Reserve Questions and Problems Solutions Manual
CHAPTER 6 DIFFUSION
Reserve Question 01: Rate of Diffusion
Diffusion by which mechanism occurs more rapidly in metal alloys?
a) Vacancy diffusion
b) Interstitial diffusion
Solution
In metal alloys, interstitial diffusion takes place more rapidly than vacancy diffusion because the
interstitial atoms are smaller and are more mobile. Also, there are more vacant adjacent interstitial
sites than there are vacancies.
Reserve Question 02: Temperature effect in diffusion
As temperature decreases, the fraction of total number of atoms that are capable of diffusive motion
a) increases.
b) decreases.
Solution
As temperature decreases, the fraction of the total number of atoms that are capable of diffusive
motion decreases.
Reserve Problem 03
A gas mixture is found to contain two diatomic A and B species for which the partial pressures of both
are 0.05065 MPa (0.5 atm). This mixture is to be enriched in the partial pressure of the A species by
passing both gases through a thin sheet of some metal at an elevated temperature. The resulting
enriched mixture is to have a partial pressure of 0.02026 MPa (0.2 atm) for gas A, and 0.01013 MPa
3
(0.1 atm) for gas B. The concentrations of A and B (CA and CB, in mol/m ) are functions of gas partial
pressures ( pA and p B , in MPa) and absolute temperature according to the following expressions:
2
2
⎛ 25.0 kJ/mol ⎞
CA = 200 pA exp ⎜ −
⎟
2
RT
⎝
⎠
⎛ 30.0 kJ/mol ⎞
CB = 1.0 × 10−3 pB exp ⎜ −
⎟
2
RT
⎝
⎠
Furthermore, the diffusion coefficients for the diffusion of these gases in the metal are functions of
the absolute temperature as follows:
⎛ 15.0 kJ/mol ⎞
DA (m2 /s) = 4.0 × 10−7 exp ⎜ −
⎟
RT
⎝
⎠
⎛ 24.0 kJ/mol ⎞
DB (m2 /s) = 2.5 × 10−6 exp ⎜ −
⎟
RT
⎝
⎠
Is it possible to purify the A gas in this manner? If so, specify a temperature at which the process
may be carried out, and also the thickness of metal sheet that would be required. If this procedure is
not possible, then state the reason(s) why.
Solution
This problem calls for us to ascertain whether or not an A2-B2 gas mixture may be enriched with
respect to the A partial pressure by allowing the gases to diffuse through a metal sheet at an elevated
temperature. If this is possible, the temperature and sheet thickness are to be specified; if such is not
possible, then we are to state the reasons why. Since this situation involves steady-state diffusion, we
employ Fick’s first law, Equation 6.2. Inasmuch as the partial pressures on the high-pressure side of
the sheet are the same, and the pressure of A2 on the low-pressure side is 2.0 times that of B2, and
concentrations are proportional to the square root of the partial pressure, the diffusion flux of A, JA,
is the square root of 2.0 times the diffusion flux of nitrogen JB—i.e.
J A = 2.0 J B
Thus, equating the Fick’s law expressions incorporating the given equations for the diffusion
coefficients and concentrations in terms of partial pressures leads to the following
JA
=
(200 )
(
0.05065 MPa −
1
×
Δx
⎛ 25.0 kJ ⎞
⎛ 15.0 kJ ⎞
0.02026 MPa exp ⎜ −
4.0 × 10 −7 m 2 /s ) exp ⎜ −
(
⎟
⎟
⎜
⎜
RT ⎟⎠
RT ⎟⎠
⎝
⎝
)
2.0 J B
=
(1.0 × 103 )
(
0.05065 MPa −
2.0
×
Δx
)
⎛ 30.0 kJ ⎞
⎛ 24.0 kJ ⎞
−6
2
0.01013 MPa exp ⎜ −
⎟ (2.5 × 10 m /s) exp ⎜ −
⎟
RT ⎠
RT ⎠
⎝
⎝
The Δx's cancel out, which means that the process is independent of sheet thickness. Now solving
the above expression for the absolute temperature T gives
T = 401 K (128°C)
Thus, it is possible to carry out this procedure at 401 K or 128°C.
