14:635:407:02
Homework III Solutions
4.1 Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K).
Assume an energy for vacancy formation of 0.55 eV/atom.
Solution:
In order to com pute the fraction of atom sites that are vacan t in lead at 600 K, we m ust employ Equation
4.1. As stated in the problem, Qv = 0.55 eV/atom. Thus,


 Q 
0.55 eV / atom
Nv
= exp  v = exp 

5
 kT 
N
eV / atom - K) (600 K) 
 (8.62  10
= 2.41  10-5
4.15 T he co ncentration o f carb on i n an i ron-carbon al loy i s 0 .15 w t%. What i s the c oncentration i n
kilograms of carbon per cubic meter of alloy?
Solution:
In order to compute the concentration in kg/m3 of C in a 0.15 wt% C-99.85 wt% Fe alloy we must employ
Equation 4.9 as
CC" =
CC
 10 3
CC
CFe

C
Fe
From inside the front cover, densities for carbon and iron are 2.25 and 7.87 g/cm3, respectively; and, therefore
CC" =
0.15
 10 3
0.15
99.85

2.25 g/cm3
7.87 g/cm3
= 11.8 kg/m3
4.16 Determine the approximate density of a high- leaded brass that has a co mposition of 64.5 wt% Cu, 33.5
wt% Zn, and 2.0 wt% Pb.
Solution:
In order to solve this problem, Equation 4.10a is modified to take the following form:
 ave =
100
CCu
CZn
C

 Pb
Cu
 Zn
Pb
And, using the density values for Cu, Zn, and Pb—i.e., 8.94 g/cm3, 7.13 g/cm3, an d 11.35 g/cm3—(as taken from
inside the front cover of the text), the density is computed as follows:
 ave =
100
64.5 wt%
33.5 wt%
2.0 wt%


8.94 g / cm3
7.13 g / cm 3
11.35 g / cm3
= 8.27 g/cm3
Substitution of the concentration values (i.e., CA = 12.5 wt% and CB = 87.5 wt%) as well as v alues for the
other parameters given in the problem statement, into the above equation gives




100

(3.95  10-8 nm)3 (6.022  1023 atoms/mol)
12.5
wt%
87.5 wt% 




6.35 g/cm3 
4.27 g/cm3
n =




100


 12.5 wt%  87.5 wt% 


125.7 g/mol 
61.4 g/mol
= 2.00 atoms/unit cell
Therefore, on the basis of this value, the crystal structure is body-centered cubic.
4.20 Gold forms a su bstitutional sol id s olution w ith silver. Compute the number of gold atom s per cubic
centimeter for a silver-gold alloy that contains 10 wt% Au and 90 wt% Ag. The densities of pure gold and
silver are 19.32 and 10.49 g/cm3, respectively.
Solution:
To solve this problem, employment of Equation 4.18 is necessary, using the following values:
C1 = CAu = 10 wt%
1 = Au = 19.32 g/cm3
2 = Ag = 10.49 g/cm3
A1 = AAu = 196.97 g/mol
Thus
N Au =
=
N AC Au
CAu AAu
A
 Au (100  CAu )
 Au
 Ag
(6.022  10 23 atoms / mol) (10 wt%)
196.97 g / mol
(10 wt%)(196.97 g / mol)

