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) 7
8.2x1012 atoms/mm2
4.95x10
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
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 3.0810 10
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 structure is . The (100) plane is the
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 (100)
plane is
2 2 1
(100)
a2 4R 4R2
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 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)
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
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) 7
8.2x1012 atoms/mm2
4.95x10
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
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 3.0810 10
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 structure is . The (100) plane is the
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 (100)
plane is
2 2 1
(100)
a2 4R 4R2
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 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)
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
