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CHAPTER 3

STRUCTURES OF METALS AND CERAMICS

PROBLEM SOLUTIONS

Point Coordinates

3.22 List the point coordinates of both the sodium and chlorine ions for a unit cell of the sodium chloride

crystal structure (Figure 3.5).

Solution

Here we are asked list point coordinates for both sodium and chlorine ions for a unit cell of the sodium

chloride crystal structure, which is shown in Figure 3.5.

In Figure 3.5, the chlorine ions are situated at all corners and face-centered positions. Therefore, point

coordinates for these ions are the same as for FCC, as presented in the previous problem—that is, 000, 100, 110,

010, 001, 101, 111, 011,

1

2

1

20,

1

2

1

21,

11

2

1

2,

01

2

1

2,

1

20

1

2, and

1

21

1

2.

Furthermore, the sodium ions are situated at the centers of all unit cell edges, and, in addition, at the unit

cell center. For the bottom face of the unit cell, the point coordinates are as follows:

1

200,

11

20,

1

210,

01

20.

While, for the horizontal plane that passes through the center of the unit cell (which includes the ion at the unit cell

center), the coordinates are

001

2,

1 01

2,

1

2

1

2

1

2,

1 11

2, and

011

2. And for the four ions on the top face

1

201,

11

21,

1

21 1, and

01

21.


3.26 What are the indices for the direction indicated by the vector in the sketch below?

Solution

We are asked for the indices of the direction sketched in the figure. The projection on the x-axis is a/2,

while projections on the y- and z-axes are 0b and c, respectively. This is a [102] direction as indicated in the

summary below.

x y z

Projections a/2 0b c

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

Reduction to integers 1 0 2

Enclosure [102]


3.27 Within a cubic unit cell, sketch the following directions:

(a)

[102 ] , (c)

[2 12] ,

(b)

[31 3] , (d) [301].

Solution

The directions asked for are indicated in the cubic unit cell shown below.


3.28 Determine the indices for the directions shown in the following cubic unit cell:

Solution

Direction A is a

[331 ] 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 a b –

c

3

Projections in terms of a, b, and c 1 1 –

1

3

Reduction to integers 3 3 –1

Enclosure

[331 ]

Direction B is a

[4 03 ] 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 –

2a

3 0b –

c

2

Projections in terms of a, b, and c –

2

3 0 –

1

2

Reduction to integers –4 0 –3

Enclosure

[4 03 ]

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 –

a

2 b

c

6


1

2 1

1

6

Reduction to integers –3 6 1

Enclosure

[3 61]

Direction D is a

[1 11 ] 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 –

a

2

b

2 –

c

2


1

2

1

2 –

1

2

Reduction to integers –1 1 –1

Enclosure

[1 11 ]


3.29 For tetragonal crystals, cite the indices of directions that are equivalent to each of the following

directions:

(a) [011]

(b) [100]

Solution

For tetragonal crystals a = b ≠ c and = = = 90; therefore, projections along the x and y axes are

equivalent, which are not equivalent to projections along the z axis.

(a) Therefore, for the [011] direction, equivalent directions are the following: [101],

[1 01 ],

[1 01],

[101 ],

[011 ] ,

[01 1] , and

[01 1 ] .

(b) Also, for the [100] direction, equivalent directions are the following:

[1 00] , [010], and

[01 0].


3.31 Determine the indices for the two directions shown in the following hexagonal unit cell:

Solution

For direction A, projections on the a1, a2, and z axes are a, 0a, and c/2, or, in terms of a and c the

projections are 1, 0, and 1/2, which when multiplied by the factor 2 become the smallest set of integers: 2, 0, and 1.

This means that

u’ = 2

v’ = 0

w’ = 1

Now, from Equations 3.7, the u, v, t, and w indices become

u 1

3(2u' v' )

1

3(2)(2) 0

4

3

v 1

3(2vÕ uÕ)

1

3(2)(0) (2)

2

3

t (u v) 4

3

2

3

2

3

w w' 1

Now, in order to get the lowest set of integers, it is necessary to multiply all indices by the factor 3, with the result

that the direction A is a

[42 2 3] direction.

For direction B, projections on the a1, a2, and z axes are –a, 0a, and 0c, or, in terms of a and c the

projections are –1, 0, and 0. This means that

u’ = –1

v’ = 0


w’ = 0

Now, from Equations 3.7, the u, v, t, and w indices become

u 1

3(2u' v)

1

3(2)(1) 0

2

3

v 1

3(2v' u' )

1

3(2)(0) (1)

1

3

t (u v) 2

3

1

3

1

3

w w' 0

Now, in order to get the lowest set of integers, it is necessary to multiply all indices by the factor 3, with the result

that the direction B is a

[2 110] direction.


3.35 Sketch within a cubic unit cell the following planes:

(a) (012), (c)

(101 ) ,

(b)

(31 3) , (d)

(21 1) .

Solution

The planes called for are plotted in the cubic unit cells shown below.


