Comprehensive Solutions Manual for "Semiconductor Physics and Devices: Basic Principles" 4th Edition by Donald A. Neamen

All Chapters Covered
k k




SOLUTION MANUAL
k

, SemiconductorkPhysicskandkDevices:kBasickPrinciples,k3rdkedition Chapterk1
SolutionskManual ProblemkSolutions


Chapter 1
F 4 r I
k
3k
Problem Solutions k k k




1.1
4 atoms per cell, so atom vol.  4G
k k k
H 3 JK k k k k k
kk


(a) fcc:k8kcornerkatomsk k 1/8k=k1katomk Then
6kfacekatomsk k ½ =k3katoms F4r IJ
4G
3



H3 K  100%  Ratio  74%
k

Totalkofk4katomskperkunitkcell
Ratio  k k k kk k k

(b) bcc:k8kcornerkatomsk k 1/8k=k1katomk
3
16k k 2krk
1kenclosedkatom =k1katom (c) Body-centeredkcubicklattice
Totalkofk2katomskperkunitkcell 4
dk k 4rk k a 3 kak  r
(c)k Diamond:k8kcornerkatomsk k 1/8k=k1katom
6kfacekatomsk k ½ =k3katoms k4
3
F I 3




3 K F 4 r I
4kenclosedkatoms
Unitkcellkvol.k k ak  r
3k
=k4katoms k
k k
3k

kTotalkofk8katomskperkunitkcell


H
1.2 2katomskperkcell,ksokatomkvol.k k2 GH 3 JK
F 4 r I
kk
(a)k 4kGakatomskperkunitkcell
4 Then 3k
Densityk   k


2G
H 3 JK
k



b 5.65x10
8k
g
3


3
k k

DensitykofkGakkk 2.22kx10 Ratiok k68%
22k
Ratio 
F4r I  100% 
cm k k k
3

4kAskatomskperkunitkcell,ksokthatk
3 k
DensitykofkAsk kk2.22kx10 cm
22k

(d)k Diamondklattice
(b) 8
Bodykdiagonalk kdk k8rk k3a 3 kak 
FI
8kGekatomskperkunitkcell r
Densityk  8 k8rk
3

b5.65x10 g k
8k 3


3
Unit cell vol.  a 
k k
H 3 K F 4 r I
k k
3k
k k
DensitykofkGekkk4.44kx10
22k 3k
cm k k




1.3
8 atoms per cell, so atom vol. 8G
k k k
H 3 JK k k k k
kk



8G 4r J
kcellkvolk k ak k 2r
k2rk8rk3 Then

HF3 K I 100% Ratio 34%
(a) Unit
Simple kcubicklattice;
3k k ak 3k 3



F 4 r I
k

3k
k k k k k


1 atom per cell, so atom vol.  1G J
Ratio
k k k
H 3 K
k k k k k k
F8r I 3k
  

H 3K
k k


Then
FG4r IJ 3k



H 3 K  100%  Ratio  52.4%
k


k
k
k
1.4
Ratio  k k k k k FromkProblemk1.3,kpercentkvolumekofkfcckatomsk
3
8rk isk74%;k Thereforekafterkcoffeekiskground,
(b) Face-centeredkcubicklattice Volumek k0.74kcm
3

d
dk k 4rk k a 2 k k ak  k 2k k 2k r
2

Unitkcellkvolk kak k
3k
c2 2 rh  16 2 r
k k
3k
k k
3
k k




3

,SemiconductorkPhysicskandkDevices:kBasickPrinciples,k3rdkedition Chapterk1
SolutionskManual ProblemkSolutions

 Thenkmasskdensitykis
23
1.5  4.85x10
kk
b g
8 
(a)
ak k5.43k A
 Fromk1.3d,k ak  r 2.8x10
8k 3k

3
k k2.21kgmk/kcm
3

ak k 3 5.43k 3 
sokthatk rk   k1.18kA
8 8
Centerkofkoneksiliconkatomktokcenterkofknearest 1.8
(a) a 3kk22.2kk21.8kk8k A


neighbork k 2rk  2.36k A
sokthat 
(b) Numberkdensity 
ak k4.62k A
b g
k 8  3
Densityk k5x10 cm
3k 22k
8k
5.43x10 1 22 3
DensitykofkAk   1.01x10k cm
(c) Masskdensity

b
28.09 g b4.62 x10 g k
8 3



NkkAt.Wt.
22k
5x10 1
k k    1.01x10 cm
22k 3




b g 
23
NA 6.02kx10 DensitykofkBk k 8k
4.62kx10 k
k k2.33kgramsk/kcm (b) Samekask(a)
3

