SOLUTIONS
, Chapter 1 Solutions
Prob. 1.1
Which semiconductor in Table 1-1 has the largest Eg? the smallest? What is the
corresponding A? How is the column III component related to Eg?
largest Eg : ZnS, 3.6 eV
1 24
X = ^=- = 0.344um
3.6
smallest E g : InSb, 0.18 eV
1 24
1=^=^- = 6.89um
0.18
Al compounds Eg > corresponding Ga compounds E g > the corresponding In
compounds Eg
Prob. 1.2
Find packing fraction of fee unit cell.
5-\/2
nearest atom separation = A = 3.54A
tetrahedral radius = 1.77A
volume of each atom =23.14A 3
number of atoms per cube =6--5- + 8 • | = 4 atoms
23 lA 3 -4
packing fraction = —'- ;— = 0.74 = 74%
(5A) 3
,Prob. 1.3
Label planes.
(6 4 3) (212)
x y z x y z
2 3 4 2 4 2
1/2 1/3 1/4 1/2 1/4 1/2
6 4 3 2 1 2
Prob. 1.4
Calculate densities of Si and GaAs.
The atomic weights of Si, Ga, and As are 28.1, 69.7, and 74.9, respectively.
Si: a = 5.43-10"8 cm, 8 atoms/cell
8 atoms
3 5-1022^
8
(5.43-10" cm) 3 cm
5-10 22 -L - 2 8 . 1 ^
density = = 2.33-¾
6.02-1023 1
mol
GaAs: a = 5.65 -10"8 cm, 4 each Ga, As atoms/cell
-=2.22-10 2 2 ^
s
(5.65-10' cm)
2.22-1022 -V(69.7 + 74.9) mol
density = cnr V / = 5.33 A
6.02-1023 1
mol
,Prob. 1.5
For InSb, find lattice constant, primitive cell volume, (110) atomic density.
^ = 1 . 4 4 A + 1.36A = 2.8A
4
a=6.47A
a3
FCC unit cell has 4 lattice points /.volume of primitive cell = — = 61.lK?
area of (110) plane = V2a2
4.1+2.1 ^/2
density of In atoms = —%=—-2
= - ^2- = 3.37-10 1 4 ^
V2a a
same number of Sb atoms = 3.37-1014 1
cm
Prob. 1.6
Find density ofsc unit cell.
nearest atom separation = 2 • 2.5A = 5A
number of atoms per cube = 8 • | = 1 atom
5 42 -^-
mass of one atom = : — ^ — = 9-10~24-§-
6.02-10 ^ f
density = * * ™ ' ^ i & r = QQ
(5A)3
Prob. 1.7
Draw <110> direction of diamond lattice.
This view is tilted slightly from (110) to show the
Y alignment of atoms. The open channels are
1 hexagonal along this direction.
,Prob. 1.8
Show bcc lattice as interpenetrating sc lattices.
p -p
0- -a
O •O •O -o
•a
view direction
iar
The shaded points are one sc lattice.
6- O- O The open points are the interpenetrating
sc lattice located a/2 behind the plane of
the front shaded points.
Prob. 1.9
(a) Find number of Si atoms/cm on (100) surface.
fee lattice with a = 5.43A
I number atoms per (100) surface = 4 - | + 1 = 2 atoms
a=5.43A
2
I atoms per (100) surface area ,- = 6 . 7 8 - 1 0 1 4 ^
2
(5.43A)
(b) Find the nearest neighbor distance in InP.
fee lattice with a = 5.87A
nearest neighbor distance = *.^i!ZA.V2-4.15A
,Prob. 1.10
Find Na CI density.
Na : atomic weight 23g/mol, radius
CI': atomic weight 35.5g/mol, radius 1.
unit cell with a = 2.8A by hard sphere approximation
1 atoms , Qg g _|_ 1 atoms . g g g g
Vi Na and lA CI atoms per unit cell = - 2 ^ ^ ^ '-^ = 4.86 -10"23 -\
23
6.02-10 ^gf ^
density = 4.86-10 2 3 ^ =22^_
(2.8-10-^)3^
The hard sphere approximation is comparable with the measured 2.17-^ density.
