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Resolução Callister Capítulo 3 Engenharia dos Materiais

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 3-24 
 3.21 A unit cell for the face-centered orthorhombic crystal structure is presented below. 
 
 
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 3-25 
 Point Coordinates 
 
 3.22 This problem asks that we list the point coordinates for all of the atoms that are associated with the 
FCC unit cell. From Figure 3.1b, the atom located of the origin of the unit cell has the coordinates 000. 
Coordinates for other atoms in the bottom face are 100, 110, 010, and 
 
1
2
1
2
0. (The z coordinate for all these points 
is zero.) 
 For the top unit cell face, the coordinates are 001, 101, 111, 011, and 
 
1
2
1
2
1. (These coordinates are the 
same as bottom-face coordinates except that the “0” z coordinate has been replaced by a “1”.) 
 Coordinates for those atoms that are positioned at the centers of both side faces, and centers of both front 
and back faces need to be specified. For the front and back-center face atoms, the coordinates are 
 
1 1
2
1
2
 and 
 
0 1
2
1
2
, 
respectively. While for the left and right side center-face atoms, the respective coordinates are 
 
1
2
0 1
2
 and 
 
1
2
1 1
2
. 
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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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 3-26 
 3.23 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 12.2. 
 In Figure 12.2, 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
2
0, 
 
1
2
1
2
1, 
 
1 1
2
1
2
, 
 
0 1
2
1
2
, 
 
1
2
0 1
2
, and 
 
1
2
1 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
2
00, 
 
1 1
2
0, 
 
1
2
10, 
 
0 1
2
0. 
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 
 
00 1
2
, 
 
1 0 1
2
, 
 
1
2
1
2
1
2
, 
 
1 1 1
2
, and 
 
01 1
2
. And for the four ions on the top face 
 
1
2
01, 
 
1 1
2
1, 
 
1
2
1 1, and 
 
0 1
2
1. 
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 3-27 
 3.24 This problem calls for us to list the point coordinates of both the zinc and sulfur atoms for a unit cell 
of the zinc blende structure, which is shown in Figure 12.4. 
 First of all, the sulfur atoms occupy the face-centered positions in the unit cell, which from the solution to 
Problem 3.22, are as follows: 000, 100, 110, 010, 001, 101, 111, 011, 
 
1
2
1
2
0, 
 
1
2
1
2
1, 
 
1 1
2
1
2
, 
 
0 1
2
1
2
, 
 
1
2
0 1
2
, and 
 
1
2
1 1
2
. 
 Now, using an x-y-z coordinate system oriented as in Figure 3.4, the coordinates of the zinc atom that lies 
toward the lower-left-front of the unit cell has the coordinates 
 
3
4
1
4
1
4
, whereas the atom situated toward the lower-
right-back of the unit cell has coordinates of 
 
1
4
3
4
1
4
. Also, the zinc atom that resides toward the upper-left-back of 
the unit cell has the 
 
1
4
1
4
3
4
 coordinates. And, the coordinates of the final zinc atom, located toward the upper-right-
front of the unit cell, are 
 
3
4
3
4
3
4
. 
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 3-28 
 3.25 A tetragonal unit in which are shown the 
 
1 1 1
2
 and 
 
1
2
1
4
1
2
 point coordinates is presented below. 
 
 
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 3-29 
 3.26 First of all, open the “Molecular Definition Utility”; it may be found in either of “Metallic Crystal 
Structures and Crystallography” or “Ceramic Crystal Structures” modules. 
 In the “Step 1” window, it is necessary to define the atom type, a color for the spheres (atoms), and specify 
an atom size. Let us enter “Sn” as the name of the atom type (since “Sn” the symbol for tin). Next it is necessary to 
choose a color from the selections that appear in the pull-down menu—for example, “LtBlue” (light blue). In the 
“Atom Size” window, it is necessary to enter an atom size. In the instructions for this step, it is suggested that the 
atom diameter in nanometers be used. From the table found inside the front cover of the textbook, the atomic radius 
for tin is 0.151 nm, and, therefore, the atomic diameter is twice this value (i.e., 0.302 nm); therefore, we enter the 
value “0.302”. Now click on the “Register” button, followed by clicking on the “Go to Step 2” button. 
 In the “Step 2” window we specify positions for all of the atoms within the unit cell; their point 
coordinates are specified in the problem statement. Now we must enter a name in the box provided for each of the 
atoms in the unit cell. For example, let us name the first atom “Sn1”. Its point coordinates are 000, and, therefore, 
we enter a “0” (zero) in each of the “x”, “y”, and “z” atom position boxes. Next, in the “Atom Type” pull-down 
menu we select “Sn”, our only choice, and the name we specified in Step 1. For the next atom, which has point 
coordinates of 100, let us name it “Sn2”; since it is located a distance of a units along the x-axis the value of 
“0.583” is entered in the “x” atom position box (since this is the value of a given in the problem statement); zeros 
are entered in each of the “y” and “z” position boxes. We next click on the “Register” button. This same procedure 
is repeated for all 13 of the point coordinates specified in the problem statement. For the atom having point 
coordinates of “111” respective values of “0.583”, “0.583”, and “0.318” are entered in the x, y, and z atom position 
boxes, since the unit cell edge length along the y and z axes are a (0.583) and c (0.318 nm), respectively. For 
fractional point coordinates, the appropriate a or c value is multiplied by the fraction. For example, the