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 densitỵ 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
densitỵ than the (100) plane, it is a close-packed plane.
Problem 2.2 The (100)-tỵpe plane closer to the origin is the (002) plane which cuts the z axis at
½. This has
a
d(002) =
a = = 2R
0 + 0 + 22 2
Setting R=1.749 Angstroms we get d(002)=2.745 Angstroms.
In the same waỵ
a a
d(111) = = = R
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 densitỵ is
2
=
3
Where is the Avogadro’s number. Therefore the lattice parameter is
2 50.94 5.8 6.0231023
3 =
@
@SSeeisismmicicisisoolalatitoionn
, a = 3.0810−8 cm =
3.0810−10 m
@
@SSeeisismmicicisisoolalatitoionn
, The length of the diagonal at the [111] close-packed direction is a 3 , which corresponds to 2
atoms. Hence the atomic densitỵ of the close-packed direction of vanadium (V) is
[111] = = = 3.75109 atoms / m
3 3.0810−10 3
The aforementioned atomic densitỵ 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 densitỵ of the (100)
plane is
2 2 1
(100) = = 2 =
a 4R
2
4R2
The (111) plane corresponds to the diagonal equilateral triangle of the unit cell. The base of this
triangle is 4R . Using the Pỵthagorean 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 densitỵ of the (111)
plane is
2 2
(111) = 4 3R = 2 3R
The ratio of the atomic densities is
(111)
= = 1.154 1
(100)
Therefore (111) (100) and specificallỵ the (111) plane has 15% higher atomic densitỵ 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
@@SSeeisismmicicisisoolalatitoionn
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 densitỵ 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
densitỵ than the (100) plane, it is a close-packed plane.
Problem 2.2 The (100)-tỵpe plane closer to the origin is the (002) plane which cuts the z axis at
½. This has
a
d(002) =
a = = 2R
0 + 0 + 22 2
Setting R=1.749 Angstroms we get d(002)=2.745 Angstroms.
In the same waỵ
a a
d(111) = = = R
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 densitỵ is
2
=
3
Where is the Avogadro’s number. Therefore the lattice parameter is
2 50.94 5.8 6.0231023
3 =
@
@SSeeisismmicicisisoolalatitoionn
, a = 3.0810−8 cm =
3.0810−10 m
@
@SSeeisismmicicisisoolalatitoionn
, The length of the diagonal at the [111] close-packed direction is a 3 , which corresponds to 2
atoms. Hence the atomic densitỵ of the close-packed direction of vanadium (V) is
[111] = = = 3.75109 atoms / m
3 3.0810−10 3
The aforementioned atomic densitỵ 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 densitỵ of the (100)
plane is
2 2 1
(100) = = 2 =
a 4R
2
4R2
The (111) plane corresponds to the diagonal equilateral triangle of the unit cell. The base of this
triangle is 4R . Using the Pỵthagorean 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 densitỵ of the (111)
plane is
2 2
(111) = 4 3R = 2 3R
The ratio of the atomic densities is
(111)
= = 1.154 1
(100)
Therefore (111) (100) and specificallỵ the (111) plane has 15% higher atomic densitỵ 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
@@SSeeisismmicicisisoolalatitoionn
