文档介绍：Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 1 Solutions Manual Problem Solutions 3 Chapter 1 Problem Solutions (a) fcc: 8 corner atoms × 1/8 = 1 atom 6 face atoms × ? = 3 atoms Total of 4 atoms per unit cell (b) bcc: 8 corner atoms × 1/8 = 1 atom 1 enclosed atom = 1 atom Total of 2 atoms per unit cell (c) Diamond: 8 corner atoms × 1/8 = 1 atom 6 face atoms × ? = 3 atoms 4 enclosed atoms = 4 atoms Total of 8 atoms per unit cell (a) 4 Ga atoms per unit cell Density =?? 4 565 10 8 3 . x bg Density of Ga = ? 10 22 3 xcm 4 As atoms per unit cell, so that Density of As = ? 10 22 3 xcm (b) 8 Ge atoms per unit cell Density =?? 8 565 10 8 3 . x bg Density of Ge = ? 10 22 3 xcm (a) Simple cubic lattice; ar = 2 Unit cell vol == = () arr 3 3 3 28 1 atom per cell, so atom vol. = () F H G I K J 1 4 3 3 π r Then Ratio r r =×? F H G I K J 4 3 8 100% 3 3 π Ratio =% (b) Face-centered cubic lattice dra== ? 42 a d r == 2 22 Unit cell vol == =arr 3 3 3 22 162 ch 4 atoms per cell, so atom vol. = F H G I K J 4 4 3 3 π r Then Ratio r r =×? F H G I K J 4 4 3 16 2 100% 3 3 π Ratio =74% (c) Body-centered cubic lattice dra a r== ?= 43 4 3 Unit cell vol. == F H I K ar 3 3 4 3 2 atoms per cell, so atom vol. = F H G I K J 2 4 3 3 π r Then Ratio r r =×? F H G I K J F H I K 2 4 3 4 3 100% 3 3 π Ratio =68% (d) Diamond lattice Body diagonal === ?=dra a r 83 8 3 Unit cell vol. == F H I K a r 3 3 8 3 8 atoms per cell, so atom vol. 8 4 3 3 π r F H G I K J Then Ratio r r =×? F H G I K J F H I K 8 4 3 8 3 100% 3 3 π Ratio =34% Fr