Reserve Problem 04
If [m] atoms of helium pass through a [a] square meter plate area every [t] hours, and if this flux is
constant with time, compute the flux of helium in units of atoms per square meter per second.
Solution:
The definition of flux is the quantity of something (which could be in terms of particle count, mass,
etc.) that passes through a specified unit of area in a specified unit of time.
In this case, we are provided an atom count that passes through the known area each hour.
Therefore, accounting for the seconds in each hour:
J=
M
[m]
=
2
At ([a] m )([t ] hr(3600 s/ hr))
Reserve Problem 05
2
If water molecules pass through a membrane with a steady state flux of [j] mole/(m day), how long
will it take, in hours, for [m] kg of water to pass through a [a] square centimeter of the membrane?
Solution:
The definition of flux is the quantity of something (which could be in terms of particle count, mass,
etc.) that passes through a specified unit of area in a specified unit of time.
In this case, we are provided the flux of water molecules and are asked to compute the time required
for a certain mass of water molecules to transit a specific area of membrane.
Therefore, accounting for the number of hours in each day:
⎛ 1000 g ⎞ ⎛ 1 mole ⎞
[m] kg ⎜
⎟⎜
⎟
⎛ 24 hr ⎞
M
⎝ 1 kg ⎠ ⎝ 18 g ⎠
t=
=
⎜
⎟
AJ
⎛ 1 m2 ⎞⎛
mol ⎞ ⎝ 1 day ⎠
2
([a] cm ) ⎜
[ j]
⎜ 1002 cm2 ⎟⎜
⎟ ⎜ m2 day ⎟⎟
⎠
⎝
⎠⎝
Reserve Problem 06a (question pool)
The cornea is the transparent outer layer of the human eye. Because it must be transparent to light,
it does not normally contain blood vessels. Therefore, it must receive its nutrients via diffusion.
Oxygen from the surrounding air diffuses to the cornea through the surface tears whereas other
nutrients diffuse to the cornea from the inner parts of the eye, such as the vitreous humor and lens.
During operation, the cornea produces waste in the form of CO2 gas that must be expelled to keep
the eye healthy and functioning. This is accomplished by the simultaneous diffusion of CO2 from the
cornea to the surrounding atmosphere, which generally features a low CO2 concentration.
It is therefore critical that modern contact lens materials allow sufficient diffusion rates of oxygen
and carbon dioxide. Without oxygen, the cornea will warp, loose transparency, and become
susceptible to scarring. The body may also react by growing additional blood vessels into the eye,
which can damage the cornea.
If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired,
which of the following contact lens / ambient condition modifications is not likely to be useful?
Note: the flow rate is equal to product of the diffusion flux and an area of interest through which
diffusion occurs.
(a) Increase the contact lens thickness
(b) Increase the diffusivity of oxygen gas by decreasing the contact lens porosity
(c) Increase the ambient temperature
(d) Increase the ambient partial pressure of oxygen gas
(e) All of the suggestions (a-d) are useful for increasing the flow rate of oxygen
Solution:
Note: Since the area of the contact lens (A) doesn’t change, the flow rate of oxygen molecules (J*A)
increases whenever the diffusion flux of oxygen (J) increases.
(a) If the contact lens thickness is increased and all other factors are preserved, the concentration
gradient decrease and the oxygen diffusion flux should decrease.
(b) If the lens features more voids, we expect the oxygen to permeate the membrane faster since
oxygen should be able to transit voids faster than the bulk lens material.
(c) Increasing the temperature should increase the diffusivity and therefore increase the diffusion
flux and flow rate.
(d) Increasing the concentration of oxygen in the ambient environment will increase the
concentration gradient, thereby increasing the diffusion flux of oxygen.
Reserve Problem 07: Non steady-state – Specific concentration at different T
For a steel alloy it has been determined that a carburizing heat treatment of 16 h duration at 757°C
will raise the carbon concentration to 0.5 wt% at a point 2.3 mm from the surface. Estimate the time
necessary to achieve the same concentration at a 8 mm position for an identical steel and at a
-5
2
carburizing temperature of 1130°C. Assume that D0 is 4.6 × 10 m /s and Qd is 104 kJ/mol.
