Chapters 2 - 10 Covered
SOLUTIONS
, Chapter 2
Problem 2.1 In FCC the relation between the lattice parameter and the atomic radius is
4R
, then α=4.95 Angstroms. On the cube phase (100) correspond 2 atoms (4x1/4+1). Then
2
the density of the (100) plane is
2 12 2
atoms/mm
(100 ) 8.2x10
4.95x10 7
In the (111) plane there are 3/6+3/2=2 atoms. The base of the triangle is 4R and the height 2 3R
After some math we get ρ(111)=9.5x1012 atoms/mm2. We see that the (111) plane has higher density than
the (100) plane, it is a close-packed plane.
Problem 2.2 The (100)-type plane closer to the origin is the (002) plane which cuts the z axis at
½. This has
a a 2R
d(002) 0 0 2 2 2 2
Setting R=1.749 Angstroms we get d(002)=2.745 Angstroms.
In the same way
a a 4R
d(111)
11 1 3 6
and d(111)=2.85 Angstroms. We see that the close-packed planes have a larger interplanar spacing.
Problem 2.3. The structure of vanadium is BCC. In this structure, the close-packed direction is [111], which
corresponds to the diagonal of the cubic unit cell where there is a consecutive contact of spheres (in the
model of hard spheres). Furthermore, the number of atoms per unit cell
for the BCC structure is 2. The first step is to find the lattice parameter α. The density is
2
3
Where is the Avogadro’s number. Therefore the lattice parameter is
3
2 50.94 a 3.08 10 8 cm 3.08 10 10 m
5.8 6.023 10 23
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,The length of the diagonal at the [111] close-packed direction is a 3 , which corresponds to 2
atoms. Hence the atomic density of the close-packed direction of vanadium (V) is
[111]
2 2 3.75 109 atoms / m
3 3.081010 3
The aforementioned atomic density result translates to 3750 atoms/μm or 3.75 atoms/nm.
4R
. The (10 0) plane is the
Problem 2.4. The lattice parameter for the FCC structure is 2
face of the unit cell. The face comprises ¼ of atoms at each corner plus 1 atom at the center of
the face. Hence the face consists of 4 () 1 2 atoms. The atomic density of the (10 0)
plane is
2 2
(100) 2 2 1
a 4R 4R 2
2
The (111) plane corresponds to the diagonal equilateral triangle of the unit cell. The base of this triangle is
4R . Using the Pythagorean Theorem, we can calculate the height of the triangle which
is 2 3R . Thus the area of the triangle is (base height / 2) 4 3R 2 . The equilateral triangle
comprises 6 of the atoms at each corner and ½ of the atoms at the middle of each side. Thus the
equilateral triangle consists of 3 (1/ 6) 3 () 2 atoms. The atomic density of the (111)
plane is
(111)
2 1
4 3R2 2 3R2
The ratio of the atomic densities is
(111)
2
(100)
1.154 1
3
Therefore (111) (100) and specifically the (111) plane has 15% higher atomic density than the
(10 0) plane. This is important since the plastic deformation of metals (Al, Cu, Ni, γ-Fe, etc.) is
accomplished with dislocation glide on the close-packed planes.
Problem 2.5. The ideal c/a ratio in HCP structure results when the atoms of this structure have an
arrangement as dense as the atoms of the FCC structure. The distance between the (0 0 01) bases of
the HCP structure is c. Using the fact that the (0 0 01) planes of HCP structure
correspond to the (111) planes of the FCC structure, we get
c 2 d(111)
FCC
Where d(111) is the distance between the (111) close-packed planes. We find that
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, a
d a a
(111) 2
h k 2
l 2
12
12 1 2
3 4R
d(111) FCC
4R 6
a 2
Thus,
8R
c c
6 4 1.63
HCP: a 2 R a 6
Therefore, the ideal ratio c/a for close packing in HCP structure is equal to 1.63. The c/a ratio for zinc (Zn)
is 1.86 while for titanium (Ti) is 1.59 (see Table 7.1, Book). This means that the distance between
the (0 0 01) planes is longer in Zn than in Ti. This fact affects the plastic
deformation in these metals, since the slip on (0 0 01) planes is easier in Zn than in Ti. Indeed the critical
shear stress of Zn is only 0.18 MPa, while of Ti is 110 MPa. Due to this, the plastic deformation in Ti
is performed on (10 1 0) plane, where the critical shear stress is approximately
49 MPa. Thus in Ti the slip is not performed on the close-packed planes of the crystal structure. For more
details look at the 7.3 paragraph of the book (plastic deformation of single crystals with slip).
