ts1201s06 - EMSE 201 — Introduction to Materials...

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Unformatted text preview: EMSE 201 — Introduction to Materials Science & Engineering 15 February 2006 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; 150 points; 6 questions; 6 pages; 15% of course grade Partial credit will be given for correct set-ups and reasoning. Give units on numerical answers where appropriate. 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) For the element uranium (U) the lattice parameters ( a , b , c , α , β , γ ), density ρ , atomic weight AW , and atomic radius r U are given below. a = 0.2854 nm b = 0.5870 nm c = 0.4955 nm α = β = γ = 90° ρ = 19.05 g cm 3 AW = 238.03 g mol –1 r U = 0.1385 nm a) (3 points) What is the crystal system for U? orthorhombic b) (3 points) For this system, specify three directions that belong to the <111> family: Any 3: [111] ( ≡ [ 1 1 1 ]) [ 1 11] ( ≡ [1 1 1 ]) [1 1 1] ( ≡ [ 1 1 1 ]) [11 1 ] ( ≡ [ 1 1 1]) 1 pt each c) (3 points) For this system, which of the following planes belong(s) to the {110} family? 0.5 pt for each correctly selected or unselected item: ( Circle all that apply:) (100) (110) (101) (011) (1 1 0) (111) d) (6 points) How many atoms are there per unit cell in U? Show your work. Denote number of atoms per unit cell by n and volume of the unit cell by V C . Rearranging eq. 3.5 in Callister: n = ρ V C N A / AW . ( 1 pt ) V C = a × b × c ( 2 pts ) = 0.2854 × 0.5870 × 0.4955 nm 3 = 8.301 × 10 –2 nm 3 = 8.301 × 10 –23 cm 3 ⇒ n = 19.05[ g • cm –3 ] ! 6.023 ! 10 23 [ mol –1 ] ! 8.301 ! 10 23 [ cm 3 ] 238.03[ g • mol –1 ] = 4.00 ( 3 pts ) e) (3 points) Based on your answers for parts a) and d), suggest a possible Bravais lattice for U: face-centered orthorhombic f) (6 points) Compute the atomic packing factor of U. Show your work. APF = n 4 3 ! r U 3 V C ( 3 pts ) = 53.6% ( 3 pts ) EMSE 201 — Introduction to Materials Science & Engineering 15 February 2006 Department of Materials Science and Engineering 2 of 6 Case Western Reserve University 2) a) (6 points) Silicon (Si) has the diamond cubic crystal structure, with a lattice parameter of 0.5431 nm. From this information, calculate the atomic radius of silicon. Show your work. 8 r Si = a Si 3 ( 3 pts ) ⇒ r Si = 0.5431 3 /8 = 0.1176 nm ( 3 pts ) b) (8 points) Ceria (cerium oxide, CeO 2 ) has the fluorite (CaF 2 ) crystal structure, with a lattice parameter of 0.5411 nm Given that the ionic radius of oxygen, r O 2– = 0.140 nm, calculate the ionic radius r Ce 4+ of Ce 4+ in ceria. Assume that the Ce 4+ and O 2– ions touch. Show your work....
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ts1201s06 - EMSE 201 — Introduction to Materials...

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