Question

Chapter 03, Problem 3.63 Using appropriate data in Table 3.1, compute the interplanar spacing for the (110) set of planes for gold Crystal Structure FCC HCP BCC HCP FCC FCC BCC FCC Atomic Radius im 0.1431 0.1490 0.1249 0.1253 0.1278 0.1442 0.1241 0.1750 Atomic Radius (nm) 0.1363 0.1246 0.1387 0.1445 0.1430 0.1445 0.1371 0.1332 Crystal Structure Metal Metal Aluminum Cadmium Chromium Cobalt Copper Gold Iron (a) Lead Molybdenum Nickel Platinum Silver Tantalum Titanium (a) Tungsten Zinc BCC FCC FCC FCC BCC HCP BCC HCP FCC face-centered cubic; HCP hexagonal close-packed; BCC body-centered cubic. A nanometer (nm) equals 10-9 m; to convert from nanometers to angstrom units (Å), multiply the nanometer value by 10. nm the tolerance is +/-2% Click if you would like to Show Work for this question: Open Show Work

Answer 1

Gold rightarrow FCC Atomic radius rightarrow 0.1442nm. for FCC retaliation b/w edge length and atomic radius is 4r = squareroot 2a. 4 times (0.1442) squareroot 2 times a. a = 0.41nm. distance between interplanar distance is calculated by alpha = a/squareroot h^2 + k^2 + l^2 where h, k, l = perpendicular, L, 0 rightarrow given (1 1 0). alpha = 0.41/squareroot 1 + 1 + 0 = > 0.41/squareroot 2 = 0.41/1.414 alpha = 0.289nm