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¹ÝµµÃ¼¹°¼º°ú ¼ÒÀÚ 3ÆÇ ¼Ö·ç¼Ç D. Neamen

¹ÝµµÃ¼¹°¼º°ú ¼ÒÀÚ 3ÆÇ ¼Ö·ç¼Ç D. Neamen

¹ÝµµÃ¼¹°¼º°ú ¼ÒÀÚ 3ÆÇ ¼Ö·ç¼Ç D. Neamen¹°¼º°ú ¼ÒÀÚ 3ÆÇ ¼Ö·ç¼Ç D. NeamenSemiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual

Chapter 1 Problem Solutions

Chapter 1
Problem Solutions
1.1 (a) fcc: 8 corner atoms ¡¿ 1/8 = 1 atom 6 face atoms ¡¿ ¨ö = 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 ¡¿ ¨ö = 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 atoms per cell, so atom vol. = 4 Then 4 Ratio =

FG 4¥ðr IJ H3K
3

16 2 r (c) Body-centered cubic lattice 4 d = 4r = a 3 ¢¡ a = r 3 Unit cell vol. = a =
3

FG 4¥ðr IJ H 3 K ¡¿ 100% ¢¡
3 3

Ratio = 74%

F 4 rI H 3K

3

2 atoms per cell, so atom vol. = 2 Then

FG 4¥ðr IJ H3K
3

b

4
?8

5.65 x10

g

3

¢¡
22 ?3

Density of Ga = 2.22 x10 cm
22

4 As atoms per unit cell, so that Density of As = 2.22 x10 cm (b) 8 Ge atoms per unit cell 8 ¢¡



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