Problem 2.6. The cell volume of HCP structure is the product of the base area (hexagon)
multiplied by the height c. The base of hexagon is A 6 R2 3 and the height is c 8R . As a
result, the cell volume is 6
V 24 2 R 3
The number of atoms per unit cell for the HCP structure is 6, thus the atomic packing factor is
4
6 R3
APFHCP 3 0.74
24 2 R 3 3 2
Regarding the BCC structure, the number of atoms per unit cell is 2 and the cell volume is where a a3 ,
4 R . Therefore the atomic packing factor of BCC structure is
3
4 3
2 R 3
APFBCC 3 8 0.68
3
4R
3
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,Since the atomic packing density of BCC is less than that of the HCP, the diffusion in BCC (movement of
atoms inside the lattice) is faster.
Problem 2.7 The density is
mass of cell atoms
cell volume
The structure of copper (Cu) is FCC and the number of atoms per unit cell is 4. The atomic mass
is , where is the atomic weight and N A is the Avogadro’s number. The cell volume is
NA
3
3 4R , thus the density is
a
2
63.57
4 23 9gr / cm3
6.023 10
4 1.276 10 8 3
2
Problem 2.8 Notice that the atomic volume is not 3
3
4 R ! It is the corresponding volume of
every atom of the structure plus the empty surrounding space inside the cell. Due to the fact that the
structure of gold (Au) is FCC, the number of atoms per unit cell is 4. Therefore the atomic volume is
3
4
For the FCC structure we get that 4R , hence
2
3
4R
2
4
After replacing the value of the atomic radius of gold R , we find that
o
10
1.7 10 29 m3 . Since
1 10 m , the atomic volume of gold is
o
17 ( )3
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,The molar volume Vm is the volume corresponding to one mole of gold and is obtained by
multiplying the atomic volume by the Avogadro’s number. Thus the molar volume is
V N 1.02 10 5 m 3 / mol
m A
Problem 2.9
For the atomic radius of the iron atom, RFCC=1.270 and RBCC=1.241 Angtroms (A) FCC has 4
atmos/cell while BCC has 2 atoms/cell.
In FCC a 4R 2 3.591A,VFCC a 3 46.34 A3
In BCC a 4R 3 2.865A,VBCC 23.51A 3
Taking 4 atoms as a reference, this corresponds to 1 FCC cell and 2 BCC cells, then
V 2VBCC VFCC
0.0144
V 2VBCC
or 1.44% volume increase.
Assume that the initial volume is V and the final volume is Vt . Hence the volume change is
V V V
V Vt V t 1 t 1
V V V V V
3
Vt ( L L) L 3
(1 )
V L3 L
Using the two previous relations, we get that the respective length change is
(1 L / L)3 1 V / V L/ L 3
1V / 1 0.00477
Therefore there is a 0.477% increase in length. V
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, Chapter 3
Problem 3.1 We consider n= number of divacancies, N= number of atomic positions
AB and AC divacancies are not distinct, while AB and AD divacancies are distinct. If the
coordination number is , then the number of the distinct divacancies is N . The change in
Gibbs free energy after the formation of divacancies is:
G n E 2v nT S 2v T S2v conf
S2v conf k ln p (configurational entropy Boltzmann)
P number of combinations for n divacancies distribution at x N positions.
x!
P
n! x n!
ln P ln x! ln n! ln x n !
x ln x x n ln n n x n ln x n x n
ln P n n
ln ln
n x n N n
n n
Since n, then
N n N
n S2v E2v 2 E1v E2v
e E2vb
X 2v exp exp . Using the following relations:
N k kT b
S2v 2 S1v S 2v
2 S2 v E 2v
X e
exp exp , where 6 for FCC structure.
We get X e b b
2v 1v k kT
Problem 3.2 Because the energies are given in kJ/mol we use RT instead of kT in the fraction
calculations.
S Ev
X exp vib exp
v
k RT
2 S E
X X exp 2v bexp 2v from
b previous problem.
