0% (0 out of 5 correct) The questions marked with symbol have not been graded.

of 15 0% (0 out of 5 correct) The questions marked with symbol have not been graded. Responses to questions are indicated by the symbol. 1. Which of the following materials may form crystalline solids? A. Polymers B. Metals C. Ceramics D. All of the above E. None of the above Under normal solidification conditions, all three of these material types may form crystalline solids; these include all metals, many ceramics, and some polymers (which will be semicrystalline). 2. The drawing below represents the unit cell for which crystal structure? A. Simple cubic B. Face-centered cubic C. Body-centered cubic D. Hexagonal close-packed Since there is one atom situated at each of the eight corners of the cube as well as one atom at the center of each of the cube faces, this is the unit cell for the face-centered cubic crystal structure. 3. Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths? a = b c

of 15 A. Cubic B. Hexagonal C. Triclinic D. Monoclinic E. Rhombohedral F. Orthorhombic G. Tetragonal H. Both C and E I. Both B and G For both hexagonal and tetragonal crystal systems, two of the unit cell edge lengths are equal to one another, but unequal to the third edge length. 4. Which crystal system(s) listed below has (have) the following interaxial angle relationship? = = = 90 A. Cubic B. Tetragonal C. Hexagonal D. Triclinic E. Orthorhombic F. Both A and D G. A, B, and E Cubic, tetragonal, and orthorhombic crystal systems have the three interaxial angles equal to 90 C. 5. If the atomic radius of a metal that has the face-centered cubic crystal structure is 0.137 nm, calculate the volume of its unit cell (in nm^3). 0.0582 nm^3

of 15 The unit cell for the face-centered cubic crystal structure is Since the atoms touch each other along the face diagonal, the length of this diagonal is four times the atomic radius. Hence, the edge length, a, may be determined as follows: or Thus, the volume of the cubic unit cell may be calculated as follows: 6. For a metal that has the body-centered cubic crystal structure, calculate the atomic radius (in nm) if the metal has a density of 7.25 g/cm 3 and an atomic weight of 50.99 g/mol. 0.124 nm The density of a metal may be calculated using the following equation:

of 15 Now, for the body-centered cubic crystal structure there are two atoms associated with each unit cell (i.e., n = 2), and the atomic radius and unit cell edge length are related as Since the unit cell for the body-centered cubic crystal structure has cubic symmetry, V C = a 3. Substitution of the last two relationships into the first equation leads to and solving for R yields = 1.24 x 10-8 cm = 0.124 nm 7. For the face-centered cubic crystal structure: (a) How many atoms are associated with each unit cell? (b) What is the coordination number? (c) What is the atomic packing factor? (a) 4 (b) 12 (c) 0.740 For the face-centered cubic crystal structure, atoms are positioned at each of the eight cube

of 15 corners, as well as at the center of each of the unit cell faces. Its unit cell is as follows: (a) There are eight corner atoms, and one-eighth of each of these is assigned to the unit cell (for the equivalence of 1 atom); also, there is one atom at each of the six faces, one-half of which is assigned to the unit cell (for the equivalence of 3 atoms). Therefore, there are four atoms per unit cell. (b) An atom has four atoms neighboring atoms in the same horizontal plane, four nearestneighbor atoms above it, and four nearest-neighbor atoms below it; therefore, the coordination number is 12. (c) Since the unit cell edge length and the atomic radius are related as the atomic packing factor is calculated as follows: 8. A hypothetical metal has an orthorhombic unit cell for which the a, b, and c lattice parameter 0.413 nm, 0.665 nm, and 0.876 nm, respectively. (a) If there are 8 atoms per unit cell and the atomic packing factor is 0.536, determine the ato radius (in nm). (b) If the density is 3.99 g/cm 3, calculate the metal's atomic weight (in g/ mol). (a) 0.157 nm (b) 72.27 g/mol (a) The atomic packing factor, APF, is defined as:

of 15 For an orthorhombic unit cell, the unit cell volume is equal to the product of the unit cell edg as The total sphere volume is equal to the volume per sphere times the number of atoms per uni (i.e., 8) as Substitution of the above expressions into the APF equation yields And solving for the atomic radius leads to the following: = 0.157 nm (b) It is possible to solve for the atomic weight by rearrangement of the equation that express density as a function of the number of atoms per unit cell, the atomic weight and the unit cell that is Incorporation of values for a, b, c, n, and provided in the problem statement, we have = 72.27 g/mol

of 15 9. What are the indices for the direction represented by the vector that has been drawn within a unit cell? Indicate a negative index with a "-" in front of the index number. Otherwise, express the direction using the conventional notation. [120] The vector projections on the x- and y-axes are a/2 and b, respectively, while the projection on the z-axis is zero (since the vector lies in the x-y plane). This is a [120] direction as indicated in the summary below. x y z Projections a/2 b 0c Projections in terms of a, b, and c 1/2 1 0 Reduction to integers 1 2 0 Enclosure [120] 10. What are the indices for the direction represented by the vector that has been drawn within a unit cell? Indicate a negative index with a "-" in front of the index number. Otherwise, express the direction using the conventional notation.

