SOLUTIONS MANUAL FOR SEMICONDUCTOR PHYSICS AND DEVICES BASIC PRINCIPLES 4TH EDITION BY NEAMEN A+ VERIFIED

All Chapters Covered
ch ch




SOLUTION MANUAL
ch

, SemiconductorchPhysicschandchDevices:chBasicchPrinciples,ch3rdchedition
SolutionschManual ProblemchChapter
Solutions
ch

1
Chapter 1
7
ch


Problem Solutions ch J4r
hc ch
3ch


4chatomschperchcell,chsochatomchvol.c h

=ch4 jH 3c h c h
jJ
1.1
(a) fcc:ch8chcornerchatomsc h c h 1/8ch=ch1ch
Then
atomch6chfacechatomsc h c h ½
=ch3chatom J 4r
4 j
7j ch
3


s
Totalchofch4chatomschperchunitchcell
H3 J
Ratioch= ch100%c h c h c h Ratioch =ch74%
3
(b) bcc:ch8chcornerchatomsc h c h 1/8ch=ch1ch 16c h c h 2chrch
atomch1chenclosedchatom (c) Body-centeredchcubicchlattice
=ch1chato 4
m 3
Totalchofch2chatomschperchunitchcell 3
dc h =c h 4rc h =c h a chac h = r
(c)c h Diamond:ch8chcornerchatomsc h c h 1/8ch=ch1chatom
6chfacechatomsc h c h ½ =ch3chatoms J4 ch 7 3



4chenclosedchatoms
Unitchcellchvol.c h =chach r
J J4r 7
3c
=ch4chatom 3 hc ch
3ch

schTotalchofch8chatomschperchunitchcell
H
h
=ch

1.2
= 2j
2chatomschperchcell,chsochatomchvol.c h
jJ
(a)c h 4chGachatomschperchunitch
H ch 3c h c h

7
cell
Densityc h =
4

Then
J4r
hc ch
3ch



2j
( 5.65x10
DensitychofchGachch−3=ch 2.22chx10
)
−8c h 3


22c h
c
H 3 jJ ch c h

Ratioch =ch68%
J 4r 7  100%
Ratio = ch
m ch ch
 3

4chAschatomschperchunitchcell,chsocht
ch
hatchDensitychofchAsc h ch=ch2.22chx10
22c h −3
(d)c h Diamondchlattice
(b) cm 8
Bodychdiagonalc h =chdc h =ch8rc 3 chac h = r
h =ch a 3
7
8chGechatomschperchunitchcell
Density
= ch
8

ch8rc h
3
J
(5.65x10 ) chch
−8c h 3

Unit cell vol. = a =
ch
3H J J4r 7
ch ch ch
3c h
ch ch
3ch
DensitychofchGechch=ch4.44chx10
22c h
hc ch
−3
cm
8 atoms per cell, so atom vol.

8j
ch ch
jJ
ch ch ch ch



1.3
(a) Simplechcubicchlattice;c h ac h =ch2
Hch 3 chch




8j
JH 3 j7
h r Then 3
=ch(2r) 4r
3c
chvolc h =c h ach
3c h
=ch8r
3
Unit chcell
ch
ch



J4rhc ch
3ch
7 Ratio
ch ch c h
100% Ratio 34%
J
1chatomchperchcell,chsochatomchvol.c h =ch(1) j =c h
J 8r3 7
  =
jJ
3c h


H
ch ch




Then
3
H3 ch c h

, SemiconductorchPhysicschandchDevices:chBasicchPrinciples,ch3rdchedition
SolutionschManual ProblemchChapter
Solutions
ch



H
1 3
J
Jj4r ch
3ch

hc
1.4

7j
Ratioch= Ratioch=ch52.4 FromchProblemch1.3,chpercentchvolumechofchfccchato
ch100%ch % mschisch74%;c h Thereforechafterchcoffeechischground
 ,
3
8rch
Volumec h =ch0.74chcm
3

(b) Face-
centeredchcubicchlattice
d
dc h =ch 4rc h =c h a2 c h c h ac h = =ch 2c h c h
2 2 r
3c h
Unitchcellchvolc h =chach =ch 2c ( h 2chr ) 3c h
=ch16c h
3
2chrch




4

, SemiconductorchPhysicschandchDevices:chBasicchPrinciples,ch3rdchedition
SolutionschManual ProblemchChapter
Solutions
ch

1
Thenchmasschdensitychis
1.5 4.85x10
−23




( )
8 ch=ch 
(a) −8ch 3c h
ach =ch5.43ch Fromch1.3d,c h ac h = r 2.8x10
 3
A c h =ch2.21chgmch/chcm
3

ac h c
sochthatc h rc h = =
(5.43)c h
=ch1.18chA


38 3 8
Centerchofchonechsiliconchatomchtochcenterchofchnea 1.8
rest
(a) a 3ch=ch2(2.2)ch+ch2(1.8)ch=ch8c h A


neighborc h =ch 2rc h 2.36c h A
 sochtha
t ac h =ch4.62c h A

(b) Numberchdensit
y
8
( )
=c h −3
Densityc h =ch5x10
22c h
5.43x10
−8ch
3c
cm
 1 22 −3



( )
h
3

DensitychofchAc h =  1.01x10c h cm
−8
(c) Masschdensity
Nch(chAt.Wt.) (5x10 )(28.0
22c h
4.62chx10
1
9)
=c h c h = =  1.01x10
22c h
cm
−3



(4.62 x10 )
23
NA 6.02chx10 DensitychofchBc h =c h −8ch
ch hc



c h =ch2.33chgramsch/chc (b) Samechasch(a)
m
3 (c) Samechmaterial

1.6 1.9
(a) ac h =ch2 =ch2(1.02)ch=ch2.04c h (a) Surfacechdensity

rA A 1 1
Now 2 c h

=a 2 = ch
ch

2rc h +ch2rc h =ch3a 3 −ch2.04
ch2rc h =ch 2
.04
A B B

sochthatc h cBh rc h =ch0.747c h A 3.31x10 cm
14c h −2


(b) A- SamechforchAchatomschandchBchatoms
type;ch1chatomchperchunitchcell
1
(b) Samechasch(a)
( )
Densityc h
= 2.04chx10
−8ch 3c h (c) c h Samechmaterial


23c h
Density(A)ch=ch1.18x10 cm 1.10
−3
1
B- (a) Volchdensity =
3
ao
type:ch1chatomchperchunitchcell
2 1
,chsochDensity(B)ch=ch1.18x10
2
3c h
cm
−3 acoh 2

1.7 
Na:chDensityc h =
(b) (c)
ach=ch1.8ch+ch1.0ch 1 ch2ch
ch 5