Foundations of Materials Science and Engineering 6th Edition By William Smith – Test Bank
To Purchase this Complete Test Bank with Answers Click the link Below
If face any problem or
Further information contact us At tbzuiqe@gmail.com
Sample Test
Copyright © McGraw-Hill Education. All rights reserved. No
reproduction or distribution without
the prior written
consent of McGraw-Hill Education.
53
Chapter 3, Problem 1
Define the following
terms: (a) crystalline solid, (b) long-range order, (c) short-range order, and
(d) amorphous.
Chapter 3, Solution 1
(a) a solid that has
an organized, repeating, 3D positioning of atoms or ions in its structure;
(b) the positioning of
atoms is periodic and repeating anywhere in the crystal. All atoms have the
same
geometric environment;
(c) there is some order (organized positioning of atoms), but it is local or
limited to
certain regions of the
material; (d) there is no organized positioning of atoms. All atom positions in
the
structure are random.
Chapter 3, Problem 2
Define the following
terms: (a) crystal structure. (b) space lattice, (c) lattice point, (d) unit
cell, (e) motif, and
(f) lattice constants
Chapter 3, Solution 2
(a) a regular, 3D
pattern of atoms or ions in space; (b) a three-dimensional array of points each
of
which has the same
geometric environment; (c) one point in the array in a space lattice;
(d) the smallest,
repeating unit of a space lattice; (e) a group of atoms that are organized
relative to
each other and are
associated with a lattice point; (f) length dimensions or angles that
characterize
the geometry of a unit
cell.
Chapter 3, Problem 3
What are the 14
Bravais unit cells?
Chapter 3, Solution 3
The fourteen Bravais
lattices are: simple cubic, body-centered cubic, face-centered cubic, simple
tetragonal,
body-centered tetragonal, simple orthorhombic, base-centered orthorhombic,
body-centered
orthorhombic,
face-centered orthorhombic, simple rhombohedral, simple hexagonal, simple
monoclinic,
base-centered
monoclinic, and simple triclinic.
Chapter 3, Problem 4
What are the three
most common metal crystal structures? List five metals that have each of these
crystal
structures.
Chapter 3, Solution 4
The three most common
crystal structures found in metals are: body-centered cubic (BCC),
face-centered
cubic (FCC), and
hexagonal close-packed (HCP). Examples of metals having these structures
include the
following.
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
54
BCC: a-iron, vanadium,
tungsten, niobium, and chromium.
FCC: copper, aluminum,
lead, nickel, and silver.
HCP: magnesium,
a-titanium, zinc, beryllium, and cadmium.
Chapter 3, Problem 5
For a BBC unit cell,
(a) how many atoms are there inside the unit cell, (b) what is the coordination
number
for the atoms, (c)
what is the relationship between the length of the side a of the BCC unit cell
and the
radius of its atoms,
and (d) what is the atomic packing factor? APF = 0.68 or 68%
Chapter 3, Solution 5
(a) A BCC crystal
structure has two atoms in each unit cell. (b) A BCC crystal structure has a
coordination
number of eight. (c)
In a BCC unit cell, one complete atom and two atom eighths touch each other
along
the cube diagonal.
This geometry translates into the relationship 3a = 4R. (d) the atomic packing
factor is:
volume of atoms in FCC
unit cell
Atomic packing factor
volume of the FCC unit
cell
=
3
Vunit cell =a
The volume of atoms
associated with this unit cell is the number of atoms in the unit cell
multiplied by the
volume of each atom
3
4 3 4 3
2 2
3 3 4 atoms
a
V pR p
é ù é æ ö ù ê ú ê çç
÷÷ ú = ê ú = ê ç ÷÷ ú ë û ê çè ÷ø ú êë úû
Therefore, the APF for
BCC is:
3
3
4 3
2
3 4
0.68
a
APF
a
p
é æ ö ù ê çç ÷÷ ú ê ç
÷÷ ú ê çè ÷ø ú = êë úû =
Chapter 3, Problem 6
For an FCC unit cell,
(a) how many atoms are there inside the unit cell, (b) What is the coordination
number
for the atoms, (c)
what is the relationship between the length of side a of the FCC unit cell and
the radius of
its atom, and (d) what
is the atomic packing factor?
