semisolpr01

# semisolpr01 - Semiconductor Physics and Devices Basic...

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Semiconductor Physics and Devices: Basic Principles, 3 rd edition Chapter 1 Solutions Manual Problem Solutions 3 Chapter 1 Problem Solutions 1.1 (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 1.2 (a) 4 Ga atoms per unit cell Density =⇒ 4 565 10 8 3 . x bg Density of Ga = 2.22 10 22 3 xc m 4 As atoms per unit cell, so that Density of As = 2.22 10 22 3 m (b) 8 Ge atoms per unit cell Density 8 8 3 . x Density of Ge = 4.44 10 22 3 m 1.3 (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 Ratio = 52.4% (b) Face-centered cubic lattice dr a 42 a d r 2 22 Unit cell vol = r 3 3 3 22 1 62 ch 4 atoms per cell, so atom vol. = F H G I K J 4 4 3 3 r Then Ratio r r F H G I K J 4 4 3 16 2 100% 3 3 Ratio = 74% (c) Body-centered cubic lattice a a r = 43 4 3 Unit cell vol. F H I K 3 3 4 3 2 atoms per cell, so atom vol. = F H G I K J 2 4 3 3 r Then Ratio r r F H G I K J F H I K 2 4 3 4 3 100% 3 3 Ratio = 68% (d) Diamond lattice Body diagonal == = ⇒= a a r 83 8 3 Unit cell vol.

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## This note was uploaded on 05/07/2008 for the course ECE E-302 taught by Professor Nabet during the Spring '08 term at Drexel.

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semisolpr01 - Semiconductor Physics and Devices Basic...

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