Probabilidad y Estadistica para Ingenieria y Ciencias - Devore 7th
Probabilidad y Estadistica para Ingenieria y Ciencias - Devore 7th
Probabilidad y Estadistica para Ingenieria y Ciencias - Devore 7th
DIFFUSION
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1. DIFFUSION
At any temperature different from absolute zero all atoms, irrespective of their state of
aggregation (gaseous, liquid or solid), are constantly in motion. Since the movement of
particles is associated with collisions, the path of a single particle is a zigzag one.
higher to places of lower concentration (fig. 1). For this reason diffusion is known as a
transport phenomenon.
In each diffusion reaction (heat flow, for example, is also a diffusion process), the flux
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from the above considerations that this diffusion constant (D) reflects the mobility of the
diffusing species in the given environment and accordingly assumes larger values in
!C
distance in the x-direction. If a
rate of change of the concentration with distance (dc/dx), which is the same as the
slope of a graph of concentration vs. position (!c/!x) (see fig. 2).
Diffusion processes may be divided into two types: (a) steady state and (b) nonsteady
state. Steady state diffusion takes place at a constant rate - that is, once the process
starts the number of atoms (or moles) crossing a given interface (the flux) is constant
with time. This means that throughout the system dc/dx = constant and dc/dt = 0.
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Nonsteady state diffusion is a time dependent process in which the rate of diffusion is a
function of time. Thus dc/dx varies with time and dc/dt # 0. Both types of diffusion are
described quantitatively by Fick’s laws of diffusion. The first law concerns both steady
state and nonsteady state diffusion, while the second law deals only with nonsteady
state diffusion.
On the basis of the above considerations, Fick’s First Law may be formulated as:
J ! " D dc
dx
# $ In words: The diffusive flux is
proportional to the
existing concentration
gradient.
The negative sign in this relationship indicates that particle flow occurs in a “down”
gradient direction, i.e. from regions of higher to regions of lower concentration. The flux
#
J moles
cm 2s
$ ! " D #dc
dx
$ # moles % cm $
cm
"3
Thus: D = cm2/s
Like chemical reactions, diffusion is a thermally activated process and the temperature
dependence of diffusion appears in the diffusivity as an “Arrhenius-type” equation:
D ! D oe "E a&RT
the rate constant) includes such factors as the jump distance, the vibrational frequency
of the diffusing species and so on. Selected values of D, Do and Ea are given in Table 1
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TABLE 1
O2 Air 0 0.178
H2 Air 0 0.611
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A typical application of Fick’s first law: Determine the rate at which helium (He), held at
5 atm and 200°C in a Pyrex glass bulb of 50 cm diameter and a wall thickness (x) of
0.1 cm, diffuses through the Pyrex to the outside. Assume that the pressure outside the
tube at all times remains negligible (see fig. 3). (For the diffusion of gases it is
He J
P1
200oC
P2 = 5 atm J = - K !P / !x
P2
P1 = 0 atm
!x
Figure 3 Conditions for outdiffusion of He from a glass bulb.
customary, although not necessary, to replace the diffusion constant D with the
Permeation constant K, normally given in units of cm2/s.atm. Using the gas laws, K is
readily converted to D if so desired.)
K = 1 x 10–9 cm2/s.atm
J ! " K dP
dx
# $ (and operate with pressures instead
of concentrations)
Jdx = –KdP
x!0.1 p 1!0
J ! K 5.0
0.1
We can forego the integration since (dP/dx) = (!P/!x) and we may immediately write:
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J = 3.9 x 10–4 ??
2 atm cm 2 ! cm 3
J ! " K x !P x A ! cm cm s
!x s % atm 1
The total flux is 3.9 x 10–4 cm3/s (with the gas volume given for 0°C and 1 atm). If the
total gas flow by diffusion were to be determined for a specified time interval, the
detail but we can consider the second law qualitatively and examine some relevant
solutions quantitatively.