2v v
k RT
Svib Ei
X exp exp
i
k RT
@
@SSeeisismmicicisisoolalatitoionn
,After the calculations we get the following results:
27 oC (300) 1000 oC (1273 )
Xv 7 1015 1.4 103
X 2v 3.34 10 23 1.17 10 5
Xi 4.111067 7.131016
Problem 3.3
Xe S
exp nb exp E exp Gv
v
k kT
kT
Xv exp Gv 1 Gv
P T
kT kT P T
G Gv
Generally V , so V .
v
P T
P T
Where Vv is the volume difference per vacancy (Vv 0 ).
Xv Xv 0.
Thus, V
v
P T kT
Therefore the application of hydrostatic pressure reduces the equilibrium vacancy concentration.
Problem 3.4 If we heat a specimen of initial length Lo , then the dilatation comprises two parts:
L L
Lo tot o Lo v
Where is the change of lattice parameter at temperature T with respect to the reference
o
L
lattice parameter o at temperature To , and is the length change due to the addition of
Lo v
Lo
vacancies. We can express the length in cell number NC , so that N C .
ao
C NC Lo L
After heating we get N , so
ao a
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, Lo a
N N
L
ao
a Lo
L
ao
C C a2 ao2
Lo a 2o
1 L
1
a
N C
1L a
ao Lo ao Lo ao
2
Where a 0 L a
and . 0.
Lo ao
Since X e NC L a
3 , we get X e 3 .
v NC v
Lo ao
Problem 3.5 The vibration entropy Svib and the vacancy formation energy Ev are both
temperature dependent.
Ev Ev T H T v
Svib S T
S T
e
Hv T
So, X v exp exp
k kT
slope
e
ln X T v
Hv T 1 H T 1 S T
Thus, v (0.1)
1 k kT 1 k 1
T T P T P
P
We can observe from (1.1) relation that the slope depends on the temperature, therefore the
Arrhenius-type diagram will exhibit curvature.
H ST
Generally, dH T dS V dP T (0.2)
1 1
T P T P
Due to the (1.2) relation, the (1.1) relation becomes:
ln X e T H T E T
v
v v
1T P
k k
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SOLUTIONS
, Chapter 2
Problem 2.1 In FCC the relation between the lattice parameter and the atomic radius is
4R
, then α=4.95 Angstroms. On the cube phase (100) correspond 2 atoms (4x1/4+1). Then
2
the density of the (100) plane is
2 12 2
atoms/mm
(100 ) 8.2x10
4.95x10 7
In the (111) plane there are 3/6+3/2=2 atoms. The base of the triangle is 4R and the height 2 3R
After some math we get ρ(111)=9.5x1012 atoms/mm2. We see that the (111) plane has higher density than
the (100) plane, it is a close-packed plane.
Problem 2.2 The (100)-type plane closer to the origin is the (002) plane which cuts the z axis at
½. This has
a a 2R
d(002) 0 0 2 2 2 2
Setting R=1.749 Angstroms we get d(002)=2.745 Angstroms.
In the same way
a a 4R
d(111)
11 1 3 6
and d(111)=2.85 Angstroms. We see that the close-packed planes have a larger interplanar spacing.
Problem 2.3. The structure of vanadium is BCC. In this structure, the close-packed direction is [111], which
corresponds to the diagonal of the cubic unit cell where there is a consecutive contact of spheres (in the
model of hard spheres). Furthermore, the number of atoms per unit cell
for the BCC structure is 2. The first step is to find the lattice parameter α. The density is
2
3
Where is the Avogadro’s number. Therefore the lattice parameter is
3
2 50.94 a 3.08 10 8 cm 3.08 10 10 m
5.8 6.023 10 23
@
@SSeeisismmicicisisoolalatitoionn
,The length of the diagonal at the [111] close-packed direction is a 3 , which corresponds to 2
atoms. Hence the atomic density of the close-packed direction of vanadium (V) is
[111]
2 2 3.75 109 atoms / m
3 3.081010 3
The aforementioned atomic density result translates to 3750 atoms/μm or 3.75 atoms/nm.