Solution
For this problem the solution to Fick's second law is according to Equation 6.6b, that is
x2
= constant
Dt
For two temperatures, T1 and T2, we may write the following expression:
x12
x2
= 2
D1t1 D2t2
Furthermore, the dependence of D depends on T is represented by Equation 6.8, which means that
⎛ Q ⎞
D1 = D0 exp ⎜⎜ − d ⎟⎟
⎝ RT1 ⎠
⎛ Q ⎞
D2 = D0 exp ⎜⎜ − d ⎟⎟
⎝ RT2 ⎠
Combining these two expressions into the previous equation, leads to the following:
x12
⎛ Q ⎞
t1 exp ⎜ − d ⎟
⎝ RT1 ⎠
=
x22
⎛ Q ⎞
t2 exp ⎜ − d ⎟
⎝ RT2 ⎠
Assuming that t2 is unknown time to be determined, the above equation takes the form
⎛ Q ⎞
t1 x 22 D0 exp ⎜ − d ⎟
⎝ RT1 ⎠
t2 =
⎛ Q ⎞
x12 D0 exp ⎜ − d ⎟
⎝ RT2 ⎠
which reduces to the following
⎛ t x2 ⎞
⎡ Q ⎛ 1 1 ⎞⎤
⎜
t2 = 1 2 ⎟ exp ⎢− d ⎜⎜ − ⎟⎟⎥
⎜ x2 ⎟
⎣⎢ R ⎝ T1 T2 ⎠⎦⎥
⎝ 1 ⎠
From our problem statement:
t1 = 17 h
T1 = 898°C (= 1171 K)
-3
x1 = 3.1 mm (3.1 × 10 m)
T2 = 1010°C (= 1283 K)
-3
x2 = 5.4 mm (5.4 × 10 m)
Qd = 118,000 J/mol
-5
2
D0 = 2.5 ×10 m /s
When these values are included into the above equation, the value of t2 is computed as follows:
⎡ (17h)(5.4 × 10 −3 m) 2 ⎤
⎡ (118,000J/mol) ⎛ 1
1 ⎞⎤
−
t2 = ⎢
⎥ exp ⎢ −
⎜
⎟⎥
−3
2
⎢⎣ (3.1 × 10 m)
⎥⎦
⎣ 8.31J/mol-K ⎝ 1171K 1283K ⎠ ⎦
= 17.9 h
Reserve Problem 08a (question pool)
The figure below features diffusion profiles that developed within four separate plain-carbon steel
specimens of equivalent geometry after they separately experienced a carburization process. The
specimens were prepared such that the scenario of a one-dimensional semi-infinite solid applies as a
solution to Fick's 2nd Law of Diffusion. Each part originally contained a uniform distribution of
carbon, and each part was processed using the same carburization temperature. In the figure, x = 0
corresponds to the surfaces of the steel parts that were exposed to the carbon-rich atmosphere during
the diffusion process.
Answer True or False for each of the following statements.
(a) The carburization surface was maintained at 1.00 wt% carbon for each specimen.
(b) Comparing the finished specimens at a depth of 0.75 mm, specimen A features the largest carbon
concentration.
(c) Comparing the finished specimens as a whole, specimen A features the lowest overall amount of
carbon.
(d) Specimen B experienced a longer carburization time compared to specimen C.
(e) The initial concentration of carbon in each part (prior to carburization) was a little less than 0.25
wt% carbon.
Solution:
(a) The carbon content at x = 0 equals 1.00 wt% carbon.
(b) The concentration profile of specimen D has the greatest value of at the specified position.
(c) The concentration profile of specimen D has the highest carbon content at every position.
Therefore, it has the highest overall carbon content.
(d) Since the temperature of each carburization was the same, a longer diffusion time will result in a
profile with higher carbon content at each position. Therefore, specimen C experienced a longer
carburization time than specimen B.