(c) Samekmaterial

1.6 1.9
(a) ak k2rA k21.02kk2.04k A

(a) Surfacekdensity
Now 1 1
 2  
2rk k2rk k a 3 k2rk k 2.04 3 k2.04 ak 2 k
k k
A B B

sokthatk k rkB k0.747k A 3.31x10 cm
14k 2


(b) A-type;k1katomkperkunitkcell SamekforkAkatomskandkBkatoms
1
Densityk  (b) Samekask(a)
b g

8k 3k (c) k Samekmaterial
2.04kx10
23k 3
Density(A)k=k1.18x10 cm 1.10
B- 1
(a) Volkdensity 
type:k1katomkperkunitkcell,ks ao
3

23k
okDensity(B)k=k1.18x10 cm 1
3
2

1.7  Surfacekdensityk  ako 2

(b) 
 (b)k Samekask(a)
akk1.8kk1.0k ak k2.8k A





(c) 1.11
 1 k2k Sketch
3
Na:kDensityk  k 2.28x10
22k
cm
k

1.12
3
Cl:kDensityk(samekaskNa)k k2.28x10
22k
cm (a)
(d) FH1 , 1 ,1IK  (313)
k
k
k
k k
Na:kAt.Wt.k=k22.99kC 1 3 1
k k k




F1 1 1 I 121
l:kAt.kWt.k=k35.45 (b)
So,kmasskperkunitkcell k k k k

1 1
k22.99kk k35.45 H 4 , 2 , 4 K 
kk
k
k k
k
kk





k k2 2 23
k
k4.85x10
23
6.02kx10

4

, SemiconductorkPhysicskandkDevices:kBasickPrinciples,k3rdkedition Chapterk1
SolutionskManual ProblemkSolutions



b g
1.13 k 2katoms
2k
 14k 2
(a) Distancekbetweenknearestk(100)kplaneskis: 4.50x10
8k 9.88x10 cm

dk kak k5.63k A
(b) Distancekbetweenknearestk(110)kplaneskis: (ii) (110)kplane,ksurfacekdensity,
2katoms 2
1 a 5.63  k 6.99kx10 cm
14k

dk k ka 2   k
2 2 2
or (iii) (111)kplane,ksurfacekdensity,
dk k3.98k A

FH31
k kk  3 1
k k kk k IK 4
(c) Distancekbetweenknearestk(111)kplaneskis: 6 2k
k 
k

1 a 5.63 3 2
k

dk k ka 3   a
3 3 3 2
15k 2
or or 1.14kx10 cm

dk k3.25k A
1.15
1.14 (a)
(a) (100)kplanekofksiliconk–ksimilarktokakfcc,
Simplekcubic:k ak k4.50k A
 2katoms
surfacekdensity k
b g

8k 2k
(i) (100)kplane,ksurfacekdensity, 5.43x10
1 atom 2 2


b g
k  4.94kx10 cm
14k 14k
6.78x10 cm
8k 2k
4.50x10 (b)
(ii) (110)kplane,ksurfacekdensity, (110)kplane,ksurfacekdensity,
2 2
k 3.49kx10 cm  k k 9.59kx10 cm
14k 14k
= 1katom 4katoms

b4.50x10 g
2k
8k 2
k



(iii) (111) plane, surface density, (c)

3 F Iatoms
k k k

(111)kplane,ksurfacekdensity,
HK
1 1
4katoms 2
 k k 7.83x10k k cm
14
k k
6k  2  1

ca 2 hx
1 2 k 2
1 ak 3 3ak
k k k
kak k 2k
2 2 2
1 1.16
2
  2.85x10 cm
14k

k dk k 4rk k a 2
 then
(b) 4r 42.25

Body-centeredk cubic ak   k6.364k A
(i) (100)kplane,ksurfacekdensity, 2 2
14k 2 (a)
Samekask(a),(i);ksurfacekdensityk 4.94x10 cm
4katoms
VolumekDensityk 
b g
(ii) (110)kplane,ksurfacekdensity, 8 
6.364kx10 3
2 atoms 2


b g
  6.99kx10 cm
14

8k 2k 22k 3
2k 4.50x10 1.55x10 cm
(iii) (111)kplane,ksurfacekdensity, (b)
14k 2 Distancekbetweenk(110)kplanes,
Samekask(a),(iii),ksurfacekdensityk 2.85x10 cm
1 a 6.364
(c) k ka 2   
Facekcenteredkcubic 2 2 2
(i) (100)kplane,ksurfacekdensity or




5