Foundations of Materials Science and Engineering 6th Edition By William Smith - Test Bank

Foundations of Materials Science and Engineering 6th Edition By William Smith – Test Bank

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Sample Questions Are Posted Below

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148
Chapter 5, Problem 1
What is a thermally activated process? What is the activation energy for such a process?
Chapter 5, Solution 1
A thermally activated process is one which requires a definite amount of thermal energy, to
overcome an activation energy barrier, and enter the reactive state.
Chapter 5, Problem 2
Write an equation for the number of vacancies present in a metal at equilibrium at a particular
temperature and define each of the terms. Give the units for each term and use electron volts for
the activation energy.
Chapter 5, Solution 2
v /
v
E kT
n NC –
= е
v
v
where number of vacancies per cubic meter of metal
total number of atom sites per cubic meter of metal
activation energy to form a vacancy (eV)
absolute temperature (K)
Boltzmann’s
n
N
E
T
k
=
=
=
=
= 6
constant 8.62 10 eV/K
constantC
–
= ´
=
Chapter 5, Problem 3
Write the Arrhenius rate equation in the (a) exponential and (b) common logarithmic forms.
Chapter 5, Solution 3
/
10 10
a) Rate of reaction
b) log rate = log constant 2.303
Q RT
C
Q
RT
–
=
–
е
Chapter 5, Problem 4
Draw a typical Arrhenius plot of log 10 of the reaction rate versus reciprocal absolute temperature,
and indicate the slope of the plot. What is the physical meaning of the slope?
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149
Chapter 5, Solution 4
A typical Arrhenius plot of the logarithmic reaction rate is shown below for the SI absolute Kelvin
temperature scale. The relationship between the log10 of the reaction rate and the inverse
absolute temperature is linear with a slope of –Q/(2.303R).
The slope can be used to calculate the activation energy of the reaction
Chapter 5, Problem 5
Describe the substitutional and interstitial diffusion mechanisms in solid metals.
Chapter 5, Solution 5
During substitutional diffusion of atoms in a solid alloy crystal lattice, solute atoms move into
positions of solvent atoms in the matrix through a vacancy mechanism. In interstitial diffusion,
small solute atoms move between the interstices of the solvent lattice.
Chapter 5, Problem 6
Write the equation for Fick’s first law of diffusion, and define each of the terms in SI units.
Chapter 5, Solution 6
Fick’s first law of diffusion is given by:
2
2 3
atoms m atoms 1
or in SI unit form, s mm m
dC
J D dx s
ì üæ öï ïì ü æ öï ï ÷ï ïçï ï ï ï÷ç÷ç= – = ´ ÷í ý í ýç÷ç ÷÷÷çï ï ï ï÷è øç⋅ï ï è øî þ ï ïï ïî þ
-11 , K
T
Intercept = log 10
log 10 of reaction
= –
D
= D
10
Slope 2.303
(log rate)
(1/ )
Q
R
T
T, K
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150
where J = flux or net flow of atoms;
D = proportionality constant called the diffusivity (atomic conductivity) or
diffusion coefficient;
dC
dx = concentration gradient.
Chapter 5, Problem 7
What factors affect the diffusion rate in solid metal crystals?
Chapter 5, Solution 7
The diffusion rate in solid metal crystals is affected by five factors:
1. Type of diffusion mechanism;
2. Temperature of diffusion;
3. Concentration of the diffusion species (concentration gradient);
4. Type of crystal structure;
5. Type of crystal imperfections present.
Chapter 5, Problem 8
Write the equation for Fick’s second law of diffusion in solids, and define each of the terms.
Chapter 5, Solution 8
Fick’s second law of diffusion in solids, written for the x-direction, is:
x xdC dCd D
dt dx dx
æ ö÷ç= ÷ç ÷÷çè ø
where rate of change of the concentration of the diffusing species in the x-direction;
concentration gradient of the diffusing species in the x-direction;
diffusion coefficient of the
dCx
dt
dCx
dx
D
=
=
= diffusing species.
Chapter 5, Problem 9
Write the equation for the solution to Fick’s second law for the diffusion of a gas into the surface
of a solid metal crystal lattice.
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151
Chapter 5, Solution 9
Fick’s second law of diffusion, for the diffusion of a gas into the surface of a solid metal crystal
lattice is:
erf 2
s x
s 0
CC x
C C Dt
– =
–
æ ö÷ç ÷ç ÷ç ÷÷çè ø
where surface concentration of element in gas diffusing into the surface;
initial uniform concentration of element in solid;
concentration of element at distance from surface at time ;
s
o
x
C
C
C x t
x
=
=
=
= distance from surface;
diffusivity of diffusing solute element;
time.
D
t
=
=
Chapter 5, Problem 10
Describe the gas-carburizing process for steel parts. Why is the carburization of steel parts
carried out?
Chapter 5, Solution 10
In the gas carburizing process for steel parts, the parts are placed in a furnace in contact with a
gas rich in CO at about 927ºC. The carbon from the gas diffuses into the surface of the steel part
and increases the carbon content of the outer surface region of the part. The higher carbon
concentration at the surface makes the steel harder in this region. A steel part can thus be
produced with a hard outer layer and a tough low carbon steel inner core. This duplex structure
is important, for example, for many types of gears.
Chapter 5, Problem 11
(a) Calculate the equilibrium concentration of vacancies per cubic meter in pure copper at 850ºC.
Assume that the energy of formation of a vacancy in pure copper is 1.0 eV. (b) What is the
vacancy fraction at 800ºC?