The difference between steady state and nonsteady state diffusion conditions can
readily be visualized (fig. 4). In the first case we have, for example, the diffusion of gas
p2 = const.
p2
membrane !x !x
!p/!x = constant !p/!x = f (time)
Figure 4 Steady state and Non-steady state diffusion
from an infinite volume (P1 const) through a membrane into an infinite volume (P2
const). The pressure gradient across the membrane remains constant as does the
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diffusive flux. In the second case we deal with diffusion from a finite volume through a
membrane into a finite volume. The pressures in the reservoirs involved change with
(You are not required to be familiar with the following derivation of the Second Fick’s
element equals the rate of accumulation of the material into this volume element:
J x " J x(dx ! )c dx
)t
[c is the average concentration in the volume element and cdx is the total amount of the
diffusing material in the element at time (t).]
)J x ) 2J x dx 2
J x(dx ! J x ( dx ( ( ...
)x )x 2 2
Accordingly, as dx " 0:
)
)x
#D dx
)c$ ! )c
)t
and if D does not vary with x (which is normally the case) we have the formulation of
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In physical terms this relationship states that the rate of compositional change is
proportional to the “rate of change” of the concentration gradient rather than to the
The solutions to Fick’s second law depend on the boundary conditions imposed by the
(a) c2 (b)
Solid (1)
Solid
conc. of comp.2
c = f (t,x)
c = c2 erfc x / 2 Dt
c2'
c2' = 0
x x
Figure 6 Diffusion at constant surface concentration ; bulk concentration of component
2 at start of diffusion is c2' (in a) and is zero (in b).
for the generation of junction devices (p-n junctions, junction transistors)]. The
boundary conditions are: the concentration of component (2) at the surface of the solid
phase (x=0) remains constant at c2 and the concentration of component (2) in the solid
prior to diffusion is uniformly c2$ (a). Under these boundary conditions the solution to
c2 " c
c 2 " c 2* ! erf # $ x
+
2 Dt
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If no component (2) is originally in the solid matrix (1) (b), the above solution is simpler:
c2 " c
c2 ! erf # $ x
+
2 Dt
1 " cc ! erf
2
# $ x
+
2 Dt
c
c 2 ! 1 " erf# $ x
+
2 Dt
c
c 2 ! erfc # $
x
+
2 Dt
c ! c 2 erfc # $
x
+
2 Dt
An analysis shows that the last form of the solution to Fick’s law relates the
concentration (c) at any position (x) (depth of penetration into the solid matrix) and time
(t) to the surface concentration (c2) and the diffusion constant (D). The terms erf and
erfc stand for error function and complementary error function respectively - it is the
Another look at the above solution to the diffusion equation shows that the
concentration (c) of component (2) in the solid is expressed in terms of the error
function of the argument x&2 +Dt. To determine at what depth a particular concentration
(c*) of (2) will appear, we substitute this concentration for (c) and obtain:
c * ! erfc
c2 # $
x
+
2 Dt
! K
Since the error function is a constant, its argument must also be a constant:
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c2
conc. of component 2
Solid (1)
c*
cx
t1 t2 t3
x x
Figure 7 Advance of the concentration front (c*)
as a function of trime
x ! K*
+
2 Dt
x ! K* 2 +Dt
x ! K* * +Dt
x , +t
4. SELF-DIFFUSION
and as such will lead to local displacements of individual atoms. This random
movement of atoms within a lattice (self-diffusion), which is not associated with any
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Ni 58 60 61 62 64
28, Ni 28, Ni 28, Ni 28 and Ni 28
If Ni 58
28 is irradiated with neutrons in a nuclear reactor, it will capture a neutron and
become Ni 59
28 which is radioactive (a radio-isotope).
