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Solved Problems in Solid State Physics -I Chapter No. 1 and 2 Dr. Ram Chand Physics Department, University of Hail 6 Feb, 2019 Ram C (UoH) Department of Physics, UoH 6 Feb, 2019 1/8 Chapter No. 1 Crystal Structure Ram C (UoH) Department of Physics, UoH 6 Feb, 2019 2/8 Problem 1: (Tetrahedral Angles) The angle between the tetrahedral bonds of diamond are the same as the angles between the body diagonals of a cube, as shown in Fig. 10. Use elemnetary vector analysis to …nd the value of the angle. Ram C (UoH) Department of Physics, UoH 6 Feb, 2019 3/8 Solution to Problem 1: (Tetrahedral Angles) Referring to Fig. 10, the angles between the tetrahedral bonds of diamond are equal to those between ! a1 and ! a2 . ! a2 and ! a3 , or ! a3 and ! a1 .So 1 2 a ( 1 1 +1 ) a1 a2 = 14a2 (12 +12 +12 ) = ka1 kka2 k 4 = cos 1 ( 1/3) = 109.49o cos θ = θ Ram C (UoH) Department of Physics, UoH 1 3 6 Feb, 2019 4/8 Problem 2: (Indice of Planes) Consider the planes with indices (100) and (001); the lattice is fcc, and the indices refer to the conventional cubic cell. What are the indices of these planes when referred to the primitive axes of Fig. 11? Ram C (UoH) Department of Physics, UoH 6 Feb, 2019 5/8 Solution to Problem 2: (Indice 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 a2 axis) and 2 a3 respectively. 21 : 0 : 21 = 1 : 0 : 1. The index referred to the primitive axes ! a1 , ! a2 , ! a3 is then (1, 0, 1). Similarly, the plan with index (001) 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 (011). Ram C (UoH) Department of Physics, UoH 6 Feb, 2019 6/8 Problem 3: (hcp Structure) Show that the c/a ratio for an ideal hexagonal close-packed structure is 8 1/2 = 1.633. If c/a is signi…cantly larger than this value, the crystal 3 structure may be thought of as composed of planes of closely packed atoms, the planes being lossely packed. Ram C (UoH) Department of Physics, UoH 6 Feb, 2019 7/8 Solution to Problem 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 a1 or a2 = 2r ( the two atoms touch). Also, from ther geometry the distance between the center layer atom and top atom is pa 3 2 + c 2 2 = 2r (the two atoms touch)= a, so we obtain c 2 a = a2 c2 a2 3 + 4 = q 8 c 8 3 or a = 3 = 1.633 q 1.633, the atoms on the top do not touch the atoms on the If 83 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. Ram C (UoH) Department of Physics, UoH 6 Feb, 2019 8/8