Chapter 6

Diffusion in

Solids

Diffusion in solids

In general- Diffusion is a process by which: -- a matter is transported through another matter. OR -- atoms are moved through solids.

- Occurs at an elevated temperature, where atoms tend to move from high conc. region to low conc. region..- Atoms diffuse in solids if: 1. vacancies or other crystal defects are present. 2. there is enough activation energy.

- Atoms move into the vacancies present.

- Normally, at higher

temperature; -- more vacancies are

created. -- diffusion rate is

higher.

Example of diffusion… 1. Movement of smoke particles in air: -- very fast.2. Movement of dye in water: -- relatively slow.3. Movement of atom in solid (solid state reactions): -- very restricted movement due to bonding.

Diffusion in solids

type (classification)

mechanism (method)

state (condition)

industrial application

self-

diff

usi

on

inte

r-d

iffu

sio

n @

im

pu

rity

diff

usi

on

vaca

ncy

@

sub

stitu

tion

al

inte

rstit

ial

ste

ad

y st

ate

d

iffu

sio

n

no

n-s

tea

dy

sta

te

diff

usi

on

case

ha

rde

nin

g

do

pin

g s

ilico

n

Example: If atom ‘A’ has sufficient activation energy, it moves into the vacancy.

Activation energy of diffusion

Activation energy to form a vacancy

Activation energy to move a atom

= +

diffusing species

solid solid

diffusing speciesheat

Diffusion process…

Diffusion in solids

type (classification)

self-

diff

usi

on

Mec

hani

sm (

met

hod)

A

B

C

D

After some time

A

B

C

D

Initially After some timeInitially

inte

rstit

ial d

iffu

sio

n - atoms move from one interstitial

site to another.- the diffusing atoms that move must be much smaller than the host atom.- i.e: carbon interstitially diffuses into α-iron (BCC) @ -iron (FCC).

vaca

ncy

diff

usi

on

- atoms exchange with vacancies. - applies to substitutional

impurities atoms. - Rate of diffusion depends on: -- number of vacancies (vacancy concentration).. -- activation energy to exchange (frequency of jumping).

- atoms migrate itself in solid. -- change its position.

- diffusion occurs in 2 different solids.- atoms migrate between 2 solids.

Example: Cu-Ni alloy1. Join 2 metals (as diffusion couple).2. Heat (below Tm) & cool.3. After heat-treated, the couple: -- contains alloyed diffusion region. -- occurs diffusion process.

inte

r-d

iffu

sio

n

Interstitial diffusion more rapid than vacancy diffusion

Diffusion in solids

state (condition)

Steady state diffusion

How do we quantify the amount or rate of diffusion?

dt

dM

A

l

At

MJ

Measured empirically:1. make thin film (membrane) of

known surface area.2. impose concentration gradient.3. measure how fast atoms or

molecules diffuse through the membrane.

J = moles (or mass) diffusing surface area x time

molcm2s

kgm2s

unit: @

Rate of diffusion @ flux, J

M

t

J slope

- rate of diffusion independent of time

(@ does not change with time).- concentration of diffusing species depends with position.

C1

C2

x

C1

C2

x1 x2

concentration gradient

rate of diffusion

J dCdx

- apply Fick’s 1st law

J dCdx

= -D

D diffusion coefficientC surface concentrationx position

12

12 linear ifxx

CC

x

C

dx

dC

Example: 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?Given: D in butyl rubber = 110 x10-8 cm2/s surface concentrations:

C2 = 0.02 g/cm3

C1 = 0.44 g/cm3

Dtb 6

2

gloveC1

C2

skinpaintremover

x1 x2

Answer:assuming linear conc. gradient

J = 1.16 x 10-5 g/cm2s

J dCdx

= -D C2 – C1

x2 – x1

= -D

(0.02 g/cm3 – 0.44 g/cm3) (0.04 cm – 0)

= -(110x10-8 cm2/s)J

Diffusion in solids

state (condition)

Non-steady state diffusion

How do we quantify the amount or rate of diffusion?

dt

dM

A

l

At

MJ

Measured empirically:1. make thin film (membrane) of

known surface area.2. impose concentration gradient.3. measure how fast atoms or

molecules diffuse through the membrane.