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
55
Chapter 3, Solution 6
(a) Each unit cell of
the FCC crystal structure contains four atoms. (b) The FCC crystal structure
has a
coordination number of
twelve. (c) the geometry can be used to find the relationship of
4
2
R
a =
(d) By definition, the
atomic packing factor is given as:
volume of atoms in FCC
unit cell
Atomic packing factor
volume of the FCC unit
cell
=
These volumes,
associated with the four-atom FCC unit cell, are
4 3 16 3
4
3 3 Vatoms pR pR
é ù
= ê ú = êë úû
and 3
Vunit cell =a
where a represents the
lattice constant. Substituting
4
2
R
a = ,
3
3
unit cell
64
2 2
R
V =a =
The atomic packing
factor then becomes,
3
3
16 1 2 2
APF (FCC unit cell)
3 32 6
R
R
=æçç p ÷÷öæçç ÷÷ö= p
çç ÷÷÷çç ÷÷÷ è øè ø
= 0.74
Chapter 3, Problem 7
For an HCP unit cell
(consider the primitive cell), (a) how many atoms are there inside the unit
cell, (b) What
is the coordination
number for the atoms, (c) what is the atomic packing factor, (d) what is the
ideal c/a ratio
for HCP metals, and
(e) repeat (a) through (c) considering the “larger” cell.
Chapter 3, Solution 7
The primitive cell has
(a) two atoms/unit cell; (b) The coordination number associated with the HCP
crystal
structure is twelve.
(c) the APF is 0.74 or 74%; (d) The ideal c/a ratio for HCP metals is 1.633;
(e) all answers
remain the same except
for (a) where the new answer is 6.
Chapter 3, Problem 8
How are atomic
positions located in cubic unit cells?
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
56
Chapter 3, Solution 8
Atomic positions are
located in cubic unit cells using rectangular x, y, and z axes and unit
distances along
the respective axes.
The directions of these axes are shown below.
Chapter 3, Problem 9
List the atom
positions for the eight corner and six face-centered atoms of the FCC unit
cell.
Chapter 3, Solution 9
The atom positions at
the corners of an FCC unit cell are:
(0, 0, 0), (1, 0, 0),
( 1, 1, 0), (0, 1, 0), (0, 0, 1), (1, 0, 1), (1, 1, 1), (0, 1, 1)
On the faces of the
FCC unit cell, atoms are located at:
(½, ½, 0), (½, 0, ½),
(0, ½, ½), (½, ½, 1), (1, ½, ½), (½, 1, ½)
Chapter 3, Problem 10
How are the indices
for a crystallographic direction in a cubic unit cell determined?
Chapter 3, Solution 10
For cubic crystals,
the crystallographic direction indices are the components of the direction
vector,
resolved along each of
the coordinate axes and reduced to the smallest integers. These indices are
designated as [uvw].
Chapter 3, Problem 11
What are the
crystallographic directions of a family or form? What generalized notation is
used to indicate
them?
Chapter 3, Solution 11
A family or form has
equivalent crystallographic directions; the atom spacing along each direction
is
identical. These
directions are indicated by uvw .
+z
−z
−x
−y
+x
y
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
57
Chapter 3, Problem 12
How are the Miller
indices for a crystallographic plane in a cubic unit cell determined? What
generalized
notation is used to
indicate them?
Chapter 3, Solution 12
The Miller indices are
determined by first identifying the fractional intercepts which the plane makes
with
the crystallographic
x, y, and z axes of the cubic unit cell. Then, all fractions must be cleared
such that the
smallest set of whole
numbers is attained. The general notation used to indicate these indices is
(hkl),
where h, k, and l
correspond to the x, y and z axes, respectively.
Chapter 3, Problem 13
What is the notation
used to indicate a family or form of cubic crystallographic planes?
Chapter 3, Solution 13
A family or form of a
cubic crystallographic plane is indicated using the notation {hkl}.
Chapter 3, Problem 14
How are
crystallographic planes indicated in HCP unit cells?
Chapter 3, Solution 14
In HCP unit cells,
crystallographic planes are indicated using four indices which correspond to
four axes:
three basal axes of
the unit cell, a1, a2, and a3 , which are separated by 120º; and the vertical c
axis.
Chapter 3, Problem 15
What notation is used
to describe HCP crystal planes?
Chapter 3, Solution 15
HCP crystal planes are
described using the Miller-Bravais indices, (hkl).
