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Chapter 5-

ISSUES TO ADDRESS...

• How does diffusion occur?

• Why is it an important part of processing?

• How can the rate of diffusion be predicted forsome simple cases?

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• How does diffusion depend on structureand temperature?

CHAPTER 5:DIFFUSION IN SOLIDS

Chapter 5-

Diffusion

Diffusion - Mass transport by atomic motion

Mechanisms• Gases & Liquids – random (Brownian) motion• Solids – vacancy diffusion or interstitial

diffusion

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Chapter 5-

Why Study Diffusion ?

• Diffusion plays a crucial role in…– Alloying metals => bronze, silver, gold

– Strengthening and heat treatment processes• Hardening the surfaces of steel

– High temperature mechanical behavior

– Phase transformations• Mass transport during FCC to BCC

– Environmental degradation• Corrosion, etc.

Chapter 5- 2

• Glass tube filled with water.• At time t = 0, add some drops of ink to one end

of the tube.• Measure the diffusion distance, x, over some time.• Compare the results with theory.

to

t1

t2

t3

xo x1 x2 x3

time (s)

x (mm)

DIFFUSION DEMO

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Chapter 5-

How do atoms move in Solids ?Why do atoms move in Solids ?

• Diffusion, simply, is atoms moving from one lattice site to another in a stepwise manner– Transport of material by moving atoms

• Two conditions are to be met:– An empty adjacent site

– Enough energy to break bonds and cause lattice distortions during displacement

• What is the energy source ?– HEAT !

• What else ?– Concentration gradient !

Chapter 5-

• Case Hardening:--Diffuse carbon atoms

into the host iron atomsat the surface.

--Example of interstitialdiffusion is a casehardened gear.

• Result: The "Case" is--hard to deform: C atoms

"lock" planes from shearing.--hard to crack: C atoms put

the surface in compression.

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Fig. 5.0, Callister 6e.(Fig. 5.0 iscourtesy ofSurface Division, Midland-Ross.)

PROCESSING USING DIFFUSION

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Chapter 5-

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

3. Result: Dopedsemiconductorregions.

silicon

Processing Using Diffusion

magnified image of a computer chip

0.5mm

light regions: Si atoms

light regions: Al atoms

2. Heat it.

1. Deposit P richlayers on surface.

silicon

Adapted from chapter-opening photograph, Chapter 18, Callister 7e.

Chapter 5-

• Interdiffusion: In an alloy, atoms tend to migratefrom regions of high conc. to regions of low conc.

Initially

Adapted from Figs. 5.1 and 5.2, Callister 7e.

Diffusion

After some time

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Chapter 5-

Chapter 5-

Diffusion Mechanisms (I)

Energy is needed to generate a vacancy, break bonds, cause distortions. Provided by HEAT , kT !Atom moves in the opposite direction of the vacancy !

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Chapter 5-

Diffusion Mechanisms (II)

Much faster than vacancy diffusion, why ? Smaller atoms like B, C, H, O. Weaker interaction with the larger atoms. More vacant sites, no need to create a vacancy !

Interstitial Diffusion

Chapter 5-

Diffusion• How do we quantify the amount or rate of diffusion?

• Measured empirically– Make thin film (membrane) of known surface area– Impose concentration gradient– Measure how fast atoms or molecules diffuse through the

membrane

sm

kgor

scm

mol

timearea surface

diffusing mass) (or molesFlux

22J

dt

dM

A

l

At

MJ M =

massdiffused

time

J slope

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Chapter 5-

• Concentration Profile, C(x): [kg/m3]

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• Fick's First Law:

Concentration

of Cu [kg/m 3]Concentration

of Ni [kg/m 3]

Position, x

Cu flux Ni flux

• The steeper the concentration profile, the greater the flux!Concentration gradient is the DRIVING FORCE !

Adapted from Fig. 5.2(c), Callister 6e.

J x D

dC

dx

Diffusion coefficient [m 2/s]

concentration

gradient [kg/m 4]

flux in x-dir.

[kg/m 2-s]

CONCENTRATION PROFILES & FLUX

Chapter 5-

Temperature Dependency !

Atom has to break bonds and “squeeze” thru => activation energy, Em ≈ 1 eV .

What is the probability to find a vacancy at a nearest site ?

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Chapter 5-

Temperature Effect !

The diffusion depends on temperature because;a- # of vacancies in the vicinityb- thermally activated successful jumps

Chapter 5-

Diffusion 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]

D

Do

Qd

R

T

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Chapter 5-

Diffusion and Temperature

Adapted from Fig. 5.7, Callister 7e. (Date for Fig. 5.7 taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)

D has exponential dependence on T

Dinterstitial >> Dsubstitutional

C in -FeC in -Fe

Al in AlFe in -FeFe in -Fe

1000K/T

D (m2/s)

0.5 1.0 1.510-20

10-14

10-8T(C)

1500

1000

600

300

Chapter 5-

• Diffusivity increases with T.

