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Solutions to Homework #2 材導 陳智 10/90 3.20 This problem asks that we calculate the unit cell volume for Re which has an HCP crystal structure. In order to do this, it is necessary to use a result of Problem 3.7, that is 2 V = 6R c 3 C The problem states that c = 1.615a, and a = 2R. Therefore 3 V = (1.615)(12 3 )R C -8 3 -23 3 -2 3 = (1.615)(12 3 )(1.37 x 10 cm) = 8.63 x 10 cm = 8.63 x 10 nm _ 3.26 (a) This portion of the problem asks that a [01 1] direction be drawn within a monoclinic unit cell (a ≠ b ≠ c, and = = 90≠ ). One such unit cell with its origin at point O is sketched below. For this direction, there is no projection along the x-axis since the first index is zero; thus, the direction lies in the y-z plane. We next move from the origin along the minus y-axis b units (from point O to point R). Since the final index is a one, _ move from point R parallel to the z-axis, c units (to point P). Thus, the [01 1] direction corresponds to the vector passing from the origin to point P, as indicated in the figure. z - c P [011] y O a -y R b x (b) A (002) plane is drawn within the monoclinic cell shown below. We first remove the parentheses and take the reciprocals of the indices; this gives and 1/2. Thus, the , , (002) plane parallels both x- and y-axes, and intercepts the z-axis at c/2, as indicated in the drawing. z c y O a b x _ 3.30 Direction A is a [4 30] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x y z 2a b Projections - 3 2 0c Projections in terms of a, b, 2 1 and c - 3 2 0 Reduction to integers -4 3 0 _ Enclosure [4 30] _ Direction B is a [23 2] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x y z 2a 2c Projections 3 -b 3 Projections in terms of a, b, 2 2 and c 3 -1 3 Reduction to integers 2 -3 2 _ Enclosure [23 2] __ Direction C is a [133 ] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x y z a Projections 3 -b -c Projections in terms of a, b, 1 and c 3 -1 -1 Reduction to integers 1 -3 -3 __ Enclosure [133 ] _ Direction D is a [136 ] direction, which determination is summarized as follows. We first of all position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system x y z a b Projections 6 2 -c Projections in terms of a, b, 1 1 and c 6 2 -1 Reduction to integers 1 3 -6 _ Enclosure [136 ] _ _ 3.36 The (11 01) and (112 0) planes in a hexagonal unit cell are shown below. z z a a 2 2 a a 3 3 a a 1 1 _ _ (1101) (1120) 3.45 Planar density, PD, is defined as A c PD = A p where A is the total plane area within the unit cell and A is the circle plane area within p c this same plane. For the (100) plane in BCC, in terms of the atomic radius, R, and the unit cell edge length a 2 4R 2 16R A =a = = 3 2 p 3 Also, upon examination of that portion of the (100) plane within a single unit cell, that there resides a single equivalent atom--one-fourth from each of the four corner atoms. Therefore, 2 A = R c Hence 2 R PD = 2 = 0.59 16R 3 That portion of a (110) plane that passes through a BCC unit cell forms a rectangle as shown below. R 4R 3 4R 2 3 In terms of the atomic radius R, the length of the rectangle base is 4R 2 / 3 , whereas 4R the height is a = . Therefore, the area of this rectangle, which is just A is 3 p 2 4R 24R = 16R 2 A = p 3 3 3 Now for the number equivalent atoms within this plane. One-fourth of each corner atom and the entirety of the center atom belong to the unit cell. Therefore, there is an equivalent of 2 atoms within the unit cell. Hence 2 A = 2(R ) c and 2 2R PD = 2 = 0.83 16R 2 3 3.49 Although each individual grain in a polycrystalline material may be anisotropic, if the grains have random orientations, then the solid aggregate of the many anisotropic grains will behave isotropically. 3.50 From the table, molybdenum has a BCC crystal structure and an atomic radius of 0.1363 nm. Using Equation (3.3), the lattice parameter a may be computed as 4R (4)(0.1363 nm) a= = = 0.3148 nm 3 3 Now, the interplanar spacing d maybe determined using Equation (3.10) as 111 a 0.3148 d = = = 0.1818 nm 111 2 2 2 3 (1) + (1) + (1)