Ni 58 59 59
28 ( n - Ni 28 - Co 27 ( " ( #
Nickel 59 is characterized by its instability which leads to the emission of & and '
radiation, with a half-life of 8 x 104 years. Since this radiation can be measured by
appropriate radiation detectors, it is possible to use Ni59 as a “tracer element” for
studies of self-diffusion.
The radioactive nickel (which is identical to ordinary nickel with the exception of its
subsequently placed into a furnace and heated up to close to its melting point for an
extended period of time. After removing the specimen, it is sectioned into slices parallel
to the surface which contained the radioactive tracer element. With the aid of a
radiation detector it can now be shown that the Ni59, which originally was only at the
surface, has diffused into the bulk material while simultaneously some bulk nickel has
counter-diffused in the other direction. If this sample is heat-treated for a much longer
time, sectioning and counting will reveal a completely uniform distribution of the
radio-tracer element. It can thus be shown that self-diffusion does occur in solids, and
5. DIFFUSION MECHANISMS
The diffusion process in interstitial solid solutions, like that of carbon in iron, can readily
readily explained. The magnitude of the observed activation energy indicates that a
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mechanism whereby atoms simply change places with each other has to be excluded.
More reasonable mechanisms were suggested by Frenkel and Schottky. They
mechanism by which atoms can move (diffuse) within a crystal. The concentration of
such vacancies, as you recall, can be calculated from simple statistical calculations.
In most solids we are not dealing with single crystals but rather with polycrystalline
volume diffusion (Dvolume < Dg-boundary). FInally, surface diffusion, which takes place on
all external surfaces, is even higher (Dvolume < Dg-boundary < Dsurface). The respective
activation energies for diffusion are:
Diffusion in Non-Metals
systems. Oxygen, for example, diffuses through many oxides by vacancy migration. In
crystalline oxides and in silicate glasses as well, it is found that oxygen diffuses much
more rapidly than the metallic ion. In glasses containing alkali atoms (Na+, K+), the
requires the motion of large molecules since intramolecular bonding is much stronger
than intermolecular bonding. This fact explains that the diffusion rates in such materials
Some gases, like hydrogen and helium, diffuse through some metals with ease even at
room temperature. Helium, for example, will diffuse through quartz and steel and limits
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TABLE 2
z erf(z) z erf(z)
0 0 0.85 0.7707
0.025 0.0282 0.90. 0.7970
0.05 0.0564 0.95 0.8209
0.10 0.1125 1.0 0.8427
0.15 0.1680 1.1 0.8802
0.20 0.2227 1.2 0.9103
0.25 0.2763 1.3 0.9340
0.30 0.3286 1.4 0.9523
0.35 0.3794 1.5 0.9661
0.40 0.4234 1.6 0.9763
0.45 0.4755 1.7 0.9838
0.50 0.5205 1.8 0.9891
0.55 0.5633 1.9 0.9928
0.60 0.6039 2.0 0.9953
0.65 0.6420 2.2 0.9981
0.70 0.6778 2.4 0.9993
0.75 0.7112 2.6 0.9998
0.80 0.7421 2.8 0.9999
Works Agency, Works Projects Administration, New York, 1941. A discussion of the
evaluation of erf(z), its derivatives and integrals, with a brief table is given by
H. Carslaw and J. Jaeger, in Appendix 2 of “Conduction of Heat in Solids”, Oxford
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1. Aluminum is to be diffused into a silicon single crystal. At what temperature will the
diffusion constant (D) be 5 x 10–11 cm2/s? (Given: EA = 315 kJ/mole and Do =
1.70 cm2/s.)
2. The diffusion constant of lithium in silicon is 10–5 cm2/s at 1100°C and 10–6 cm2/s
at 692°C. What are the values of EA and Do for diffusion of Li in silicon?
3. We know that in some nuclear reactors the probability of “fission” increases with
decreasing velocity of neutrons. For this reason we use, for example, graphite to
“moderate” (slow down) neutrons; the neutrons diffuse through the graphite matrix
and by collision are gradually slowed down. Do you expect the diffusion constant
(D) of neutrons in graphite to be small (~10–15) or large (~10–5) as in liquids, or
even larger? Explain your answer.
