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ts1201f09

# ts1201f09 - EMSE 201 Introduction to Materials Science...

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EMSE 201 — Introduction to Materials Science & Engineering 23 September 2009 Name: S O L U T I O N Department of Materials Science and Engineering 1 of 6 Case Western Reserve University Test #1 — 75 minutes; 100 points; 6 questions; 6 pages; 10% of course grade Partial credit will be given for correct set-ups and reasoning. Give units on numerical answers where appropriate and use the correct number of significant figures. Please write all answers on these pages; use the backs if needed. Boltzmann’s constant, k B = 1.381 × 10 -23 J K -1 = 8.620 × 10 -5 eV K -1 Avogadro’s number, N A = 6.023 × 10 23 mol -1 gas constant, R : 8.314 J mol -1 K -1 1) The sketch at right shows one unit cell of a compound of titanium (Ti, black circles) and oxygen (O, white circles). The coordinates of the atoms are as follows: Ti: 0, 0, 0 1/2, 1/2, 1/2 O: x 1 , x 1 , 0 1– x 1 , 1– x 1 , 0 1– x 2 , x 2 , 1/2 x 2 , 1– x 2 , 1/2 (0 < ( x 1 , x 2 ) < 1, with specific values that depend on the ionic radii of Ti and O.) a) (2 points) Determine the number of Ti atoms per unit cell. net 1 atom from the corners + 1 atom in the interior = 2 b) (2 points) Determine the number of O atoms per unit cell. 4 atoms on faces for a net of 2 atoms + 2 atoms in the interior = 4 c) (2 points) The coordination number of titanium in this compound is 6, i.e. every titanium atom is surrounded by 6 oxygen atoms. What is the coordination number of oxygen? By inspection, the interior oxygens have 3 nearest Ti neighbors. Or reasoning from the number of each atom per unit cell, CN O = (2/4) CN Ti = 3 . The values of the lattice parameters are a = b = 0.45933 nm, c = 0.29592 nm, α = β = γ = 90°. d) (2 points) To what crystal system does this structure belong? tetragonal e) (5 points) Calculate the atomic packing fraction in this compound. For the radius of oxygen use 0.136 nm. The radius of titanium is given on the next page. APF = V atoms / V cell ( 1 pt ) = (4 V O + 2 V Ti )/( abc ) = 4 3 π 4 r O 3 + 2 r Ti 3 ( ) a 2 c = 4 3 π 4(0.136) 3 + 2(0.0605) 3 ( ) (0.45933) 2 (0.29592) = 70.5% ( 2 pts ) ( 1 pt ) ( 1 pt ) continued on next page x y z c a a

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EMSE 201 — Introduction to Materials Science & Engineering 23 September 2009 Department of Materials Science and Engineering 2 of 6 Case Western Reserve University 1) f) (5 points) In the space below, sketch the (001) plane of this structure. Label the atoms and (where appropriate) the cell edges. g) (5 points) In the space below, sketch the (110) plane of this structure. Label the atoms and (where appropriate) the cell edges. A compound of tin (Sn) and oxygen also has the structure shown on page 1. h) (10 points) Briefly state the rules for solid solution formation, as they would be applied to oxides. Evaluate each rule for these Ti-O and Sn-O compounds. Given the data below, predict whether these two compounds would be expected to form a complete solid solution. Ti Sn cation radius, nm 0.0605 0.0690 cation electronegativity 1.54 1.96 cation oxidation numbers +4 , +3 +4 , +2 1 pt per correctly stated criterion; 1 pt per correct evaluation; 2 pts conclusion.
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