Chapter 5, Solution 11
a) In general, the equilibrium number of vacancies is v /
v
E kT
n NCe-
= . For copper,
23 6 3 28 3Cu (6.02 10 atoms/at. mass)(8.96 10 g/m ) 8.49 10 atoms/m
at. mass Cu (63.54 g/at. mass)
oN
N r ´ ´
= = = ´
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152
Substituting and assuming Ev = 1.00 eV at 1123 K,
u
ì üé ùï ïï ïê ú= ´ -í ýê úï ï´ê úï ïë ûî þ
=
28 3
-5
1.00 eV
(8.49 10 atoms/m ) exp (8.62 10 eV/K)(1123 K)
/
n
24 3
2.77 10 vacancies m´
b) The vacancy fraction at 1073 K is,
u –
é ù-ê ú= = =ê ú´ê úë û
10.81
-5
1.00 eV
exp /
(8.62 10 eV/K)(1073 K)
n
N
-5
2.02 10 vacancies atom´е
Chapter 5, Problem 12
(a) Calculate the equilibrium concentration of vacancies per cubic meter in pure silver at 750ºC.
Assume that the energy of formation of a vacancy in pure silver is 1.10 eV. (b) What is the vacancy
fraction at 700ºC?
Chapter 5, Solution 12
a) The equilibrium number of vacancies is calculated as v /
v .E kT
n NC –
= е Thus for silver,
r
= .
Nn Ag
N at mass Ag =
23 6 3 28 3(6.02×10 atoms/at.mass)(10.5×10 g/m )
= 5.86×10 atoms/m
(107.870 g/at.mass)
Substituting and assuming Ev = 1.10 eV for vacancies formed at 1023 K,
u
ì üé ùï ïï ïê ú= ´ -í ýê úï ï´ê úï ïë ûî þ
=
28 3
-5
1.10 eV
(5.86 10 atoms/m ) exp (8.62 10 eV/K)(1023 K)
/
n
23 3
2.24 10 vacancies m´
b) The vacancy fraction at 973 K is,
13.12
-5
1.10 eV
exp /
(8.62 10 eV/K)(973 K)
n
N
u –
é ù-ê ú= = = ´ê ú´ê úë û
-6
2.01 10 vacancies atomе
Chapter 5, Problem 13
Determine the diffusion flux of zinc atoms in a solid solution of zinc in copper between two points
A and B, 20 μm apart, at 500°C. C A = 1026 atoms/m 3 and C B = 1024 atoms/m 3
.
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153
Chapter 5, Solution 13
C C J
J D X X D
D D
= –  = –
D D
24 3 26 3 25 3
10 / 10 / 9.9 10 /B AC C C atoms m atoms m atoms mD = – = – = – ´
6
20 20 10X m mm –
D = = ´
At 500C, D for zinc in copper is 18 2
4 10 /m s-
´
25 3
18 2 13 2
6
9.9 10 /
(4 10 / ) 1.98 10 /
20 10
atoms m
J m s atoms m s
m
–
–
– ´
= – ´ = ´
´
Chapter 5, Problem 14
The diffusion flux of copper solute atoms in aluminum solvent from point A to point B, 10 μm apart,
is 4 × 1017 atoms/ (m2 · s) at 500°C. Determine (a) the concentration gradient and (b) difference in
the concentration levels of copper between the two points.
Chapter 5, Solution 14
Copper in Aluminum T = 500°C 17 2
J 4 10 atoms/m s= ´
A J B
10μm
a)
C C J
J D X X D
D D
= –  = –
D D
14 2
@500 CD 4 10 m /s-
 = ´
æ ö÷ç ÷ç ´ ÷ç ÷ç ÷- = – = – ´ç ÷ç ÷÷ç ÷ç ´ ÷ç ÷çè ø
2
17
31 4
-14 2
atoms
4 10 s
J m 1 10 atoms/m
D 10 m
4 s
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154
b)
( ) -D = – ´  D = – ´ ´
D
31 31 4 6C 1 10 C 1 10 atoms/m (10 10 m)
X
 D = – = – ´ /26 3
B AC C C 1 10 atoms m
Chapter 5, Problem 15
Consider the gas carburizing of a gear of 1018 steel (0.18 wt %) at 927ºC (1700ºF). Calculate the time
necessary to increase the carbon content to 0.35 wt % at 0.40 mm below the surface of the gear.
Assume the carbon content at the surface to be 1.15 wt % and that the nominal carbon content of
the steel gear before carburizing is 0.18 wt %. D (C in γ iron) at 927ºC = 1.28 × 10-11 m2
/s.
Chapter 5, Solution 15
The time required for this diffusion process is calculated using Fick’s second law,
erf 2
s x
s 0
C C x
C C Dt
æ ö- ÷ç ÷ç= ÷ç ÷ç ÷- è ø
4 11 2
927
where: 1.15% 0.18% 0.35%
0.40 mm = 4 10 m 1.28 10 m /s
s 0 x
C
C C C
x D- –
= = =
= ´ = ´
4
11 2
1.15 0.35 4 10 m
Substituting erf
1.15 0.18 2 (1.28 10 m /s)
55.90
0.8247 erf erf
t
z
t
–
–
é ù
– ´ê ú
= ê ú
– ê ú´ë û
é ù
ê ú= =
ê úë û
Interpolating from Table 5.3,
erf z z
0.8209 0.95
0.8247 x
0.8427 1.0
0.8247 0.8209 0.95 0.959
0.8427 0.8209 1.0 0.95
Thus,
55.90 0.959
3397.7 s =
x x
z t
t
– –
= =
– –
= =
= 56.6 min.
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155
Chapter 5, Problem 16
The surface of a steel gear made of 1022 steel (0.22 wt % C) is to be gas-carburized at 927ºC
(1700ºF). Calculate the time necessary to increase the carbon content to 0.30 wt % at 0.030 in. below
the surface of the gear. Assume the carbon content of the surface to be 1.20 wt %. D (C in γ iron) at
927ºC = 1.28 × 10-11 m2
/s.
