# Solutions to Homework _2 by wuxiangyu

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```									                     Solutions to Homework #2

3.20 This problem asks that we calculate the unit cell volume for Re which has an HCP crystal
structure. In order to do this, it is necessary to use a result of Problem 3.7, that is

2
V = 6R c 3
C

The problem states that c = 1.615a, and a = 2R. Therefore
3
V = (1.615)(12 3 )R
C

-8      3            -23   3           -2  3
= (1.615)(12 3 )(1.37 x 10        cm) = 8.63 x 10     cm = 8.63 x 10 nm

_
3.26 (a) This portion of the problem asks that a [01 1] direction be drawn within a monoclinic

unit cell (a ≠ b ≠ c, and  =  = 90≠ ). One such unit cell with its origin at point O is
sketched below. For this direction, there is no projection along the x-axis since the first
index is zero; thus, the direction lies in the y-z plane. We next move from the origin
along the minus y-axis b units (from point O to point R). Since the final index is a one,
_
move from point R parallel to the z-axis, c units (to point P). Thus, the [01 1] direction

corresponds to the vector passing from the origin to point P, as indicated in the figure.

z

-                                                              c
P        [011]

y
           

O
                      a

-y          R
b
x
(b) A (002) plane is drawn within the monoclinic cell shown below. We first remove the
parentheses and take the reciprocals of the indices; this gives   and 1/2. Thus, the
, ,
(002) plane parallels both x- and y-axes, and intercepts the z-axis at c/2, as indicated in
the drawing.

z

c

y
           

O
                        a

b
x

_
3.30 Direction A is a [4 30] direction, which determination is summarized as follows. We first

of all position the origin of the coordinate system at the tail of the direction vector; then in
terms of this new coordinate system

x                 y                z
2a              b
Projections                                   - 3               2                 0c

Projections in terms of a, b,
2               1
and c                                - 3               2                 0

Reduction to integers                          -4               3                 0
_
Enclosure                                                     [4 30]

_
Direction B is a [23 2] direction, which determination is summarized as follows.

We first of all position the origin of the coordinate system at the tail of the direction vector;
then in terms of this new coordinate system
x                 y                z
2a                                 2c
Projections                                    3                -b               3

Projections in terms of a, b,
2                                   2
and c                                 3                 -1                3

Reduction to integers                          2                -3                2
_
Enclosure                                                     [23 2]

__
Direction C is a [133 ] direction, which determination is summarized as follows.

We first of all position the origin of the coordinate system at the tail of the direction vector;
then in terms of this new coordinate system
x                 y                z
a
Projections                                   3                 -b                -c

Projections in terms of a, b,
1
and c                                 3                 -1                -1

Reduction to integers                          1                -3                -3
__
Enclosure                                                     [133 ]

_
Direction D is a [136 ] direction, which determination is summarized as follows.

We first of all position the origin of the coordinate system at the tail of the direction vector;
then in terms of this new coordinate system

x                 y                z

a                 b
Projections                                   6                 2                 -c

Projections in terms of a, b,
1                 1
and c                                 6                 2                 -1

Reduction to integers                          1                 3                -6
_
Enclosure                                                     [136 ]
_            _
3.36 The (11 01) and (112 0) planes in a hexagonal unit cell are shown below.

z                                           z

a                                          a
2                                          2

a                                          a
3                                          3                          a
a
1                                      1
_                                            _
(1101)                                      (1120)

3.45 Planar density, PD, is defined as

A
c
PD = A
p

where A is the total plane area within the unit cell and A is the circle plane area within
p                                                  c
this same plane. For the (100) plane in BCC, in terms of the atomic radius, R, and the unit
cell edge length a

2
4R 2 16R
A =a =   = 3
2
p      3

Also, upon examination of that portion of the (100) plane within a single unit cell, that there
resides a single equivalent atom--one-fourth from each of the four corner atoms.
Therefore,

2
A = R
c

Hence
2
R
PD =          2 = 0.59
16R
3
That portion of a (110) plane that passes through a BCC unit cell forms a rectangle
as shown below.

R                   4R
3

4R 2
3

In terms of the atomic radius R, the length of the rectangle base is 4R 2    /   3 , whereas
4R
the height is a =      . Therefore, the area of this rectangle, which is just A is
3                                                         p

2
4R 24R = 16R 2
A =       
p     3  3       3

Now for the number equivalent atoms within this plane. One-fourth of each corner atom
and the entirety of the center atom belong to the unit cell.          Therefore, there is an
equivalent of 2 atoms within the unit cell. Hence

2
A = 2(R )
c

and
2
2R
PD =       2   = 0.83
16R 2
3

3.49 Although each individual grain in a polycrystalline material may be anisotropic, if the
grains have random orientations, then the solid aggregate of the many anisotropic grains
will behave isotropically.

3.50 From the table, molybdenum has a BCC crystal structure and an atomic radius of 0.1363
nm. Using Equation (3.3), the lattice parameter a may be computed as
4R   (4)(0.1363 nm)
a=      =                = 0.3148 nm
3           3

Now, the interplanar spacing d         maybe determined using Equation (3.10) as
111

a                    0.3148
d    =                            =          = 0.1818 nm
111         2       2        2          3
(1) + (1) + (1)

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