(100  10 wt%)
10.49 g / cm3
19.32 g / cm 3
= 3.36  1021 atoms/cm3
4.27 For an FCC single cr ystal, would you e xpect the surface energy for a (100) plane to be greater or less
than that for a (111) plane? Why? (Note: You may want to consult the solution to Problem 3.54 at the end of
Chapter 3.)
Solution:
The surface e nergy for a c rystallographic plane will depend on its pac king density [i.e., the planar density
(Section 3.11)]—that is, the higher the packing density, the greater the number of nea rest-neighbor atoms, and t he
more ato mic bonds in that plane that are satisfied, an
d, consequently, the lowe r th e surface ene rgy. From the
1
1
solution to Problem 3.54, planar densities for FCC (100) and (111) planes are
and
, respectively—that
2
2
4R
2R 3
0.25
0.29
is
and
(where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have the
2
R
R2
lower surface energy.
4.28 For a B CC single crystal, would you expect the surface energy for a (100) plane to be greater or less
than that for a (110) plane? Why? (Note: You may want to consult the solution to Problem 3.55 at the end of
Chapter 3.)
Solution:
The surface e nergy for a c rystallographic plane will depend on its pac king density [i.e., the planar density
(Section 3.11)]—that is, the higher the packing density, the greater the number of nea rest-neighbor atoms, and t he
more ato mic bonds in that plane that are satisfied, an
d, consequently, the lowe r th e surface ene rgy. From the
3
3
solution to Problem 3.55, the planar densities for BCC (100) and (110) are
and
, respectively—that is
2
2
8R 2
16R
0.19
0.27
and
. Thus, since the planar density for (110) is greater, it will have the lower surface energy.
2
R
R2
4.29 (a) For
F a given material,
m
wo uld
d you expect the surface en
nergy to be ggreater than, tthe same as, oor less
than the grain
g
boundary energy? Wh
hy?
(b
b) The grain boundary
b
en ergy
e
of a sma
all-angle grain
n boun dary iss less than fo r a high-anglle one .
Why is thiis so?
Solution:
(aa) The s urfacee energy will b e greater than the grain bounndary energy. For grain bouundaries, some atoms
on one side of a boundarry will bond to
o atoms on the other side; suuch is no t the ccase for surfacce atoms. Therrefore,
b fewer unsatiisfied bonds allong a grain bo
oundary.
there will be
(b
b) The sm all-aangle grain bo undary energy
y is lower thann for a high-a nngle one beca uuse more atomss bond
across the boundary
b
for th
he small-anglee, and, thus, theere are fewer uunsatisfied bondds.
4.30 (a) Briefly
B
describe a twin and a twin bounda
ary.
(b
b) Cite the diffference between mechanica
al and annealin
ng twins.
Solution:
(aa) A twin bou
undary is an in
nterface such th
hat atoms on oone side are l oocated at mirroor image positiions of
those atom
ms situated on th
he other bound
dary side. The region on one side of this booundary is calleed a twin.
(b
b) Mechanicall twins are prod
duced as a resu
ult of mechaniical deformatioon and generallly occur in BC
CC and
HCP metalls. Annealing twins form durring annealing heat treatmentts, most often iin FCC metals..
4.31 F or each of the fo
ollowing stack
king sequencees found in F C
CC metals, ciite the type off planar defecct that
exists:
(a
a) . . . A B C A B C B A C B A . . .
(b
b) . . . A B C A B C B C A B C . . .
Now, copy
y the stacking sequences and
d indicate the position(s) off planar defectt(s) with a verrtical dashed lline.
Solution:
(a) The in
nterfacial defec
ct that exists fo
or this stacking
g sequence iss a twin bound
dary, which occcurs at the ind
dicated
position.
The stackin
ng sequence on
n one side of th
his position is mirrored
m
on thee other side.
b) Th e interfacial defect th at
a exists within
n this FCC sta ccking sequencee is a stackin g fault, which ooccurs
(b
between th
he two lines.
Within thiss region, the staacking sequencce is HCP.
14:440:407 Section 02 Fall 2010 SOLUTION OF HOMEWORK 04
Question 5.3: (a) Compare interstitial and vacancy atomic mechanisms for diffusion.
(b) Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion.
Solution of 5.3:
(a) With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy. Self-diffusion and the
diffusion of substitutional impurities proceed via this mechanism. On the other hand, atomic motion is from
interstitial site to adjacent interstitial site for the interstitial diffusion mechanism.
(b) Interstitial diffusion is normally more rapid than vacancy diffusion because: (1) interstitial atoms, being smaller,
are more mobile; and (2) the probability of an empty adjacent interstitial site is greater than for a vacancy adjacent
to a host (or substitutional impurity) atom.
Question 5.4: Briefly explain the concept of steady state as it applies to diffusion.
Solution of 5.4:
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.
Question 5.6: The purification of hydrogen gas by diffusion through a palladium sheet was discussed in Section 5.3.
Compute the number of kilograms of hydrogen that pass per hour through a 5-mm-thick sheet of palladium having
an area of 0.20 m2 at 500C. Assume a diffusion coefficient of 1.0  10-8 m2/s, that the concentrations at the highand low-pressure sides of the plate are 2.4 and 0.6 kg of hydrogen per cubic meter of palladium, and that steadystate conditions have been attained.
Solution of 5.6:
This problem calls for the mass of hydrogen, per hour, that diffuses through a Pd sheet. It first becomes necessary to
employ both Equations 5.1a and 5.3. Combining these expressions and solving for the mass yields
M = JAt =  DAt
C
x
0.6  2.4 kg / m3 
=  (1.0  10 -8 m2 /s)(0.20 m2 ) (3600 s/h) 