Prob. 1.11
Find packing fraction, B atoms per unit volume, and A atoms per unit area.
AhJ V rt A D
A rt
) Note: The atoms are the same size and touch each
other by the hard sphere approximation.
radii of A and B atoms are then lA
number of A atoms per unit cell = 8 • \ = 1
4A number of B atoms per unit cell = 1
volume of atoms per unit cell = l - ^ - ( l A ) 3 + l-^-(lA) 3 = f^A3
volume of unit cell = (4A)3 = 64A3
M 3
3
A 71
packing fraction = . , = — = 0.13 = 13%
64A3 24
B atoms volume density = ^- = 1.56-1022 ~
64A3
number of A atoms on (100) plane = 4-1 = 1
1 a
A atoms (100) aerial density = ^ ° m = 6.25-1014 -½
(4A)2
,Prob. 1.12
Find atoms/cell and nearest neighbor distance for sc, bcc, andfee lattices.
sc: atoms/cell = 8-j = 1
nearest neighbor distance = a
bcc: atoms/cell = 8 - 1 + 1 = 2
L->/3
nearest neighbor distance =
fee: atoms/cell = 8 - | + 6 - f = 4
i-V2
nearest neighbor distance =
Prob. 1.13
Draw cubes showing four {111} planes andfour {110} planes.
{111} planes
{110} planes
,Prob. 1.14
Findfraction occupiedfor sc, bcc, and diamond lattices.
sc: atoms/cell = 8-1 = 1
nearest neighbour = a -> radius = —
2
4TE Tt-a
atom sphere volume = (-)3
\2)
unit cell volume = a3
1 7c-a
ft 71
fraction occupied = ^— = — = 0.52
3
a 6
bcc: atoms/cell = 8 • { + 1 = 2
•V3
nearest neighbour = -> radius = -
4% (- >/3K\I TI-S-
atom sphere volume
v 4
y 16
unit cell volume = a3
fraction occupied = j ^ = = 0.68
a 8
diamond: atoms/cell = 4 (fee) + 4 (offset fee) = 8
nearest neighbour = — : — -> atom radius =
^a/2
4 8
471 <*•&* 7C-V3 •;
atom sphere volume =
128 -
v. " y
3
unit cell volume —= ar ,
128 Tt-A/3
fraction occupied = 0.34
16
,Prob. 1.15
Calculate densities ofGe andlnP.
The atomic weights of Ge, In, and P are 72.6,114.8, and 31, respectively.
Ge: a = 5.66-10"8 cm, 8 atoms/cell
8 atoms
8
= 4.41-1022 ^ r
a (5.66-l(r cm)
4.41-10 2 2 ^-72.6-!
mol
density = 23 = 5.32-^
6.02-10 _J_
mol
GaAs: a = 5.87 -10~8 cm, 4 each In, P atoms/cell
8
-=1.98-1022 ^ r
(5.87-10" cm)
1.98-10^^-(114.8+31)-^
J
density = , "* \ „, -^- = 4 . 7 9 ^
6-02-10 2 3 ^
Prob. 1.16
Sketch diamond lattice showing four atoms of interpenetratingfee in unit cell..
Full Interpenetrating Lattice Four Interpenetrating Atoms in Unit Cell
, Prob. 1.17
FindAlSbxAsi-x to lattice match InP and give band gap.
Lattice constants of AlSb, AlAs, and InP are 6.14k, 5.66k, and 5.87A, respectively from
Appendix III. Using Vegard's Law,
6.14A-x + 5.66A-(l-x) = 5.87A -> x = 0.44
AlSbo.44Aso.56 lattice matches InP and has Eg=1.9eV from Figure 1-13.
Find InxGai-x P to lattice match GaAs and give band gap.
