PHY 140A Solid State Physics Solution to Homework 1 TA Xun Jia1 October 2 2006 1 Email jiaxun physics ucla edu Fall 2006 c Xun Jia October 2 2006 Physics 140A Problem 1 In each of the following cases indicate whether the structure is a Bravais lattice If it is give 3 primitive vectors if it is not describe it as Bravais lattice with as small as possible a basis z o x y a b c d Figure 1 a Coordinate system b Base centered cubic c Sidecentered cubic d Edge centered cubic a Base centered cubic simple cubic with additional points in the centers of the horizontal faces of the cubic cell b Side centered cubic simple cubic with additional points in the centers of the vertical faces of the cubic cell c Edge centered cubic simple cubic with additional points at the midpoints of the lines join nearest neighbors Solution Construct the coordinate as illustrated in Fig 1 a Let the lattice constant for the corresponding simple cubic be a Denote the unit vector along x y and z directions by x y and z respectively 1 Fall 2006 c Xun Jia October 2 2006 Physics 140A a It is a Bravais lattice since from whichever point the lattice is viewed it appears exactly the same Indeed if we take three primitive vectors as a1 a2 x a2 y a2 a2 x a2 y a3 az then every lattice point will be at R n1 a1 n2 a2 n3 a3 where n1 n2 n3 are integers b It is not a Bravais lattice since there are lattice points at R1 a2 x a2 z and R2 a2 y a2 z but there is no point at R1 R2 a2 x a2 y az If we consider points at R1 R2 and R3 0 as a basis then the lattice becomes a Bravais lattice c It is not a Bravais lattice since there are lattice points at R1 a2 x and R2 a y but there is no point at R1 R2 a2 x a2 y If we consider points at R1 2 R2 R3 a2 z and R4 0 as a basis then the lattice becomes a Bravais lattice Those three types of lattices are shown in Fig 1 Problem 2 What is the Bravais lattice formed by all points with Cartesian coordinates n1 n2 n3 if a The ni are all even b The ni are all even or all odd c The sum of the ni is required to be even Solution a If the ni are all even the lattice will be simple cubic with lattice constants a 2 b If the ni are all even or all odd the lattice will be body centered cubic with lattice constants a 2 Indeed if we shift the lattice in part a as n1 n2 n3 n1 1 n2 1 n3 1 we will get a lattice with ni all odd and each new lattice point is in the center of the cubic of old lattice hence we obtain a body centered cubic lattice c If the sum of the ni is required to be even the lattice will be face centered cubic with lattice constants a 2 Indeed if we shift the lattice in part a as n1 n2 n3 n1 1 n2 1 n3 n1 n2 n3 n1 n2 1 n3 1 and n1 n2 n3 n1 1 n2 n3 1 the sum of ni are still even and each new lattice point is in the center of each square of the old lattice hence we obtain a face centered cubic lattice 2 Fall 2006 c Xun Jia October 2 2006 Physics 140A Problem 3 The face centered cubic is the most dense and the simple cubic is the least dense of the three cubic Bravais lattices The diamond structure is less dense than any of these One measure of this is that the coordination numbers are fcc 12 bcc 8 sc 6 diamond 4 Another is the following suppose identical solid spheres are distributed through space in such a way that their centers lie on the points of each of these four structures and spheres on nearest neighboring points just touch without overlapping Such an arrangement of spheres is called a close packing arrangement Show that the ratio of the volume contained in the spheres to the total volume the packing fraction for each of the four structures is a fcc 2 6 0 74 b bcc 3 8 0 68 c sc 6 0 52 d diamond 3 16 0 34 Solution As in Fig 2 Let the radius of the spheres be r and the length of the cube be a then consider the side diagonal as in Fig 2 a we have 2a 4r a For fcc lattice thus a 4r 2 Moreover there are 8 spheres at the corners and 1 8 of each is contained in the cube there are 6 spheres on the sides and 1 2 of each is contained in the cube thus the number of spheres in the cube is 8 81 6 12 4 Therefore the packing fraction is 4 r3 3 2 0 74 6 4r 3 2 4 1 b For bcc lattice consider the body diagonal as in Fig 2 b we have 3a 4r thus a 4r 3 Moreover there are 8 spheres at the corners and 1 8 of each is contained in the cube there are 1 spheres at the center which is totally contained in the cube thus the number of spheres in the cube is 8 81 1 1 2 Therefore the packing fraction is 4 r3 2 3 3 0 68 8 4r 3 3 3 2 Fall 2006 c Xun Jia October 2 2006 Physics 140A c For sc lattice consider the side as in Fig 2 c we have a 2r Moreover there are 8 spheres at the corners and 1 8 of each is contained in the cube thus the number of spheres in the cube is 8 81 1 Therefore the packing fraction is 4 r3 3 0 52 2r 3 6 1 3 d For diamond lattice it requires thinking carefully Consider the body diagonal as in Fig 2 d From simple geometry we have 3a 8r thus a 8r 3 In fact if you take a plane containing the side diagonals on the top surface and bottom surface you will find why those 5 big spheres in Fig 2 d are packing along the body diagonal This actually follows from the similarity among triangles Moreover there are 8 spheres at the corners and 1 8 of each is contained in the cube there are 6 spheres on the sides and 1 2 of each is contained in the cube there are 4 spheres at the center of each tetrahedron a b c d Figure 2 a Face centered cubic b Body centered cubic c Simple cubic d Diamond 4 Fall 2006 c Xun Jia October 2 2006 Physics 140A which are all contained in the cube thus the number of spheres in the cube is 8 18 6 …