 5  103 m 
= 2.6  10-3 kg/h
Question 5.11: Determine the carburizing time necessary to achieve a carbon concentration of 0.45 wt% at a
position 2 mm into an iron–carbon alloy that initially contains 0.20 wt% C. The surface concentration is to be
maintained at 1.30 wt% C, and the treatment is to be conducted at 1000C. Use the diffusion data for -Fe in Table
5.2.
Solution of 5.11:
In order to solve this problem it is first necessary to use Equation 5.5:
C x  C0
 x 
= 1  erf 

2 Dt 
Cs  C0
wherein, Cx = 0.45, C0 = 0.20, Cs = 1.30, and x = 2 mm = 2  10-3 m. Thus,
 x 
0.45  0.20
Cx  C0
=
= 0.2273 = 1  erf 

2 Dt 
1.30  0.20
Cs  C0
or
 x 
erf 
= 1  0.2273 = 0.7727
2 Dt 
By linear interpolation using data from Table 5.1
z
erf(z)
0.85
0.7707
z
0.7727
0.90
0.7970
0.7727  0.7707
z  0.850
=
0.900  0.850 0.7970  0.7707
From which
z = 0.854 =
x
2 Dt
Now, from Table 5.2, at 1000C (1273 K)


148,000 J/mol
D = (2.3  10 -5 m2 /s) exp 

 (8.31 J/mol- K)(1273 K) 
= 1.93  10-11 m2/s
Thus,
0.854 =
2  103 m
(2) (1.93  1011 m2 /s) (t)
Solving for t yields
t = 7.1  104 s = 19.7 h
Question 5.12: An FCC iron-carbon alloy initially containing 0.35 wt% C is exposed to an oxygen-rich and
virtually carbon-free atmosphere at 1400 K (1127C). Under these circumstances the carbon diffuses from the alloy
and reacts at the surface with the oxygen in the atmosphere; that is, the carbon concentration at the surface position
is maintained essentially at 0 wt% C. (This process of carbon depletion is termed decarburization.) At what position
will the carbon concentration be 0.15 wt% after a 10-h treatment? The value of D at 1400 K is 6.9  10-11 m2/s.
Solution of 5.12:
This problem asks that we determine the position at which the carbon concentration is 0.15 wt% after a 10-h heat
treatment at 1325 K when C0 = 0.35 wt% C. From Equation 5.5
 x 
0.15  0.35
Cx  C0
=
= 0.5714 = 1  erf 

2 Dt 
Cs  C0
0  0.35
Thus,
 x 
erf 
 = 0.4286
2 Dt 
Using data in Table 5.1 and linear interpolation
z
erf (z)
0.40
0.4284
z
0.4286
0.45
0.4755
0.4286  0.4284
z  0.40
=
0.45  0.40 0.4755  0.4284
And,
z = 0.4002
Which means that
x
= 0.4002
2 Dt
And, finally
x = 2(0.4002) Dt = (0.8004)
(6.9  1011 m2 /s)( 3.6  10 4 s)
= 1.26  10-3 m = 1.26 mm
Note: this problem may also be solved using the “Diffusion” module in the VMSE software. Open the “Diffusion”
module, click on the “Diffusion Design” submodule, and then do the following:
1. Enter the given data in left-hand window that appears. In the window below the label “D Value” enter
the value of the diffusion coefficient—viz. “6.9e-11”.
2. In the window just below the label “Initial, C0” enter the initial concentration—viz. “0.35”.
3. In the window the lies below “Surface, Cs” enter the surface concentration—viz. “0”.
4. Then in the “Diffusion Time t” window enter the time in seconds; in 10 h there are (60 s/min)(60
min/h)(10 h) = 36,000 s—so enter the value “3.6e4”.
5. Next, at the bottom of this window click on the button labeled “Add curve”.
6. On the right portion of the screen will appear a concentration profile for this particular diffusion
situation. A diamond-shaped cursor will appear at the upper left-hand corner of the resulting curve. Click and drag
this cursor down the curve to the point at which the number below “Concentration:” reads “0.15 wt%”. Then read
the value under the “Distance:”. For this problem, this value (the solution to the problem) is ranges between 1.24
and 1.30 mm.
Question 5.15: For a steel alloy it has been determined that a carburizing heat treatment of 10-h duration will raise
the carbon concentration to 0.45 wt% at a point 2.5 mm from the surface. Estimate the time necessary to achieve the
same concentration at a 5.0-mm position for an identical steel and at the same carburizing temperature.
Solution of 5.15:
This problem calls for an estimate of the time necessary to achieve a carbon concentration of 0.45 wt% at a point 5.0
mm from the surface. From Equation 5.6b,
x2
= constant
Dt
But since the temperature is constant, so also is D constant, and
x2
= constant
t
or
x12
t1
=
x22
t2
Thus,
(2.5 mm) 2
(5.0 mm) 2
=
10 h
t2
from which
t2 = 40 h
Question 5.16: Cite the values of the diffusion coefficients for the interdiffusion of carbon in both α-iron (BCC) and
γ-iron (FCC) at 900°C. Which is larger? Explain why this is the case.
Solution 5.16:
We are asked to compute the diffusion coefficients of C in both  and  iron at 900C. Using the data in Table 5.2,


