HW 3 Solution - Spring 2018 Homework 3 Answers, Professor: Dr. Cogan PDF

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Spring 2018 Homework 3 Answers, Professor: Dr. Cogan...

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3.27 List the point coordinates of both the zinc (Zn) and sulfur (S) atoms for a unit cell of the zinc blende (ZnS) crystal structure (Figure 12.4).

Solution The ZnS unit cell, Figure 12.4, is shown below on which is superimposed an x-y-z coordinate axis system. Also, each Zn atom in the unit cell is labeled with an uppercase letter; S atoms are labeled with numbers. Of course, because the unit cell is cubic, the unit cell edge length along each of the x, y, and z axes is a.

Coordinates for each of these points is determined in a manner similar to that demonstrated in Example Problem 3.6. For the S atom labeled 7, we determine its point coordinates by using rearranged forms of Equations 3.9a, 3.9b, and 3.9c as follows: The lattice position referenced to the x axis is a = qa The lattice position referenced to the y axis is a/2 = ra The lattice position referenced to the x axis is a/2 = sa Solving these expressions for the values of q, r, and s leads to q=1

r

s=

Therefore, the q r s coordinates for this point are

The S atom labeled C, has lattice position indices referenced to the x, y, and z axes of a/4, a/4, and 3a/4, respectively. Therefore, The lattice position referenced to the x axis is a/4 = qa The lattice position referenced to the y axis is a/4 = ra The lattice position referenced to the x axis is 3a/4 = sa Thus, values for values of q, r, and s are as follows: q = 1/4

r = 1/4

s = 3/4

and the coordinates for the location of this atom are

.

This same procedure is carried out for determine point coordinates for both Zn and S atoms in the unit cell. Indices for all these points are listed in the following two tables.

S atom point number

q

r

s

1 2 3 4

0 1 1 0

0 0 1 1

0 0 0 0

5

0

6 7

0 1

8

1

9

0

10 11 12 13

0 1 1 0

14

0 0 1 1

1 1 1 1 1

Zn atom point letter

q

r

s

B C

3.34 Within a cubic unit cell, sketch the following directions: (a)

[101]

(e) (b) [211]

(f)

(c)

(g)

(d)

(h) [301] Solution (a) For the [101] direction, it is the case that u =1

v=0

w=1

lf we select the origin of the coordinate system as the position of the vector tail, then x 1 = 0a

y 1 = 0b

z1 = 0 c

It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows:

Thus, the vector head is located at a, 0b, and c, and the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.]

(b) For a [211] direction, it is the case that u =2

v=1

w=1

lf we select the origin of the coordinate system as the position of the vector tail, then x 1 = 0a

y 1 = 0b

z1 = 0 c

It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows:

Thus, the vector head is located at 2a, b, and c, and the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.]

(c) For the u =1

direction, it is the case that v=0 w = 2

lf we select the origin of the coordinate system as the position of the vector tail, then x 1 = 0a

y 1 = 0b

z1 = 0 c

It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows:

Thus, the vector head is located at a, 0b, and c. However, in order to reduce the vector length, we have divided these coordinates by ½, which gives the new set of head coordinates as a/2, 0b, and c; the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.]

(d) For the u=3

direction, it is the case that v = 1 w=3

lf we select the origin of the coordinate system as the position of the vector tail, then x 1 = 0a y 1 = 0b z1 = 0 c It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows:

Thus, the vector head is located at 3a, b, and 3c. However, in order to reduce the vector length, we have divided these coordinates by 1/3, which gives the new set of head coordinates as a, b/3, and c; the direction vector having these head coordinates is plotted below.

[Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.]

(e) For the u = 1

direction, it is the case that v=1 w = 1

lf we select the origin of the coordinate system as the position of the vector tail, then x 1 = 0a y 1 = 0b z 1 = 0c It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows:

Thus, the vector head is located at a, b, and c, and the direction vector having these head coordinates is plotted below.

[Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.]

(f) For the u = 2

direction, it is the case that v=1 w=2

lf we select the origin of the coordinate system as the position of the vector tail, then x 1 = 0a

y 1 = 0b

z1 = 0 c

It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows:

Thus, the vector head is located at 2a, b, and 2c. However, in order to reduce the vector length, we have divided these coordinates by 1/2, which gives the new set of head coordinates as a, b/2, and c; the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.]

(g) For the u =3

direction, it is the case that v = 1 w=2

lf we select the origin of the coordinate system as the position of the vector tail, then x 1 = 0a

y 1 = 0b

z1 = 0 c

It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows:

Thus, the vector head is located at 3a, b, and 2c. However, in order to reduce the vector length, we have divided these coordinates by 1/3, which gives the new set of head coordinates as a, b/3, and 2c/3; the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.]

(g) For the [301] direction, it is the case that u =3

v=0

w=1

lf we select the origin of the coordinate system as the position of the vector tail, then x 1 = 0a

y 1 = 0b

z1 = 0 c

It is now possible to determine values of x2, y2, and z2 using rearranged forms of Equations 3.10a through 3.10c as follows:

Thus, the vector head is located at 3a, 0b, and c. However, in order to reduce the vector length, we have divided these coordinates by 1/3, which gives the new set of head coordinates as a, 0b, and c/3; the direction vector having these head coordinates is plotted below. [Note: even though the unit cell is cubic, which means that the unit cell edge lengths are the same (i.e., a), in order to clarify construction of the direction vector, we have chosen to use b and c to designate edge lengths along y and z axes, respectively.]

3.50 Cite the indices of the direction that results from the intersection of each of the following pairs of planes within a cubic crystal: (a) the (110) and (111) planes, (b) the (110) and (1)

planes, and (c) the

and

planes. Solution (a) In the figure below is shown (110) and (111) planes, and, as indicated, their intersection results in a , or equivalently, a

direction.

(b) In the figure below is shown (110) and [001], or equivalently, a

direction.

planes, and, as indicated, their intersection results in a

(c) In the figure below is shown

and (001) planes, and, as indicated, their intersection results in a

, or equivalently, a

3.51Sketch the atomic packing of (a) the (100) plane for the FCC crystal structure, and (b) the (111) plane for the BCC crystal structure (similar to Figures 3.12b and 3.13b). Solution

(a) An FCC unit cell, its (100) plane, and the atomic packing of this plane are indicated below.

(b) A BCC unit cell, its (111) plane, and the atomic packing of this plane are indicated below....