2. a.) Draw an HCP unit cell. b.) What is the ratio of cell edge height (c) to cell edge length (a) in an HCP structure?

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2 A Draw An Hcp Unit Cell B What Is The Ratio Of Cell Edge Height C To Cell Edge Length A In An Hcp Structure 1 (47.9 KiB) Viewed 125 times

2 A Draw An Hcp Unit Cell B What Is The Ratio Of Cell Edge Height C To Cell Edge Length A In An Hcp Structure 2 (157.62 KiB) Viewed 125 times

2 A Draw An Hcp Unit Cell B What Is The Ratio Of Cell Edge Height C To Cell Edge Length A In An Hcp Structure 3 (259.11 KiB) Viewed 125 times

2. a.) Draw an HCP unit cell. b.) What is the ratio of cell edge height (c) to cell edge length (a) in an HCP structure? c.) If you were to assume that in an HCP unit cell, a = 2R, determine the cell edge length (a) and cell edge height (c) for Cadmium (Cd). (Hint: Use Table 3.1 in supplemental materials) d.) If you were to assume that the unit cell volume (V...) of an HCP unit cell equals 6R2cV3, and that the atomic weight of Cd equals 112.41 g/mol, calculate the theoretical density () (Hint: If struggling to visualize how these values were derived, refer to Example Problem 3.3 in supplemental materials)
Table 3.1 Crystal Atomic Radius Crystal Atomic Metal Structure (nm) Metal Structure Radius (nm) Aluminum FCC 0.1431 Molybdenum BCC 0.1363 Cadmium HCP 0.1490 Nickel FCC 0.1246 Chromium BCC 0.1249 Platinum FCC 0.1387 Cobalt HCP 0.1253 Silver FCC 0.1445 Copper FCC 0.1278 Tantalum BCC 0.1430 Gold FCC 0.1442 Titanium (a) HCP 0.1445 Iron (a) BCC 0.1241 Tungsten BCC 0.1371 Lead FCC 0.1750 Zinc HCP 0.1332 "FCC = face-centered cubic: HCP = hexagonal close-packed: BCC = body-centered cubic. "A nanometer (nm) equals 10m; to convert from nanometers to angstrom units (Å), multiply the nanometer value by 10.
EXAMPLE PROBLEM 3.3 Determination of HCP Unit Cell Volume (a) Calculate the volume of an HCP unit cell in terms of its a and clattice parameters (b) Now provide an expression for this volume in terms of the atomic radius, R. and the clattice parameter Solution (a) We use the adjacent reduced-sphere HCP unit cell to solve this problem. Now, the unit cell volume is just the prod. uct of the base area times the cell height, This base area is just three times the area of the parallelepiped ACDE shown below. (This ACDE parallelepiped is also labeled in the above unit cell.) The area of ACDE is just the length of times the height BC. But CD is just a, and BC is equal to BC = cos(30) - Thus, the base area is just 60 4V3 2R AREA = (3/TDXBC) – (310164 3) - Bopy G2R 2 Again, the unit cell volume V is just the product of the AREA and thus Vc - AREA) - (Birma) es 30 V3 (3.7a) 2 (b) For this portion of the problem, all we need do is realize that the lattice parameter a is related to the atomic radius Ras - 2R Now making this substitution for a in Equation 3.7a gives 3(2RV3 Vc 2 -6R V3 (3.7b)