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```of i is

i = − (h + k) = −[0 + (−1)] = 1

Hence, this is a (01 10) 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 3.7, the value of i is

i = − (h + k) = − (−1 − 1) = 2

Hence, this is a (1 1 22) plane.

(c) 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 3.7, the value of i is

i = − (h + k) = − (2 − 1) = −1
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3-59

Hence, this is a (21 1 0) plane.

(d) 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 3.7, the value of i is

i = − (h + k) = − (−1 + 1) = 0

Therefore, this is a (1 102) plane.
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
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3-60
3.50 This problem asks that we draw (011 1) and (21 1 0) planes within hexagonal unit cells.
For (011 1) the reciprocals of h, k, i, and l are, respectively, ∞, 1, –1, and 1; thus, this plane is parallel to
the a1 axis, and intersects the a2 axis at a, the a3 axis at –a, and the z-axis at c. The plane having these intersections
is shown in the figure below

For (21 1 0) 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.

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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
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3-61
Linear and Planar Densities

3.51 (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 = x
2 + y2
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
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3-62

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 + (2 R 2)2 = 24 R2 = 2 R 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 nm−1 = 2.76 × 109 m−1

While for the [111] direction

LD111(Cu) =
1
2 R 6
=
1
(2)(0.128 nm) 6
= 1.59 nm−1 = 1.59 × 109 m−1
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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
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3-63
3.52 (a) In the figure below is shown a [110] direction within a BCC unit cell.

For this [110] direction there is one atom at each of the two unit cell corners, 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 x in this figure, which
is equal to

x = z
2 − y2

where y is the unit cell edge length, which, from Equation 3.3 is equal to

4 R
3
. Furthermore, z is the length of the
unit cell diagonal, which is equal to 4R Thus, using the above equation, the length x may be calculated as follows:

x = (4R)2 − 4 R
3
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
2
=
32 R2
3
= 4 R 2
3

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

LD110 =
number of atoms centered on [110] direction vector
length of [110] direction vector

=
1 atom
4 R 2
3
=
3
4 R 2

A BCC unit cell within which is drawn a [111] direction is shown below.

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students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted
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3-64

For although the [111] direction vector shown passes through the centers of three atoms, there is an equivalence of
only two atoms associated with this unit cell—one-half of each of the two atoms at the end of the vector, in addition
to the center atom belongs entirely to the unit cell. Furthermore, the length of the vector shown is equal to 4R, since
all of the atoms whose centers the vector passes through touch one another. Therefore, the linear density is equal to

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

=
2 atoms
4R
=
1
2R```