(e) The concentration profiles approach a value slightly less than 0.25 wt% carbon as the x position
increases.
Reserve Problem 09
The diffusion coefficient for aluminum in silicon is DAl in Si = 3x10
is about room temperature).
-16
2
cm /s at 300 K (note that 300 K
What is a reasonable value for DAl in Si at 600 K ?
Note: Rather than performing a specific calculation, you should be able to justify your answer from
the options below based on the mathematical temperature dependence of the diffusion coefficient.
(a) D < 3x10
-16
(b) D = 3x10
(c) D = 6x10
-16
-16
(d) D = 1.5x10
(e) D > 6x10
2
cm /s
2
cm /s
-16
-16
(f) D = 6x10
2
cm /s
-17
2
cm /s
2
cm /s
2
cm /s
Solution:
We expect the diffusion coefficient to increase if the temperature of this system is increased.
Therefore, options (a), (b), (d), and (f) are eliminated.
Furthermore, we expect that since the diffusion coefficient is exponentially dependent on
temperature, the diffusivity should increase by more than a factor of two if the absolute temperature
is doubled.
Reserve Problem 10a (question pool)
Depicted below are five different steady-state concentration profiles for the same gas across five
separate and identical plastic membranes at the same temperature. Which concentration profile
results in the lowest diffusion flux through the membrane?
Solution:
Since we are comparing diffusion profiles for the same diffuser and host in each case, and the
temperature is the same for each of the cases, only the concentration gradient needs to be compared.
Since the diffusion flux is directly proportional to the concentration gradient, the case featuring the
lowest magnitude slope should exhibit the lowest magnitude diffusion flux.
Specifically, since case (a) has no apparent gradient (a uniform concentration is implied), there is no
driving force for chemical diffusion.
Reserve Problem 11
One integrated circuit design calls for the diffusion of arsenic into silicon wafers; the background
20
3
concentration of As in Si is 2.5 × 10 atoms/m . The predeposition heat treatment is to be conducted
26
3
at 1000°C for 45 minutes, with a constant surface concentration of 8 × 10 As atoms/m . At a drivein treatment temperature of 1100°C, determine the diffusion time required for a junction depth of
-3
2
1.2 μm. For this system, values of Qd and D0 are 4.10 eV and 2.29 × 10 m /s, respectively.
Solution
This problem asks that we compute the drive-in diffusion time for arsenic diffusion in silicon. It is
first necessary to determine the value of Q0 using Equation 6.12. But before this is possible, the value
of Dp at 1000°C must be computed with the aid of Equation 6.8. Thus,
⎛ Q
D p = D0 exp ⎜ − d
⎜ kT p
⎝
⎞
⎟
⎟
⎠
⎡
⎤
4.10 eV
= (2.29 × 10 −3 m2 /s) exp ⎢ −
⎥
−5
⎣⎢ (8.62 × 10 eV/atom-K)(1000°C + 273 K) ⎦⎥
= 1.36 × 10 −19 m2 /s
Now for the computation of Q0 using Equation 6.12:
Q0 = 2Cs
= (2)(8 × 1026 atoms/m3 )
Dpt p
π
(1.36 × 10−19 m2 /s)(45 min)(60 s/min)
π
= 1.73 × 1019 atoms/m2
We now desire to calculate td in Equation 6.13. Algebraic manipulation and rearrangement of this
expression leads to
⎛ x2
j
exp⎜
⎜ 4Dd td
⎝
⎞
Q0
⎟=
⎟ CB π Dd td
⎠
At this point it is necessary to determine the value of Dd (at 1100°C). Thus, using Equation 6.8
⎡
⎤
4.10 eV
Dd = (2.29 × 10 −3 m2 /s) exp ⎢ −
⎥
−5
⎣⎢ (8.62 × 10 eV/atom-K)(1100°C + 273 K) ⎦⎥
= 2.06 × 10 −18 m2 /s
And incorporation of values of all parameters except td in the above expression yields
⎡
(1.2 × 10 −6 m)2
exp ⎢
⎢⎣ (4)(2.06 × 10 −18 m2 /s)td
⎤
1.73 × 1019 atoms/m2
⎥=
⎥⎦ (2.5 × 1020 atoms/m3 ) (π )(2.06 × 10 −18 m2 /s)t
d
which expression reduces to
⎛ 1.75 × 10 5
exp ⎜
⎜
td
⎝
s ⎞ 2.72 × 10 7 s1/2
⎟=
⎟
td
⎠
Solving for td is not a simple matter. One possibility is to use a graphing technique. Let us take the
logarithm of both sides of the above equation, which gives
⎛ 2.72 × 107 s1/2 ⎞
1.75 × 105 s
⎟
= ln ⎜
⎜
⎟
td
t
d
⎝
⎠
Now if we plot the terms on both left and right hand sides of this equation versus td, the value of td at
the point of intersection of the two resulting curves is correct answer. Below is such a plot:
As noted, the two curves intersect at about 13,900 s, which corresponds to td = 3.86 h.
