Chapter 5: Diffusion
Chapter 5: Diffusion
ISSUES TO ADDRESS...
• What is diffusion?
• By what atomic mechanisms does diffusion occur?
• What are examples of diffusion in materials processing?
• What equations do we use to solve diffusion problems?
• How does the rate of diffusion depend on
temperature?
Chapter 5 - 1
Diffusion
Diffusion - Mass transport by atomic motion
• ______________________
- Gases & Liquids – random (___________) motion
- Solids – ______ diffusion and _________ diffusion
• Interdiffusion - diffusion of atoms of one material into
another material
• Self-diffusion – atomic migration in a pure metal
Chapter 5 - 2
Diffusion
• Atoms tend to _________ from regions of ____ concentration to
regions of ____ concentration.
Before diffusion
After diffusion
Figs. 5.1,
Callister &
Rethwisch 10e.
Concentration Profiles
Chapter 5 - 3
1
Chapter 5: Diffusion
Diffusion
• Self-diffusion: Migration of __________ in pure metals
Locations of 4 labeled
atoms before diffusion
Locations of 4 labeled
atoms after diffusion
C
C
A
D
A
D
B
B
Chapter 5 - 4
Diffusion Mechanism I
Vacancy Diffusion
• ___________________ exchange positions
• applies to host and ___________________ impurity atoms
• diffusion rate depends on:
-- number of ___________
-- activation _________ to exchange.
increasing elapsed time
Chapter 5 - 5
Diffusion Mechanism II
____________ Diffusion
• Small, _____________ atoms move from one
interstitial position to an adjacent one
Fig. 5.2 (b), Callister & Rethwisch 10e.
More rapid than ___________ diffusion
Chapter 5 - 6
2
Chapter 5: Diffusion
Processing Using Diffusion
Case hardened
region
• Case _______________:
- Example of ____________ diffusion
- Outer surface selectively hardened by
diffusing carbon atoms into surface
- Presence of C atoms makes
iron (steel) harder
• Example: __________________ gear
- Case hardening improves wear
resistance of gear
- Resulting residual compressive
stresses improve resistance to
fatigue failure
Chapter-opening photograph, Chapter 5,
Callister & Rethwisch 10e. (Courtesy of
Surface Division, Midland-Ross.)
Chapter 5 - 7
Processing Using Diffusion
Diffusion in Semiconducting Devices
• _______ – Diffusion of very small concentrations of atoms
of an impurity (e.g., P) into the semiconductor silicon.
• Process:
1. Deposit ___ rich
layers on surface
2. __________
the sample to
drive in P
3. Result is P doped
______________
regions
silicon
silicon
Chapter 5 - 8
Rate of Diffusion
• Diffusion is a time-dependent process.
• Rate of Diffusion- expressed as diffusion flux, J
mass of diffused species M ⎛ kg ⎞
J ≡ Flux ≡
=
⎜
⎟
(area)(time)
At ⎝ m2 -s ⎠
• Measured experimentally
– Use thin sheet (______________) – cross-sectional area A
– Impose _________________ gradient across sheet
– Measure mass of diffusing species (M) that passes through
the sheet over time period (t)
J=
M
l dM
=
At A dt
M=
mass
diffused
J ∝ slope
time
Chapter 5 - 9
3
Chapter 5: Diffusion
Steady-State Diffusion
Rate of ___________ (or flux) independent of time
Flux (J) proportional to ___________ gradient: J ∝
dC
dx
C = concentration
x = diffusion direction
C1
C1
____________ of diffusion
C
C2
x1
if linear
x
C2
J = −D
x2
dC ΔC C2 − C1
≅
=
dx Δx x 2 − x1
dC
dx
D = diffusion coefficient
Chapter 5 - 10
Diffusion Example
Chemical Protective Clothing (CPC)
• ________________ 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.
• Lets investigate whether butyl rubber gloves (_____ cm
thick) commonly found in the _______ can be used as
protective gloves.