Chapter 3, Problem 16
What is the difference
in the stacking arrangement of close-packed planes in (a) the HCP crystal
structure
and (b) the FCC
crystal structure?
Chapter 3, Solution 16
Although the FCC and
HCP are both close-packed lattices with APF = 0.74, the structures differ in
the
three dimensional
stacking of their planes:
(a) the stacking order
of HCP planes is ABAB… ;
(b) the FCC planes
have an ABCABC… stacking sequence.
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
58
Chapter 3, Problem 17
What are the
closest-packed directions in (a) the BCC structure, (b) the FCC structure and (c)
the HCP
structure?
Chapter 3, Solution 17
(a) The closest packed
directions in BCC lattice are < 1 1 1 > directions.
(b) The closest-packed
directions in the FCC lattice are the 1 10 directions.
(c) The closest-packed
directions in the HCP lattice are the 1120 directions.
Chapter 3, Problem 18
Identify the
close-packed planes in (a) the BCC structure, (b) the FCC structure, and (c)
the HCP structure.
Chapter 3, Solution 18
(a) Does not have a
close-packed plane; (b) {111} planes; (c) {0001} planes
Chapter 3, Problem 19
What is polymorphism
with respect to metals?
Chapter 3, Solution 19
A metal is considered
polymorphic if it can exist in more than one crystalline form under different
conditions of
temperature and pressure.
Chapter 3, Problem 20
What are X rays, and
how are they produced?
Chapter 3, Solution 20
X-rays are
electromagnetic radiation having wavelengths in the range of approximately 0.05
nm to
0.25 nm. These waves
are produced when accelerated electrons strike a target metal.
Chapter 3, Problem 21
Draw a schematic
diagram of an x-ray tube used for x-ray diffraction, and indicate on it the
path of the
electrons and X rays.
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
59
Chapter 3, Solution 21
See Figure 3.25 of
textbook.
Figure 3.25
Chapter 3, Problem 22
What is the
characteristic x-ray radiation? What is its origin?
Chapter 3, Solution 22
Characteristic
radiation is an intense form of x-ray radiation which occurs at specific
wavelengths for a
particular element.
The Ka radiation, the most intense characteristic radiation emitted, is caused
by excited
electrons dropping
from the second atomic shell
(n = 2) to the first
shell (n = 1). The next most intense radiation, Kb, is caused by excited
electrons dropping
from the third atomic
shell (n = 3) to the first shell (n = 1).
Chapter 3, Problem 23
Distinguish between
destructive interference and constructive interference of reflected x-ray beams
through crystals.
Chapter 3, Solution 23
Destructive
interference occurs when the wave patterns of x-ray beams, reflected from a
crystal, are out of
phase. Conversely,
when the wave patterns leaving a crystal plane are in phase, constructive
interference
occurs and the beam is
reinforced.
Copper X rays Vacuum
Electrons
Cooling water
Target
Beryllium window X
rays Metal focusing cup
To transformer
Tungsten
filament lass
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written
consent of McGraw-Hill Education.
60
Application and
Analysis Problems
Chapter 3, Problem 24
Tungsten at 20°C is
BCC and has an atomic radius of 0.137 nm. Calculate a value for its lattice
constant
a in nanometers. (b)
Calculate the volume of the unit cell.
Chapter 3, Solution 24
Letting a represent
the edge length of the BCC unit cell and R the tungsten atomic radius,
4 4 ( )
3 4 or 0.137nm
3 3
a = R a = R = = 0.316
nm
Since BCC has a cubic
structure, the length of each side is a. Therefore its volume should be
V = a3 = 0.316nm3 =
0.0316 nm3
Chapter 3, Problem 25
Lead is FCC and has an
atomic radius of 0.175 nm. (a) Calculate a value for its lattice constant
a in nanometers. (b)
Calculate the volume of the unit cell in nm3 .
Chapter 3, Solution 25
Letting a represent
the FCC unit cell edge length and R the palladium atomic radius,
4 4 ( )
2 2
2a = 4R or a = R =
0.175 nm = 0.495 nm
Since FCC has a cubic
structure, the length of each side is a. Therefore its volume should be
V = a3 = 0.495nm3 =
0.121 nm3
Chapter 3, Problem 26
Verify that the atomic
packing factor for the FCC structure is 0.74.
