Condensed Matter Physics - Solutions for Homework I | PHYS 460, Assignments of Physics

Download Condensed Matter Physics - Solutions for Homework I | PHYS 460 and more Assignments Physics in PDF only on Docsity! Solutions for Homework 1 September 18, 2006 1 Tetrahedral angles Referring to Fig10, the angles between the tetrahedral bonds of diamond are equal to those between −→a1 and −→a2,−→a2 and −→a3, or −→a3 and −→a1. So cosθ = a1 × a2 ‖a1‖‖a2‖ = 1 4a 2(−1− 1 + 1) 1 4a 2(12 + 12 + 12) = −1 3 θ = cos−1(−1 3 ) ≈ 109.470 . 2 Indices of Planes Referring to Fig 11, the plane with index (100) is the plane which parallel to y-z plane abd cuts x-axis at x=a. and this plane intercepts −→a1, −→a2, −→a3 axes at 2−→a1,∞−→a2 (does not intercept −→a2axis) and 2−→a3 respectively. 12 :0: 12=1:0:1. The index referred to the primitive axes −→a1,−→a2,−→a3 is then (1,0,1). Similarly, the plan with index (0,0,1) referred to cubic cell. The plane is parallel to x-y plane and cuts z-axis at z=a. And this plane intercepts −→a1,−→a2,−→a3 axes at ∞−→a1,2−→a2,2−→a3.Hence the index referred to the primitive axes is (0,1,1). 3 Hcp Structure Suppose the radius of an atom is r. Since it’s an ideal hexagonal close-packed structure, see Fig 21, c=2r ( the two atoms touch) and a1ora2 = 2r( the two atoms touch). Also, from the geometry the distance between the center layer atom and top atom is √ ( a√ 3 )2 + ( c2 ) 2=2r (the two atoms touch)=a, so we obtain a2 3 + c2 4 = a2 ⇒ ( c a )2 = 8 3 or c a = √ 8 3 ∼= 1.633 1 If ca À √ 8 3 , the atoms on the top do not touch the atoms on the center layer. And this means, the crystal structure is composed of planes of closely packed atoms ( atoms on the each layer still touch each other), the plane being loosely stacked. 4 Problem 4 • For ideal close packed hcp, r = a 2 , c = √ 8 3 a The volume of conventional unit cell=[( √ 3 4 a 2)× 6]× √ 8 3a = 3 √ 2a3. Since there are 6 atoms in a unit cell, the volume occupied by those 6 atoms= 6× 43π(a2 )3 = πa3. Therefore packing fraction= πa 3 3 √ 2a3 = √ 2π 6 ∼= 0.74 • For close packed fcc, 4r = √ 2a ⇒ r = √ 2 4 a and there are 4 atoms per conventional cell, therefore packing fraction= 4× 43 πr3 a3 = 16 3 π( √ 2 4 ) 3a3 a3 = √ 2 6 π ∼= 0.74 • For ’close packed’ bcc, 4r = √3a (body diagonal), and there are only two atoms per unit cell. Therefore packing fraction=2× 4 3 πr 3 a3 = √ 3 8 π ∼= 0.68 5 Problem 5 Suppose the plane intercepts x,y,z axes at x1−→a1, x2−→a2, x3−→a3 respectively.Then x1 : x2 : x3 = 1h : 1 k : 1 l . (a) Prove that the reciprocal lattice vector −→G = h−→b1 + k−→b2 + l−→b3 is perpen- dicular to this plane. The normal vector to the plane is (−x1−→a1 + x2−→a2)× (−x1−→a1 + x3−→a3) = x1x3−→a3 ×−→a1 + x1x2−→a1 ×−→a2 + x2x3−→a2 ×−→a3 = x1x2x3( 1 x1 −→a2 ×−→a3 + 1 x2 −→a3 ×−→a1 + 1 x3 −→a1 ×−→a2) ∼ h−→b1 + k−→b2 + l−→b3 Therefore −→G = h−→b1 + k−→b2 + l−→b3 is perpendicular to this plane. 2