Material Science II
Today’s topic: Diffusion of materials
Compiled by : Dr N.N. Nyangiwe
Learning outcomes
After completing this learning unit you should be able to:
• Understand and explain various mechanisms of diffusion
• Understand and explain the factors affecting diffusion
• Understand and explain Fick’s first and second laws
• Discuss and calculate impurity diffusion,
• Activation energy,
• Pre-exponential factor in material alloys.
• Understand how diffusion is used in materials processing.
Assessment criteria
•
Diffusion and its applications are discussed
•
The various mechanisms of diffusion are compared.
•
Fick's first law is formulated.
•
Factors affecting diffusion are discussed.
•
Fick's second law is formulated.
•
Various material processing methods are discussed.
•
Use expressions of diffusion rate to describe the influence of structure and
temperature.
Outline
•
•
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•
•
•
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Diffusion mechanisms
Vacancy diffusion
Interstitial diffusion
Impurities
The mathematics of diffusion
Steady-state diffusion (Fick’s first law)
Nonsteady-State Diffusion (Fick’s second law)
Factors that influence diffusion
Diffusing species
Host solid
Temperature
Microstructure
Have you wondered
• Aluminum oxidizes more easily than iron, so why do we say aluminum normally does not “rust?”
• What kind of plastic is used to make carbonated beverage bottles?
• How are the surfaces of certain steels hardened?
• Why do we encase optical fibers using a polymeric coating?
• Who invented the first contact lens?
• How does a tungsten filament in a light bulb fail?
Diffusion
• What is diffusion?
• How does it occur?
• Why is it an important part of processing?
• How can the rate of diffusion be predicted for some simple cases?
• How does diffusion depend on structure and temperature?
• Be able to perform diffusion calculations
• Calculate number of vacancies in a solid and the diffusion coefficient
using Arrhenius equation.
Diffusion
Diffusion - is the movement
of particles from an area of
higher concentration to an
area of lower concentration
Molecules spread out or move from
areas where there are many of them
to areas where there are fewer of
them in order to reach equilibrium
Diffusion Animation
Diffusion
• Concentration on the left is higher than on the right of the imaginary
barrier.
• Many of the molecules on the left can pass to the right, but few can
pass from right to left.
• There is a net movement from the higher concentration to the lower
concentration
Diffusion
Diffusion
• Diffusion: Mass transport by atomic motion
• Gases & Liquids – random (Brownian) motion
• Solids – Two mechanisms:
Vacancy diffusion or Interstitial diffusion
Osmosis
•
Osmosis is the movement of water from a region
where its concentration is high, across a
selectively permeable membrane, into a region
where its concentration is lower.
•
A selectively permeable membrane is one that
allows passage of some molecules, but not others
Aims of studying diffusion
• Overview of Fick’s laws that describe the diffusion process.
•
How the different crystal structures and the size of atoms or ions,
temperature, and concentration of diffusing species affect the
rate at which diffusion occurs.
• We will discuss specific examples of how diffusion is used in the
synthesis and processing of advanced materials as well as
manufacturing of components using advanced materials
Imperfections in atomic and ionic arrangements
Imperfections
Point defects
Line Defects
(Dislocation)
Surface
Defects
Point defects
• Point defects - Imperfections, such as vacancies, that are located
typically at one (in some cases a few) sites in the crystal.
• Extended defects - Defects that involve several atoms/ions and thus
occur over a finite volume of the crystalline material (e.g.,
dislocations, stacking faults, etc.).
• Vacancy - An atom or an ion missing from its regular crystallographic
site.
• Interstitial defect - A point defect produced when an atom is placed
into the crystal at a site that is normally not a lattice point.
• Substitutional defect - A point defect produced when an atom is
removed from a regular lattice point and replaced with a different
atom, usually of a different size.
Imperfections in atomic and ionic arrangements
• These imperfections only represent defects or deviations from the
perfect or ideal atomic or ionic arrangements expected in a given
crystal structure. The material is not considered defective from a
technological viewpoint.
•
In many applications, the presence of such defects is useful. For
example, defects known as dislocations are useful for increasing
the strength of metals and alloys; however, in single crystal silicon,
used for manufacturing computer chips, the presence of
dislocations is undesirable.
• Often the “defects” may be created intentionally to produce a
desired set of electronic, magnetic, optical, or mechanical
properties.
Point defects
Point defects are localized disruptions in otherwise
perfect atomic or ionic arrangements in a crystal structure.
Even though we call them point defects, the disruption
affects a region involving several atoms or ions.
These imperfections, shown in Figure 1-1, may be
introduced by movement of the atoms or ions when they
gain energy by heating, during processing of the material,
or by the intentional or unintentional introduction of
impurities.
Impurities and dopants
• Impurities are elements or compounds that are present from raw
materials or processing. For example, silicon crystals grown in
quartz crucibles contain oxygen as an impurity.
• Dopants, on the other hand, are elements or compounds that are
deliberately added, in known concentrations, at specific locations
in the microstructure, with an intended beneficial effect on
properties or processing.
• In general, the effect of impurities is harmful, whereas the effect of
dopants on the properties of materials is useful.
• Phosphorus (P) and boron (B) are examples of dopants that are
added to silicon crystals to improve the electrical properties of pure
silicon (Si).
Vacancies
• A vacancy is produced when an atom or an ion is missing from its
normal site in the crystal structure as. When atoms or ions are
missing (i.e., when vacancies are present), the overall randomness
or entropy of the material increases, which increases the
thermodynamic stability of a crystalline material.
• All crystalline materials have vacancy defects. Vacancies are
introduced into metals and alloys during solidification, at high
temperatures, or as a consequence of radiation damage.
• Vacancies play an important role in determining the rate at which
atoms or ions move around or diffuse in a solid material, especially
in pure metals.
Vacancies
At room temperature (298 K), the concentration of vacancies is
small, but the concentration of vacancies increases exponentially
as the temperature increases, as shown by the following
Arrhenius type behavior:
nv = n exp [-Qv/RT]
where
nv is the number of vacancies per cm3;
n is the number of atoms per cm3;
Qv is the energy required to produce one mole of vacancies, in cal/mol
or Joules/ mol;
R is the gas constant, 1.987 cal/mol.K or 8.31J/mol.K
T is the temperature in degrees Kelvin.
The Effect of Temperature on Vacancy concentrations
(a) Calculate the concentration of vacancies in copper at room
temperature (25oC). Assume that 20,000 cal are required to produce a
mole of vacancies in copper.
(b) What temperature will be needed to heat treat copper such that the
concentration of vacancies produced will be 1000 times more than the
equilibrium concentration of vacancies at room temperature
Example 1 SOLUTION
The lattice parameter of FCC copper is 0.36151 nm. The basis is 1,
therefore, the number of copper atoms, or lattice points, per cm3 is:
n 
4 atoms/cell
22
3

8
.
47

10
copper
atoms/cm
8
3
(3.6151  10 cm)
Example 1 SOLUTION continued
At room temperature, T = 25 + 273 = 298 K:
 Q 
n  n exp 

