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
(100) 8.2x1012 atoms/mm2
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 22 2 2
Setting R=1.749 Angstroms we get d(002)=2.745 Angstroms.
In the same
way
a 4R
d(111)
a 6
1 1 1 3
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
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, 3
Where is the Avogadro’s number. Therefore the lattice parameter is
2 50.94
3
a 3.08 10 8 cm 3.08 10 10 m
5.8
6.023 1023
,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
2 2
[111] 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
Problem 2.4. The lattice parameter for the FCC . The (100) plane is the
2
structure is
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
(100)
plane is
2 2 1
(100)
a2 2
4R 4R2
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 3R2 . 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 () 3 () 2 atoms. The atomic density of the
(111)
plane is
2 1
(111)
4 3R2 2 3R2
The ratio of the atomic densities is
(111) 2
1.154 1
(100)
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,Therefore (111) (100) and specifically the (111) plane has 15% higher atomic density
than the
(100) 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 (0001) bases of the HCP structure is c. Using the fact that the
(0001) 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
, a
d a a
(111)
h2 k 2 l 2 12 12 12 3 4R
d (111)
4R 6
a FCC
2
Thus
,
8R c 4
c 1.63
6
a 6
HCP: a 2R
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 (0001) planes is longer in Zn than in Ti. This fact
affects the plastic
deformation in these metals, since the slip on (0001) 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 A 6 3 and the height is c . As a
R2 8R
hexagon is
6
result, the cell volume is
V 24 R3
The number of atoms per unit cell for the HCP structure is 6, thus the atomic packing
factor is
4
6 R3
APF 3 0.74
HCP
24 R3
Regarding the BCC structure, the number of atoms per unit cell is 2 and the cell
volume is
4R
where a . Therefore the atomic packing factor of BCC structure is
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,a3 ,
4
2 R3
3 3 0.68
APFBCC
8
3
4R
,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 N is the Avogadro’s number. The cell volume is
A
and
NA
4R
3
, thus the density is
a3
2
63.57
4
6.023 1023
9gr / cm3
3
4 1.276 10 8
2
4
Problem 2.8 Notice that the atomic volume is R3 ! It is the corresponding volume of
3
not
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
4R
For the FCC structure we get that , hence
2
3
4R
2
4
After replacing the value of the atomic radius of gold R , we find that 1.7 10 29 m3 .
Since
o 1 10
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,10
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, The molar volume is the volume corresponding to one mole of gold and is obtained
Vm by
multiplying the atomic volume by the Avogadro’s number. Thus the molar volume is
V N 1.02 10 5 m3 / 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 a3 46.34 A3
In BCC a 4R 3 2.865A,V 23.51A3
BCC
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 Vt V Vt V
1 1
Vt
V V V V V
V (L L)3 L 3
t (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 1 0.00477
/L
Therefore there is a 0.477% increase in length.
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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
(100) 8.2x1012 atoms/mm2
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 22 2 2
Setting R=1.749 Angstroms we get d(002)=2.745 Angstroms.
In the same
way
a 4R
d(111)
a 6
1 1 1 3
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
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, 3
Where is the Avogadro’s number. Therefore the lattice parameter is
2 50.94
3
a 3.08 10 8 cm 3.08 10 10 m
5.8
6.023 1023
,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
2 2
[111] 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
Problem 2.4. The lattice parameter for the FCC . The (100) plane is the
2
structure is
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
(100)
plane is
2 2 1
(100)
a2 2
4R 4R2
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 3R2 . 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 () 3 () 2 atoms. The atomic density of the
(111)
plane is
2 1
(111)
4 3R2 2 3R2
The ratio of the atomic densities is
(111) 2
1.154 1
(100)
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,Therefore (111) (100) and specifically the (111) plane has 15% higher atomic density
than the
(100) 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 (0001) bases of the HCP structure is c. Using the fact that the
(0001) 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
, a
d a a
(111)
h2 k 2 l 2 12 12 12 3 4R
d (111)
4R 6
a FCC
2
Thus
,
8R c 4
c 1.63
6
a 6
HCP: a 2R
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 (0001) planes is longer in Zn than in Ti. This fact
affects the plastic
deformation in these metals, since the slip on (0001) 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 A 6 3 and the height is c . As a
R2 8R
hexagon is
6
result, the cell volume is
V 24 R3
The number of atoms per unit cell for the HCP structure is 6, thus the atomic packing
factor is
4
6 R3
APF 3 0.74
HCP
24 R3
Regarding the BCC structure, the number of atoms per unit cell is 2 and the cell
volume is
4R
where a . Therefore the atomic packing factor of BCC structure is
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,a3 ,
4
2 R3
3 3 0.68
APFBCC
8
3
4R
,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 N is the Avogadro’s number. The cell volume is
A
and
NA
4R
3
, thus the density is
a3
2
63.57
4
6.023 1023
9gr / cm3
3
4 1.276 10 8
2
4
Problem 2.8 Notice that the atomic volume is R3 ! It is the corresponding volume of
3
not
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
4R
For the FCC structure we get that , hence
2
3
4R
2
4
After replacing the value of the atomic radius of gold R , we find that 1.7 10 29 m3 .
Since
o 1 10
@LECTJULIESOLUTIONSSTUVIA
,10
m
,
t
h
e
a
t
o
m
ic
v
o
l
u
m
e
o
f
g
o
l
d
is
o
17 ( )3
, The molar volume is the volume corresponding to one mole of gold and is obtained
Vm by
multiplying the atomic volume by the Avogadro’s number. Thus the molar volume is
V N 1.02 10 5 m3 / 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 a3 46.34 A3
In BCC a 4R 3 2.865A,V 23.51A3
BCC
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 Vt V Vt V
1 1
Vt
V V V V V
V (L L)3 L 3
t (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 1 0.00477
/L
Therefore there is a 0.477% increase in length.
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