Chapter 3, Solution 26
By definition, the
atomic packing factor is given as:
volume of atoms in FCC
unit cell
Atomic packing factor
volume of the FCC unit
cell
=
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
61
These volumes,
associated with the four-atom FCC unit cell, are
4 3 16 3
4
3 3 Vatoms pR pR
é ù
= ê ú = êë úû
and 3
Vunit cell = a
where a represents the
lattice constant. Substituting
4
2
R
a = ,
3
3
unit cell
64
2 2
R
V = a =
The atomic packing
factor then becomes,
3
3
16 1 2 2
APF (FCC unit cell)
3 32 6
R
R
=æçç p ÷÷öæçç ÷÷ö= p =
çç ÷÷÷çç ÷÷÷ è øè ø
0.74
Chapter 3, Problem 27
Calculate the volume
in cubic nanometers of the cobalt crystal structure unit cell (use the larger
cell).
Cobalt is HCP at 20°C
with a = 0.2507 nm and c = 0.4069 nm.
Chapter 3, Solution 27
For a hexagonal prism,
of height c and side length a, the volume is given by:
( )( ) ( )( )
( )( )
( )( )( )
o
o
2
2
3
Area of Base Height 6
Equilateral Triangle Area Height
3 sin60
3 0.2507 sin 60 0.4069
0.0664
V
a c
nm nm
nm
= = éê ´ ùú ë û
=
=
=
Chapter 3, Problem 28
Consider a 0.05-mm
thick, 500 mm2 (about three times the area of a dime) piece of aluminum foil.
How
many unit cells exist
in the foil? If the density of aluminum is 2.7 g/cm3, what is the mass of each
cell?
Chapter 3, Solution 28
A = 500mm2 t = 0.05mm
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
62
Thickness = 0.05 mm
Area = 500 mm2
a) The lattice
constant for aluminum is a = 0.405nm
a = 0.405 nm = 0.405´10-6
mm
volume of the sheet =
(A)(t) = (500)(0.05) = 25 mm3
volume of a single
cell = a3 = ( 0.405´ 10-6 )3
Þ a3
=6.64´ 10-20 mm3
number of unit cell in
the sheet
3
20 3
25 mm
6.64 10- mm =
´
number of cells =
3.76´ 1020 ( ~ 2000 times smaller than Avogadro’s number)
b)
m
V
r =
m = rV =
20 3 3 3
3 3
2.7 gr 6.64 10 mm 10
cm
cm mm
æçç ÷÷öççæ ´ – ´ – ÷÷ö
çè ÷øçè ÷ø
m = 1.8 ´10-22 gr
Chapter 3, Problem 29
Draw the following directions in a BCC unit cell and list the position
coordinates of
the atoms whose centers
are intersected by the direction vector: Determine the repeat distance in terms
of
the lattice constant
in each direction.
(a) [010] (b) [011]
(c) [111] (d) Find the angle between directions in (b) and (c)
Chapter 3, Solution 29
(a) Position Coordinates:
(b) Position Coordinates: (c) Position Coordinates:
(0, 0, 0), (0, 1, 0)
(0, 0, 0), (0, 1, 1) (0, 0, 0), (1, 1, 1),
The repeat distance is
defined as the distance between two consecutive lattice points on
a specified direction
z
y
(0, 0, 0) (0, 0, 0)
(0, 0, 0)
(1, 1, 1)
(0,1,0)
(0,1,1)
x
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
63
(a) repeat distance:
The distance between
the centers of the atoms corner atoms is a, so repeat distance is a
(b) repeat distance:
The distance between
the centers of the corner atoms is 2a, so the repeat distance is 2a
(c) repeat distance:
The distance between
the center of the corner atom, and the center of the BCC center atom is
( 3 /2)a, so the
repeat distance is ( 3/2)a
(d) the angle between
two directions can be found using:
1 2 1 2 1 2
2 2 2 2 2 2
1 1 1 2 2 2
h h k k l l
cos
h k l h k l
q
+ +
=
+ + + +
q = 35.26 Degrees
Chapter 3, Problem 30
Draw direction vectors
in an FCC unit cell for the following cubic directions, and list the position
coordinated of the
atoms whose centers are intersected by the direction vector. Determine the
repeat
distance in terms of
the lattice constant in each direction
(a)é1 1 1ù (b)é10 1 ù
(c)é21 1ù (d)é1 3 1ù (e)Find the angle between directions in(b)and(d).
