materials science

Materials Science

ME 274

Dr Yehia M. Youssef

1Copyright © YM Youssef, 4-Oct-10 Materials Science

Chapter 5: Diffusion in Solids

ISSUES TO ADDRESS...

• How does diffusion occur?

• Why is it an important part of processing?

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

• How does diffusion depend on structureand temperature?


Diffusion

Diffusion M t t b t i tiDiffusion - Mass transport by atomic motion

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


Diffusion

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

Initially After some time

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


Diffusion

• Self-diffusion: In an elemental solid, atomsalso migrate.also migrate.

Label some atoms After some timeC

AC

A

C

D

BD

AB


Diffusion MechanismsVacancy Diffusion:

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

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

increasing elapsed time


Diffusion Simulation

• Simulation of Simulation of interdiffusionacross an interface:

• Rate of substitutionaldiffusion depends on:diffusion depends on:--vacancy concentration--frequency of jumping.

(Courtesy P.M. Anderson)


Diffusion Mechanisms

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

Adapted from Fig 5 3 (b) Callister 7e

More rapid than vacancy diffusionAdapted from Fig. 5.3 (b), Callister 7e.


Processing Using Diffusion

Adapted from• Case Hardening:

Diffuse carbon atoms Adapted from chapter-opening photograph, Chapter 5, Callister 7e. (C t f

--Diffuse carbon atomsinto the host iron atomsat the surface.

(Courtesy ofSurface Division, Midland-Ross.)

--Example of interstitialdiffusion is a casehardened gearhardened gear.

• Result: The presence of C t k i ( t l) h datoms makes iron (steel) harder.


D i ili ith h h f t i d tProcessing Using Diffusion

• Doping silicon with phosphorus for n-type semiconductors:• Process: 0.5mm

1 Deposit P rich

magnified image of a computer chip

1. Deposit P richlayers on surface.

magnified image of a computer chip

silicon

3. Result: Dopedsemiconductor

light regions: Si atoms

2. Heat it.

semiconductorregions.

siliconlight regions: Al atoms

Adapted from chapter-opening photograph,

10

Chapter 18, Callister 7e.

Copyright © YM Youssef, 4-Oct-10 Materials Science

Diffusion

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

( )( ) smkgor

scmmol

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

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

membranemembrane

dMlMM =

mass J ∝ slope

dtdM

Al

AtMJ == mass

diffusedtime

J ∝ slope


Steady-State Diffusion

Rate of diffusion independent of timeFlux proportional to concentration gradient =

dCFlux proportional to concentration gradient =

dx

dC

Fick’s first law of diffusionC1C1

dxdCDJ −=C2C2

xx1 x2

D ≡ diffusion coefficientD ≡ diffusion coefficient

12

12 linear ifxxCC

xC

dxdC

−−

=ΔΔ

≅


Example: Chemical Protective Clothing (CPC)(CPC)

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

• 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

13

2 gCopyright © YM Youssef, 4-Oct-10 Materials Science

Example (cont).

• Solution – assuming linear conc. gradient

12

12- xxCCD

dxdCDJ

−−

−≅=D

tb 6

2=

gloveC1

i t 12 xxdx −D6

C2

skinpaintremover

D = 110x10-8 cm2/s Data:x1 x2

C2 = 0.02 g/cm3

C1 = 0.44 g/cm3

x x = 0 04 cm

g10161)g/cm 44.0g/cm02.0(/ )10110( 5-33

28- −J

x2 – x1 = 0.04 cm

scmg10x 16.1

cm) 04.0()gg(/s)cm10 x 110( 2

528 =−=J


Diffusion and Temperature

• Diffusion coefficient increases with increasing T. Diffusion coefficient increases with increasing T.

QD = Do exp⎛

⎝ ⎞ ⎠

−Qd

RT

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

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

o

Qd

R= absolute temperature [K]T

15Materials ScienceCopyright © YM Youssef, 4-Oct-10

Diffusion and Temperature

D has exponential dependence on T

10-8T(°C)15

00

1000

600

300

Dinterstitial >> DsubstitutionalD (m2/s)

0

C in α-FeC in γ-Fe

Al in AlFe in α-FeFe in γ-Fe

10-14

1000K/T0 5 1 0 1 510-20

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

1000K/T0.5 1.0 1.5

16

( ) ,ed., Butterworth-Heinemann, Oxford, 1992.)


Example: At 300ºC the diffusion coefficient and ti ti f C i Siactivation 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?

transform D ln Ddata

Temp = T 1/T

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−=

101

202

1lnln and 1lnlnTR

QDDTR

QDD dd

Temp = T 1/T

⎠⎝⎠⎝ 12

⎟⎟⎠

⎞⎜⎜⎝

⎛−−==−∴

121

212

11lnlnln TTR

QDDDD d

17

⎠⎝ 121 TTRDCopyright © YM Youssef, 4-Oct-10 Materials Science

Example (cont.)