Chapter 5, Solution 16
4 11 2
927
Given: 1.20% 0.22% 0.30%
ft
0.03 in.(0.3048 m/ft) = 7.62 10 m 1.28 10 m /s
12 in.
s 0 x
C
C C C
x D- –
= = =
æ ö÷ç= ´ = ´÷ç ÷÷çè ø 
4
11 2
1.20 0.30 7.62 10 m
erf
1.20 0.22 2 (1.28 10 m /s)
106.49
0.9184 erf erf
s x
s 0
C C
C C t
z
t
–
–
é ù
– – ´ê ú
= = ê ú
– – ê ú´ë û
é ù
ê ú= =
ê úë û
Interpolating from Table 5.3,
erf z z
0.9103 1.2
0.9184 x
0.9340 1.3
2 2
0.9184 0.9103 1.2 1.234
0.9340 0.9103 1.3 1.2
Thus,
106.49 106.49 7,446.6 s =
1.234
x x
t z
– –
= =
– –
é ù é ù
ê ú ê ú= = =
ê ú ê úë û ë û
124 min.
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156
Chapter Problem 17
A gear made of 1020 steel (0.20 wt % C) is to be gas-carburized at 927ºC (1700ºF). Calculate the
carbon content at 0.90 mm below the surface of the gear after a 4.0-hour carburizing time. Assume
the carbon content at the surface of the gear is 1.00 wt %. D (C in γ iron) at 927ºC = 1.28 × 10-11 m2
/s.
Chapter 5, Solution 17
D (C in λ iron) at 927ºC = 1.28 × 10 –11 m2 /s.
4 11 2
927
Given: 1.00% 0.20% ? 4 h = 14,400 s
0.09 mm = 9.0 10 m 1.28 10 m /s
s 0 x
C
C C C t
x D- –
= = = =
= ´ = ´
4
11 2
1.00 9.0 10 m
erf
1.00 0.20 2 (1.28 10 m /s)(14,400 s)
1.25(1 ) erf (1.0482)
s x x
s 0
x
C C C
C C
C
–
–
é ù
– – ´ê ú
= = ê ú
– – ê ú´ë û
– =
Interpolating from Table 5.3,
Chapter 5, Problem 18
A gear made of 1020 steel (0.20 wt % C) is to be gas-carburized at 927ºC (1700ºF). Calculate the
carbon content at 0.040 in. below the surface of the gear after a 7.0-hour carburizing time. Assume
the carbon content at the surface of the gear is 1.15 wt %. D (C in γ iron) at 927ºC = 1.28 × 10-11 m2
/s.
Chapter 5, Solution 18
D (C in λ iron) at 927ºC = 1.28 ´ 10 -11 m2 /s.
erf z z
0.8427 1.00
x 1.0482
0.8802 1.10
1.0482 1.00 0.8427 0.8608
1.10 1.00 0.8802 0.8427
Thus, 0.8608 erf( 1.0482)
Substituting,
1.25(1 ) 0.8608x x
x x
C C
– –
= =
– –
=
– = = 0.311 wt %
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157
0.89798 0.85 0.7707 0.7959
0.90 0.85 0.7970 0.7707
Thus, 0.7959 erf ( 0.89798)
Substituting,
1.15 0.7959
0.95
x x
x x
C C
– –
= =
– –
=
– = = 0.394 wt %
4 11 2
927
Given: 1.00% 0.20% ? 4 h = 14,400 s
0.09 mm = 9.0 10 m 1.28 10 m /s
s 0 x
C
C C C t
x D- –
= = = =
= ´ = ´
3
11 2
1.15 1.02 10 m
erf
1.15 0.20 2 (1.28 10 m /s)(25,200 s)
1.15 erf (0.89798)
0.95
s x x
s 0
x
C C C
C C
C
–
–
é ù
– – ´ê ú
= = ê ú
– – ê ú´ë û
– =
Interpolating from Table 5.3,
Chapter 5, Problem 19
The surface of a steel gear made of 1018 steel (0.18 wt % C) is to be gas-carburized at 927ºC.
Calculate the time necessary to increase the carbon content to 0.35 wt % at 1.00 mm below the
surface of the gear after an 8.0 hour carburizing time. Assume the carbon content at the surface
of the gear is 1.20 wt %. D (C in γ iron) at 927ºC = 1.28 × 10-11 m2 /s.
Chapter 5, Solution 19
3 11 2
927
Given: 1.20% 0.18% 0.35%
1.0 mm = 1.0 10 m 1.28 10 m /s
s 0 x
C
C C C
x D- –
= = =
= ´ = ´
3
11 2
1.20 0.35 1.00 10 m
erf
1.20 0.18 2 (1.28 10 m /s)
139.75
0.8333 erf erf
s x
s 0
C C
C C t
z
t
–
–
é ù
– – ´ê ú
= = ê ú
– – ê ú´ë û
é ù
ê ú= =
ê úë û
erf z z
0.7707 0.85
x 0.89798
0.7970 0.90
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158
Interpolating from Table 5.3,
erf z z
0.8209 0.95
0.8333 x
0.8427 1.0
Chapter 5, Problem 20
The surface of a steel gear made of 1020 steel (0.20 wt % C) is to be gas-carburized at 927ºC.
Calculate the carbon content at 0.95 mm below the surface of the gear after an 8.0-hour
carburizing time. Assume the carbon content at the surface of the gear is 1.25 wt %. D (C in γ iron)
at 927ºC = 1.28 × 10-11 m2 /s.