êë úû êë úû êë úû êë
úû
Chapter 3, Solution 30
z
x
y
a) (1 1 1)
c) (211)
b)
1/3
1/3
1/3
d) (131)
(1 0 1)
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
64
Position coordinates
a) (0,0,0), (-1, -1,
1) b) (0,0,0) (1 ,0, -1) (1/2, 0, -1/2) c) (0,0,0,) (1, -1/2, -1/2) (2, -1, -1)
d) (0,0,0) (-1, 3, -1)
Repeat distances:
a) The distance
between two lattice points (in this case, the corner atoms) is ( 3)a
b) The distance
between two lattice points (in this case, the corner atom the face-centered
atom)
is ( 2/2)a
c) The distance
between two lattice points (in this case, the corner atom and the face-centered
atom
is ( 3 / 2)a
d) The distance
between two lattice points (in this case, the corner atom and the corner atom
two lattice
cells over) is ( 11)a
e) The angle between
(b) and (d) can be calculated using the following equation:
1 2 1 2 1 2
2 2 2 2 2 2
1 1 1 2 2 2
h h k k l l
cos
h k l h k l
q
+ +
=
+ + + +
q =64.76 Degrees
Chapter 3, Problem 31
Draw direction vectors
in unit cells for the following cubic directions:
(a) éêë1 12ùúû (c)
éêë331ùúû (e) éêë212ùúû (g) éêë 101ùúû (i) éë321ùû (k) éêë122ùúû
(b) éêë123ùúû (d)
éêë021ùúû (f) éêë233ùúû (h) éêë121 ùúû (j) éêë103ùúû (l) éêë223ùúû
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
65
Chapter 3, Solution 31
z
y
½ x
⅓
⅔
⅔
⅓
½
½
½
1
1
1
1
1
(a) Dividing
1
1
1
[ 1 1 2 ] by 2, (b)
Dividing [ 123] by 3, (c) Dividing [331] by 3,
x = , 1
2
x = , 2
3
y = , 1
2
,
1
2
z = 1 z = 1
3
x = , x = 1, y =1, 1
3
y = , 2
3
z = 1
(d) Dividing [021] by
2, (e) Dividing [212] by 2, (f) Dividing [233] by 3,
z = 1
2
x =0 , y = 1, x =1, y
= z= 1 y = 1, z= 1
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
66
Figure P3.32
Chapter 3, Problem 32
What are the indices
of the directions shown in the unit cubes of Fig. P3.32?
1 1
1
1
1 1
1
½
½
½
⅓
⅓
⅔
⅔
⅔
(g) For [101], (h)
Dividing [ 1 2 1 ] by 2, (i) Dividing [321] by 3,
x = 1, y = 0, z =1 x
=1,
z = 1 y= 1 z = 1
x = , , 1
2
y = , 2
3
z =1
3
z = 1
2
x= 2
3
y=1
x = , , 1
3
x , , , y= 2
3
= , 1
2
y=0 z=1
( j) Dividing [103] by
3, (k) Dividing [ 122] by 2, (l) Dividing [223] by 3,
z
a
e
h
f
g
c
b
1
3
y
x
(a)
z
y
x
(b)
d
3
4
1
2
1
2
1
4
1
4
1
4
1
4
3
4
3
4
2
3
1
3
1
2
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
67
Chapter 3, Solution 32
a
b
c
½
a. Vector components:
Direction indices: [ 1
1 0 ]
x = −1, y = 1, z = 0
b. Moving direction
vector c. Moving direction vector fordown
¼, vector components
ward ½, vector components
are: x = 1, y = −1, z
= are: x = − , y = 1, z = 1
½
⅓
1 6
1 6
New 0
New 0
¼
¼
¼Direction indices:
[441] Direction indices: [166]
d e
f
½ ¼
¼
⅔
⅓
⅓
¼
d. Moving direction
vector e. Vector components are: f. Moving direction vector up
left ¼, vector
components ⅓, vector components are:
are: x = 1, y = ½, z =
1
x = −¾, y = −1, z = 1
x = −1, y = 1, z =−⅓
New 0
New 0
Direction indices:
[212]
−
Direction indices:
[344] Direction indices: [331]
x
z
y
z
y
a
¼
½ ¼
¾
½
⅓
b
c
d
(a) (b)
¼
¾
½
⅓
¼
¾
⅔
f
g
h
e
x
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
68
Chapter 3, Problem 33
A direction vector
passes through a unit cube from the
3 1
,0,
4 4
æç ÷öçç ÷÷÷ è ø
to the
1
, 1,0
2
æç ÷öçç ÷÷÷ è ø
positions. What are
its
direction indices?