4R
. The (10 0) plane is the
Problem 2.4. The lattice parameter for the FCC structure is 2
face of the unit cell. The face comprises ¼ of atoms at each corner plus 1 atom at the center of
the face. Hence the face consists of 4 () 1 2 atoms. The atomic density of the (10 0)
plane is
2 2
(100) 2 2 1
a 4R 4R 2
2
The (111) plane corresponds to the diagonal equilateral triangle of the unit cell. The base of this triangle is
4R . Using the Pythagorean Theorem, we can calculate the height of the triangle which
is 2 3R . Thus the area of the triangle is (base height / 2) 4 3R 2 . The equilateral triangle
comprises 6 of the atoms at each corner and ½ of the atoms at the middle of each side. Thus the
equilateral triangle consists of 3 (1/ 6) 3 () 2 atoms. The atomic density of the (111)
plane is
(111)
2 1
4 3R2 2 3R2
The ratio of the atomic densities is
(111)
2
(100)
1.154 1
3
Therefore (111) (100) and specifically the (111) plane has 15% higher atomic density than the
(10 0) plane. This is important since the plastic deformation of metals (Al, Cu, Ni, γ-Fe, etc.) is
accomplished with dislocation glide on the close-packed planes.
Problem 2.5. The ideal c/a ratio in HCP structure results when the atoms of this structure have an
arrangement as dense as the atoms of the FCC structure. The distance between the (0 0 01) bases of
the HCP structure is c. Using the fact that the (0 0 01) planes of HCP structure
correspond to the (111) planes of the FCC structure, we get
c 2 d(111)
FCC
Where d(111) is the distance between the (111) close-packed planes. We find that
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, a
d a a
(111) 2
h k 2
l 2
12
12 1 2
3 4R
d(111) FCC
4R 6
a 2
Thus,
8R
c c
6 4 1.63
HCP: a 2 R a 6
Therefore, the ideal ratio c/a for close packing in HCP structure is equal to 1.63. The c/a ratio for zinc (Zn)
is 1.86 while for titanium (Ti) is 1.59 (see Table 7.1, Book). This means that the distance between
the (0 0 01) planes is longer in Zn than in Ti. This fact affects the plastic
deformation in these metals, since the slip on (0 0 01) planes is easier in Zn than in Ti. Indeed the critical
shear stress of Zn is only 0.18 MPa, while of Ti is 110 MPa. Due to this, the plastic deformation in Ti
is performed on (10 1 0) plane, where the critical shear stress is approximately
49 MPa. Thus in Ti the slip is not performed on the close-packed planes of the crystal structure. For more
details look at the 7.3 paragraph of the book (plastic deformation of single crystals with slip).
Problem 2.6. The cell volume of HCP structure is the product of the base area (hexagon)
multiplied by the height c. The base of hexagon is A 6 R2 3 and the height is c 8R . As a
result, the cell volume is 6
V 24 2 R 3
The number of atoms per unit cell for the HCP structure is 6, thus the atomic packing factor is
4
6 R3
APFHCP 3 0.74
24 2 R 3 3 2
Regarding the BCC structure, the number of atoms per unit cell is 2 and the cell volume is where a a3 ,
4 R . Therefore the atomic packing factor of BCC structure is
3
4 3
2 R 3
APFBCC 3 8 0.68
3
4R
3
@
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,Since the atomic packing density of BCC is less than that of the HCP, the diffusion in BCC (movement of
atoms inside the lattice) is faster.
Problem 2.7 The density is
mass of cell atoms
cell volume
The structure of copper (Cu) is FCC and the number of atoms per unit cell is 4. The atomic mass
is , where is the atomic weight and N A is the Avogadro’s number. The cell volume is
NA
3
3 4R , thus the density is
a
2
63.57
4 23 9gr / cm3
6.023 10
4 1.276 10 8 3
2
Problem 2.8 Notice that the atomic volume is not 3
3
4 R ! It is the corresponding volume of
every atom of the structure plus the empty surrounding space inside the cell. Due to the fact that the
structure of gold (Au) is FCC, the number of atoms per unit cell is 4. Therefore the atomic volume is
3
4
For the FCC structure we get that 4R , hence
2
3
4R
2
4
After replacing the value of the atomic radius of gold R , we find that
o
10
1.7 10 29 m3 . Since
1 10 m , the atomic volume of gold is
o
17 ( )3
@
@SSeeisismmicicisisoolalatitoionn
,The molar volume Vm is the volume corresponding to one mole of gold and is obtained by
multiplying the atomic volume by the Avogadro’s number. Thus the molar volume is
V N 1.02 10 5 m 3 / mol
m A
Problem 2.9
For the atomic radius of the iron atom, RFCC=1.270 and RBCC=1.241 Angtroms (A) FCC has 4
atmos/cell while BCC has 2 atoms/cell.