RT



22
  8.47  10

cal


 20,000


atoms 
mol

. exp 
3 
cal
cm 
 1.987
 298K 


mol  K


 1.815  108 vacancies /cm 3
We could do this by heating the copper to a temperature at which this number of
vacancies forms:
 Q 
n  1.815  1011  n exp 

RT


 (8.47  10 ) exp( 20,000 /(1.987  T )), T  102 C
22
o
Diffusion-how do atoms move through solids
Diffusion mechanisms
Vacancy diffusion
Interstitial diffusion
Impurities
The mathematics of diffusion
Steady-state diffusion (Fick’s first law)
Nonsteady-State Diffusion (Fick’s second law)
Factors that influence diffusion
Diffusing species
Host solid
Temperature
Microstructure
Applications of diffusion
•
•
•
•
•
•
Nitriding - Carburization for Surface Hardening of Steels.
p-n junction - Dopant Diffusion for Semiconductor
Devices.
Manufacturing of Plastic Beverage Bottles/Balloons.
Sputtering, Annealing - Magnetic Materials for Hard
Drives.
Hot dip galvanizing - Coatings and Thin Films
Thermal Barrier Coatings for Turbine Blades
Interdiffusion
Interdiffusion: In an alloy, atoms tend to migrate from regions of
high concentration to regions of low concentration
Initially
After some time
Interdiffusion
Hea
t
Diffusion
Diffusion
Self-diffusion: In an elemental solid, atoms also migrate.
Labelled atoms
C
A
After some time
C
D
A
D
B
B
Self-diffusion
• Self-diffusion: Defects known as vacancies exist in materials.
The disorder that vacancies create minimizes the free energy
and increases the thermodynamic stability of a crystalline
material.
• In materials having vacancies atoms move or “jump” from one
lattice position to another. This process is known as selfdiffusion.
• It can be detected by using radioactive tracers. If we introduce
a radioactive isotope of gold (Au198) onto the surface of
standard gold (Au197). After a period of time, the radioactive
atoms would move into the standard gold.
• Eventually, the radioactive atoms would be uniformly
distributed throughout the entire standard gold sample.
Diffusion mechanisms
• From an atomic perspective, diffusion is just the stepwise
migration of atoms from lattice site to lattice site. In fact,
the atoms in solid materials are in constant motion,
rapidly changing positions. For an atom to make such a
move, two conditions must be met:
(1) there must be an empty adjacent site, and
(1) the atom must have sufficient energy to break
bonds with its neighbor atoms and then cause
some lattice distortion during the displacement.
Diffusion mechanism types
1. Interstitial
Diffusion
2. Vacancy
Diffusion
Diffusion mechanism 2
Interstitial Diffusion When a small interstitial atom or ion is
present in the crystal structure, the atom or ion moves from
one interstitial site to another. No vacancies are required for
this mechanism. Partly because there are many more
interstitial sites than vacancies, interstitial diffusion occurs
more easily than vacancy diffusion. Interstitial atoms that
are relatively smaller can diffuse faster.
Diffusion mechanism 2
Interstitial diffusion: smaller atoms can diffuse between atoms.
More rapid than vacancy diffusion
Interstitial diffusion mechanism
Before
After
Interstitial diffusion is generally faster than vacancy
diffusion because bonding of interstitials to the
surrounding atoms is normally weaker and there are
many more interstitial sites than vacancy sites to
jump to.
Requires small impurity atoms (e.g. C, H, O) to
fit into interstices in host.
Diffusion mechanism 1
Vacancy Diffusion In self-diffusion and diffusion involving
substitutional atoms, an atom leaves its lattice site to fill a
nearby vacancy (thus creating a new vacancy at the original
lattice site). As diffusion continues, we have large number of
atoms and vacancies, called vacancy diffusion. The number
of vacancies, which increases as the temperature increases,
influences the extent of both self-diffusion and diffusion of
substitutional atoms.
Diffusion mechanism 1
Vacancy Diffusion: atoms exchange with vacancies
increasing elapsed time
Applies to substitution (impurity) atoms
Rate of vacancy diffusion depends on:
• number of vacancies
• activation energy to exchange
Vacancy diffusion mechanism
Befor
e
Afte
r
The direction of flow of atoms is opposite the
vacancy flow direction.
Stability of Atoms and Ions
Imperfections are present or deliberately introduced into a material;
atoms or ions in their normal positions in the crystal structures are
not stable or at rest. Instead, the atoms or ions possess thermal
energy, and they will move.
An atom may move from a normal crystal structure location to
occupy a nearby vacancy. An atom may also move from one
interstitial site to another. Atoms or ions may jump across a grain
boundary, causing the grain boundary to move.
The ability of atoms and ions to diffuse increases as the
temperature, or thermal energy possessed by the atoms and ions,
increases.
Stability of Atoms and Ions
The rate of atom or ion movement is related to temperature or
thermal energy by the Arrhenius equation:
−𝑄
𝑅𝑎𝑡𝑒 = 𝑐𝑜 𝑒𝑥𝑝
𝑅𝑇
where c0 is a constant, R is the gas constant (1.987cal/(mol.K), T is
the absolute temperature(K), and Q is the activation energy
(cal/mol) required to cause Avogadro’s number of atoms or ions to
move.
This equation is derived from a statistical analysis of the probability
that the atoms will have the extra energy Q needed to cause
movement. The rate is related to the number of atoms that move.
Stability of Atoms and Ions
We can rewrite the equation by taking natural logarithms of
both sides:
ln 𝑟𝑎𝑡𝑒 = ln 𝑐0 −
𝑄
𝑅𝑇
If we plot ln(rate) versus 1/T , the slope of the line will be -Q
/R and, consequently, Q can be calculated. The constant c0
corresponds to the intercept at ln(c0) when 1/T is zero.
 Activation energy -The energy required to cause a
particular reaction to occur.
Activation Energy for Interstitial Atoms
Suppose that interstitial atoms are found to move from one site to another at the rates of
5  108 jumps/s at 500oC and 8  1010 jumps/s at 800oC. Calculate the activation energy
Q for the process.
The Arrhenius plot
of ln(rate) versus
1/T can be used to
determine the
activation energy
required for a
reaction
SOLUTION
we could write two simultaneous equations:
Rate
5x
𝑗𝑢𝑚𝑝𝑠
−𝑄
= 𝑐0 𝑒𝑥𝑝
𝑠
𝑅𝑇
𝑗𝑢𝑚𝑝𝑠
108
𝑠
=
𝑗𝑢𝑚𝑝𝑠
𝑐0
𝑠
exp
−𝑄
𝑐𝑎𝑙
1.987𝑚𝑜𝑙.𝐾
= 𝑐0 exp(−0.000651 𝑄)
𝑐𝑎𝑙
𝑚𝑜𝑙
500+273 𝐾
SOLUTION conti…
8 x 1010
−𝑄
𝑗𝑢𝑚𝑝𝑠
𝑗𝑢𝑚𝑝𝑠
= 𝑐0
exp
𝑐𝑎𝑙
𝑠
𝑠
1.987
𝑚𝑜𝑙. 𝐾
= 𝑐0 exp −0.000469 𝑄
𝑐𝑎𝑙
𝑚𝑜𝑙
800 + 273 𝐾
Now the temperatures were converted into K. Since
5 x 108
