20. In a crystalline material, planes of symmetry are labeled using triplets of integer numbers of...

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20. In a crystalline material, planes of symmetry are labeled using triplets of integer numbers of the form (hkl), e.g. (100) is equivalent to h = 1, k = 0, 1 = 0. For this question it is not important how this labeling scheme works exactly or what h, k and I mean. By performing certain experiments, it is possible to determine these (hkl) triplets for multiple planes of symmetry. From the knowledge of several such triplets the microscopic structure of the material can be inferred using the following rules: • If for all planes all positive integer values of h, k and I are possible the material has a simple cubic lattice. • If for all planes h+k+1 = 2n where n is any positive integer then the material has a body centred cubic lattice. • If for all planes h and k and I are either all even or all odd the material has a face centred cubic lattice. Note: zero is an even number During the analysis of the data from such an experiment the following four planes of symmetry were found: (111), (200), (220), (311). (a) What type of lattice did the material in these experiments have? (b) In a material with some form of cubic lattice the distance d between the planes of symmetry is given by: a /h²+k² + 1² where a is a constant length that is usually of the order of a fraction of a nanometre. Using the formula for d and the data in the table below determine the best estimate for a. d (nm) 0.224 0.195 (200) 0.137 (220) 0.117 (311) Plane (111) (c) A cube of the material has a length L in each dimension where L = Na and N is a very large number. Pressure is applied to two opposite sides of the cube. The volume of the cube remains constant but one of the cube's dimensions reduces to a length of L'= Na. What is the new length of the deformed object in the other two dimensions? [2] [3] [2] 20. In a crystalline material, planes of symmetry are labeled using triplets of integer numbers of the form (hkl), e.g. (100) is equivalent to h = 1, k = 0, 1 = 0. For this question it is not important how this labeling scheme works exactly or what h, k and I mean. By performing certain experiments, it is possible to determine these (hkl) triplets for multiple planes of symmetry. From the knowledge of several such triplets the microscopic structure of the material can be inferred using the following rules: • If for all planes all positive integer values of h, k and I are possible the material has a simple cubic lattice. • If for all planes h+k+1 = 2n where n is any positive integer then the material has a body centred cubic lattice. • If for all planes h and k and I are either all even or all odd the material has a face centred cubic lattice. Note: zero is an even number During the analysis of the data from such an experiment the following four planes of symmetry were found: (111), (200), (220), (311). (a) What type of lattice did the material in these experiments have? (b) In a material with some form of cubic lattice the distance d between the planes of symmetry is given by: a /h²+k² + 1² where a is a constant length that is usually of the order of a fraction of a nanometre. Using the formula for d and the data in the table below determine the best estimate for a. d (nm) 0.224 0.195 (200) 0.137 (220) 0.117 (311) Plane (111) (c) A cube of the material has a length L in each dimension where L = Na and N is a very large number. Pressure is applied to two opposite sides of the cube. The volume of the cube remains constant but one of the cube's dimensions reduces to a length of L'= Na. What is the new length of the deformed object in the other two dimensions? [2] [3] [2]

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a To determine the type of lattice the material has we need to apply the rules given for the hkl triplets of the planes of symmetry For a simple cubic ... View the full answer