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Resolução Callister Capítulo 3 Engenharia dos Materiais

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(b) From the table inside the front cover, the atomic radius for iron is 0.124 nm. Therefore, the linear 
density for the [110] direction is 
 
 
LD110 (Fe) =
3
4 R 2
=
3
(4)(0.124 nm) 2
= 2.47 nm−1 = 2.47 × 109 m−1 
 
While for the [111] direction 
 
 
LD111(Fe) =
1
2 R
=
1
(2)(0.124 nm)
= 4.03 nm−1 = 4.03 × 109 m−1 
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 
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 3-65 
 3.53 (a) In the figure below is shown a (100) plane for an FCC unit cell. 
 
 
 
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, while the center atom lies entirely within the unit cell. Thus, there is the equivalence of 2 atoms 
associated with this FCC (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, 2R 2 (Equation 3.1); and, thus, the area of this square is just 
 (2R 2)2 = 8R2. Hence, the planar density for this (100) plane is just 
 
 
PD100 = 
number of atoms centered on (100) plane
area of (100) plane
 
 
 
=
2 atoms
8R2
=
1
4R2
 
 
 That portion of an FCC (111) plane contained within a unit cell is shown below. 
 
 
 
There are six atoms whose centers lie on this plane, which are labeled A through F. One-sixth of each of atoms A, 
D, and F are associated with this plane (yielding an equivalence of one-half atom), with one-half of each of atoms 
B, C, and E (or an equivalence of one and one-half atoms) for a total equivalence of two atoms. Now, the area of 
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. 
 
 3-66 
the triangle shown in the above figure is equal to one-half of the product of the base length and the height, h. If we 
consider half of the triangle, then 
 
 (2 R)
2 + h2 = (4 R)2 
 
which leads to h = 2 R 3 . Thus, the area is equal to 
 
 
Area = 4 R(h)
2
=
(4 R) (2 R 3)
2
= 4 R2 3 
 
And, thus, the planar density is 
 
 
PD111 = 
number of atoms centered on (111) plane
area of (111) plane
 
 
 
=
2 atoms
4 R2 3
=
1
2 R2 3
 
 
 (b) From the table inside the front cover, the atomic radius for aluminum is 0.143 nm. Therefore, the 
planar density for the (100) plane is 
 
 
PD100 (Al) =
1
4 R2
=
1
4 (0.143 nm)2
= 12.23 nm−2 = 1.223 × 1019 m−2 
 
While for the (111) plane 
 
 
PD111(Al) =
1
2 R2 3
=
1
2 3 (0.143 nm)2
= 14.12 nm−2 = 1.412 × 1019 m−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. 
 
 3-67 
 3.54 (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 
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. 
 
 3-68 
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 = z
2 − 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
 
 
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. 
 
 3-69 
 (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 nm−2 = 1.014 × 1019 m−2 
 
While for the (110) plane 
 
 
PD110 (Mo) =
3
8 R2 2
=
3
8 (0.136 nm)2 2
= 14.34 nm−2 = 1.434 × 1019 m−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. 
 
 3-70 
 3.55 (a) A (0001) plane for an HCP unit cell is show below. 
 
 
 
Each of the 6 perimeter atoms in this plane is shared with three other unit cells, whereas the center atom is shared 
with no other unit cells; this gives rise to three equivalent atoms belonging to this plane. 
 In terms of the atomic radius R, the area of each of the 6 equilateral triangles that have been drawn is 
 R
2 3 , or the total area of the plane shown is 6 R
2 3 . And the planar density for this (0001) plane is equal to 
 
 
PD0001 =
number of atoms centered on (0001) plane
area of (0001) plane
 
 
 
=
3 atoms
6R2 3
=
1
2R2 3
 
 
 (b) From the table inside the front cover, the atomic radius for titanium is 0.145 nm. Therefore, the planar 
density for the (0001) plane is 
 
 
PD0001(Ti)