J = moles (or mass) diffusing surface area x time

molcm2s

kgm2s

unit: @

Rate of diffusion @ flux, J

M

t

J slope

- occur for most practical diffusion situations. - rate of diffusion & concentration gradient vary with time & position. - concentration of diffusing species

is a function of both time and position. C = C(x,t)- apply Fick’s 2nd law

2

2

x

CD

t

C

Copper diffuses into a bar of aluminum.

Example:

before diffusion

after some time

during diffusion

Al barCoCS

C(x,t)

- boundary conditions (B. C)

at t = 0, C = Co for 0 x

at t > 0, C = CS for x = 0 (constant surface conc.)

C = C(x,t) for x = xt = x2

C = Co for x =

Dt

x

CC

Ct,xC

os

o

2 erf1

Co = pre-existing concentration

CS = surface concentration

C(x,t) = concentration at depth x after

time t

where;

- B. C are related to equation:

Dt

xzerf

2 erf)(

x1 x2

erf (z) = error function

Note: erf(z) values are given in Table 6.1

Position, x

Co

C(x,t)

Cs

time = t2

time= t1

time = to

x

Diffusion in solids

state (condition)

Non-steady state diffusion

Table 6.1: Tabulation of Error Function Values

zerf (z) z erf (z) z erf (z)

0 0 0.55 0.5633 1.3 0.9340

0.025 0.0282 0.60 0.6039 1.4 0.9523

0.05 0.0564 0.65 0.6420 1.5 0.9661

0.10 0.1125 0.70 0.6778 1.6 0.9763

0.15 0.1680 0.75 0.7112 1.7 0.9838

0.20 0.2227 0.80 0.7421 1.8 0.9891

0.25 0.2763 0.85 0.7707 1.9 0.9928

0.30 0.3286 0.90 0.7970 2.0 0.9953

0.35 0.3794 0.95 0.8209 2.2 0.9981

0.40 0.4284 1.0 0.8427 2.4 0.9993

0.45 0.4755 1.1 0.8802 2.6 0.9998

0.50 0.5205 1.2 0.9103 2.8 0.9999

Diffusion in solids

state (condition)

Non-steady state diffusion

Example:

Consider one such alloy that initially has a uniform carbon concentration of 0.25 wt% and is to be treated at 950 oC. If the concentration of carbon at the surface is suddenly brought to & maintained at 1.20 wt%, how long will it take to achieve a carbon content of 0.80 wt% at a position 0.5 mm below the surface? The D for carbon in iron at this temperature is 1.6 x 10-11 m2/s.

Dt

x

CC

CtxC

os

o

2erf1

),(

Answer:

x = 0.5 mm = 5 x 10-4 mC(x,t) = 0.80 wt% CCs = 1.20 wt% CCo = 0.25 wt% CD = 1.6 x 10-11 m2/s

Given;

erf (z) = 0.4210

z erf (z)

0.35z0.40

0.37940.42100.4284

C(x,t) – Co = 1 – erf (z)

Cs – Co

0.80 – 0.25 = 1 – erf (z)

1.20 – 0.25

z – 0.35 = 0.4210 – 0.37940.40 – 0.35 0.4284 – 0.3794

z = 0.392

use table 6.1 to find z value

An interpolation is necessary as follows

x = z 2 (Dt)1/2

Therefore,

5 x 10-4 m = 0.392

2 [(1.6 x 10-11 m2/s)(t)]1/2

t = 25400 s = 7.1 h

- diffusion coefficient, D increases with increasing T.

- apply the formula:

= 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

D = Do exp æ

è ç

ö

ø-

Qd

RT

Diffusion in solids

Factors that influence diffusion

Temperature

D interstitial >> Dsubstitutional

C in a-Fe

C in g-Fe

Al in Al

Fe in a-Fe

Fe in g-Fe

1000 K/T

D (m2/s) C in a-Fe

C in g-Fe

Al in Al

Fe in a-Fe

Fe in g-Fe

0.5 1.0 1.510-20

10-14

10-8

T(C)

15

00

10

00

60

0

30

0

D vs T for several metals…

Diffusing species

- rearrange;

ln D = ln Do – Qd R (T)

Note: Diffusion data values are given in Table 6.2

Smaller diffusing species, diffusion faster.

