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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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