MSE280B: Introduction to Engineering Materials
Homework 2: Solutions
1.
(a)
For indium, and from the definition of the APF
Problem 3.8 from Callister (6 points)
APF =
V
S
V
C
=
n
4
3
π
R
3
a
2
c
we may solve for the number of atoms per unit cell,
n
, as
n
=
(APF)
a
2
c
4
3
π
R
3
=
(0.693)(4.59
×
10
8
cm)
2
(4.95
×
10
8
cm)
4
3
π
(1.625
×
10
8
cm
)
3
= 4.0 atoms/unit cell
(b)
In order to compute the density, we just employ Equation 3.5 as
ρ
=
nA
In
a
2
c N
A
=
(4 atoms/unit cell)(114.82 g/mol)
(4.59
×
10
8
cm)
2
(4.95
×
10
8
cm)/unit cell
[
]
(6.02
×
10
23
atoms/mol)
= 7.31 g/cm
3
Problem 3.28 from Callister (6 points)
Direction A is a
[33
1 ]
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
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y
z
Projections
a
b
–
c
3
Projections in terms of
a
,
b
, and
c
1
1
–
1
3
Reduction to integers
3
3
–1
Enclosure
[33
1 ]
Direction B is a
[
4 0
3 ]
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
Projections
–
2
a
3
0
b
–
c
2
Projections in terms of
a
,
b
, and
c
–
2
3
0
–
1
2
Reduction to integers
–4
0
–3
Enclosure
[
4 0
3 ]
Direction C is a
[
3 61]
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
Projections
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 Fall '08
 Johnson
 Crystal

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