𝑗𝑢𝑚𝑝𝑠
𝑐0 =
exp −0.000651 𝑄
𝑠
then
8x
1010
5 x 108 exp −0.000469 𝑄
=
exp −0.000651 𝑄
160 = exp 0.000651 − 0.000469 𝑄 = exp 0.000182 𝑄
ln 160 = 5.075 = 0.000182 𝑄
5.075
𝑄=
= 27,880 𝑐𝑎 𝑙 𝑚 𝑜𝑙
0.000182
The
Arrhenius
plot of
ln(rate)
versus 1/T
can be used
to
determine
the
activation
energy
required for
a reaction
Activation Energy for Diffusion
A diffusing atom must squeeze past the surrounding atoms to
reach its new site. In order for this to happen, energy must be
supplied to allow the atom to move to its new position, as
shown in the figure (SEE NEXT SLIDE). The atom is originally
in a low-energy, relatively stable location. In order to move to a
new location, the atom must overcome an energy barrier. The
energy barrier is the activation energy Q.
The thermal energy supplies atoms or ions with the energy
needed to exceed this barrier.
Note that the symbol Q is often used for activation energies for
different processes (rate at which atoms jump, a chemical
reaction, energy needed to produce vacancies, etc)
Activation Energy for Diffusion
Diffusion couple - A combination of elements involved in diffusion
studies
A high energy is
required to squeeze
atoms past one another
during diffusion. This
energy is the activation
energy Q.
Generally more energy
is required for a
substitutional atom
than for an interstitial
atom
Activation Energy for Diffusion
Less energy is required to squeeze an interstitial atom past the surrounding
atoms; hence, activation energies are lower for interstitial diffusion than for
vacancy diffusion. Typical values for activation energies for diffusion of different
atoms in different host materials are shown in Table 5-1.
Diffusion couple is used to indicate a combination of an atom of a given
element (e.g., carbon) diffusing in a host material (e.g., BCC Fe).
A low-activation energy indicates easy diffusion.
In self-diffusion, the activation energy is equal to the energy needed to create a
vacancy and to cause the movement of the atom.
Table 5-1 also shows values of D0, which is the pre-exponential term and the
constant c0,when the rate process is diffusion. We will see later that D0 is the
diffusion coefficient when 1/T = 0.
Diffusion coupe
Interstitial diffusion:
C in FCC iron
C in BCC iron
N in FCC iron
N in BCC iron
H in FCC iron
H in BCC iron
Diffusion data for selected materials
Q (cal/mol)
Do (cm2/s)
32900
20900
34600
18300
10300
3600
0.23
0.011
0.0034
0.0047
0.0063
0.0012
Self-diffusion (vacancy diffusion):
Pd in FCC Pb
Al in FCC Al
Cu in FCC Cu
Fe in FCC Fe
Zn in HCP Zn
Mg in HCP Mg
Fe in BCC Fe
W in BCC W
Si in Si (covalent)
C in C (covalent)
25900
32200
49300
66700
21800
32200
58900
143300
110000
163000
1.27
0.10
0.36
0.65
0.1
1.0
4.1
1.88
1800.0
5.0
Heterogeneous diffusion (vacancy diffusion):
Ni in Cu
Cu in Ni
Zn in Cu
Ni in FCC iron
Au in Ag
Ag in Au
Al in Cu
Al in Al203
O in Al2O3
Mg in MgO
O in MgO
57900
61500
43900
64000
45500
40200
39500
114000
152000
79000
82100
2.3
0.65
0.78
4.1
0.26
0.072
0.045
28.0
1900.0
0.249
0.000043
Steady- State Diffusion: Fick’s first law
• Fick’s first law: the diffusion flux along
direction x is proportional to the concentration
gradient.
• where D is the diffusion coefficient
Several factors affect the flux of atoms during diffusion. If
we are dealing with diffusion of ions, electrons, holes, etc.,
the units of J, D, and will reflect the appropriate species that
are being considered.
The negative sign in Equation 1 tells us that the flux of
diffusing species is from higher to lower concentrations.
Fick’s first law describes flow under steady state conditions.
ie. The concentration profile is maintained the same
throughout the flow process. Hence flux will be independent
of time.
Steady- State Diffusion: Fick’s first law
The flux during
diffusion is
defined as the
number of atoms
passing through
a plane of unit
area per unit
time
Steady- State Diffusion
•
The flux of diffusing atoms, J, is used to
quantify how fast diffusion occurs. The flux is
defined as either the
number of atoms
diffusing through unit area (A) per unit time (t)
(atoms/m2-second) or the mass of atoms (M)
diffusing through unit area per unit time,
(kg/m2- second).
•
For example, for the mass flux we can write
Steady- State Diffusion
• Steady state diffusion: the diffusion
flux does not change with time.
• Concentration profile: concentration
of atoms/molecules of interest as
function of position in the sample.
• Concentration gradient: dC/dx (Kg m4): the
slope at a particular point on
concentration profile.
©2003 Brooks/Cole, a division of Thomson Learning, Inc. Thomson Learning™ is a trademark used herein under license.
Concentration gradient
Figure 5.15
Illustration of the
concentration
gradient
Steady- State Diffusion
Diffusional processes can be either steady-state or non-steadystate. These two types of diffusion processes are distinguished
by use of a parameter called flux. It is defined as net number of
atoms crossing a unit area perpendicular to a given direction per
unit time. For steady-state diffusion, flux is constant with time,
whereas for non-steady-state diffusion, flux varies with time. A
schematic view of concentration gradient with distance for both
steady-state and non-steady-state diffusion processes are shown
in figure
Rate of Diffusion (Fick’s First Law)
 Fick’s first law - The equation relating the flux of atoms by
diffusion to the diffusion coefficient and the concentration
gradient.
 Diffusion coefficient (D) - A temperature-dependent coefficient
related to the rate at which atoms, ions, or other species diffuse.
 Concentration gradient - The rate of change of composition with
distance in a nonuniform material, typically expressed as
atoms/cm3.cm or at%/cm.
Example: Semiconductor doping
One way to manufacture transistors, which amplify electrical
signals, is to diffuse impurity atoms into a semiconductor
material such as silicon (Si). Suppose a silicon wafer 0.1 cm
thick, which originally contains one phosphorus atom for
every 10 million Si atoms, is treated so that there are 400
phosphorous (P) atoms for every 10 million Si atoms at the
surface . The lattice parameter of silicon is 5.4307 Å.
Calculate the concentration gradient
(a) in atomic percent/cm and
(b) in atoms /cm3.cm.
Example: Semiconductor doping
Solution…
a) Calculate the initial and surface compositions in
atomic percent.
1Patom
ci  7
 100  0.00001at % P
10 atoms
400 Patom
cs  7
 100  0.004at % P
10 atoms
c
0.00001  0.004at % P
at % P