Chapter 3, Solution 33
The starting point coordinates,
subtracted from the end point, give the vector components:
1 3 1 1 1
1 0 1 0
2 4 4 4 4
x = – =- y = – = z = –
=-
The fractions can then
be cleared through multiplication by 4, giving x =-1, y = 4, z =-1 . The
direction
indices are therefore
éêë 1 41ùúû .
Chapter 3, Problem 34
A direction vector
passes through a unit cube from the
3
1,0,
4
æç ÷öçç ÷÷÷ è ø
to the
1 1
, 1,
4 4
æç ÷öçç ÷÷÷ è ø
positions. What are
its
direction indices?
Chapter 3, Solution 34
Subtracting
coordinates, the vector components are:
1 3 1 3 1
1 1 0 1
4 4 4 4 2
x = – =- y = – = z = –
=-
¾
g.
h. Moving direction
vector
up ¼, vector
components
are: x = ¾, y = −1, z
= −¾
Direction indices:
¾
−¾
¾
¼
½
New 0 New 0
g h
[441]
[343]
Moving direction
vector
up ½, vector components
are: x = 1, y = −1, z
= −¼
Direction indices:
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
69
Clearing fractions
through multiplication by 4, gives x =-3, y = 4, z =-2.
The direction indices
are therefore éêë3 4 2ùúû .
Chapter 3, Problem 35
What are the
directions of the á103ñ family or form for a unit cube? Draw all the directions
in a unit cell.
Chapter 3, Solution 35
[ 103] , [301] , [310]
, [130] , [031] , [013] , [103] , [130] , [310] , [301] , [031] , [013] , [
103] , [3 10] , [ 130] ,
[301 ] , [0 13] , [031
] , [ 103] , [3 10] , [ 1 30] , [301 ] , [031 ] , [0 1 3]
Chapter 3, Problem 36
What are the
directions of the 111 family or form for a unit cube? Draw all the directions
in a BCC unit cell.
Can you identify a
special quality of these directions?
Chapter 3, Solution 36
[111] , [1 1 1] , [1 1
1] , [ 1 1 1] ,
[ 1 11] , [1 1 1] ,
[11 1] , [ 1 1 1]
These directions
intersect in the middle of the cube.
[111]
[111]
[111]
[111]
[111]
[111]
[111]
[111]
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
70
Chapter 3, Problem 37
What 110 -type
directions lie on the (111) plane of a cubic unit cell? Draw those directions
in an FCC unit
cell. Can you identify
a special quality of these directions?
Chapter 3, Solution 37
[0 1 1] , [0 1 1] , [
1 10] , [1 1 0] , [ 101] , [10 1]
These directions
connect at the corners.
Chapter 3, Problem 38
What 111 -type
directions lie on the (110) plane of a BCC unit cell? Draw those directions in
a unit cell.
Can you identify a
special quality of these directions?