The above formula can be used for steady state & non-steady state diffusion

Diffusion in solids

Factors that influence diffusion

Temperature Diffusing species

Diffusing Species

Host Metal

Do (m2/s)

Activation Energy Qd Calculated Values

kJ/mol eV/atom T (oC) D (m2/s)

Feα-Fe (BCC)

2.8 x 10-4 251 2.60 500 3.0 x 10-21

900 1.8 x 10-15

Fe-Fe (FCC)

5.0 x 10-5 284 2.94 900 1.1 x 10-17

1100 7.8 x 10-16

C α-Fe 6.2 x 10-7 80 0.83 500 2.4 x 10-12

900 1.7 x 10-10

C -Fe 2.3 x 10-5 148 1.53 900 5.9 x 10-12

1100 5.3 x 10-11

Cu Cu 7.8 x 10-5 211 2.19 500 4.2 x 10-19

Zn Cu 2.4 x 10-5 189 1.96 500 4.0 x 10-18

Al Al 2.3 x 10-4 144 1.49 500 4.2 x 10-14

Cu Al 6.5 x 10-5 136 1.41 500 4.1 x 10-14

Mg Al 1.2 x 10-4 131 1.35 500 1.9 x 10-13

Cu Ni 2.7 x 10-5 256 2.65 500 1.3 x 10-22

Table 6.2: Tabulation of diffusion data

Example 1: At 300ºC the diffusion coefficient and activation energy for Cu in Si are 7.8 x 10-11 m2/s and 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 æ

è ç

ö

ø-

Qd

RT

Diffusion in solids

Factors that influence diffusion

Temperature Diffusing species

Answer:

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 = 573 K

T2 = 273 + 350 = 623 K

D2 = 15.7 x 10-11 m2/s

Diffusion in solids

Example 2: (for non-steady state diffusion application)

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.

Dt

x

CC

CtxC

os

o

2erf1

),(

Answer:

Factors that influence diffusion

Temperature Diffusing species

)(erf12

erf120.00.1

20.035.0),(z

Dt

x

CC

CtxC

os

o

t = 49.5 h x = 4 x 10-3 mC(x,t) = 0.35 wt%Cs = 1.0 wt%Co = 0.20 wt%

Given;

erf(z) = 0.8125

We must now determine from Table 6.1 the value of z for which the error function is 0.8125. An interpolation is necessary as follows

z erf(z)

0.90 0.7970 z 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

from Table 6.2, for diffusion of Carbon in FCC Fe

Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol

)lnln( DDR

QT

o

d

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

J/mol 000,14821125

T

T = 1300 K = 1027°C

Diffusion in solids

Example 3:

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 breakthrough time (tb), i.e., how long could the gloves be used before methylene chloride reaches the hand?

Answer:

Factors that influence diffusion

Temperature Diffusing species

Given;D in butyl rubber = 110 x10-8 cm2/s

assuming linear conc. gradient

Dtb 6

2

gloveC1

C2

skinpaintremover

x1 x2

Dtb 6

2

Breakthrough time = tb

Time required for breakthrough ca. 4 mincm 0.04 12 xx

min 4 s 240/s)cm 10 x 110)(6(

cm) 04.0(28-

2bt

Diffusion in solids

Indu

stria

l app

licat

ion C

ase

har

deni

ng

Dop

ing

silic

on w

afe

r- sliding and rotating parts needs to have hard surfaces. - these parts are usually machined with low carbon steel as they are easy to machine.- their surface is then hardened by carburizing. - steel parts are placed at elevated temperature (927oC) in an atmosphere of hydrocarbon gas (CH4).- carbon diffuses into iron (steel) surface and

fills interstitial space to make it harder.

Low carbonsteel part

Diffusing carbonatoms

- Doping silicon with phosphorus, P for n-type semiconductors.

- impurities (P) are made to diffuse into silicon wafer to change

its electrical characteristics. - used in integrated circuits.- silicon wafer is exposed to vapor of impurity at 1100oC in a quartz tube furnace. - the concentration of impurity at any point depends on depth and time of exposure.

1. Deposit P rich layers on surface.

silicon silicon

2. heat

Doping process…3. Result: Doped semiconductorregions.

Diffusion FASTER for...

• open crystal structures

• materials w/secondary bonding

• smaller diffusing atoms

• lower density materials

Diffusion SLOWER for...

• close-packed structures

• materials w/covalent bonding

• larger diffusing atoms

• higher density materials

Summary

End of Chapter

6