Chapter 5, Solution 20
3 11 2
927
Given: 1.20% 0.18% 0.35%
1.0 mm = 1.0 10 m 1.28 10 m /s
s 0 x
C
C C C
x D- –
= = =
= ´ = ´
3
11 2
1.20 0.35 1.00 10 m
erf
1.20 0.18 2 (1.28 10 m /s)
139.75
0.8333 erf erf
s x
s 0
C C
C C t
z
t
–
–
é ù
– – ´ê ú
= = ê ú
– – ê ú´ë û
é ù
ê ú= =
ê úë û
2 2
0.8333 0.8209 0.95 0.978
0.8427 0.8209 1.0 0.95
Thus,
139.75 139.75
0.978
20,400 s
x x
t z
t
– –
= =
– –
é ù é ù
ê ú ê ú= =
ê ú ê úë û ë û
= = =340 min. 5.67 h
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159
0.7823 0.75 0.7112 0.7312
0.80 0.75 0.7421 0.7112
Thus, 0.7312 erf(0.7823)
Substituting,
1.25 0.7312
1.05
x x
x x
C C
– –
= =
– –
=
– = = 0.48 wt %
Interpolating from Table 5.3,
erf z z
0.7112 0.75
x 0.7823
0.7421 0.80
Chapter 5, Problem 21
A steel gear made of 1018 steel (0.18 wt % C) is to be gas-carburized at 927ºC. If the carburizing
time is 7.5 h, at what depth in millimeters will the carbon content be 0.040 wt %? Assume the
carbon content at the surface of the gear is 1.20 wt %. D (C in γ iron) at 927ºC = 1.28 × 10-11 m2 /s.
Chapter 5, Solution 21
11 2
927
Given: 1.20% 0.18% 0.40%
7.5 h = 27,000 s 1.28 10 m /s
s 0 x
C
C C C
t D –
= = =
= = ´
11 2
1.20 0.40 erf
1.20 0.18 2 (1.28 10 m /s)(27,000 s)
0.7843 erf(850.52 ) erf
s x
s 0
C C x
C C
x z
–
é ù
– – ê ú
= = ê ú
– – ê ú´ë û
= =
Interpolating from Table 5.3,
erf z z
0.7112 0.75
x 0.7823
0.7421 0.80
1 1
0.850.7843 0.7707 0.8759
0.7970 0.7707 0.90 0.85
Substituting,
0.8759 850.52
0.00103 m = 1.03 mm
x x
z x
x
— = =
– –
= =
=
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160
Chapter 5, Problem 22
If boron is diffused into a thick slice of silicon with no previous boron in it at a temperature of 1100ºC
for 5 h, what is the depth below the surface at which the concentration is 1017 atoms/cm3 if the surface
concentration is 1018 atoms/cm3
? D = 4 × 10-13 cm2
/s for boron diffusing in silicon at 1100ºC.
Chapter 5, Solution 22
18 3 17 3
4 13 2
1100
Given: 10 atoms/cm 10 atoms/cm 0.0
5.0 h 1.8 10 s 4.0 10 cm /s
s x 0
C
C C C
t D –
= = =
= = ´ = ´
18 17
18 13 2 4
4
10 10 erf
10 0 2 (4.0 10 cm /s)(1.80 10 s)
0.90 erf erf
1.697 10
s x
s 0
C C x
C C
x z
–
–
é ù
– – ê ú
= = ê ú
– ê ú- ´ ´ë û
æ ö÷ç= =÷ç ÷÷çè ø´
Interpolating from Table 5.3,
erf z z
0.8802 1.1
0.9000 x
0.9103 1.2
Chapter 5, Problem 23
If aluminum is diffused into a thick slice of silicon with no previous aluminum in it at a temperature
of 1100ºC for 6 h, what is the depth below the surface at which the concentration is 1016 atoms/cm3
if the surface concentration is 10 18 atoms/cm3
?
D = 2 × 10-12 cm2 /s for aluminum diffusing in silicon at 1100ºC.
1 1
4
4
1.10.9000 0.8802 1.166
0.9103 0.8802 1.2 1.1
Substituting,
1.166 1.697 10
1.98 10 cm
x x
x
z
x
–
–
— = =
– –
= = ´
= ´
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161
Chapter 5, Solution 23
18 3 16 3
4 12 2
1100
Given: 10 atoms/cm 10 atoms/cm 0.0
6.0 h = 2.16 10 s 2.0 10 cm /s
s x 0
C
C C C
t D –
= = =
= ´ = ´
18 16
18 12 2 4
4
10 10 erf
10 0 2 (2.0 10 cm /s)(2.16 10 s)
0.99 erf erf
4.157 10
s x
s 0
C C x
C C
x z
–
–
é ù
– – ê ú
= = ê ú
– ê ú- ´ ´ë û
æ ö÷ç= =÷ç ÷÷çè ø´
Interpolating from Table 5.3,
erf z z
0.9891 1.8
0.9900 x
0.9928 1.9
Chapter 5, Problem 24
Phosphorus is diffused into a thick slice of silicon with no previous phosphorus in it at a temperature
of 1100ºC. If the surface concentration of the phosphorus is 1 × 1018 atoms/cm3 and its concentration
at 1 μm is 1 × 1015 atoms/cm 3
, how long must be the diffusion time? D = 3.0 × 10-13 cm2 /s for P
diffusing in Si at 1100ºC.
Chapter 5, Solution 24
18 3 15 3
4 13 2
1100
Given: 10 atoms/cm 10 atoms/cm 0.0
1.0 m = 1.0 10 cm 3.0 10 cm /s
s x 0
C
C C C
x Dm – –
= = =
= ´ = ´
1 1
4
4
1.80.9900 0.9891 1.824
0.9928 0.9891 1.9 1.8
Substituting,
1.824 4.157 10
7.58 10 cm
x x
x
z
x
–
–
— = =
– –
= = ´
= ´
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162
18 15 4
18 13 2
10 10 10 cm
erf
10 0 2 (3.0 10 cm /s)
91.287
0.999 erf erf
s x
s 0
C C
C C t
z
t
–
–
é ù
– – ê ú
= = ê ú
– ê ú- ´ë û
æ ö÷ç= =÷ç ÷ç ÷è ø
Interpolating from Table 5.3,
erf z z
0.9981 2.2
0.9990 x
0.9993 2.4
Chapter 5, Problem 25
If the diffusivity in Prob. 4.53 had been 1.5 × 10-13 cm2 /s, at what depth in micrometers would the
phosphorus concentration be 1 × 1015 atoms/cm 3
?