80,000 J/mol
D = (6.2  10 -7 m2 /s) exp

 (8.31 J/mol - K)(1173 K) 
D = (2.3  10 -5
= 1.69  10-10 m2/s


148, 000 J/mol
m2 /s) exp

 (8.31 J/mol - K)(1173 K) 
= 5.86  10-12 m2/s
The D for diffusion of C in BCC  iron is larger, the reason being that the atomic packing factor is smaller
than for FCC  iron (0.68 versus 0.74—Section 3.4); this means that there is slightly more interstitial void space in
the BCC Fe, and, therefore, the motion of the interstitial carbon atoms occurs more easily.
Question 5.18: At what temperature will the diffusion coefficient for the diffusion of copper in nickel have a value
of 6.5  10-17 m2/s. Use the diffusion data in Table 5.2.
Solution of 5.18:
Solving for T from Equation 5.9a
T = 
Qd
R (ln D  ln D0 )
and using the data from Table 5.2 for the diffusion of Cu in Ni (i.e., D0 = 2.7  10-5 m2/s and Qd = 256,000 J/mol) ,
we get
T = 

256,000 J/mol
(8.31 J/mol- K) ln (6.5  10 -17 m2 /s)  ln (2.7  10 -5 m2 /s)

= 1152 K = 879C
Note: this problem may also be solved using the “Diffusion” module in the VMSE software. Open the “Diffusion”
module, click on the “D vs 1/T Plot” submodule, and then do the following:
1. In the left-hand window that appears, there is a preset set of data for several diffusion systems. Click on
the box for which Cu is the diffusing species and Ni is the host metal. Next, at the bottom of this window, click the
“Add Curve” button.
2. A log D versus 1/T plot then appears, with a line for the temperature dependence of the diffusion
coefficient for Cu in Ni. At the top of this curve is a diamond-shaped cursor. Click-and-drag this cursor down the
line to the point at which the entry under the “Diff Coeff (D):” label reads 6.5  10-17 m2/s. The temperature at
which the diffusion coefficient has this value is given under the label “Temperature (T):”. For this problem, the
value is 1153 K.
Question 5.24: Carbon is allowed to diffuse through a steel plate 15 mm thick. The concentrations of carbon at the
two faces are 0.65 and 0.30 kg C/m3 Fe, which are maintained constant. If the preexponential and activation energy
are 6.2  10-7 m2/s and 80,000 J/mol, respectively, compute the temperature at which the diffusion flux is 1.43  10-9
kg/m2-s.
Solution 5.24:
Combining Equations 5.3 and 5.8 yields
J = D
C
x
 Q 
C
=  D0
exp  d 
x
 RT 
Solving for T from this expression leads to
Q 
1
T =  d 

D
C 
R
 
ln  0

 J x 
And incorporation of values provided in the problem statement yields
80,000 J/mol 
1
= 

8.31 J/mol - K  (6.2  107 m2 /s)(0.65 kg/m3  0.30 kg/m3) 
ln 

(1.43  109 kg/m2 - s)(15  103 m) 

= 1044 K = 771C