CHAPTER 3
STRUCTURES OF METALS AND CERAMICS
PROBLEM SOLUTIONS
3.32
(a) Using the ionic radii in Table 3.4, compute the theoretical density of CsCl. (Hint: Use a
modification of the result of Problem 3.3.)
(b) The measured density is 3.99 g/cm3. How do you explain the slight discrepancy between your
calculated value and the measured one?
Solution
(a) We are asked to compute the density of CsCl. Modifying the result of Problem 3.3, we get
a =
2rCs   2rCl
3
=
2 (0.170 nm)  2 ( 0.181 nm)
3
= 0.405 nm = 4.05  10-8 cm

From Equation 3.8
 
n( ACs + ACl )
n( ACs + ACl )
=
VC N A
a3 N A
For the CsCl crystal structure, n' = 1 formula unit/unit cell, and thus

 
(1 formula unit/unit cell)(132.91 g/mol + 35.45 g/mol)
(4.05  10-8 cm) 3 /unit cell(6.022  1023 formula units/mol)

= 4.21 g/cm3
(b) This value of the density is greater than the measured density (3.99 g/cm3 ). The reason for this
discrepancy is that the ionic radii in Table 3.4, used for this computation, were for a coordination number of six, when,
in fact, the coordination number of both Cs + and Cl- is eight. The ionic radii should be slightly greater, leading to a
larger VC value, and a lower density.
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3.54 Determine the indices for the directions shown in the following cubic unit cell:
Solution
Direction A is a [430] 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

2a
b
3
2
2
1
3
2
–
Projections in terms of a, b, and c
Enclosure
y
–
Projections
Reduction to integers
x

–4



z
0c
0
3
0
[430]
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

Projections
Enclosure
y
2a
–b
3
2
Projections in terms of a, b, and c
Reduction to integers
x
3


2
z
2c
3
2
–1
–3
[23 2]
3

2

Direction C is a [13 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

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Projections
x
y
z
a
–b
–c
–1
–1
–3
–3
3
1
Projections in terms of a, b, and c
3
Reduction to integers

Enclosure

1
[13 3]
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

Projections
Projections in terms of a, b, and c
Reduction to integers

Enclosure

x
y
z
a
b
6
1
2
1
–c
6
2
1


3
–1
–6
[136 ]

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3.62 Sketch within a cubic unit cell the following planes:

(a) (01 1 ) ,
(e) (1 11 ) ,
(b) (112 ) ,
(f) (12 2 ) ,
(c) (102 ) ,


(g) (1 23 ) ,

(h) (01 3 )
(d) (13 1) ,



Solution

The planes called for are plotted in the cubic unit cells shown below.
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3.66 Cite the indices of the direction that results from the intersection of each of the following pair of
planes within a cubic crystal: (a) the (100) and (010) planes, (b) the (111) and (111 ) planes, and (c) the (101 )
and (001) planes.

Solution

(a) In the figure below is shown (100) and (010) planes, and, as indicated, their intersection results in a [001], or
equivalently, a [001 ] direction.

(b) In the figure below is shown (111) and (111) planes, and, as indicated, their intersection results in a
[1 10] , or equivalently, a [11 0] direction.



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(c) In the figure below is shown (101) and (001) planes, and, as indicated, their intersection results in a
[010], or equivalently, a [01 0] direction.


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