• Experimental Data:

1000K/T

D (m2/s) C in -Fe

C in -Fe

Al in A

l

Cu in Cu

Zn in CuFe in -Fe

Fe in -Fe

0.5 1.0 1.5 2.010 -20

10 -14

10 -8T(C)15

00

1000

600

300

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pre-exponential [m 2/s] (see Table 5.2, Callister 6e )activation energy

gas constant [8.31J/mol-K]

D Do exp QdRT

diffusivity

[J/mol],[eV/mol] (see Table 5.2, Callister 6e )

Dinterstitial >> Dsubstitutional

C in -FeC in -Fe Al in Al

Cu in Cu

Zn in Cu

Fe in -FeFe in -Fe

Adapted from Fig. 5.7, Callister 6e. (Date for Fig. 5.7 taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)

DIFFUSION AND TEMPERATURE

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Chapter 5-

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?

101

202

1lnln and

1lnln

TR

QDD

TR

QDD dd

121

212

11lnlnln

TTR

Q

D

DDD d

transform data

D

Temp = T

ln D

1/T

D Do exp

Qd

RT

Chapter 5-

Example (cont.)

K 573

1

K 623

1

K-J/mol 314.8

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

2D

1212

11exp

TTR

QDD d

T1 = 273 + 300 = 573K

T2 = 273 + 350 = 623K

D2 = 15.7 x 10-11 m2/s

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Chapter 5-

Concentration Gradient

Chapter 5-

• Steady State: the concentration profile doesn't change with time.

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• Apply Fick's First Law:

• Result: the slope, dC/dx, must be constant(i.e., slope doesn't vary with position)!

J x(left) = J x(right)

Steady State:

Concentration, C, in the box doesn’t change w/time.

J x(right)J x(left)

x

Jx D

dC

dx

dC

dx

left

dC

dx

right

• If Jx)left = Jx)right , then

STEADY STATE DIFFUSION

Why is the minus sign ?

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Chapter 5-

• Steel plate at 700º C

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• Q: How much carbon transfers from the rich to the deficient side?

J DC2 C1

x2 x1

2 .4 10 9 kg

m2s

Adapted from Fig. 5.4, Callister 6e.

C 1 = 1.2kg/m3

C 2 = 0.8kg/m3

Carbon rich gas

10mm

Carbon deficient

gas

x1 x205mm

D=3x10 -11 m2/s

Steady State = straight line!

EX: STEADY STATE DIFFUSION

Chapter 5 -

HW: Chemical Protective Clothing (CPC)

• Methylene chloride is a common ingredient of paint removers. Besides being an irritant, it also may be absorbed through skin. When using this paint remover, protective gloves should be worn.

• If butyl rubber gloves (0.04 cm thick) are used, what is the diffusive flux of methylene chloride through the glove?

• Data:

– diffusion coefficient in butyl rubber: D = 110x10-8 cm2/s

– surface concentrations:

C2 = 0.02 g/cm3

C1 = 0.44 g/cm3

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Chapter 5 -

scm

g 10 x 16.1

cm) 04.0(

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

25-

3328-

J

Example (cont).

12

12- xx

CCD

dx

dCDJ

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:

Chapter 5-

Fick’s Second Law ; Non-steady state Diffusion

• In most practical cases, J (flux) and dC/dx (concentration gradient) change with time (t).– Net accumulation or depletion of species diffusing

• How do we express a time dependent concentration?

Concentration at a point xChanging with time

? Flux, J, changes at any point x !

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Chapter 5-

How do we solve this partial differential equation ?

• Use proper boundary conditions:– t=0, C = C0, at 0 ≤ x ≤ ∞

– t>0, C = Cs, at x = 0

C = C0, at x = ∞

Chapter 5 -

Non-steady State Diffusion

Adapted from Fig. 5.5, Callister 7e.

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 conc., Co of copper atoms

Surface conc., C of Cu atoms bar

s

Cs

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Chapter 5 -

• Copper diffuses into a bar of aluminum.

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• General solution:

“Gaussian error function"

pre-existing conc., C o of copper atoms

Surface conc., Cs of Cu atoms bar

Co

Cs

position, x

C(x,t)

tot1

t2t3 Adapted from

Fig. 5.5, Callister 6e.