of 15 [0-11] The vector projection on the x-axis is zero (since the vector lies in the y-z plane), while projections on the y- and z-axes are -b, and c, respectively. This is a [0-11] direction as indicated in the summary below. x y z Projections 0a -b c Projections in terms of a, b, and c 0-1 1 Enclosure [0-11] 11. What are the indices for the direction represented by the vector that has been drawn within a unit cell? Indicate a negative index with a "-" in front of the index number. Otherwise, express the direction using the conventional notation.

of 15 [-1-43] The origin of the coordinate system must be repositioned so that the tail of the direction vector is situated at its origin. In terms of this new coordinate system, the vector projections on the x-, y-, and z-axes are -a/6, -2b/3, and c/2, respectively. This is a [-1-43] direction as indicated in the summary below. x y z Projections -a/6-2b/3 c/2 Projections in terms of a, b, and c -1/6-2/3 1/2 Reduction to integers -1-4 3 Enclosure [-1-43] 12. What are the Miller indices for the plane shown below? Indicate a negative index with a "- " in front of the index number. Otherwise, express the plane using the conventional notation. (110) The intercepts of the plane with the x-, y-, and z-axes are a, b, and c, respectively. This is a (110) plane as indicated by the summary below. x y z Intercepts a b c

of 15 Intercepts in terms of a, b, and c 1 1 Reciprocals of intercepts 1 1 0 Enclosure (110) 13. What are the Miller indices for the plane shown below? Indicate a negative index with a "- " in front of the index number. Otherwise, express the plane using the conventional notation. (40-3) The intercepts of the plane (extended) with the x-, y-, and z-axes are a/2, respectively. This is a (40-3) plane as indicated in the summary below. and -2c/3, x y z Intercepts a/2 b -2c/3 Intercepts in terms of a, b, and c 1/2-2/3 Reciprocals of intercepts 2 0-3/2 Reduction to integers 4 0-3 Enclosure (40-3) 14. What are the Miller indices for the plane shown below? Indicate a negative index with a "- " in front of the index number. Otherwise, express the plane using the conventional notation.

of 15 (2-23) Since the plane passes through the origin, the origin must be translated, in this case, one unit cell distance to the right along the y axis. The intercepts of this plane relative to the new x-, y-, and z-axes are a, -b, and 2c/3, respectively. This is a (2-23) plane as indicated in the summary below. x y z Intercepts a b 2c/3 Intercepts in terms of a, b, and c 1-1 2/3 Reciprocals of intercepts 1-1 3/2 Reduction to integers 2-2 3 Enclosure (2-23) 15. For the direction represented by the vector drawn within the hexagonal unit cell below, specify its indices (using the 3-index scheme, referenced to a 1, a 2, and z axes). Indicate a negative index with a "-" in front of the index number. Otherwise, express the direction using the conventional notation.

of 15 [221] The projections on the a 1 -, a 2 -, and z-axes are a, a, and c/2, respectively. This is a [221] direction as indicated in the summary below. a 1 a 2 z Projections a a c/2 Projections in terms of a and c 1 1 1/2 Reduction to integers 2 2 1 Enclosure [221] 16. What are the Miller-Bravais indices for the plane shown below? Indicate a negative index with a "-" in front of the index number. Otherwise, express the plane using the appropriate notation.

of 15 (-1010) The intercepts of this plane relative to the a 1 -, a 2 -, a 3 -, and z-axes are a, a, a, and c, respectively. This is a (-1010) plane as indicated in the summary below. a 1 a 2 a 3 z Intercepts a a a c Intercepts in terms of a and c 1 1 Reciprocals of intercepts -1 0 1 0 Reduction to integers Enclosure (-1010) 17. For the body-centered cubic crystal structure, in terms of the atomic radius, R, determine the distance between the centers of nearest-neighbor atoms. 2R A unit cell for the body-centered cubic crystal structure is shown below. Since corner atoms touch the central atom along unit cell diagonals, the central atom is a nearest neighbor to the corner atoms; therefore, the distance between nearest-neighbor atoms is 2R. 18. Below is shown the atomic packing of a set of planes for the simple cubic crystal

of 15 structure; atoms drawn to full size are represented by circles. From the list below select the set of indices for this plane. A. (120) B. (100) C. (111) D. (110) The atomic packing shown is for a (100) type plane. Below is presented a (100) plane that passes through a simple cubic reduced sphere unit cell, and which yields the packing arrangement indicated in the problem statement. 19. In terms of the atomic radius, R, determine the distance between the centers of adjacent atoms for the FCC crystal structure along the [100] direction. 2*R*sqrt(2) The [100] direction within an FCC unit cell is shown below.

of 15 Thus, the distance between the centers of adjacent atoms in this [100] direction is simply the edge length of the face-centered cubic unit cell, which, in terms of the atomic radius R is just Retake Test