 0.0399
x
0.1cm
cm
Solution…
b) The volume of the unit cell:
Vcell = (5.4307  10-8 cm)3 = 1.6  10-22 cm3/cell
The volume occupied by 107 Si atoms, which are
arranged in a diamond cubic (DC) structure with 8
atoms/cell, is:
V = 2  10-16 cm3
The compositions in atoms/cm3 are:
Solution…
1Patom
atoms
18
ci 
 0.005  10 P (
)
16
3
3
2  10 cm
cm
400 Patoms
atoms
18
cs 

2

10
P
(
)
16
3
3
2  10 cm
cm
atoms
18
18
0.005  10  2  10 P (
)
3
c
cm

x
0.1cm
atoms
19
 1.995  10 P
cm 3 .cm
Example: Diffusion of Nickel in Magnesium Oxide (MgO)
A 0.05 cm layer of magnesium oxide (MgO) is deposited
between layers of nickel (Ni) and tantalum (Ta) to provide a
diffusion barrier that prevents reactions between the two
metals (Figure 5.17). At 1400oC, nickel ions are created and
diffuse through the MgO ceramic to the tantalum. The diffusion
coefficient of nickel ions in MgO is 9  10-12 cm2/s, and the
lattice parameter of nickel at 1400oC is 3.6  10-8 cm.
Determine the number of nickel ions that pass through the
MgO per second.
Solution…
Solution…
The composition of nickel at the Ni/MgO interface is 100% Ni, or
c
Ni / MgO
atoms
4 Ni
22 atoms
unitcell

 8.57  10
8
3
3
(3.6  10 cm)
cm
The composition of nickel at the Ta/MgO interface is 0%
Ni. Thus, the concentration gradient is:
atoms
0  8.57  10
3
c
24 atoms
cm

 1.71  10
3
x
0.05cm
cm .cm
22
Solution…
The flux of nickel atoms through the MgO layer is:
c
12
2
24 atoms
J  D
 (9  10 cm / s)( 1.71  10
)
3
x
cm .cm
13 Niatoms
J  1.54  10
2
cm .s
The total number of nickel atoms crossing the 2 cm  2
cm interface per second is:
Total Ni atoms per second = J(Area)
= (1.54  1013 atoms/cm2.s) (2 cm)(2 cm)
= 6.16  1013 Ni atoms/s
Steady-state Diffusion: J ~ gradient of c
• Concentration Profile, C(x): [kg/m3]
Cu flux Ni flux
Concentration
of Cu [kg/m 3 ]
Adapted from Fig. 6.2(c)
Concentration
of Ni [kg/m 3 ]
Position, x
• Fick's First Law: D is a constant!
• The steeper the concentration profile, the greater the flux!
Steady-state Diffusion
• Steady State: concentration profile not changing with time.
dC
• Apply Fick's First Law:
• If Jx)left = Jx)right , then
J x  D
dx
 dC
 dC 
 
 

 dx  left  dx  right
• Result: the slope, dC/dx, must be constant
(i.e., slope doesn't vary with position)!
Modeling rate of diffusion: flux
• Flux:
• Directional Quantity
• Flux can be measured for:
- vacancies
- host (A) atoms
- impurity (B) atoms
A = Area of flow
• Empirically determined:
– Make thin membrane of known surface area
– Impose concentration gradient
–Measure how fast atoms or molecules diffuse
through the membrane
diffused
mass
M
J  slope
time
Steady-state diffusion (Fick’s first law)
Steady-state diffusion
Rate of diffusion independent of time
dC
Flux proportional to concentration gradient =
dx
C1
C1
Fick’s first law of diffusion:
C2
x1
if linear
x
x2
dC C C2  C1