Reserve Problem 12
Phosphorus atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat
19
treatments; the background concentration of P in this silicon material is known to be 5 × 10
3
atoms/m . The predeposition treatment is to be conducted at 950°C for 45 minutes; the surface
26
3
concentration of P is to be maintained at a constant level of 1.5 × 10 atoms/m . Drive-in diffusion
will be carried out at 1200°C for a period of 2.5 h. For the diffusion of P in Si, values of Qd and D0 are
-4
2
3.40 eV and 1.1 × 10 m /s, respectively.
(a) Calculate the value of Q0.
(b) Determine the value of xj for the drive-in diffusion treatment.
(c) Also for the drive-in treatment, compute the position x at which the concentration of P atoms is
24
-3
10 m .
Solution
(a) For this portion of the problem we are asked to determine the value of Q0. This is possible using
Equation 6.12. However, it is first necessary to determine the value of D for the predeposition
treatment [Dp at Tp = 950°C (1223 K)] using Equation 6.8. Thus
⎛ Q
D p = D0 exp ⎜ − d
⎜ kT p
⎝
⎞
⎟
⎟
⎠
⎡
⎤
3.40 eV
= (1.1 × 10 −4 m 2 /s) exp ⎢ −
⎥
⎢⎣ (8.62 × 10 −5 eV/atom-K)(1223 K) ⎥⎦
= 1.08 × 10 −18 m 2 /s
The value of Q0 may be determined as follows:
Q0 = 2Cs
= (2)(1.5 × 1026 atoms/m3 )
Dpt p
π
(1.08 × 10 −18 m2 /s)(45 min)(60 s/min)
π
18
= 9.14 × 10 atoms/m
2
(b) Computation of the junction depth requires that we use Equation 6.13. However, before this is
possible it is necessary to calculate D at the temperature of the drive-in treatment [Dd at 1200°C
(1473 K)]. Thus,
⎡
⎤
3.40 eV
Dd = (1.1 × 10 −4 m 2 /s) exp ⎢ −
⎥
−5
⎣⎢ (8.62 × 10 eV/atom-K)(1473 K) ⎦⎥
= 2.58 × 10 − 16 m 2 /s
Now from Equation 6.13
⎡
⎛
Q0
x j = ⎢(4Dd td )ln ⎜
⎜ C πD t
⎢
d d
⎝ B
⎣
1/2
⎞⎤
⎟⎥
⎟⎥
⎠⎦
1/2
⎧
⎡
⎤⎫
9.14 ×1018 atoms/m2
⎪
−16 2
⎢
⎥ ⎪⎬
= ⎨(4)(2.58 ×10 m /s)(9000 s) ln
⎢ (5 ×1019 atoms/m3 ) (π )(2.58 × 10−16 m2 /s)(9000 s) ⎥ ⎪
⎪⎩
⎣
⎦⎭
= 1.21×10−5 m = 12.1 μm
24
3
(c) For a concentration of 10 P atoms/m for the drive-in treatment, we compute the value of x
using Equation 6.11. However, it is first necessary to manipulate Equation 5.11 so that x is the
dependent variable. Taking natural logarithms of both sides leads to
⎛ Q
0
ln C ( x,t ) = ln ⎜
⎜ πD t
d d
⎝
⎞
x2
⎟−
⎟ 4 Dd td
⎠
Now, rearranging and solving for x leads to
⎧
⎡
Q0
⎪
x = ⎨(4 Dd td )ln ⎢
⎢⎣ C (x,t ) π Dd td
⎪⎩
1/2
⎤ ⎫⎪
⎥⎬
⎥⎦ ⎪
