EN380 Homework #5 Solution 1. Aluminum forms an FCC structure and has an atomic radius of 0:143 nm. (a) Calculate the lattice constant, a. For FCC structure: 4R 4 · 0:143 nm a = p = p = 0:404 nm 2 2 (b) Calculate the atomic packing factor, APF. For FCC structure, # Atoms = 6 faces · 1 atom per face + 8 corners · 1 atom per corner = 4 cell 2 8 # Atoms # Atoms 4π 3 4π 3 · V{ atom · R V{ atoms cell cell 3 4 · 3 (0:143 nm) AP F = = 3 = 3 = 3 = 0:740 V{ cell a a (0:404 nm) (c) Calculate the theoretical density of the metal in the unit cell. How does this compare to the actual density of aluminum? # Atoms 1 · atomic mass · 4 atoms · 26:982 g 1 mol matoms cell NA mol 6:02·1023atoms ρtheo = = 3 = 3 V{ cell a (0:404 nm) g g = 2:719 · 10−21 = 2:719 nm3 cm3 g The theoretical density is slightly higher than actual density (2:700 cm3 ) due to the presence of defects in the crystal structure (dislocations, vacancies, grain boundaries, etc.). 1 2. Find the direction indices of the vectors shown in the following figures. (a) (b) Tail is already at origin Move origin to vector tail Cube Projections Cube Projections x y z x y z 1 1 1 1 − 2 0 1 Multiply by LCF (2) All integers, can't be simplified ) [ 1 1 1 ] −1 0 2 Can't be simplified ) [ 1 0 2 ] (c) Move the origin to vector tail. This produces projections on the a1; a2; and a3 axes of -1, -1, and +2, respectively. Now we need to check that our vectors add to attain our desired endpoint and scale them if they do not: 2 The correct projects on a1; a2; and a3 axes are 2 2 2 4 therefore 3 · [−1; −1; 2] = [− 3 ; − 3 ; 3 ] Hexagonal Prism Projections a1 a2 a3 z 2 2 4 1 − 3 − 3 3 2 Multiply by LCD (6) -4 -4 8 3 Can't be simplified ) [ 4 4 8 3 ] (d) Repeat problem 2.(a) with the head and (e) Repeat problem 2.(b) with the head and the the tail of the vector reversed. tail of the vector reversed. Same as before but flip the sign of all Same as before but flip the sign of all of the projections: of the projections: Cube Projections Cube Projections x y z x y z 1 -1 -1 -1 2 0 -1 Multiply by LCD (2) All integers, can't be simplified 1 0 -2 ) [ 1 1 1 ] Can't be simplified ) [ 1 0 2 ] (f) Repeat problem 2.(c) with the head and the tail of the vector reversed. By inspection, we may simply reverse the sign of all of the components from 2.(c) just as we had for the previous two parts: Hexagonal Prism Projections a1 a2 a3 z 2 2 4 1 3 3 − 3 − 2 Multiply by LCD (6) 4 4 -8 -3 ) [ 4 4 8 3 ] 3 3. Find the Miller indices of the planes shown in the following figures. (a) (b) Plane passes through origin Plane does not pass through origin ! place the origin at the point (0,0,1) ! leave origin unchanged. (represented above with the x0, y0, z axes). Axis Intercepts Axis Intercepts x y z 0 0 2 1 x y z 1 3 2 1 1 -1 Reciprocals Reciprocals x y z 0 0 3 x y z 1 2 2 1 0 -1 Multiply by LCF (2) 2 3 4 All integers, can't be simplified ) ( 1 0 1 ) or ( 1 0 1 ) Can't be simplified ) ( 2 3 4 ) (depending on how you selected your origin) 4 (c) Plane does not pass through origin ! leave origin unchanged. Axis Intercepts a1 a2 a3 z 1 -1 1 1 Reciprocals 1 -1 0 1 All integers, can't be simplified ) ( 1 1 0 1 ) 5.