Chapter 5, Solution 25
Since 0.999, erf is still 2.35. Thus, for the same diffusion period,s x
s 0
C C z
C C
– =
–
–
é ù
ê ú
= = = ´ =ê ú
ê ú´ë û
5
-13 2 2.35, 7.07 10 cm
2 (1.5 10 cm /s)(1509 s)
x
z x 0.707 μm
2 2
0.9990 0.9981 2.2 2.35
0.9993 0.9981 2.4 2.2
Thus,
91.287 91.287
2.35
1508 s
x x
t z
t
– –
= =
– –
é ù é ù
ê ú ê ú= =
ê ú ê úë û ë û
= = 25.1 min.
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163
Chapter 5, Problem 26
Arsenic is diffused into a thick slice of silicon with no previous arsenic in it at 1100ºC. If the surface
concentration of the arsenic is 5.0 × 1018 atoms/cm3 and its concentration at 1.2 μm below the
silicon surface is 1.5 × 1016 atoms/cm3
, how long must be the diffusion time? (D = 3.0 × 10-14 cm2
/s for
As diffusing in Si at 1100ºC.)
Chapter 5, Solution 26
18 3 16 3
6 4 14 2
1100
Given: 5.0 10 atoms/cm 1.5 10 atoms/cm 0.0
1.20 10 m 1.20 10 cm 3.0 10 cm /s
s x 0
C
C C C
x D- – –
= ´ = ´ =
= ´ = ´ = ´
18 16 4
18 14 2
5.0 10 1.5 10 1.20 10 cm
erf
5.0 10 0 2 (3.0 10 cm /s)
346.4
0.9970 erf erf
s x
s 0
C C
C C t
z
t
–
–
é ù
– ´ – ´ ´ê ú
= = ê ú
– ê ú´ – ´ë û
æ ö÷ç= =÷ç ÷ç ÷è ø
Interpolating from Table 5.3,
2 2
0.9970 0.9953 2.0 2.12 and,
0.9981 0.9953 2.2 2.0
346.4 346.4 26,700 s
2.12
x x
t z
– –
= =
– –
é ù é ù
ê ú ê ú= = = =
ê ú ê úë û ë û
7.42 h
Chapter 5, Problem 27
Calculate the diffusivity D in square meters per second for the diffusion of nickel in FCC iron at
1100ºC. Use values of D0 = 7.7 × 10-5 cm2 /s; Q = 280 kJ/mol; R = 8.314 J/(mol · K).
Chapter 5, Solution 27
The diffusivity of the nickel into FCC iron at 1373 K is:
[ ]
/ 5 2
5 2 24.53
280,000 J/mol
(7.7 10 m /s) exp 8.314 J/(mol K) (1373 K)
(7.7 10 m /s)(e )
Q RT
0D D e- –
– –
ì üé ùï ï-ï ïï ïê ú= = ´ í ýê úï ï⋅ê úï ïë ûï ïî þ
= ´
= 15 2
1.71 × 10 m /s-
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164
Chapter 5, Problem 28
Calculate the diffusivity in square meters per second of carbon in HCP titanium at 700ºC. Use
D0 = 5.10 × 10-4 m2 /s; Q = 182 kJ/mol; R = 8.314 J/(mol · K).
Chapter 5, Solution 28
The diffusivity of carbon into HCP titanium is:
[ ]
/ 4 2
4 2 22.49
182,000 J/mol
(5.10 10 m /s) exp 8.314 J/(mol K) (973 K)
(5.10 10 m /s)(e )
Q RT
0D D e- –
– –
ì üé ùï ï-ï ïï ïê ú= = ´ í ýê úï ï⋅ê úï ïë ûï ïî þ
= ´
= 14 2
8.64 10 m /s-
´
Chapter 5, Problem 29
Calculate the diffusivity in square meters per second for the diffusion of zinc in copper at 350ºC. Use
D0 = 3.4 × 10-5 m2 /s; Q = 191 kJ/(mol · K), R = 8.314 J/(mol · K).
Chapter 5, Solution 29
The diffusivity of zinc into copper at 623 K is:
– –
æ öì üé ùï ï-÷ç ï ï÷ ê úç= = í ý÷ç ÷ ê úï ï÷çè ø ë ûï ïî þ
2
/ 5
0
191,000 /
3.40 10 [8.314 /( . )](623 )
Q RT m J mol
D D e x exp
s J mol k k
2
5 36.88
3.40 10 ( )
m
x s
– –
æ ö÷ç ÷ç= ÷ç ÷÷çè ø е
21 2
3.29 10 /x m s-
=
Chapter 5, Problem 30
The diffusivity of manganese atoms in the FCC iron lattice is 1.50 × 10 -14 m 2 /s at 1300ºC and
1.50 × 10 -15 m 2 /s at 400ºC. Calculate the activation energy in kJ/mol for this case in this
temperature range. R = 8.314 J/(mol · K).
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165
Chapter 5, Solution 30
The activation energy may be calculated using the Arrhenius type equation,
1300 C 2
1 2 1400 C
exp( / ) 1 1
exp
exp( / )
D Q RT Q
D Q RT R T T
é ùæ ö- – ÷çê ú÷= = -ç ÷ê úç ÷ç- è øê úë û


1 2where 400 C = 673 K and 1300 C = 1573 K. Substituting,T T= = 
14 2
15 2
4
4
1.5 10 m /s 1 1
exp 8.314 J/(mol K) 1573 K 673 K1.5 10 m /s
10 exp (1.0226 10 )
ln(10) (1.0226 10 )
22,518 J/mol
Q
Q
Q
Q
–
–
–
–
é ùæ ö´ – ÷çê ú= – ÷ç ÷÷çê úè ø⋅´ ë û
é ù= ´ê úë û
= ´
= = 22.5 kJ/mol
Chapter 5, Problem 31
The diffusivity of copper atoms in the aluminum lattice is 7.50 × 10-13 m2
/s at 600ºC and 2.50 × 10-15 m2
/s
at 400ºC. Calculate the activation energy for this case in this temperature range. R = 8.314 J/(mol · K).
