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# Chapter 3 Homework solutions (part 2).pdf

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Materials Science & Engineering

MSE101H1

Scott Ramsay

Spring

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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 10, 1 1 1 1 1 1 ,0 1 1 , 10 1 , and 111 .
2 2 2 2 2 2 2 2 2 2 2 2
Furthermore, the sodium ions are situated at the centers of all unit cell edges, and, in addition, at the unit
1 1 1 1
cell center. For the bottom face of the unit cell, the point coordinates are as follows: 00 1 0 , 10 0 0 .
2 2 2 2
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 are00 1 ,1 0 1 , 1 1 1, 11 1 , and01 1. And for the four ions on the top face 1 01,1 1 1,
2 2 2 2 2 2 2 2 2
1 1
1 1, and 0 1 .
2 2
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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]
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.27 Within a cubic unit cell, sketch the following directions:
(a) [102] , (c) [212] ,
(b) [313] , (d) [301].
Solution
The directions asked for are indicated in the cubic unit cell shown below.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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
c
Projections a b –
3
1
Projections in terms of a, b, and c 1 1 – 3
Reduction to integers 3 3 –1
Enclosure [331]
Direction B is [403] 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 c
Projections – 3 0b – 2
2 1
Projections in terms of a, b, and c – 0 –
3 2
Reduction to integers –4 0 –3
Enclosure [403]
Direction C is a361] 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
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. x y z
Projections – a b c
2 6
1 1
Projections in terms of a, b, and c – 1
2 6
Reduction to integers –3 6 1
Enclosure [361]
Direction D is a [111] direction, which determation 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 c
Projections – –
2 2 2
Projections in terms of a, b, and c – 1 1 – 1
2 2 2
Reduction to integers –1 1 –1
Enclosure [111]
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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], [101] [101] ,
[101] [011] [011] , and [011] .
(b) Also, for the [100] direction, equivalent directions are the following: [100] , [010], and [010] .
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.31 Determine the indices for the two directions shown in the following hexagonal unit cell:
Solution
For direction A, projections on the a , a , and z axes are a, 0a, and c/2, or, in terms of a and c the
1 2
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 (2u' v' ) (2)(2) 4
3 3 3
v 1 (2v uÕ) 1 (2)(0) (2) 2
3 3 3
4 2 2
t (uv)
3 3 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 a223] direction.
For direction B, projections on the a , a , and z axes are –a, 0a, and 0c, or, in terms of a and c the
1 2
projections are –1, 0, and 0. This means that
u’ = –1
v’ = 0
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. w’ = 0
Now, from Equations 3.7, the u, v, t, and w indices become
u 1 (2u' v) 1(2)(1) 0 2
3 3 3
1 1 1
v 3 (2v' u' ) 3 (2)(0) (1 3
2 1 1
t (uv)
3 3 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[2110] direction.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 3.35 Sketch within a cubic unit cell the following planes:
(a) (012), (c) (101) ,
(b) (313) , (d) (211) .
Solution
The planes called for are plotted in the cubic unit cells shown below.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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(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 11)].
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
a b
Intercepts ∞c
2 3
1 1
Intercepts in terms of a, b, and c ∞
2 3
Reciprocals of intercepts 2 3 0
Enclosure (230)
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to
students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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(211) plane, as
summarized below.
x y z
Intercepts a b – c
2
1
Intercepts in terms of a, b, and c 2 1 – 1
Reciprocals of intercepts 2 1 – 1
Reduction not necessary
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; thi(021)aplane, as summarized below.
x y z
b
Intercepts ∞a – c
2
1
Intercepts in terms of a, b, and c ∞ – 1
2

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