NON STEADY STATE DIFFUSION

C(x, t ) Co

Cs Co

1 erfx

2 Dt

C (x, t) = concentration at any time and position !

Chapter 5 -

• Concentration profile,C(x), changesw/ time.

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• To conserve matter: • Fick's First Law:

• Governing Eqn.:

Concentration, C, in the box

J (right)J (left)

dx

d C

dt= D

d 2C

dx 2

dx

d C

dtJ D

d C

dxor

J (left)J (right)

d J

dx

d C

dt

d J

dx D

d2 C

dx 2

(if D does not vary with x)

equate

NON STEADY STATE DIFFUSION

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Chapter 5 -

Solution:

C(x,t) = Conc. at point x at time t

erf (z) = error function

erf(z) values are given in Table 5.1

CS

Co

C(x,t)

Dt

x

CC

Ct,xC

os

o

2 erf1

dye yz 2

0

2

Chapter 5 -

Non-steady State Diffusion• Sample Problem: An FCC iron-carbon alloy initially containing

0.20 wt% C is carburized at an elevated temperature and in an atmosphere that gives a surface carbon concentration constant 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, determine the temperature at which the treatment was carried out.

• Solution: use Eqn. 5.5

• Determine z value from Table 5.1

Dt

x

CC

CtxC

os

o

2erf1

),(

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Chapter 5 -

Solution (cont.):

– t = 49.5 h x = 4 x 10-3 m

– Cx = 0.35 wt% Cs = 1.0 wt%

– Co = 0.20 wt%

Dt

x

CC

C)t,x(C

os

o

2erf1

)(erf12

erf120.00.1

20.035.0),(z

Dt

x

CC

CtxC

os

o

erf(z) = 0.8125

Chapter 5 -

Solution (cont.):

We must now determine from Table 5.1 the value of z for which the error function is 0.8125. An interpolation is necessary as follows

z erf(z)

0.90 0.7970z 0.81250.95 0.8209

7970.08209.0

7970.08125.0

90.095.0

90.0

z

z 0.93

Now solve for D

Dt

xz

2

tz

xD

2

2

4

/sm 10 x 6.2s 3600

h 1

h) 5.49()93.0()4(

m)10 x 4(

4

2112

23

2

2

tz

xD

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Chapter 5 -

• To solve for the temperature at which D has above value, we use a rearranged form of Equation (5.9a);

)lnln( DDR

QT

o

d

from Table 5.2, for diffusion of C in FCC Fe

Do = 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 000,14821125

T

Solution (cont.):

T = 1300 K = 1027°C

Chapter 5-

• Copper diffuses into a bar of aluminum.• 10 hours at 600C gives desired C(x).• How many hours would it take to get the same C(x)

if we processed at 500C?

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(Dt) 500ºC =(Dt) 600ºCs

C (x, t ) CoC Co

= 1 erfx

2Dt

• Result: Dt should be held constant.

• Answer:Note: valuesof D areprovided here.

Key point 1: C(x,t500C) = C(x,t600C).Key point 2: Both cases have the same Co and Cs.

t 500 (Dt )600

D500

110 hr

4.8x10 -14 m2/s

5.3 x10 -13 m2/s 10hrs

Example 5.3

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Chapter 5-

Size Impact on Diffusion

Smaller atoms diffuse faster

Chapter 5-

Fast Tracks for diffusion !eg. self-diffusion of Ag :-Areas where lattice is pre-strained can allow for faster diffusion of atoms-Less energy is needed to distort an already strained lattice !

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Chapter 5-

Important

• Temperature - diffusion rate increases with increasing temperature (WHY ?)

• Diffusion mechanism – interstitials diffuse faster (WHY ?)

• Diffusing and host species - Do, Qd is different forevery solute - solvent pair

• Microstructure - grain boundaries and dislocationcores provide faster pathways for diffusing species, hence diffusion is faster in polycrystalline vs. single crystal materials. (WHY ?)

Chapter 5- 20

Diffusion FASTER for...

• open crystal structures

• lower melting T materials

• materials w/secondarybonding

• smaller diffusing atoms

• cations

• lower density materials

Diffusion SLOWER for...

• close-packed structures

• higher melting T materials

• materials w/covalentbonding

• larger diffusing atoms

• anions

• higher density materials

SUMMARY:STRUCTURE & DIFFUSION

WHY ?

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Chapter 5-

HW 6:

Solve EXAMPLE PROBLEMS: 5.1, 5.2, 5.3, 5.4, 5.5

HW 7: Chemical Protective Clothing (CPC)

Chapter 5-

QUIZ 4

a.Compare interstitial and vacancy atomic mechanisms for diffusion.

b.Which one is more rapid?

c.Why?