⎭
24
3
Now, incorporating values for Q0 and Dd determined above and taking C(x,t) = 10 P atoms/m
yields
1/2
⎧
⎡
⎤⎫
9.14 × 1018
⎪
⎥ ⎪⎬
x = ⎨(4)(2.58 × 10 −16 )(9000) ln ⎢
24
−16
⎢
⎥
⎪⎩
⎣ (10 ) (π )(2.58 × 10 )(9000) ⎦ ⎪⎭
= 3.36 × 10 −6 m = 3.36 μm
Reserve Problem 13
Aluminum atoms are to be diffused into a silicon wafer using both predeposition and drive-in heat
19
treatments; the background concentration of Al in this silicon material is known to be 3 × 10
3
atoms/m . The drive-in diffusion treatment is to be carried out at 1050°C for a period of 4.0 h, which
gives a junction depth xj of 3.0 μm. Compute the predeposition diffusion time at 950°C if the surface
25
3
concentration is maintained at a constant level of 2 × 10 atoms/m . For the diffusion of Al in Si,
-4
2
values of Qd and D0 are 3.41 eV and 1.38 × 10 m /s, respectively.
Solution
This problem asks that we compute the time for the predeposition heat treatment for the diffusion of
Al in Si. In order to do this, it is necessary to determine the value of Q0 from Equation 6.13.
However, before doing this we must first calculate Dd, using Equation 6.8. Therefore
⎛ Q
Dd = D0 exp ⎜⎜ − d
⎝ kTd
⎞
⎟⎟
⎠
⎡
⎤
3.41 eV
= (1.38 × 10−4 m2 /s)exp ⎢−
⎥
−5
⎣⎢ (8.62 × 10 eV/atom-K)(1050°C + 273 K) ⎦⎥
= 1.43 × 10−17 m2 /s
Now, solving for Q0 in Equation 6.13 leads to
⎛ x2 ⎞
j
⎟
Q0 = CB πDd td exp⎜
⎜ 4Dd td ⎟
⎝
⎠
(
)
In the problem statement we are given the following values:
19
3
CB = 3 × 10 atoms/m
td = 4 h (14,400 s)
-6
xj = 3.0 μm = 3.0 × 10 m
Therefore, incorporating these values into the above equation yields
⎡
⎤
(3.0 × 10 −6 m)2
Q0 = ⎡ (3 × 1019 atoms/m3 ) (π )(1.43 × 10 −17 m2 /s)(14,400 s) ⎤ exp ⎢
⎥
−17
2
⎢⎣
⎥⎦
⎢⎣ (4)(1.43 × 10 m / s )(14, 400 s) ⎥⎦
= 1.34 × 1018 atoms/m 2
We may now compute the value of tp using Equation 6.12. However, before this is possible it is
necessary to determine Dp (at 950°C) using Equation 6.8. Thus
⎡
⎤
3.41 eV
D p = (1.38 × 10 −4 m 2 /s) exp ⎢ −
⎥
−5
⎣⎢ (8.62 × 10 eV/atom-K)(950°C + 273 K) ⎦⎥
= 1.24 × 10 −18 m 2 /s
Now, solving for tp in Equation 6.12 we get
tp =
πQ02
4Cs2 Dp
25
3
And incorporating the value of Cs provided in the problem statement (2 × 10 atoms/m ) as well as
values for Q0 and Dp determined above, leads to
tp =
(
π 1.34 × 1018 atoms/m2
(
)
2
)
2
(4) 2 × 1025 atoms/m3 (1.24 × 10−18 m2 /s)
= 2.84 × 103 s = 47.4 min