Chapter 3, Solution 38
[1 1 1] , [ 1 11] , [1
1 1] , [ 1 1 1]
These directions
intersect at the center of the cell
Chapter 3, Problem 39
Draw in unit cubes the
crystal planes that have the following Miller indices:
(a) (1 1 1 ) (c) (121
) (e) (321) (g) (201 ) (i) (232) (k) (312)
(b) (102) (d) (213)
(f) (302) (h) (212) (j) (133) (l) (331 )
[ 1 1 1]
[1 1 1]
[1 1 1]
[ 1 0 1]
[0 1 1 ]
[ 1 10]
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
71
Chapter 3, Solution 39
Chapter 3, Problem 40
What are the Miller
indices of the cubic crystallographic planes shown in Fig. P3.40
Figure P3.40
z
2
3
a
b
d
a
c d
b
c
(a) (b)
y
x
z
y
x
1
3
1
3
3
4
2
3
2
3
1
2
1
3
1
3
1
4
3
4
a. For ( 1 1 1)
reciprocals
are: x = 1, y = -1, z
= -1
b. For ( 102)
reciprocals
are: x = 1, y = ¥, z =
½
c. For ( 1 2 1)
reciprocals
are: x = 1, y = -½ , z
= -1
-1
-1
+1
(0, 0, 0)
(0, 0, 0)
+1
-⅓
+½
d. For (213)
reciprocals are:
x = ½ , y = 1, z = -⅓
e. For (321)
reciprocals
are: x = ⅓ , y = -½ ,
z = 1
(0, 0, 0)
+1
-½
+⅓
f. For (302)
reciprocals are:
x = ⅓ , y = ∞, z = -½
•
(0, 0, 0)
+⅓
-½
• •
•
+1
-½
(0, 0, 0) •
-1
-½ (0, 0, 0)
+1
•
z
y
x
(1 1 1) (102) (1 2 1)
(213)
(321)
(302)
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
72
Chapter 3, Solution 40
Miller Indices for
Figure P3.40(a)
Plane a based on (0,
1, 1) as origin Plane b based on (1, 1, 0) as origin
Planar Intercepts
Reciprocals of Intercepts Planar Intercepts Reciprocals of Intercepts
x = ∞
1
0
x
= x = -1
1
1
x
=-
y = -1
1
1
y
=- 5
12
y
= –
1 12
y 5
= –
1
4
z =-
1
4
z
=- z = ∞
1
0
z
=
The Miller indices of
plane a are (0 1 4). The Miller indices of plane b are (5 120) .
b
x b
z
y
a
⅔
½
⅓
¾
⅓
⅓
c
d
(a) ¾ z
y
(b) ¼
½
⅓
⅔
a
d
c
x ⅔
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
73
Plane c based on (1,
1, 0) as origin
Plane d based on (0,
0, 0) as origin
Planar Intercepts
Reciprocals of Intercepts Planar Intercepts Reciprocals of Intercepts
x = ∞
1
0
x
= x = 1
1
1
x
=
y = -1
1
1
y
=- y = 1
1
1
y
=
1
3
z =
1
3
z
=
2
3
z =
1 3
z 2
=
The Miller indices of
plane c are (0 1 3). The Miller indices of plane d are (2 2 3).
Miller Indices for
Figure P3.40(b)
Plane a based on (1,
0, 1) as origin Plane b based on (0, 1, 1) as origin
Planar Intercepts
Reciprocals of Intercepts Planar Intercepts Reciprocals of Intercepts
x = -1
1
1
x
=- x = 1
1
1
x
=
y = ∞
1
0
y
= y = -1
1
1
y
=-
1
3
z =-
1
3
z
=-
2
3
z =-
1 3
z 2
=-
The Miller indices of
plane a are (1 03) . The Miller indices of plane b are (223).
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
74
Plane c based on (0,
1, 0) as origin Plane d based on (0, 1, 0) as origin
Planar Intercepts
Reciprocals of Intercepts Planar Intercepts Reciprocals of Intercepts
x = 1
1
1
x
= x = 1
1
1
x
=
5
12
y
= –
1 12
y 5
= – y = -1
1
1
y
=-
z = ∞
1
0
z
=
1
2
z =
1
2
z
=
The Miller indices of
plane c are (5 120) . The Miller indices of plane d are (1 1 2).
Chapter 3, Problem 41
What are the {100}
family of planes of the cubic system? Draw those planes in a BCC unit cell and
show all
atoms whose centers
are intersected by the planes. What is your conclusion?
Chapter 3, Solution 41
(1 0 0) , (01 0) ,
(001) , (1 00), (01 0), (001 )
(100) (101)
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution without
the prior written
consent of McGraw-Hill Education.
75
These planes make up
the sides of the unit cell, so none of those planes intersect the
body-centered atom
Chapter 3, Problem 42
Draw the following
crystallographic planes in a BCC unit cell and list the position of the atoms
whose
centers are
intersected by each of the planes:
(a) (010) (b) (011)
(c) (111)
(010) (0 10)
(001) (100)
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written consent
of McGraw-Hill Education.
76
Chapter 3, Solution 42
Chapter 3, Problem 43
Draw the following
crystallographic planes in an FCC unit cell and list the position coordinates
of the atoms
whose centers are
intersected by each of the planes:
(a) (010) (b) (011)
(c) (111)
(010) plane.
Positions of atoms
intersected by plane:
(1,1,0), (1,1,1),
(1,1,0),
(1,1,1)
(011) plane.