dx
x
x2  x1
dC
J  D
dx
D  diffusion coefficient
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?
Data:
Diffusion coefficient in butyl rubber: D = 110 x10-8 cm2/s
surface concentrations:
C1 = 0.44 g/cm3
C2 = 0.02 g/cm3
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, 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:
– D in butyl rubber:
– surface concentrations:
D = 110 x10-8 cm2/s
C1 = 0.44 g/cm3 C2 = 0.02 g/cm3
– Diffusion distance:
glove
C1
l2
t 
x2 – x1 = 0.04 cm
b
6D
paint
remover
skin
C2
C -C
Jx= -D 2 1
x 2- 1x
= 1.16 x 10
x1 x2
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004, 2006-
-5
g
cm2 s
Example: Chemical protective clothing (CPC)
Solution: – assuming linear conc. gradient
glove
C1
paint
remover
dC
C2  C1
J -D
 D
dx
x2  x1
2
tb 
6D
skin
Data:
D = 110 x 10-8 cm2/s
C1 = 0.44 g/cm3
C2 = 0.02 g/cm3
x2 – x1 = 0.04 cm
C2
x1 x2
J   (110 x 10
-8
(0.02 g/cm 3  0.44 g/cm3 )
g
cm /s)
 1.16 x 10-5
(0.04 cm)
cm2s
2
Example: Diffusion in steel plate
• Steel plate at 7000C with geometry shown:
700 C
Knowns:
C1= 1.2 kg/m3 at 5mm
(5 x 10–3 m) below surface.
C2 = 0.8 kg/m3 at 10mm
(1 x 10–2 m) below surface.
D = 3 x10-11 m2/s at 700 C.
•Q: In steady-state, how much carbon transfers
from the rich to the deficient side?
Adapted from Fig.
5.4, Callister 6e.
Example: Diffusion of radioactive atoms
• Surface of Ni plate at 10000C contains
50% Ni63 (radioactive) and 50% Ni (non-radioactive).
• 4 microns below surface Ni63 /Ni = 48:52
• Lattice constant of Ni at 1000 C is 0.360 nm.
• Experiment shows that self-diffusion of Ni is 1.6 x 10-9 cm2/sec
What is the flux of Ni63 atoms through a plane 2 m below surface?
C1 
(4 Ni / cell )(0.5 Ni 63 / Ni )
(0.36 x 10 9 m ) 3 / cell
 42.87 x 10 27 Ni 63 / m 3
 (1.6x10 13
m / 2sec)
C2 
(4Ni /cell )(0.48Ni 63 /Ni )
(0.36x10 9 m) 3 /cell
 41.15x10 27 Ni 63 /m 3
(41.15  42.87)x10 27 Ni 63 /m 3
(4 0)x106m
 0.69x1020Ni 63 /m 2  s
How many Ni63 atoms/second through cell?
J (0.36 nm ) 2  9 Ni 63 / s
Where can we use Fick’s Law?
Fick's law is commonly used to model transport processes in
• foods,
• clothing,
• biopolymers,
• pharmaceuticals,
• porous soils,
• semiconductor doping process, etc.
Example
The total membrane surface area in the lungs (alveoli)
may be on the order of 100 square meters and have a
thickness of less than a millionth of a meter, so it is a
very effective gas-exchange interface.
CO2 in air has D~16 mm2/s, and, in water, D~ 0.0016 mm2/s
Non-Steady-State Diffusion
• Concentration profile,
C(x), changes w/ time.
• To conserve matter:
• Governing Eqn.:
• Fick's First Law:
Factors affecting diffusion
Diffusion coefficient and temperature
The diffusion
coefficient D as a
function of reciprocal
temperature for some
metals and ceramics.
In the Arrhenius plot,
D represents the rate
of the diffusion
process. A steep
slope denotes a high
activation energy
Types of Diffusion
In volume diffusion, the atoms move through the crystal from one
regular or interstitial site to another. Because of the surrounding
atoms, the activation energy is large and the rate of diffusion is
relatively slow.
Atoms can also diffuse along boundaries, interfaces, and surfaces in
the material. Atoms diffuse easily by grain boundary diffusion,
because the atom packing is disordered and less dense in the grain
boundaries. Because atoms can more easily squeeze their way
through the grain boundary, the activation energy is low (Table 5-2).
Surface diffusion is easier still because there is even less constraint
on the diffusing atoms at the surface.
Problem 7:Tungsten Thorium Diffusion
Couple
Consider a diffusion couple setup between pure tungsten and
a tungsten alloy containing 1 at.% thorium. After several
minutes of exposure at 20000C, a transition zone of 0.01 cm
thickness is established. What is the flux of thorium atoms at
this time if diffusion is due to
(a)volume diffusion,
(b) grain boundary diffusion
(c) surface diffusion? (See Table 5.2.)
(a) Volume diffusion
−120000
2
𝐷 = 1.0
𝑐𝑚
exp
𝑠
1.987
𝑐𝑎𝑙
𝑚𝑜𝑙. 𝐾
𝑐𝑎𝑙
𝑚𝑜𝑙
2273 𝐾
= 2.89 𝑋 10−12 𝑐𝑚2 /𝑠
∆𝑐
𝐽 = −𝐷∆𝑥
= − 2.89 𝑋
= 18. 2 𝑋 1010
10−12
𝑎𝑡𝑜𝑚𝑠
𝑐𝑚2 .𝑠
𝑐𝑚2
𝑠
−6.3 𝑋 1022
𝑎𝑡𝑜𝑚𝑠
𝑐𝑚3 . 𝑐𝑚
(b) Grain boundary diffusion
𝑐𝑚2
𝐷 = 0.74
exp
𝑠
−9000
1.987
𝑐𝑎𝑙
𝑚𝑜𝑙. 𝐾
𝑐𝑎𝑙
𝑚𝑜𝑙
2273 𝐾
= 1.64 𝑋 10−9 𝑐𝑚2 /𝑠
𝐽 = − 1.64 𝑋
10−9
= 10.3 𝑋 1013
𝑐𝑚2
𝑠
𝑎𝑡𝑜𝑚𝑠
𝑐𝑚2 .𝑠
−6.3 𝑋 1022
𝑎𝑡𝑜𝑚𝑠
𝑐𝑚3 . 𝑐𝑚
(c) Surface diffusion
𝐷 = 0.47
𝑐𝑚2
𝑠
−66400
exp
1.987
𝑐𝑎𝑙
𝑚𝑜𝑙. 𝐾
𝑐𝑎𝑙
𝑚𝑜𝑙
2273 𝐾
= 1.94 𝑋 10−7 𝑐𝑚2 /𝑠
2
𝑐𝑚
𝐽 = − 1.94 𝑋 10−7
𝑠
−6.3 𝑋 1022
= 12.2 𝑋 1015
𝑎𝑡𝑜𝑚𝑠
𝑐𝑚2 . 𝑠
𝑎𝑡𝑜𝑚𝑠
𝑐𝑚3 . 𝑐𝑚
Dependence on bonding and crystal structure
A number of factors influence the activation energy for
diffusion and, hence, the rate of diffusion.
Interstitial diffusion, with a low-activation energy, usually
occurs much faster than vacancy, or substitutional diffusion.
Activation energies are usually lower for atoms diffusing
through open crystal structures than for close-packed crystal
structures. Because the activation energy depends on the
strength of atomic bonding, it is higher for diffusion of atoms in
materials with a high melting temperature (Figure 5-17).
Dependence on bonding and crystal structure
The smaller size, cations (with a positive charge) often
have higher diffusion coefficients than those for anions
(with a negative charge).
In sodium chloride, for instance, the activation energy for
diffusion of chloride ions (Cl-) is about twice that for
diffusion of sodium ions (Na+).
Diffusion of ions also provides a transfer of electrical
charge. The electrical conductivity of ionically bonded
ceramic materials is related to temperature by an Arrhenius
equation. As the temperature increases, the ions diffuse
more rapidly, electrical charge is transferred more quickly,
and the electrical conductivity is increased. Some ceramic
materials are good conductors of electricity.
Time: Diffusion requires time. The units for flux are
atoms/cm2.s . If a large number of atoms must diffuse to
produce a uniform structure, long times may be required,
even at high temperatures. Times for heat treatments may
be reduced by using higher temperatures or by making the
diffusion distances (related to Δx) as small as possible.
Remarkable structures and properties are obtained if we
prevent diffusion. Steels are quenched rapidly from high
temperatures to prevent diffusion, they form nonequilibrium
structures and provide the basis for sophisticated heat
treatments.
Steady state diffusion and nonsteady state diffusion
Steady State Diffusion:
If the diffusion flux does not change with time a steady state
condition exist. Steady-state diffusion is the situation wherein
the rate of diffusion into a given system is just equal to the rate
of diffusion out, such that there is no net accumulation or
depletion of diffusing species--i.e., the diffusion flux is
independent of time.
Non Steady State Diffusion: If the diffusion flux and the
concentration gradient at some particular point in a solid varies
with time there will be a depletion /accumulation of diffusing
species. Then there is a non steady state of diffusion.
Steady-state diffusion and non-steady-state diffusion
Fick’s Second Law
 Fick’s second law - The partial differential equation
that describes the rate at which atoms are
redistributed in a material by diffusion.
 Interdiffusion - Diffusion of different atoms in
opposite directions.
 Kirkendall effect - Physical movement of an
interface due to unequal rates of diffusion of the
atoms within the material.
 Purple plague - Formation of voids in goldaluminum welds due to unequal rates of diffusion
of the two atoms; eventually failure of the weld
can occur.
Nonsteady-State Diffusion: Fick’s second law
• In many real situations the concentration
profile and the concentration gradient are
changing with time. The changes of the
concentration profile can be described in this
case by a differential equation, Fick’s
second law.
• Solution
of
concentration
C(x,t):
this
equation
is
profile as function
of
time,
Gas diffusion into a Solid
Non- Steady-state diffusion (Fick’s second law)
The concentration of diffusing species is a function of both
time and position C = C(x,t)
C
 2C
D 2
In this case Fick’s Second Law is used
t
x
C s
Non-Steady-State Diffusion : C = c (x,t)
concentration of diffusing species is a function of both time and position






c
Fick's Second "Law" c  D
t
x

2 


2

• Copper diffuses into a bar of aluminum.
Cs
B.C. at t = 0, C = Co for 0  x  
at t > 0, C = C S for x = 0 (fixed surface conc.)
C = Co for x = 
Adapted from Fig. 6.5,
Callister & Rethwisch 3e.
Non-Steady-State Diffusion
• Cu diffuses into a bar ofAl.
CS
C(x,t)
Fick's Second "Law":
c
t