Chapter 5, Solution 31
The activation energy associated with copper diffusing into aluminum for this temperature range is,
600 C
2 1400 C
1 2
1 1
exp
where 400 C = 673 K and 600 C = 873 K. Substituting,
D Q
D R T T
T T
é ùæ ö- ÷çê ú÷= -ç ÷ê úç ÷çè øê úë û
= =


 
13 2
15 2
5
5
7.5 10 m /s 1 1
exp 8.314 J/(mol K) 873 K 673 K2.5 10 m /s
300 exp (4.094 10 )
ln(300) (4.094 10 )
139,320 J/mol
Q
Q
Q
Q
–
–
–
–
é ùæ ö´ – ÷çê ú= – ÷ç ÷÷çê úè ø⋅´ ë û
é ù= ´ê úë û
= ´
= = 139.3 kJ/mol
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166
Chapter 5, Problem 32
The diffusivity of iron atoms in the BCC iron lattice is 4.5 × 10-23 m2
/s at 400ºC and 5.9 × 10-16 m2
/s
at 800ºC. Calculate the activation energy in kJ/mol for this case in this temperature range.
R = 8.314 J/(mol · K).
Chapter 5, Solution 32
The activation energy associated with iron diffusing into BCC iron for this temperature range is,
800 C
2 1400 C
1 2
1 1
exp
where 400 C = 673 K and 800 C = 1073 K. Substituting,
D Q
D R T T
T T
é ùæ ö- ÷çê ú÷= -ç ÷ê úç ÷çè øê úë û
= =


 
16 2
23 2
5
5
5.9 10 m /s 1 1
exp 8.314 J/(mol K) 1073 K 673 K4.5 10 m /s
13,111,111.1 exp (6.662 10 )
ln(13,111,111.1) (6.662 10 )
246,007 J/mol
Q
Q
Q
Q
–
–
–
–
é ùæ ö´ – ÷çê ú= – ÷ç ÷÷çê úè ø⋅´ ë û
é ù= ´ê úë û
= ´
= = 246 kJ/mol
Chapter 5, Problem 33
The concentration of Manganese (Mn) at 500°C on the surface of an iron sample is 0.6 a%. At a
distance of 2 mm below the surface, the concentration is 0.1 at%. Determine the flux of Mn atoms
between the surface and plane 2 mm deep. Hint: convert a% to atoms/m 3 using information in
Table 3.2.
Chapter 5, Solution 33
Convert atom % (a%) to atoms/m 3
.
For FCC iron, lattice constant a = 0.351 nm.
( )
= = =
4 0.124 nm4R
a 0.351 nm
2 2
There are 4 atoms per unit cell.
Thus,
( )-
=
´ 3
9
atoms 4 atoms
unit volume 0.351 10 m
= 28 3
9.24 10 atoms m´
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167
C
J D X
D
= – D ; 24 2
@500 CD 3 10 m s-
 = ´ (metastable FCC ion)
X 0.002mD =
Surface ( )28 3 26 3
sC 0.006 9.24 10 atoms m 5.55 10 atoms m= ´ ´ = ´
Surface ( )28 3 25 3
2 mmC 0.001 9.24 10 atoms m 9.24 10 atoms m= ´ ´ = ´
( ) 2mm s24 2 5 2
C C
J 3 10 m s 6.939 10 atoms m s
0.002 m
– –
 = – ´ = ´
Chapter 5, Problem 34
The concentration of carbon on the surface of a 1018 steel gear is 0.8 wt% at 1000°C. Determine
the flux of carbon atoms from the surface to a plane 25 mm below the surface where carbon
concentration is unaffected by the surface concentration. Hint: convert wt% to atoms/m 3 using
information in Table 3.2.
Chapter 5, Solution 34
At 100°C, the structure of iron will be FCC.
Convert wt % to atoms/m 3
. (assume 100 grams)
(1)
23 226.02 10
0.8 4.01 10
12.01 g
´
 ´ = ´ atoms of carbon in 0.8 g
23 246.02 10
99.2 1.07 10
55.85 g
´
´ = ´ atoms of iron in 99.2 g
(2) For FCC iron,
( )
28 3
3
9
4 atoms 9.25 10 atoms m
0.351 10 m-
= ´
´
Note
( )4 0.124nm4R
a 0.351nm
2 2
= = =
From (1) atom % carbon
( )
22
22 24
4.01 10 100
4.01 10 1.07 10
´
= ´
´ + ´ = 3.61 %
 carbon concentration on the surface ( )28 3
sC 0.036 9.25 10 atoms m@ ´
@ ´ 27 3
3.33 10 atoms msC
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168
C
J D X
D
 = – D ; 11
@ 1000°CD 3 10-
= ´ (carbon in FCC iron)
Carbon concentration 25μm below the surface is unaffected. Thus,
( )28
25μmC 0.18 wt% 0.0018 9.25 10= = ´
26 3
25μmC 1.66 10 atoms m= ´
26 27
25μm s11 11
6
C C 1.66 10 3.33 10
J 3 10 3 10
X 25 10
– –
–
æ öæ ö- ´ – ´ ÷÷ çç ÷÷ ç= – ´ = – ´ç ÷÷ çç ÷÷ç ÷çD ´è ø è ø
= ´ 21 2
3.79 10 atoms m sJ
Chapter 5, Problem 35
A copper-zinc alloy (85wt% Cu-10wt% zinc) is coupled with pure copper (interfaced). The diffusion
couple is then heated to a temperature of 1000°C. (a) How long will it take for the concentration
of zinc to reach 0.2%, 2.5 mm below the interface? (b) How much will the zinc concentration at
the same point be in twice the time calculated in part a?
