Solutions Manual for Semiconductor Physics And Devices Basic Principles 4th Edition by Neamen (1)

All Chapters Covered
M M




SOLUTION MANUAL
M

, SemiconductorMPhysicsMandMDevices:MBasicMPrinciples,M3rdMedition
SolutionsMManua ProblemMChapter
Solution
M
l1 s
Chapter 1
F 4r I
M
3M
ProblemMSolutions M M
4 atoms per cell, so atom vol.
M


 4G
M M M
JK M M



1.1
(a) fcc:M8McornerMatomsM M 1/8M=M1M
M
H
M 3 MM



Then
atomM6MfaceMatomsM M ½
=M3Mato F4r IJ
4G
3



H3 K
M

ms
TotalMofM4MatomsMperMunitMcell
RatioM M100%M M M RatioM M 74%
(b) bcc:M8McornerMatomsM M 1/8M=M1
3
16M M 2MrM
MatomM1 Menclosed Matom (c) Body-centeredMcubicMlattice
=M1Mato 4
m 3
TotalMofM2MatomsMperMunitMcell 3
dM M 4rM M a MaM  r
(c)M Diamond:M8McornerMatomsM M 1/8M=M1Matom
6MfaceMatomsM M ½ =M3Matoms F4 I M
3




3 K F 4r I
4MenclosedMatoms
Unit cell vol.  a r 3
=M4Mato M M M M M
M M
3M

msMTotalMofM8MatomsMperMunitMcell

H
M
M



1.2
2 atoms per cell, so atom vol.
M


 2G
M M
JKM M M


(a)M 4MGaMatomsMperMunitM M
H M 3 MM



F 4 r I
cell Then
4 3M
DensityM   M M

2G
b g H 3 JK
8M 3
5.65x10
M M
DensityMofMGaMM3M 2.22Mx10 Ratio  68%
22M
Ratio 
F4r I
M M M
cm  100% M M
 3

4MAsMatomsMperMunitMcell,MsoM
M
thatMDensityMofMAsM MM2.22Mx
22M 3
(d)M DiamondMlattice
(b) 10 cm 8
BodyMdiagonalM MdM M8r 3 MaM  r
M M a 3
F I
8MGeMatomsMperMunitMcell 3
8 M8rM

DensityM

b5.65x10 g M
8M 3

Unit cell vol.  a 
M M
3 H KF 4r I
M M M
3M
M
3M
MofMGeMMM 4.44Mx10
22
M
Density
3
M M
cm
8 atoms per cell, so atom vo
M


l. 8G
M M
JK M M M


1.3
(a) SimpleMcubicMlattice;M aM M2
M
H 3 MM




8G
FH43r JI
Then
M MvolM M aM M 2r
3
Mr 3M 3
MM8rcell
3
Unit M


F 4 r I
3M


K
M M M M M
Ratio 100% Ratio 34%

1MatomMperMcell,MsoMatomMvol.M M1
GH JK

F 8r3 I   
M
M M
3M





Then
3
H3 M M

, SemiconductorMPhysicsMandMDevices:MBasicMPrinciples,M3rdMedition
SolutionsMManua ProblemMChapter
Solution
M



H K
l1 3 s


FG 4r
M
M
3M


1.4

IJ
RatioM RatioMM52.4 FromMProblemM1.3,MpercentMvolumeMofMfccMat
M100%M % omsMisM74%;M ThereforeMafterMcoffeeMisMgrou
 nd,
3
8rM
VolumeM M0.74Mcm
3

(b) Face-
centeredMcubicMlattice
d
dM M 4rM M a2 M M aM  M 2M M
2 2 r

UnitMcellMvolM MaM M
3M
c2 M 2Mr h
3M
M16M 2MrM
3




4

, SemiconductorMPhysicsMandMDevices:MBasicMPrinciples,M3rdMedition
SolutionsMManua ProblemMChapter
Solution
M
l1 s
 ThenMmassMdensityMis
23
1.5  4.85x10
MM
b g
8 
8M 3M
aM M5.43M FromM1.3d,M aM  r
(a) 2.8x10
 3
A M M2.21MgmM/Mcm
3

aM M 5.43M 
soMthatM rM   M1.18MA
38 3 8
CenterMofMoneMsiliconMatomMtoMcenterMofMn 1.8
earest
(a) a 3MM22.2MM21.8MM8M A


neighborM M 2rM 2.36M A
soMtha 
(b) NumberMdensit 
t aM M4.62M A
y

b g
M 8 3
DensityM M5x10 cm
22M
8M 3
 5.43x10 1 22 3



b g
M
3


 DensityMofMAM   1.01x10M cm
8
(c) MassMdensity
NMMAt.Wt. b5x10 g28.09
22M
4.62Mx10
1


M M    1.01x10
22M 3
cm
b4.62 x10
23
NA 6.02Mx10 DensityMofMBM M 8M
M




M M2.33MgramsM/M
g(b)Same as (a)
M
M M

cm
3 (c) SameMmaterial

1.6 1.9
(a) aM M 2 M21.02MM2.04M (a) SurfaceMdensity

rA A 1 1
Now M M
a
2 M 2  
2rM M2rM M a3 3 M2.0
M2rM M 2. 4
04
A B B

soMthatM MB rM M0.747M A 3.31x10 cm
14M 2

(b) A- SameMforMAMatomsMandMBMatoms
type;M1MatomMperMunitMcell
1
(b) SameMasM(a)
b g
DensityM
8M 3M (c) M SameMmaterial
 2.04Mx10


23M
Density(A)M=M1.18x10 c 1.10
3
m 1
(a) VolMdensity 
B- ao
3


type:M1MatomMperMunitMcel 1
l,MsoMDensity(B)M=M1.18x1
23M 3 a
2 M 2
0 cm o


1.7 (c) 
(b) 1 M2M 
aMM1.8MM1.0 Na:MDensityM  aM M2.8M A

M
M





5