• Note: The maximum allowable flux for a 150 lb person is
less than _______________________
• Compute the diffusion flux of methylene chloride
through the gloves.
Chapter 5 - 11
CPC Example (cont.)
• Solution – diffusion flux of methylene chloride
assume __________ conc. gradient
glove
C1
paint
remover
2
tb =
6D
skin
C2
x1 x2
J = − (110 x 10-8 cm2 /s)
J = −D
C − C1
dC
≅ −D 2
dx
x 2 − x1
Data:
D = ____________
C1 = _____________
C2 = 0.02 g/cm3
x2 – x1 = 0.04 cm
(0.02 g/cm3 − 0.44 g/cm3 )
g
= 1.16 x 10-5
(0.04 cm)
cm2 -s
Note: this is more than 30 times the allowable flux.
Unsafe to use these gloves for paint removal.
Chapter 5 - 12
4
Chapter 5: Diffusion
Influence of Temperature on Diffusion
• Diffusion coefficient increases with increasing _____
D = Do exp -
Qd
RT
D = __________ coefficient [m2/s]
Do = pre-exponential [m2/s]
Qd = __________ energy [J/mol]
R = gas constant [____________]
T = absolute temperature [K]
Chapter 5 - 13
Influence of Temperature on Diffusion
(cont.)
D
transform
data
ln D
Temp = T
1/T
⎛ Q ⎞
D = D0 exp ⎜ − d ⎟
⎝ RT ⎠
take natural log
of both sides
lnD = lnD0 −
Qd
RT
Chapter 5 - 14
Influence of Temperature on Diffusion
(cont.)
300
600
1000
10-8
1500
D has _________________ dependence on T
T(°C)
C
in
D (m2/s)
C in α-Fe
C in γ-Fe
Al
1.0
Dinterstitial >> Dsubstitutional
in
0.5
α-F
e
Al
-Fe
10-20
Fe
α-
γ
in
Fe
n
Fe Fe i
γ-
10-14
C in
1.5
Al in Al
Fe in α-Fe
Fe in γ-Fe
1000 K/T
Adapted from Fig. 5.6, Callister & Rethwisch 10e.
(Data for Fig. 5.7 taken from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals
Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)
Chapter 5 - 15
5
Chapter 5: Diffusion
VMSE: Screenshot of Diffusion Plots
Chapter 5 - 16
Influence of Temperature on Diffusion (cont.)
Derive an equation relating the ________ coefficients at
two temperature T1 and T2 using the ________ derived
on slide 14.
Q "1%
Q "1%
lnD2 = lnD0 − d $$ '' and lnD1 = lnD0 − d $$ ''
R # T2 &
R # T1 &
______________ equation at T1 from equation at T2
lnD2 − lnD1 = ln
Q ⎛ 1 1⎞
D2
= − d ⎜⎜ − ⎟⎟
D1
R ⎝ T2 T1 ⎠
Take the ___________ of each side to get the final equation
⎡ Q ⎛ 1 1 ⎞⎤
D2 = D1 exp ⎢− d ⎜⎜ − ⎟⎟⎥
⎢⎣ R ⎝ T2 T1 ⎠⎥⎦
Chapter 5 - 17
Example Diffusion Problem
At 300°C the diffusion coefficient and _______________
for Cu in Si are
D1(300°C) = 7.8 x 10-11 m2/s
Qd = 41.5 kJ/mol
Compute the _____________________ D2 at 350°C?
⎡ Q ⎛ 1 1 ⎞⎤
D2 = D1 exp ⎢− d ⎜⎜ − ⎟⎟⎥
⎢⎣ R ⎝ T2 T1 ⎠⎥⎦
T1 = 273 + 300 = 573 K
T2 = 273 + 350 = 623 K
⎡ −41,500 J/mol ⎛ 1
1 ⎞⎤
D2 = (7.8 x 10−11 m2 /s) exp ⎢
−
⎜
⎟⎥
8.314
J/mol-K
623
K
573
K ⎠⎦
⎝
⎣
D2 = _____________
Chapter 5 - 18
6
Chapter 5: Diffusion
Non-steady State Diffusion
• The ___________ of diffusing species is a function of
both time and position C = C(x,t)
• For non-steady state diffusion, we seek solutions to
____________________
____________ Law
∂C
∂2C
=D 2
∂t
∂x
This form of the equation assumes D is independent
of concentration
Chapter 5 - 19
Non-steady State Diffusion
• Consider the diffusion of copper into a bar of aluminum
Surface conc.,
Cs of Cu atoms
bar
pre-existing conc., Co of copper atoms
Cs
Fig. 5.4,
Callister &
Rethwisch 10e.