Chapter 5, Solution 35
Copper is solvent, Zn is solute (non-steady diffusion)
interface
85% wt % Cu- pure copper
1% wt Zn
2.5 mm
S X
S 0
C C X
erf
C C 2 Dt
æ ö- ÷ç= ÷ç ÷ç ÷- è ø
sC 10 wt %= (amount of zinc present at the surface)
XC 0.2 wt %= (concentration @ 2.5mm)
0 0 %C = (pure copper)
3
X 2.5 10 m-
= ´ ; 13 2
@1000 CD 5 10 m s-
 = ´ (table 5.2)
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169
a) 10% 0.2% x
erf erf(z)
10% 0% 2 Dt
æ ö- ÷ç= =÷ç ÷ç ÷- è ø
0.98 = erf (z)
from table 5.3 using interpolation
z erf (z)
1.6 0.9763
z 0.9800
1.7 0.9838
1 0.9763 0.9800.6 1.65
1.6 1.7 0.9763 0.9838
z z
—
 =  =
– –
3
13
2.5 10 m
1.65
2 5 10 t
–
–
´
 =
´
t 318 hrs =
b)
3
23 6
2.5 10 1.25
2 5 10 2.29 10
z
–
–
´
= =
´ ´ ´
 erf (z) = 0.916 from table 5.3
X X
10 wt% C
0.916 C (0.1) (0.916)(0.1)
10 wt% 0
–
=  = –
–
XC 0.84 wt % = (increases by more than 4 times)
Chapter 5, Problem 36
A bar of pure nickel is coupled with a bar of pure iron (interfaced). The diffusion couple is then
heated to a temperature of 1000°C. (a) How long will it take for the concentration of nickel to
reach 0.1wt%, 1.0 um below the interface? (b) How long will it take for the concentration of nickel to
reach 0.1wt%, 1.0 mm below the interface? (c) What does the comparison of the two answers show?
Chapter 5, Solution 36
Iron is solvent, Ni is solute (non-steady diffusion); T = 1000°C
Ni Fe
1μm
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170
æ ö- ÷ç= ÷ç ÷ç ÷- è ø
S X
S 0
C C X
erf
C C 2 Dt
SC 100 %= (pure nickel)
XC 0.1 wt %=
0C 0%= (pure iron)
16 2
@1000 CD 2 10 m s-
 = ´ (table 5.2)
a) 100 0.1 X
erf erf(z)
100 0 2 Dt
æ ö- ÷ç= =÷ç ÷ç ÷- è ø
erf (z) = 0.999  z = 2.8
( )
6
16 2
X 1 10 m
2.8 2.8
2 Dt 2 2 10 m s .(t)
–
–
´
=  =
´
6 2
2 16
(1 10 )
t 159 s
(2.8) (4) (2 10 )
–
–
´
 = =
´
b) For concentration of 0.1 wt % at 1.0mm
3 2 8
2 16
(1 10 )
t 1.59 10 s
(2.8) (4) (2 10 )
–
–
´
= = ´
´
The conclusion is that diffusion is a slow process. Also, a small change in Δx (by a factor of 1000)
will increase the time by a factor of 10 6
.
Chapter 5, Problem 37
In Table 5.1, propose a diffusion mechanism for metals and nonmetals listed, and justify your
answer. Why are the activation energies for self-diffusion of nonmetals Si and C significantly
higher than those of metals in this table?
Chapter 5, Solution 37
For the metals listed in Table 5.1, the diffusion mechanisms is substitutional diffusion. This is
because metals will always have some defects present at equilibrium, and these defects enable
atoms to move into them, and thus undergo substitutional diffusion.
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171
For the non-metals listed in Table 5.1, the diffusion mechanism is interstitial diffusion. This is
because the atomic packing factor of these structures is small (there is more space in each unit
cell), and the condition for interstitial diffusion is that the diffusing atoms need to be small
compared to the matrix.
The activation energy is higher for the non-metals since these atoms need enough energy to
squeeze between matrix atoms.
Chapter 5, Problem 38
In Figure 5.14, without any calculations, can you identify the highest and lowest activation
energies in the given systems? What does a high activation energy imply as compared to a low
activation energy?
Chapter 5, Solution 38
For Arrhenius plots, the slope can be used to find the activation energy (slope=-Ea/R). So, a larger
slope will result in a larger activation energy. With this in mind, we can say that “Mn in FCC Fe”
has the highest activation energy, and “Ag along Ag boundaries” has the lowest.
A high activation energy implies that more energy will be needed for diffusion to occur.
Chapter 5, Problem 39
The activation energy of nickel atoms in FCC iron is 280 kJ/mol and carbon atoms in FCC iron is
142 kJ/mol. (a) What does this tell you about the comparative diffusion of nickel and carbon in
iron? (b) Can you explain why the activation energies are so drastically different? (c) Find a way to
qualitatively explain how much energy is 142 kJ to a non-engineer or a non-scientist.
Chapter 5, Solution 39
a) Since QC-Fe < QNi-Fe , it can be concluded that carbon has an easier time diffusing in Fe
than Ni atoms do in Fe.
b) The difference in activation energies is due to differences in mechanisms (interstitial
diffusion for C-Fe and substitution for Ni-Fe).
c) You would need 4.2 J of heat to increase the temperature of one gram of water by one
degree C. Thus, 142000 J would raise the temperature of 33,809 g (142000/4.2) of water
by 1°C. Other ways to explain:
336 J to boil one gram of water
142000 J to boil 422 grams of water (or 0.422 kg)
Note that the activation energy is per mol. Therefore, the actual energy for practical
applications will be significantly higher and depends on the number of moles.
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172
Chapter 5, Problem 40
The melt temperatures of copper and aluminum are 1083°C and 657°C respectively. Compare the
diffusivities of copper in copper and copper in aluminum at 500°C (use Table 5.2). Can you
explain why a drastic difference exists?
Chapter 5, Solution 40
Copper has a higher melting temperature than aluminum. This means that the bonds between
copper atoms are stronger and, therefore, harder to break. Diffusion will likely be more difficult in
copper than aluminum. A second reason could be that, although both metals are FCC, the void
spaces in aluminum are larger (see example 4.3) since r Al > rCu .