D
c
x 
2


2 
Co

• Solution:
"error function” Values calibrated in Table 6.1
erf (z) 
MatSE 280: Introduction to Engineering Materials
2


z
e
y 2
dy
0
©D.D. Johnson 2004, 2006-
15
Non-Steady-State Diffusion : another look
• Concentration profile,
C(x), changes w/ time.
• Rate of accumulation C(x)
C
dx  J x  J x dx
t
• Using Fick’s Law:
• If D is constant:
C
J x
J x

dx  J x  (J x 
dx)  
dx
t
x
x
C
Jx

C 

   D
t
x
x 
x 
Fick's Second "Law"
c 

t x
Fick’s
2nd Law





D
c








D
x x

c
2


2 

Example: Non-Steady-State Diffusion
FCC iron-carbon alloy initially containing 0.20 wt% C is carburized at an elevated
temperature and in an atmosphere that gives a surface C content 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, what temperature was treatment done?
Solution

C(x,t )  Co 0.35  0.20
x

 1 erf
  1 erf(z)
Cs  Co
1.0  0.20
 2 Dt 
Using Table 6.1 find z where erf(z) = 0.8125. Use interpolation.
z  0.90 0.8125  0.7970

0.95  0.90 0.8209  0.7970
Now solve for D
z
So, z  0.93
x
2 Dt
D
 x2 
(4 x 103 m)2
1h
D  

2
2
 4z t  (4)(0.93) (49.5 h) 3600 s
 erf(z) = 0.8125
z
erf(z)
0.90
z
0.95
0.7970
0.8125
0.8209
x2
4z2 t
 2.6 x 1011 m2 /s
Solution (cont.):
•
To solve for the temperature at which D
has the above value, we use a rearranged
form of Equation (6.9a);
D=D0 exp(-Qd/RT)
Qd
T
R(lnDo  lnD)
• From Table 6.2, for diffusion of C in FCC Fe
D  2.6 x 1011 m2 /s
Do = 2.3 x 10-5 m2/s Qd = 148,000J/mol

T
148,000 J/mol
(8.314 J/mol-K)(ln 2.3x105 m2 /s  ln 2.6x1011 m2 /s)
T = 1300 K = 1027°C
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004, 2006-
Example: Processing
• Copper diffuses into a bar of aluminum.
• 10 hours processed at 600 C gives desired C(x).
• How many hours needed to get the same C(x) at 500 C?
Key point 1: C(x,t500C) = C(x,t600C).
Key point 2: Both cases have the same Co and Cs.
• Result: Dt should be held constant.
C(x,t)  C o
C s C o
• Answer:
 x 
 1 erf 