4. It is desired to diffuse In into pure silicon such that the In concentration at a depth
of 3 x 10–4 cm from the surface will be one-half of the surface concentration. How
long should the silicon be in contact with In at 1600K in order to accomplish this
diffusion? (D = 8 x 10–12 cm2/s)
5. A sample of Fe which has been “carburized” (carbon was diffused into it) at 930°C
for ten hours has a “carburized” depth of 0.04 cm. How long must this same Fe
sample be carburized (exposed to carbon) to produce a carburized depth of
0.08 cm?
6. Suppose the average grain size of a solidified Cu-Ni alloy is 10–2 cm. Each grain
exhibits a lower Cu composition in the center than on the periphery. Approximately
how long will it take to eliminate the “coring phenomenon” (make the alloy
homogeneous) through high temperature solid state diffusion (by annealing) at
1100°C. Given: Do = 2.7 cm2/s and EA = 235 kJ/mole.
7. Indium (In) is to be diffused into pure silicon such that the In concentration at a
depth (x) of 2.3 x 10–4 cm below the surface will be one-half of the surface
concentration (maintained constant throughout diffusion). How long must diffusion
be conducted at 1600K if D = 8 x 10–12 cm2/s?
8 Ga is diffused into a thick slice of silicon at a temperature of 1100°C for three (3)
hours. What is, after that time, the Ga concentration (Ga/cm3) at a depth of:
(a) 1 x 10–4 cm and (b) 3 x 10–4 cm? (A constant Ga source is used at the
surface, Co = 1018/cm3; D1100°C = 7 x 10–13 cm2/s.)
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10. Indium diffusion into silicon is to proceed at 1600°C until the In concentration at a
depth of 10–3 cm is one-half of Co (the constant surface concentration). What is
the time of diffusion if D1600°C = 8 x 10–12 cm2/s?
11. For small values of X the Gaussian error function (erf) can be approximated as
erf(X) . X. Suppose a silicon wafer is exposed to aluminum vapor at 1300°C and
the aluminum begins to diffuse into the silicon. How long will it take for the
concentration of Al at a point 0.01 cm below the surface to be 0.35 of that at the
surface? (D1300°C = 10–10 cm2/s.)
12. A sample of steel which has been carburized at 930°C for ten (10) hours has a
carburized depth of 0.04 cm. How long must this same steel be carburized at the
same temperature to produce a carburized depth of 0.08 cm?
13. It is intended to market a new steel for automotive applications. This steel, to be
effective, is to have 20% of the surface concentration of Cr at a depth of 0.02 cm
below the surface, to be achieved by diffusion at 1000°C. Determine if it is
economically feasible to produce this steel, knowing that Do = 0.47 cm2/s and
EA = 332 kJ/mole.
14. Gallium is to be diffused for three hours at a temperature of 1375K into a thick
slice of silicon. Given a constant surface concentration of gallium (Co = 1018/cm3)
(D1375 = 5 x 10–13 cm2/sec), after that time what will be the Ga concentration at a
depth of (a) 1 x 10–4 cm and (b) 3 x 10–4 cm?
15. A metal exposed to air is found to oxidize according to a linear rate law. Upon
addition of aluminum (Al) as an alloying constituent, the same metal oxidizes
according to a logarithmic rate law. Explain the oxidation behavior of the pure and
the alloyed metal.
16. To increase its corrosion resistance, chromium is diffused into steel at 980°C. If
during diffusion the surface concentration (cs) of chromium remains constant at
100%, how long will it take (in days) to achieve a Cr concentration of 1.8% at a
depth of 0.002 cm below the steel surface? (Do = 0.54 cm2/sec; EA = 286 kJ/mole)
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