Positions of atoms
intersected by plane:
(1,1,0) (1,0,1)
(0,1,0), (0,0,1)
(0,0,1) (1/2,1/2,1/2)
(111) plane.
Positions of atoms
intersected by plane:
(1,0,0) (0,1,0)
(0,0,1)
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
77
Chapter 3, Solution 43
Chapter 3, Problem 44
A cubic plane has the
following axial intercepts:
1 2 1
, ,
3 3 2
a = b =- c = . What
are the Miller indices of this
plane?
Chapter 3, Solution 44
Given the axial
intercepts of (⅓,-⅔, ½), the reciprocal intercepts are:
1 1 3 1
3, , 2.
x y 2 z
= =- =
Multiplying by 2 to
clear the fraction, the Miller indices are (6 3 4).
Chapter 3, Problem 45
A cubic plane has the
following axial intercepts:
1 1 2
, ,
2 2 3
a =- b =- c = . What
are the Miller indices of
this plane?
(010) plane.
Positions of atoms
intersected by plane:
(1,1,0) (1,1,1)
(0,1,0), (0,1,1)
(1/2, 1, ½)
(011) plane.
Positions of atoms
intersected by plane:
(1,1,0) (1,0,1)
(1,1,0) (0,0,1)
(111) plane.
Positions of atoms
intersected by plane:
(1,0,0) (0,1,0)
(0,0,1) (1/2, ½, 0)
(½, 0, ½) (0, ½, ½)
Copyright © McGraw-Hill
Education. All rights reserved. No reproduction or distribution without
the prior written
consent of McGraw-Hill Education.
78
Chapter 3, Solution 45
Given the axial
intercepts of (-½,-½, ⅔), the reciprocal intercepts are:
1 1 1 3
2 , 2 ,
x y z 2
=- =- = .
Multiplying by 2, the
Miller indices are (4 4 3).
Chapter 3, Problem 46
Determine the Miller
indices of the cubic crystal plane that intersects the following position
coordinates:
( 1 ) ( 1 3 ) ( 1 )
2 2 4 2 1, ,1; ,0, ;
1,0, .
Chapter 3, Solution 46
After locating the
three position coordinates, connect points b and c and extend the line to point
d.
Complete the plane by
connecting point d to a and a to c. Using
(1, 0, 1) as the plane
origin, x = -1, y = ½ and z = -½. The intercept reciprocals then become
1 1 1
1, 2, 2.
x y z
=- = =- The Miller
indices are(1 2 2).
Chapter 3, Problem 47
Determine the Miller
indices of the cubic crystal plane that intersects the following position
coordinates:
( 1 ) ( ) ( 1 1 )
2 24 0,0, ; 1,0,0 ; ,
,0 .
(1, 0, 1)
b
a
• c
• •
(0, 0, 0)
(½, 0, ¾)
• •
(1, 0, ½,) •
(1, ½, 1)
d
Copyright ©
McGraw-Hill Education. All rights reserved. No reproduction or distribution
without
the prior written
consent of McGraw-Hill Education.
79
Chapter 3, Solution 47
After locating the three
position coordinates, connect points b and c and extend the line to point d.
Complete the plane by
connecting point d to a and a to b. Using
(0, 0, 0) as the plane
origin, x = 1, y = ½ and z = ½. The intercept reciprocals are thus
1 1 1
1, 2, 2
x y z
= = = .
The Miller indices are
therefore (1 2 2).
Chapter 3, Problem 48
Rodium is FCC and has
a lattice constant a of 0.38044 nm. Calculate the following interplanar
spacings:
(a) d111 (b) d200 (c)
d220
Chapter 3, Solution 48
(a) 111 2 2 2
0.38044 nm 0.38044 nm
1 1 1 3
d = = =
+ +
0.220 nm
(b) 200 2 2 2
0.38044 nm 0.38044 nm
2 0 0 4
d = = =
+ +
0.190 nm
(c) 220 2 2 2
0.38044 nm 0.38044 nm
2 2 0 8
d = = =
+ +
0.135 nm
Chapter 3, Problem 49
Tungsten is BCC and
has a lattice constant a of 0.31648 nm. Calculate the following interplanar
spacings:
(a) d110 (b) d220 (c)
d310
d
b
a
c
•
(0, ½, 0) •
(0, 0, 0)
(1, 0, 0)
Comments
Post a Comment