⎥⎤

⎢⎡

⎟⎟⎞

⎜⎜⎛

−−=11exp QDD d ⎥

⎦⎢⎣

⎟⎟⎠

⎜⎜⎝

=12

12 exp TTR

DD

T1 = 273 + 300 = 573KT2 = 273 + 350 = 623K

⎥⎤

⎢⎡

⎟⎞

⎜⎛−

= − 11J/mol 500,41exp/s)m10x87( 211D

2

⎥⎦

⎢⎣

⎟⎠

⎜⎝

−=K 573K 623K-J/mol 314.8

exp/s)m10 x 8.7(2D

D2 = 15.7 x 10-11 m2/s


Non-steady State Diffusion

• The concentration of diffucing species is a function of• The concentration of diffucing species is a function of both time and position C = C(x,t)

• In this case Fick’s Second Law is used

2

2

xCD

tC

∂∂

=∂∂Fick’s Second Law

xt ∂∂



• Copper diffuses into a bar of aluminum.Surface conc.,

barpre-existing conc., Co of copper atoms C of Cu atoms bars

Cs

Adapted from Fig. 5.5,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 = C for x = ∞20

C = Co for x = ∞Copyright © YM Youssef, 4-Oct-10 Materials Science

Solution

( )⎟⎞

⎜⎛−=

− xCt,xC o erf1

C(x t) = Conc at point x at

⎟⎠

⎜⎝− DtCC os 2

erf1

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

erf (z) = error function

CS

( )

C(x,t)dye yz 2

0

2 −∫=

erf(z) values are given in Table 5 1

Co

0∫π

Table 5.1



• Sample Problem: An FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated g %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 carriedthe temperature at which the treatment was carried out.

• Solution: use Eqn. 5.5 ⎟⎠

⎞⎜⎝

⎛−=−

−Dtx

CCCtxC

os

o

2erf1),(

⎠⎝ Dtos 2


Solution (cont.): ⎟⎠

⎞⎜⎝

⎛−=−

−Dtx

CCC)t,x(C

os

o

2erf1

t 49 5 h 4 10 3

⎠⎝ Dtos 2

– t = 49.5 h x = 4 x 10-3 m– Cx = 0.35 wt% Cs = 1.0 wt%

C 0 20 t%– Co = 0.20 wt%

200350)( xCtxC o ⎟⎞

⎜⎛−− )(erf1

2erf1

20.00.120.035.0),( z

Dtx

CCCtxC

os

o −=⎟⎠

⎞⎜⎝

⎛−=

−=

−

∴ erf(z) = 0.8125


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 followsp y

z erf(z)79700820907970.08125.0

90095090.0 −

=−z

0.90 0.7970z 0.81250.95 0.8209

7970.08209.090.095.0 −−

z = 0.930.95 0.8209

Now solve for D x xD2

Dtxz

2=

tzxD 24

=

/sm 10 x 6.2s3600

h 1h)549()930()4(

m)10 x 4(4

2112

23

2

2−

−==⎟

⎟⎠

⎞⎜⎜⎝

⎛=∴

tzxD

24

s3600h)5.49()93.0()4(4 ⎟⎠

⎜⎝ tz


Solution (cont.):

• To solve for the temperature at which D has above value, we )lnln( DDR

QT d−

=c as abo e a ue, euse a rearranged form of Equation (5.9a);

)lnln( DDR o −

from Table 5.2, for diffusion of C in FCC Fe

D = 2 3 x 10-5 m2/s Qd = 148 000 J/molDo 2.3 x 10 m /s Qd 148,000 J/mol

J/mol000148/s)m 10x6.2 ln/sm 10x3.2 K)(ln-J/mol 314.8(

J/mol 000,14821125 −− −

=T∴

T = 1300 K = 1027°C


Example: Chemical Protective Clothing (CPC)(CPC)

• Methylene chloride is a common ingredient of paint removers. Besides being an irritant it also may be absorbed through skinBesides being an irritant, it also may be absorbed through skin. When using this paint remover, protective gloves should be worn.If b t l bb l (0 04 thi k) d h t i th• If butyl rubber gloves (0.04 cm thick) are used, what is the breakthrough time (tb), i.e., how long could the gloves be used before methylene chloride reaches the hand?

• Data (from Table 22.5)– diffusion coefficient in butyl rubber:

D 110 10 8 2/D = 110x10-8 cm2/s


Example (cont).

• Solution – assuming linear conc. gradientglove

C1

i t Dtb 6

2= Equation 22.24

C2

skinpaintremover

D6

cm0 0412 == xxx1 x2cm0.0412 =−= xx

D = 110x10-8 cm2/s

min 4 s 240/ )10110)(6(

cm) 04.0(28-

2===bt

Time required for breakthrough ca. 4 min

/s)cm10x 110)(6( 28

27

q gCopyright © YM Youssef, 4-Oct-10 Materials Science

Summary

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

• open crystal structures • close-packed structures

• materials w/secondarybonding

• materials w/covalentbonding

• smaller diffusing atoms • larger diffusing atoms

• lower density materials • higher density materials


materials science

Documents