In FCC a 4R 2 3.591A,VFCC a 3 46.34 A3
In BCC a 4R 3 2.865A,VBCC 23.51A 3
Taking 4 atoms as a reference, this corresponds to 1 FCC cell and 2 BCC cells, then
V 2VBCC VFCC
0.0144
V 2VBCC
or 1.44% volume increase.
Assume that the initial volume is V and the final volume is Vt . Hence the volume change is
V V V
V Vt V t 1 t 1
V V V V V
3
Vt ( L L) L 3
(1 )
V L3 L
Using the two previous relations, we get that the respective length change is
(1 L / L)3 1 V / V L/ L 3
1V / 1 0.00477
Therefore there is a 0.477% increase in length. V
@
@SSeeisismmicicisisoolalatitoionn
, Chapter 3
Problem 3.1 We consider n= number of divacancies, N= number of atomic positions
AB and AC divacancies are not distinct, while AB and AD divacancies are distinct. If the
coordination number is , then the number of the distinct divacancies is N . The change in
Gibbs free energy after the formation of divacancies is:
G n E 2v nT S 2v T S2v conf
S2v conf k ln p (configurational entropy Boltzmann)
P number of combinations for n divacancies distribution at x N positions.
x!
P
n! x n!
ln P ln x! ln n! ln x n !
x ln x x n ln n n x n ln x n x n
ln P n n
ln ln
n x n N n
n n
Since n, then
N n N
n S2v E2v 2 E1v E2v
e E2vb
X 2v exp exp . Using the following relations:
N k kT b
S2v 2 S1v S 2v
2 S2 v E 2v
X e
exp exp , where 6 for FCC structure.
We get X e b b
2v 1v k kT
Problem 3.2 Because the energies are given in kJ/mol we use RT instead of kT in the fraction
calculations.
S Ev
X exp vib exp
v
k RT
2 S E
X X exp 2v bexp 2v from
b previous problem.
2v v
k RT
Svib Ei
X exp exp
i
k RT
@
@SSeeisismmicicisisoolalatitoionn
,After the calculations we get the following results:
27 oC (300) 1000 oC (1273 )
Xv 7 1015 1.4 103
X 2v 3.34 10 23 1.17 10 5
Xi 4.111067 7.131016
Problem 3.3
Xe S
exp nb exp E exp Gv
v
k kT
kT
Xv exp Gv 1 Gv
P T
kT kT P T
G Gv
Generally V , so V .
v
P T
P T
Where Vv is the volume difference per vacancy (Vv 0 ).
Xv Xv 0.
Thus, V
v
P T kT
Therefore the application of hydrostatic pressure reduces the equilibrium vacancy concentration.
Problem 3.4 If we heat a specimen of initial length Lo , then the dilatation comprises two parts:
L L
Lo tot o Lo v
Where is the change of lattice parameter at temperature T with respect to the reference
o
L
lattice parameter o at temperature To , and is the length change due to the addition of
Lo v
Lo
vacancies. We can express the length in cell number NC , so that N C .
ao
C NC Lo L
After heating we get N , so
ao a
@
@SSeeisismmicicisisoolalatitoionn
, Lo a
N N
L
ao
a Lo
L
ao
C C a2 ao2
Lo a 2o
1 L
1
a
N C
1L a
ao Lo ao Lo ao
2
Where a 0 L a
and . 0.
Lo ao
Since X e NC L a
3 , we get X e 3 .
v NC v
Lo ao
Problem 3.5 The vibration entropy Svib and the vacancy formation energy Ev are both
temperature dependent.
Ev Ev T H T v
Svib S T
S T
e
Hv T
So, X v exp exp
k kT
slope
e
ln X T v
Hv T 1 H T 1 S T
Thus, v (0.1)
1 k kT 1 k 1
T T P T P
P
We can observe from (1.1) relation that the slope depends on the temperature, therefore the
Arrhenius-type diagram will exhibit curvature.
H ST
Generally, dH T dS V dP T (0.2)
1 1
T P T P
Due to the (1.2) relation, the (1.1) relation becomes:
ln X e T H T E T
v
v v
1T P
k k
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