Boundary/Initial Conditions
at t = 0, C = Co for 0 ≤ x ≤ ∞
at t > 0, C = CS for x = 0 (constant surface conc.)
C = Co for x = ∞
Chapter 5 - 20
Non-steady State Diffusion (cont.)
( )
C x,t − Co
Cs − Co
⎛ x ⎞
= 1− erf ⎜
⎟
⎝ 2 Dt ⎠
C(x,t) = Conc. at point x at
time t
erf(z) = error ___________
z and erf(z) values are given
in Table _______
Fig. 5.5, Callister & Rethwisch 10e.
Chapter 5 - 21
7
Chapter 5: Diffusion
Non-steady State Diffusion
Example Problem
An FCC iron-carbon alloy initially containing ______
% C is carburized at an elevated temperature and in
an atmosphere in which the surface carbon
concentration is maintained at 1.0 wt%. If, after
_____, the concentration of carbon is 0.35 wt% at a
position ________ below the surface, determine the
temperature at which the treatment was carried out.
Chapter 5 - 22
Example Problem (cont.):
Solution: use Eqn. 5.5
⎛ x ⎞
C(x,t) − Co
= 1− erf ⎜
⎟
Cs − Co
⎝ 2 Dt ⎠
Data for problem tabulated as follows:
– t = 49.5 h
x = ________
– Cx = 0.35 wt%
Cs = 1.0 wt%
– Co = __________
⎛ x ⎞
C(x,t) − Co 0.35 − 0.20
=
= 1− erf ⎜
⎟ = 1− erf(z)
Cs − Co
1.0 − 0.20
⎝ 2 Dt ⎠
erf(z) = ______
Chapter 5 - 23
Example Problem (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
z
0.95
0.7970
0.8125
0.8209
Now solve for D
z − 0.90
0.8125 − 0.7970
=
0.95 − 0.90 0.8209 − 0.7970
z = 0.93
z=
x
2 Dt
D=
x2
4z 2t
⎛ x2 ⎞
(4 x 10−3 m)2
1h
∴D = ⎜⎜ 2 ⎟⎟ =
= 2.6 x 10−11 m2 /s
2
⎝ 4z t ⎠ (4)(0.93) (49.5 h) 3600 s
Chapter 5 - 24
8
Chapter 5: Diffusion
Example Problem (cont.):
• To solve for the temperature at
which D has the above value,
we use a rearranged form of
Equation 5.9a
}
T=
Qd
R(lnDo − lnD)
From Table 5.2, for diffusion of C in FCC Fe
Do = 2.3 x 10-5 m2/s Qd = 148,000 J/mol
T=
148,000 J/mol
(8.314 J/mol-K)[ln (2.3 x 10−5 m2 /s) − ln (2.6 x 10−11 m2 /s)]
T = 1300 K = _______
Chapter 5 - 25
Summary
• Solid-state diffusion is mass transport within solid
materials by stepwise atomic motion
• Two diffusion mechanisms
- Vacancy diffusion
- Interstitial diffusion
dC
dx
• Fick’s First Law of Diffusion
J = −D
• Fick’s Second Law of Diffusion
- non-steady state diffusion
∂C
∂2C
=D 2
∂t
∂x
• Diffusion coefficient
- Effect of temperature
⎛ Q ⎞
D = D0 exp ⎜⎜ − d ⎟⎟
⎝ RT ⎠
Chapter 5 - 26
9