Chapter 1
Problem Solutions
1.1 (a) fcc: 8 corner atoms?1/8?1atom 6 face atoms ?1/2?3atoms
Total of 4 atoms per unit cell (b) bcc: 8 corner atoms?1/8?1atom
1 enclosed atom =1 atom
Total of 2 atoms per unit cell (c) Diamond: 8 corner atoms?1/8?1atom 6 face atoms?1/2?3atoms
4 enclosed atoms = 4 atoms
Total of 8 atoms per unit cell
_______________________________________ 1.2 (a) Simple cubic lattice: a?2r
Unit cell vol ?a3??2r?3?8r3 1 atom per cell, so atom vol ??1???4?r3????3?? Then
??4?r3?? Ratio ??3???8r3?100%?52.4% (b) Face-centered cubic lattice
d?4r?a2?a?d??2?22?r
Unit cell vol ?a3?22?r3?162?r3 4 atoms per cell, so atom vol
??4???4?r3????3? ? Then
?4???4?r3??? Ratio ??3??162?r3?100%?74% (c) Body-centered cubic lattice
d?4r?a3?a?43?r 系统教学#
3 Unit cell vol ?a3???4??r???3?? 2 atoms per cell, so atom vol
??2???4?r3??3?? ??
Then
???4?r32??? Ratio ??3????100%?68%??4r?3???3??(d) Diamond lattice Body diagonal
?d?8r?a3?a?83?r
3 Unit cell vol ?a3???8r????3?? 8 atoms per cell, so atom vol ??8???4?r3????3?? Then
?8???4?r3?? Ratio ??3?? ??100%?34% ??8r?3???3??_______________________________________ 1.3
(a)
a?5.43Ao; From Problem 1.2d, a?83?r
Then r?a3?5.43?3o8?8?1.176A
Center of one silicon atom to center of
nearest neighbor ?2r?2.35Ao (b) Number density
?8?5.43?10?8?3?5?1022cm?3
(c)
Mass density
???N?At.Wt.??5?1022??28.09?N?6.02?1023 A ???2.33 grams/cm3
_______________________________________ 1.4 (a) 4 Ga atoms per unit cell
Number density ?4?5.65?10?8?3
?Density of Ga atoms
?2.22?1022cm?3
4 As atoms per unit cell ?Density of As atoms ?2.22?1022cm?3 (b) 8 Ge atoms per unit cell
Number density ?8?5.65?10?8?3
?Density of Ge atoms ?4.44?1022cm?3
_______________________________________ 1.5
From Figure 1.15
(a)
d???a???3???2????2???0.4330?a ? ??0.4330??5.65??d?2.447Ao (b)
d???a??2??2??0.7071?a
??0.7071??5.65??d?3.995Ao _______________________________________ 1.6
asin????222??2???a?3?2?54.74?
23 ???109.5?
_______________________________________ 1.7
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(a) Simple cubic: a?2r?3.9Ao
(b) fcc: a?4r2?5.515Ao
(c) bcc: a?4r3?4.503Ao (d) diamond: a?2?4r??9.007Ao3
_______________________________________ 1.8
(a)
2?1.035?2?2?1.035??2rB roB?0.4287A (b) a?2?1.035??2.07Ao
(c)
A-atoms: # of atoms ?8?18?1 Density ?1?2.07?10?8?3
?1.13?1023cm?3
B-atoms: # of atoms ?6?12?3
Density ?3?2.07?10?8?3
?3.38?1023 cm?3 _______________________________________ 1.9 (a)
a?2r?4.5Ao
# of atoms ?8?18?1 Number density ?1?4.5?10?8?3
?1.097?1022cm?3
Mass density ???N?At.Wt.?N A ??1.0974?1022??12.5?6.02?1023 ?0.228gm/cm3
(b)
a?4r3?5.196Ao
# of atoms 8?18?1?2
Number density ?2?5.196?10?8?3
?1.4257?1022cm?3
Mass density ????1.4257?1022??12.5?6.02?1023
?0.296gm/cm3
_______________________________________ 1.10
From Problem 1.2, percent volume of fcc atoms is 74%; Therefore after coffee is ground,
Volume = 0.74 cm3
_______________________________________ 1.11
(b)a?1.8?1.0?2.8Ao (c)
Na: Density ??1/2??2.8?10?8?3
?2.28?1022cm?3
Cl: Density ?2.28?1022cm?3 (d) Na: At. Wt. = 22.99
Cl: At. Wt. = 35.45
So, mass per unit cell ??1???2???22.99????1??2???35.45?6.02?1023?4.85?10?23 Then mass density
??4.85?10?23?2.8?10?8?3?2.21 grams/cm3
_______________________________________ 1.12 (a)
a3?2?2.2??2?1.8??8Ao
Then a?4.62Ao Density of A:
?1?4.62?10?8?3?1.01?1022cm?3
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Density of B:
?1?10?8?3?1.01?1022cm?34.62?
(b) Same as (a) (c) Same material
_______________________________________ 1.13
a?2?2.2??2?1.8?o3?4.619A
(a) For 1.12(a), A-atoms Surface density ?11a2??4.619?10?8?2 ?4.687?1014cm?2
For 1.12(b), B-atoms: a?4.619Ao Surface density ?1a2?4.687?1014cm?2 For 1.12(a) and (b), Same material
(b) For 1.12(a), A-atoms; a?4.619Ao Surface density
?1?3.315?1014a22cm?2
B-atoms;
Surface density
?1?3.315?1014cm?2a22 For 1.12(b), A-atoms; a?4.619Ao Surface density
?1?3.315?1014cm?2a22
B-atoms;
Surface density
?122?3.315?1014cm?2a For 1.12(a) and (b), Same material
半导体物理与器件第四版课后习题答案(完整教资)
