of copper atoms

Surface conc., C of Cu atoms bars

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Solution: Solution:

CC((xx,,tt) = Conc. at point ) = Conc. at point xx at time at time tt

erferf ((zz) = Gaussian ) = Gaussian error function error function

CS

Co

C(x,t)

( )⎟⎠

⎞⎜⎝

⎛−=−

−Dtx

CCCt,xC

os

o

2 erf1

dye yz 2

0

2 −∫π=

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NonNon--steady State Diffusionsteady State Diffusion

Sample Problem: An FCC ironSample Problem: An FCC iron--carbon alloy initially carbon alloy initially containing 0.20 wt% C is carburized at an elevated containing 0.20 wt% C is carburized at an elevated temperature and in an atmosphere that gives a temperature and in an atmosphere that gives a surface carbon concentration constant at 1.0 wt%. surface carbon concentration constant at 1.0 wt%. If after 49.5 h the concentration of carbon is 0.35 If after 49.5 h the concentration of carbon is 0.35 wt% at a position 4.0 mm below the surface, wt% at a position 4.0 mm below the surface, determine the temperature at which the treatment determine the temperature at which the treatment was carried out.was carried out.

Solution: use equationSolution: use equation ⎟⎠

⎞⎜⎝

⎛−=−

−Dtx

CCCtxC

os

o

2erf1),(

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Solution (cont.):

t = 49.5 h x = 4 x 10-3 mCx = 0.35 wt% Cs = 1.0 wt%Co = 0.20 wt%

⎟⎠

⎞⎜⎝

⎛−=−

−Dtx

CCC)t,x(C

os

o

2erf1

)(erf12

erf120.00.120.035.0),( z

Dtx

CCCtxC

os

o −=⎟⎠

⎞⎜⎝

⎛−=

−−

=−

−

∴ erf(z) = 0.8125

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Solution (cont.):

We must now determine from table the value of z for which the error function is 0.8125.

z erf(z)0.90 0.7970z 0.81250.95 0.8209

7970.08209.07970.08125.0

90.095.090.0

−−

=−

−z

z = 0.93

Now solve for D

Dtxz

2=

tzxD 2

2

4=

/sm 10 x 6.2s 3600

h 1h) 5.49()93.0()4(

m)10 x 4(4

2112

23

2

2−

−==⎟

⎟⎠

⎞⎜⎜⎝

⎛=∴

tzxD

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To solve for the temperature To solve for the temperature at which at which DD has above value, has above value, we use a rearranged form ofwe use a rearranged form of

)lnln( DDRQTo

d−

=

for diffusion of C in FCC FeDo = 2.3 x 10-5 m2/s Qd = 148,000 J/mol

/s)m 10x6.2 ln/sm 10x3.2 K)(ln-J/mol 314.8(J/mol 000,148

21125 −− −=T

T = 1300 K = 1027°C

D = Do exp⎛

⎝⎜

⎞

⎠⎟−

Qd

RT