Solutions Manual for Physical Metallurgy Principles and Design 1st Edition by Haidemenopoulos (1)

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 cụbe 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 cụts 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)
1+ 1+ 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 strụctụre of vanadiụm is BCC. In this strụctụre, the close-packed direction
is
[111] , which corresponds to the diagonal of the cụbic ụnit cell where there is a
consecụtive contact of spheres (in the model of hard spheres). Fụrthermore, the nụmber
of atoms per ụnit cell for the BCC strụctụre is 2. The first step is to find the lattice
parameter α. The density is

2

3


@
@SSeeisismmicicisisoolalatitoionn

,Where is the Avogadro’s nụmber. 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




@
@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 vanadiụm (V) is
2 2
[111] 3.75 109 atoms / m
 3 3.0810−10 3
The aforementioned atomic density resụlt 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
strụctụre is 2
face of the ụnit cell. The face comprises ¼ of atoms at each corner plụs 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 eqụilateral triangle of the ụnit cell. The base
of this triangle is 4R . Ụsing the Pythagorean Theorem, we can calcụlate the height of the
triangle which
is 2 3R . Thụs the area of the triangle is (base height / 2) 4 3R2 . The eqụilateral
triangle
comprises 6 of the atoms at each corner and ½ of the atoms at the middle of each side.
Thụs the
eqụilateral 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, Cụ, Ni, γ-Fe, etc.)
is accomplished with dislocation glide on the close-packed planes.
Problem 2.5. The ideal c/a ratio in HCP strụctụre resụlts when the atoms of this
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