Chapter 5, Problem 41
The self-diffusion of iron atoms in BCC iron is significantly higher than in FCC iron (See Table 5.2).
Explain why.
Chapter 5, Solution 41
The packing factor for BCC metals is lower than that of FCC metals. As a result, there is more
space for diffusion in BCC iron (APF = 0.68) than in FCC iron (APF = 0.74).
Chapter 5, Problem 42
Would you expect the diffusion rate of copper (self-diffusion) to be lower or higher in copper with
ASTM grain size 4 than in copper with ASTM grain size 8? Explain your answer.
Chapter 5, Solution 42
There are more grains and grain boundaries in fine grained copper (the one with n=8). Grain
boundaries are less efficiently packed, and they allow for higher diffusion rates. Thus, copper
with an ASTM grain size of 8 will have higher self-diffusion.
Chapter 5, Problem 43
Would you expect the diffusion rate of copper (self-diffusion) to be lower in a pure copper sample
that is loaded with dislocations or in a pure copper sample that is free of dislocations (refer to
Sec. 4.4.2 for characteristics of dislocations)? Explain your answer.
Chapter 5, Solution 43
Metals loaded with dislocation possess higher stored energy. The atoms are forced out of their
normal position, and this causes lattice strain. These atoms have a higher state of energy than in
an unstrained array, and the extra energy could help with the diffusion process.
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173
Chapter 5, Problem 44
In NaCl, would you expect the activation energy of cation (Na+) to be higher or the anion (Cl-)?
Why?
Chapter 5, Solution 44
Cations are smaller in size than their bonding anions. This offers an advantage for higher rates of
diffusion for cations. Na + will have the lower activation energy.
Chapter 5, Problem 45
Is the non-steady diffusion process more sensitive to temperature or time? Explain using
appropriate equations.
Chapter 5, Solution 45
In the non-steady diffusion equation,
æ ö- ÷ç= ÷ç ÷ç ÷- è ø
S X
S 0
C C X
erf
C C 2 Dt
Note that time, t, appears in the denominator under the radical sign. Temperature, T, also
appears in the same place through “D”.
Q
RT
0D D e-
=
Small changes in temperature, T, will cause large changes in D. Thus, cutting the temperature by
half will have a more significant impact on the diffusion process than cutting the time by half.
Chapter 5, Problem 46
Show, using equations only, that as time increases in the gas carburization process, the
concentration Cx increases.
Chapter 5, Solution 46
In the non-steady diffusion equation,
æ ö- ÷ç= ÷ç ÷ç ÷- è ø
S X
S 0
C C X
erf
C C 2 Dt
as time increases the denominator in the argument of erf also increases. As the denominator
increases, the argument X
2 Dt will decrease. As the argument becomes smaller, the X
erf 2 Dt
æ ö÷ç ÷ç ÷ç ÷è ø
will also become smaller (see table 5.3). As X
erf 2 Dt
æ ö÷ç ÷ç ÷ç ÷è ø becomes smaller, the differences between
CS and CX also become smaller. Thus CX is approaching CS (becoming larger).
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174
Chapter 5, Problem 47
If hydrogen diffuses in ferrous alloys, it will make the material significantly more brittle and
susceptible to fracture. The activation energy of hydrogen in steel is 3.6 Kcal/mol. Should we
worry about hydrogen embrittlement of steels (is it very likely to occur)? Explain.
Chapter 5, Solution 47
Compare the activation energy of hydrogen in iron (3.6 kcal/mol) to that of carbon in iron (29 to
34 kcal/mol) in table 5.4. It is clear that Q H-Fe is 10 times smaller than Q C-Fe . Now, consider the
equation for diffusivity:
Q
RT
0D D e-
=
As Q is decreased by a factor of 10, -Q/RT also decreases. This results in a large increase in
Q
RTe-
. Therefore, we should be concerned with diffusion of hydrogen in ferrous alloys.
Chapter 5, Problem 48
In Figure 5.14, compare the diffusivity of silver in silver inside the grain with silver in silver along
the grain boundaries at any given temperature. What is your conclusion? How do you explain the
difference?
Chapter 5, Solution 48
Figure 5.14 shows the Arrhenius plots of the diffusivity data. Looking at this, we can see that the
slope is larger for “Ag in Ag” compared to “Ag along Ag boundaries”, which tells us that “Ag in Ag”
has the higher activation energy. This difference can be explained by the fact that high energy
areas with a higher defect concentrations make it easier for diffusion to occur. Grain boundaries
would have higher energy than inside the silver grain.
Chapter 5, Problem 49
Calculate the diffusivity D in square meters per second for the diffusion of nickel in FCC iron at
1100°C. Use values of D0 = 7.7 × 10−5 cm2 /s; Q = 280 kJ/mol; R = 8.314 J/(mol · K).
Chapter 5, Solution 49
The diffusivity of the nickel into FCC iron at 1373 K is:
[ ]
/ 5 2
5 2 24.53
280,000 J/mol
(7.7 10 m /s) exp 8.314 J/(mol K) (1373 K)
(7.7 10 m /s)(e )
Q RT
0D D – –
– –
ì üé ùï ï-ï ïï ïê ú= = ´ í ê úï ï⋅ê úï ïë ûï ïî 
= ´
= 15 2
1.71 10 m /s-
´
е
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175
Chapter 5, Problem 50
Figure 5.15 shows the diffusion coefficient as a function of temperature for diffusion of various
impurities in silicon, including aluminum, gallium, boron, arsenic, and antimony. By examining the
figure, can you determine which impurity diffuses more easily across the silicon lattice? Can you
explain why?
Chapter 5, Solution 48
For an Arrhenius plot, the larger slope will have a higher activation energy. Thus, Antimony that
has the largest slope diffuses more easily across the silicon lattice.