 2 Dt 
Note
D(T) are T dependent!
Values of D are provided.
Diffusion of atoms into the surface of a material illustrating
the use of Fick’s second law
Graph showing the
argument and value of
error function
encountered in Fick’s
second law
Diffusion down the concentration gradient
• Why do the random jumps of atoms result in a flux of
atoms from regions of high concentration towards the
regions of low concentration?
Diffusion in this
direction
• As a result there is a net flux of atoms from left to right.
Factors that influence diffusion
1- Diffusing Species
The diffusing species as well as the host material
influence the diffusion coefficient (Table 5.2). For
example, there is a significant difference in
magnitude between self-diffusion (D = 3 x 10-21)
and carbon interdiffusion in iron at 500oC (D= 2.4
x 10-12). This comparison also provides a contrast
between
rates of diffusion via vacancy and
interstitial modes.
Self-diffusion occurs by a vacancy
mechanism, whereas carbon diffusion in
iron is interstitial.
Diffusing species
Problem: Design of a Carburizing treatment
The surface of a 0.1% C steel gears is to be hardened by
carburizing. In gas carburizing, the steel gears are placed in an
atmosphere that provides 1.2% C at the surface of the steel at a
high temperature (Figure 5.1). Carbon then diffuses from the surface
into the steel. For optimum properties, the steel must contain 0.45%
C at a depth of 0.2 cm below the surface. Design a carburizing heat
treatment that will produce these optimum properties. Assume that
the temperature is high enough (at least 900oC) so that the iron has
the FCC structure.
Figure 5.1 Furnace
for heat treating
steel using the
carburization
process. (Courtesy
of Cincinnati Steel
Treating).
The exact combination of temperature and time will
depend on the maximum temperature that the heat
treating furnace can reach, the rate at which parts
must be produced, and the economics of the trade
offs between higher temperatures versus
longer times.
Another factor to consider is changes in
microstructure that occur in the rest of the material.
For example, while carbon is diffusing into the surface,
the rest of the microstructure can begin to experience
grain growth or other changes.
Problem: Design of a more economical heat
treatment
We find that 10 h are required to successfully carburize a
batch of 500 steel gears at 900oC, where the iron has the
FCC structure. We find that it costs $1000 per hour to
operate the carburizing furnace at 900oC and $1500 per
hour to operate the furnace at 1000oC. Is it economical to
increase the carburizing temperature to 1000oC? What
other factors must be considered?
Notice, we do not need the value of the pre-exponential term D0, since it
canceled out.
Considering the cost of operating the furnace, by increasing the
temperature
• reduces the heat-treating cost of the gears and increases the production
rate.
• could cause some microstructural or other changes. For example, would
increased temperature cause grains to grow significantly? If this is the
case, we will be weakening the bulk of the material.
• affect the life of the other equipment such as the furnace itself and any
accessories
• How long would the cooling take? Will cooling from a higher temperature
cause residual stresses?
• Would the product still meet all other specifications?
solution we propose is not only technically sound and economically
sensible, it should recognize and make sense for the system as a whole. A
good solution is often simple, solves problems for the system, and does
not create new problems.
Problem: Silicon Device Fabrication
Devices such as transistors (Figure 5.2) are made by doping
semiconductors with different dopants to generate regions that have p- or
n-type semiconductivity. The diffusion coefficient of phosphorus (P) in Si is
D = 65  10-13 cm2/s at a temperature of 1100oC. Assume the source
provides a surface concentration of 1020 atoms/cm3 and the diffusion time
is one hour. Assume that the silicon wafer contains no P to begin with.
(a) Calculate the depth at which the concentration of P will be 1018
atoms/cm3. State any assumptions you have made while solving this
problem.
(b) What will happen to the concentration as we cool the Si wafer
containing P?
(c) What will happen if now the wafer has to be heated again for boron (B)
diffusion for creating a p-type region?
Figure 5.2 Schematic of a n-p-n transistor. Diffusion plays a
critical role in formation of the different regions created in the
semiconductor substrates. The creation of millions of such
transistors is at the heart of microelectronics technology
The initial carbon content of the steel is 0.20 wt%,
whereas the surface concentration is to be
maintained at 1.00 wt%. In order for this
treatment to be effective, a carbon content of 0.60
wt% must be established at a position 0.75 mm
below the surface. Specify an appropriate heat
treatment in terms of temperature and time for
temperatures between 900C and 1050C. Use data
in Table 6.2 for the diffusion of carbon in -iron.
For some applications, it is necessary to harden the surface of a steel
above the hardness of its interior. One way this may be accomplished
is by increasing the surface concentration of carbon in a process
termed carburizing; the steel piece is exposed, at an elevated
temperature, to an atmosphere rich in a hydrocarbon gas, such as
methane (CH4).
Consider one such alloy that initially has a uniform carbon
concentration of 0.25 wt% and is to be treated at 950◦C If the
concentration of carbon at the surface is suddenly brought to and
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
diffusion coefficient for carbon in iron at this temperature is 1.6 × 10−11
m2/s; assume that the steel piece is semi-infinite.
Factors that influence diffusion
2- Temperature
• Diffusion coefficient is the measure of
mobility of diffusing species.
•
D0 – temperature-independent preexponential (m2/s)
•
Qd – the activation energy for diffusion (J/mol or eV/atom)
•
R – the gas constant (8.31 J/mol-K or 8.62×10-5 eV/atom-K)
•
T – absolute temperature (K)
•
We can find D0 And Qd By drawing Arrhenius plots (lnD versus 1/T or logD
versus 1/T)
Diffusion and Temperature
• Diffusivity increases with T exponentially
(so does Vacancy conc.).
 Qd 
D  Do exp  
RT
D = diffusion coefficient [m2/s]
Do = pre-exponential [m2/s]
(see Table 6.2)
Qd = activation energy [J/mol or eV/atom]
R = gas constant [8.314 J/mol-K]
T = absolute temperature [K]
Q  1
d
Note: ln D  ln D 
0 R T 
log D  log D 
0
Q  1
d 
2.3R T 
Factors that influence diffusion
Y= b + a
X
Factors that influence diffusion
Graph of log D vs. 1/T has slope of –Qd/2.3 R, and intercept of log D0
Reciprocal of temperature =
1/T
Diffusion and Temperature
10-8
300
600
1000
Adapted from Fig. 6.7,
1500
• Experimental Data:
T(C)
D (m2/s)
D interstitial >> D substitutional
C in -Fe
C in -Fe
10-14
10-20
0.5
1.0
1.5
1000K/T
Adapted from Fig. 6.7, Callister & Rethwisch 3e.
(Data for Fig. 6.7 from E.A. Brandes and G.B. Brook (Ed.) Smithells Metals
Reference Book, 7th ed., Butterworth-Heinemann, Oxford, 1992.)
Cu in Cu
Al in Al
Fe in -Fe
Fe in -Fe
Zn in Cu
Example: Comparing Diffuse in Fe
D0=2.3x10–5(m2/2)
Q=1.53 eV/atom
T = 900 C D=5.9x10–12(m2/2)
• bcc-Fe:
D0=6.2x10–7(m2/2)
Q=0.83 eV/atom
T = 900 C D=1.7x10–10(m2/2)
10-8
300
600
1000
(Table 6.2)
• fcc-Fe:
1500
Is C in fcc Fe diffusing faster than C in bcc Fe?
T(C)
D (m2/s)
10-14
10-20
0.5
1.0
1.5
1000K/T
• FCC Fe has both higher activation energy Q and D0 (holes larger in FCC).
• BCC and FCC phase exist over limited range of T (at varying %C).
• Hence, at same T, BCC diffuses faster due to lower Q.
• Cannot always have the phase that you want at the %C and T you want!
which is why this is all important.
Factors that influence diffusion
3- Interstitial and vacancy diffusion
mechanisms
• Diffusion of interstitials is typically faster as
compared to the vacancy diffusion mechanism
(self- diffusion or diffusion of substitutional
atoms).
• Smaller atoms cause less distortion of the
lattice during migration and diffuse more
readily than big ones (the atomic diameters
decrease from C to N to H).
• Diffusion is faster in open lattices or in
open directions
Factors that influence diffusion
Factors that influence diffusion
Factors that influence diffusion
4- Role of the
microstructure
• Self-diffusion
coefficients for Ag
depend on the
diffusion path. In
general the diffusivity
is greater through less
restrictive structural
regions – grain
boundaries, dislocation
cores, external
surfaces.
Factors that influence diffusion
• The plots are from the computer
simulation by T. Kwok, P. S. Ho,
and S. Yip. Initial atomic positions
are shown by the circles,
trajectories of atoms are shown
by lines. We can see the
difference between atomic mobility
in the bulk crystal and in the grain
boundary region.
Example: Diffusion in nanocrystalline materials
Lin et al. J. Phys. Chem. C 114, 5686,
2010
Arrhenius plots for 59Fe diffusivities in nanocrystalline Fe and
other alloys compared to the crystalline Fe (ferrite). [Wurschum
et al. Adv. Eng. Mat. 5, 365, 2003]
Factors that influence diffusion: Summary
• Temperature - diffusion rate increases very
rapidly with increasing temperature
• Diffusion mechanism – diffusion by interstitial
mechanism is usually faster than by vacancy
mechanism.
• Diffusing and host species – D0, Qd are different for
every solute, solvent pair.
• Microstructure diffusion is
faster in
polycrystalline materials compared to single
crystals because of the accelerated diffusion
along grain boundaries.
Diffusion in material processing 1
• Case Hardening:
Hardening the surface of a
metal by exposing it to
impurities that diffuse into
the surface region and
increase surface hardness.
• Common example of case
hardening is carburization of
steel. Diffusion of carbon
atoms (interstitial
mechanism) increases
concentration of C atoms
and makes iron (steel)
harder.
Processing using diffusion
• Case Hardening:
-Diffuse carbon atoms into the
host iron atoms at the surface.
-Example of interstitial diffusion
is a case hardened gear.
• Result: The "Case" is
-hard to deform: C atoms
"lock" planes from shearing.
-hard to crack: C atoms put
the surface in compression.
Processing using diffusion
• Doping Silicon with P for n-type semiconductors:
• Process:
1. Deposit P rich
layers on surface.
silicon
2. Heat it.
3. Result: Doped
semiconductor
regions.
silicon
Fig. 18.0,
Callister 6e.
Diffusion in material processing 2
• Doping silicon with phosphorus for
n-type semiconductors.
• Process of doping:
Demonstration of diffusion
The mechanism of diffusion can be demonstrated by considering a diffusion
couple. A diffusion couple is formed by joining bars of two different materials, in
this case, copper (Cu) and nickel (Ni) is considered.
1. The couple is heated to high temperatures for a period of time (below
melting point) and cooled at room temperature.
2. Chemical analysis will show pure copper and pure nickel on the left
and right respectively, separated by a copper-nickel alloy.
3. This shows that copper atoms have diffused into the nickel and the
nickel atoms have diffused into the copper.
4. This process whereby atoms of one metal diffuse into another is
termed interdiffusion or impurity diffusion.
5. Diffusion also occurs in pure metals, this is termed self diffusion.
Stability of Atoms and Ions
• Imperfections are present or deliberately introduced into a
material; atoms or ions in their normal positions in the crystal
structures are not stable or at rest. Instead, the atoms or ions
possess thermal energy, and they will move.
• An atom may move from a normal crystal structure location to
occupy a nearby vacancy. An atom may also move from one
interstitial site to another.
• Atoms or ions may jump across a grain boundary, causing the
grain boundary to move.
• The ability of atoms and ions to diffuse increases as the
temperature, or thermal energy possessed by the atoms and
ions, increases.
Diffusion Analysis
• The experiment: we recorded combinations of
t and x that kept C constant.
x 
C(x i ,t i )  C o
i