3.36 Determine the Miller indices for the planes shown in the following unit cell:

Solution

For plane A we will leave the origin at the unit cell as shown. If we extend this plane back into the plane of

the page, then it is a

(111 ) plane, as summarized below.

x y z

Intercepts a b – c

Intercepts in terms of a, b, and c 1 1 – 1

Reciprocals of intercepts 1 1 – 1

Reduction not necessary

Enclosure

(111 )

[Note: If we move the origin one unit cell distance parallel to the x axis and then one unit cell distance parallel to the

y axis, the direction becomes

(1 1 1) ].

For plane B we will leave the origin of the unit cell as shown; this is a (230) plane, as summarized below.

x y z

Intercepts

a

2

b

3 ∞c

Intercepts in terms of a, b, and c

1

2

1

3 ∞

Reciprocals of intercepts 2 3 0

Enclosure (230)


3.37 Determine the Miller indices for the planes shown in the following unit cell:

Solution

For plane A since the plane passes through the origin of the coordinate system as shown, we will move the

origin of the coordinate system one unit cell distance vertically along the z axis; thus, this is a

(211 ) plane, as

summarized below.

x y z

Intercepts

a

2 b – c

Intercepts in terms of a, b, and c

1

2 1 – 1



Enclosure

(211 )

For plane B, since the plane passes through the origin of the coordinate system as shown, we will move the

origin one unit cell distance vertically along the z axis; this is a

(021 ) plane, as summarized below.

x y z

Intercepts ∞a

b

2 – c

Intercepts in terms of a, b, and c ∞

1

2 – 1



Enclosure

(021 )


3.38 For each of the following crystal structures, represent the indicated plane in the manner of Figures

3.26 and 3.27, showing both anions and cations:

(a) (111) plane for the diamond cubic crystal structure, and

(b) (110) plane for the fluorite crystal structure.

Solution

(a) A (111) plane for the diamond cubic crystal structure would appear as

(b) A (110) plane for the fluorite crystal structure would appear as


3.41 Determine the indices for the planes shown in the hexagonal unit cells below:

Solution

(a) For this plane, intersections with the a1, a2, and z axes are a/2, –a, and c (the plane parallels the z

axis). In terms of a and c these intersections are 1/2, –1, and , the respective reciprocals of which are 2, –1, and 0.

This means that

h = 2

k = –1

l = 0

Now, from Equation 3.8, the value of i is

i (h k) (2 1) 1

Hence, this is a

(21 1 0) plane.

(b) For this plane, intersections with the a1, a2, and z axes are –a, a, and c/2, respectively. In terms of a

and c these intersections are –1, 1, and 1/2, the respective reciprocals of which are –1, 1, and 2. This means that

h = –1

k = 1

l = 2

Now, from Equation 3.8, the value of i is

i (h k) (1 1) 0

Therefore, this is a

(1 102) plane.


3.42 Sketch the

(21 1 0) plane in a hexagonal unit cell.

Solution

For this

(21 1 0) plane, the reciprocals of h, k, i, and l are, respectively, 1/2, –1, –1, and ; thus, this plane

is parallel to the c axis, and intersects the a1 axis at a/2, the a2 axis at –a, and the a3 axis at –a. The plane having

these intersections is shown in the figure below.


Linear and Planar Densities

3.43 (a) Derive linear density expressions for FCC [100] and [111] directions in terms of the atomic

radius R.

(b) Compute and compare linear density values for these same two directions for copper.

Solution

(a) In the figure below is shown a [100] direction within an FCC unit cell.

For this [100] direction there is one atom at each of the two unit cell corners, and, thus, there is the equivalent of 1

atom that is centered on the direction vector. The length of this direction vector is just the unit cell edge length,

2R 2 (Equation 3.1). Therefore, the expression for the linear density of this plane is

LD100 = number of atoms centered on [100] direction vector

length of [100] direction vector

1 atom

2 R 2

1

2 R 2

An FCC unit cell within which is drawn a [111] direction is shown below.


For this [111] direction, the vector shown passes through only the centers of the single atom at each of its ends, and,

thus, there is the equivalence of 1 atom that is centered on the direction vector. The length of this direction vector is

denoted by z in this figure, which is equal to

z x2 y2

where x is the length of the bottom face diagonal, which is equal to 4R. Furthermore, y is the unit cell edge length,

which is equal to

2R 2 (Equation 3.1). Thus, using the above equation, the length z may be calculated as follows:

z (4R)2 (2R 2)2 24R2 2R 6

Therefore, the expression for the linear density of this direction is

LD111 = number of atoms centered on [111] direction vector

length of [111] direction vector

1 atom

2 R 6

1

2 R 6

(b) From the table inside the front cover, the atomic radius for copper is 0.128 nm. Therefore, the linear

density for the [100] direction is

LD100 (Cu) 1

2 R 2

1

(2)(0.128 nm) 2 2.76 nm1 2.76 109 m1

While for the [111] direction

LD111(Cu) 1

2 R 6

1

(2)(0.128 nm) 6 1.59 nm1 1.59 109 m1


3.44 (a) Derive planar density expressions for BCC (100) and (110) planes in terms of the atomic radius

R.

(b) Compute and compare planar density values for these same two planes for molybdenum.

Solution

(a) A BCC unit cell within which is drawn a (100) plane is shown below.