 1 erf 

Cs C o
 2 Dt i 
= (constant here)
• Diffusion depth given by:
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004, 2006-
17
Data from Diffusion Analysis
• Experimental result: x ~ t0.58
• Theory predicts x ~ t0.50 from
• Close agreement.
Connecting Holes, Diffusion and Stress
z-axis
Red
octahedral fcc
Green + Red octahedral bcc
FCC represented as a BCT cell, with relevant holes.
Other types of diffusion (Beside Atomic)
• Flux is general concept: e.g. charges, phonons,…
• Charge Flux –
1 dq 1 d(Ne)
jq 

A dt A dt
dV
Defining conductivity 
j  
(a material property)
dx
I 1V
V
Ohm’s Law j q 

A AR
• Heat Flux –
(by phonons)
e = electric chg.
N = net # e- cross A
Solution: Fick’s 2nd Law
Vx  Vs  erf  x  
V V
2 t


0
s
l
Q
1 dH 1 d(N  )

A dt
A dt
Defining thermal conductivity 
(a material property)
Or w/ Thermal Diffusivity: h 
Q  
dT
dx

cV 
N = # of phonons with
avg. energy 
Q  cVT
Solution: Fick’s 2nd Law
Tx  T s
 erf
T0 Ts
x 


 2 ht 
Inter-diffusion (diffusion couples)
interface
Cu flux Ni flux
C1
Concentration
of Cu [kg/m 3 ]
Concentration
of Ni [kg/m 3 ]
C0
C2
Position, x
Assuming D A = DB:
For C x2
C1x C 0
x
 erf   
 2 Dt 
C1  C 0
C0= (C +C
/2
1
2)
Curve is symmetric about C ,0 erf z  erf z
MatSE 280: Introduction to Engineering Materials
©D.D. Johnson 2004, 2006-
Kirkendall Effect: What is D A > D B ?
• Kirkendall studied Mo markers in Cu-brass (i.e., fcc Cu70Zn30).
• Symmetry is lost: Zn atoms move more readily in one direction
(to the right) than Cu atoms move in the other (to the left).
Cu70Zn30
Cu
Cu
concentration
t=0
Zn
Mo wire markers
After diffusion
• When diffusion is asymmetric, interface moves away markers,
i.e., there is a net flow of atoms to the right past the markers.
• Analyzing movement of markers determines DZn and DCu.
• Kirkendall effect driven by vacancies, effective for T > 0.5 Tmelt.
Diffusion in Compounds: Ionic Conductors
• Unlike diffusion in metals, diffusion in compounds involves
second-neighbor migration.
• Since the activation energies are high, the D’s are low unless
vacancies are present from non-stoichiometric ratios of atoms.
e.g., NiO
There are
Schottky defects
Ni2+ O2–
Ni2+
O
Ni
O
Ni
O
Ni
O
• The two vacancies cannot accept neighbors because they have
wrong charge, and ion diffusion needs 2nd neighbors with high
barriers (activation energies).
Diffusion in Compounds: Ionic Conductors
• D’s in an ionic compound are seldom comparable because of size, change
and/or structural differences.
• Two sources of conduction: ion diffusion and via e- hopping from ions of variable
valency, e.g., Fe2+ to Fe3+, in applied electric field.
e.g., ionic
• In NaCl at 1000 K, DNa+~ 5DCl– ,whereas at 825 K DNa+ ~ 50DCl–!
• This is primarily due to size rNa+ = 1 A vs rCl–=1.8 A.
e.g., oxides
• In uranium oxide, U4+(O2–)2, at 1000 K (extrapolated), D O ~ 107 DU.
• This is mostly due to charge, i.e. more energy to activate 4+ U ion.
• Also, UO is not stoichiometric, having U3+ ions to give UO2-x, so that the
anion vacancies significantly increase O2- mobility.
e.g., solid-solutions of oxides (leads to defects, e.g., vacancies)
• If Fe1-xO (x=2.5-4% at 1500 K, 3Fe2+ -> 2Fe3+ + vac.) is dissolved in MgO
under reducing conditions, then Mg2+ diffusion increases.
• If MgF2 is dissolved in LiF (2Li+ -> Mg2+ + vac.), then Li+ diffusion increases.
All due to additional vacancies.
Ionic Conductors: related to fuel cells
• Molten salts and aqueous electrolytes conduct charge when placed in
electric field, +q and –q move in opposite directions.
• The same occurs in solids although at much slower rate.
• Each ion has charge of Ze (e = 1.6 x 10–19 amp*sec),
so ion movement induces ionic conduction
• Conductivity
  n Ze is related to mobility, , which is
related to D via the Einstein equations:   ZeD / kBT
• Hence
nZ 2 e 2
nZ 2 e 2
Doe Q/ RT
 ionic 
D
kBT
kBT
log10 
ionic
 nZ 2e2 
Q
~ ln 
Do 
 k BT
 2.3RT
So, electrical conduction can be used
determine diffusion data in ionic solids.
e.g., What conductivity results by Ca2+ diffusion in CaO at 2000 K?
• CaO has NaCl structure with a= 4.81 A, with D(2000 K)~10 -14m2/s, and Z=2.
nCa2
4
cell
cell (4.81x10 10
m)
3
 3.59x10 28/m3
1.3x10 5
nZ 2 e2

D~
kBT
ohm  cm
Example: Solid-oxide fuel cell (SOFC)
• SOFC is made up of four layers, three of
which are ceramics (hence the name).
• A single cell consisting of these four layers
stacked together is only a few mm thick.
• Need to stack many, many together to
have larger DC current.
Overall: H 2 + 1/2O2  H2O
H2  2H++2e–
O2– + 2H + + 2e–  H2O
S. Haile’s SOFC (2004 best): Thin-film of Sm-doped Ceria electrolyte (CeO2, i.e. SmxCe1-xO2-x/2)
and BSCF cathode (Perovskite Ba0.5Sr0.5Co0.8Fe0.2O3-d) show high power densities – over 1
W/cm2 at 600 C – with humidified H 2 as the fuel and air at the cathode.
Ceramic Compounds: Al2O3
Holes for diffusion
Unit cell defined by Al ions: 2 Al + 3 O
Summary: Structure and Diffusion
Diffusion FASTER for...
Diffusion SLOWER for...
• open crystal structures
• close-packed structures
• lower melting T materials
• higher melting T materials
• materials w/secondary
bonding
• materials w/covalent
bonding
• smaller diffusing atoms
• larger diffusing atoms
• cations
• anions
• lower density materials
• higher density materials
Summary
• Solid-state diffusion is mass transport at the
atomic level.
• Diffusion occurs using imperfections in solids i.e.
Vacancies and interstitials.
• The coefficient of diffusion is a function of
temperature.