For this (100) plane there is one atom at each of the four cube corners, each of which is shared with four adjacent

unit cells. Thus, there is the equivalence of 1 atom associated with this BCC (100) plane. The planar section

represented in the above figure is a square, wherein the side lengths are equal to the unit cell edge length,

4 R

3

(Equation 3.3); and, thus, the area of this square is just

4R

3

2

=

16 R2

3. Hence, the planar density for this (100)

plane is just

PD100 = number of atoms centered on (100) plane

area of (100) plane

1 atom

16 R2

3

3

16 R2

A BCC unit cell within which is drawn a (110) plane is shown below.


For this (110) plane there is one atom at each of the four cube corners through which it passes, each of which is

shared with four adjacent unit cells, while the center atom lies entirely within the unit cell. Thus, there is the

equivalence of 2 atoms associated with this BCC (110) plane. The planar section represented in the above figure is a

rectangle, as noted in the figure below.

From this figure, the area of the rectangle is the product of x and y. The length x is just the unit cell edge length,

which for BCC (Equation 3.3) is

4 R

3. Now, the diagonal length z is equal to 4R. For the triangle bounded by the

lengths x, y, and z

y z2 x2

Or

y (4 R)2 4R

3

2

4 R 2

3

Thus, in terms of R, the area of this (110) plane is just


Area (110) xy 4 R

3

4 R 2

3

16 R2 2

3

And, finally, the planar density for this (110) plane is just

PD110 = number of atoms centered on (110) plane

area of (110) plane

2 atoms

16 R2 2

3

3

8 R2 2

(b) From the table inside the front cover, the atomic radius for molybdenum is 0.136 nm. Therefore, the

planar density for the (100) plane is

PD100 (Mo) 3

16 R2

3

16 (0.136 nm)2 10.14 nm2 1.014 1019 m2

While for the (110) plane

PD110 (Mo) 3

8 R2 2

3

8 (0.136 nm)2 2 14.34 nm2 1.434 1019 m2


3.46 Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangement of

O2–

ions.

(a) Which type of interstitial site will the Fe2+

ions occupy? Why?

(b) Which type of interstitial site will the Ti4+

ions occupy? Why?

(c) What fraction of the total tetrahedral sites will be occupied?

(d) What fraction of the total octahedral sites will be occupied?

Solution

(a) We are first of all asked to cite, for FeTiO3, which type of interstitial site the Fe2+ ions will occupy.

From Table 3.4, the cation-anion radius ratio is

rFe2

rO2

= 0.077 nm

0.140 nm= 0.550

Since this ratio is between 0.414 and 0.732, the Fe2+ ions will occupy octahedral sites (Table 3.3).

(b) Similarly, for the Ti4+ ions

rTi4

rO2

= 0.061 nm

0.140 nm= 0.436

Since this ratio is between 0.414 and 0.732, the Ti4+ ions will also occupy octahedral sites.

(c) Since both Fe2+ and Ti4+ ions occupy octahedral sites, no tetrahedral sites will be occupied.

(d) For every FeTiO3 formula unit, there are three O2- ions, and, therefore, three octahedral sites; since

there is one ion each of Fe2+ and Ti4+, two-thirds of these octahedral sites will be occupied.


X-ray Diffraction: Determination of Crystal Structures

3.47 Determine the expected diffraction angle for the first-order reflection from the (310) set of planes for

BCC chromium when monochromatic radiation of wavelength 0.0711 nm is used.

Solution

We first calculate the lattice parameter using Equation 3.3 and the value of R (0.1249 nm) cited in Table

3.1, as follows:

a =4 R

3=

(4) (0.1249 nm)

3= 0.2884 nm

Next, the interplanar spacing for the (310) set of planes may be determined using Equation 3.15 according to

d310 = a

(3)2 + (1)2 + (0)2=

0.2884 nm

10= 0.0912 nm

And finally, employment of Equation 3.14 yields the diffraction angle as

sin =n

2d310

=(1)(0.0711 nm)

(2)(0.0912 nm)= 0.390

Which leads to

= sin-1(0.390) = 22.94

And, finally

2 = (2)(22.94) = 45.88


3.49 The metal niobium has a BCC crystal structure. If the angle of diffraction for the (211) set of planes

occurs at 75.99° (first-order reflection) when monochromatic x-radiation having a wavelength of 0.1659 nm is used,

compute (a) the interplanar spacing for this set of planes, and (b) the atomic radius for the niobium atom.

Solution

(a) From the data given in the problem, and realizing that 75.99 = 2, the interplanar spacing for the (211)

set of planes for Nb may be computed using Equation 3.14 as follows:

d211 =n

2 sin =

(1)(0.1659 nm)

(2) sin75.99

2

= 0.1347 nm

(b) In order to compute the atomic radius we must first determine the lattice parameter, a, using Equation

3.15, and then R from Equation 3.3 since Nb has a BCC crystal structure. Therefore,

a= d211 (2)2 + (1)2 + (1)2 (0.1347 nm)( 6) 0.3300 nm

And, from Equation 3.3

R a 3

4=

(0.3300 nm) 3

4 0.1429 nm

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