Lattice constant of InP, GaP, and GaAs are 5.87A, 5.45A, and 5.65A, respectively
from Appendix IE. Using Vegard's Law,
5.87A-x + 5.45A-(l-x) = 5.65A -> x = 0.48
Ino.48Gao.52P lattice matches GaAs and has Eg=2.0eV from Figure 1-13.
Prob. 1.18
Find weight of As (kd=0.3) added to lkg Si in Czochralski growth for 1015 cm'3 doping.
atomic weight of As = 7 4 . 9 ^
15
1015 -J-
C =k d - ( ^ = 1 0 ^ ->CL = ^- = 3.33-1015-½
assume As may be neglected for overall melt weight and volume
i ° ° 2 « ® - 429.2cm' Si
2.33A
cm
3.33-10 15 -^ • 429.2cm3 = 1.43-1018As atoms
cm
1.43-1018atoms • 74.9-^
mol _ 1 0 1 A - 4 „ A „ - 1 0 1 A-7
= 1.8-10^g As = 1.8-10"'kg As
6.02-1023^
, Chapter 1 Solutions
Prob. 1.1
Which semiconductor in Table 1-1 has the largest Eg? the smallest? What is the
corresponding A? How is the column III component related to Eg?
largest Eg : ZnS, 3.6 eV
1 24
X = ^=- = 0.344um
3.6
smallest E g : InSb, 0.18 eV
1 24
1=^=^- = 6.89um
0.18
Al compounds Eg > corresponding Ga compounds E g > the corresponding In
compounds Eg
Prob. 1.2
Find packing fraction of fee unit cell.
5-\/2
nearest atom separation = A = 3.54A
tetrahedral radius = 1.77A
volume of each atom =23.14A 3
number of atoms per cube =6--5- + 8 • | = 4 atoms
23 lA 3 -4
packing fraction = —'- ;— = 0.74 = 74%
(5A) 3
,Prob. 1.3
Label planes.
(6 4 3) (212)
x y z x y z
2 3 4 2 4 2
1/2 1/3 1/4 1/2 1/4 1/2
6 4 3 2 1 2
Prob. 1.4
Calculate densities of Si and GaAs.
The atomic weights of Si, Ga, and As are 28.1, 69.7, and 74.9, respectively.
Si: a = 5.43-10"8 cm, 8 atoms/cell
8 atoms
3 5-1022^
8
(5.43-10" cm) 3 cm
5-10 22 -L - 2 8 . 1 ^
density = = 2.33-¾
6.02-1023 1
mol
GaAs: a = 5.65 -10"8 cm, 4 each Ga, As atoms/cell
-=2.22-10 2 2 ^
s
(5.65-10' cm)
2.22-1022 -V(69.7 + 74.9) mol
density = cnr V / = 5.33 A
6.02-1023 1
mol
,Prob. 1.5
For InSb, find lattice constant, primitive cell volume, (110) atomic density.
^ = 1 . 4 4 A + 1.36A = 2.8A
4
a=6.47A
a3
FCC unit cell has 4 lattice points /.volume of primitive cell = — = 61.lK?
area of (110) plane = V2a2
4.1+2.1 ^/2
density of In atoms = —%=—-2
= - ^2- = 3.37-10 1 4 ^
V2a a
same number of Sb atoms = 3.37-1014 1
cm
Prob. 1.6
Find density ofsc unit cell.
nearest atom separation = 2 • 2.5A = 5A
number of atoms per cube = 8 • | = 1 atom
5 42 -^-
mass of one atom = : — ^ — = 9-10~24-§-
6.02-10 ^ f
density = * * ™ ' ^ i & r = QQ
(5A)3
Prob. 1.7
Draw <110> direction of diamond lattice.
This view is tilted slightly from (110) to show the
Y alignment of atoms. The open channels are
1 hexagonal along this direction.
,Prob. 1.8
Show bcc lattice as interpenetrating sc lattices.
p -p
0- -a
O •O •O -o
•a
view direction
iar
The shaded points are one sc lattice.
6- O- O The open points are the interpenetrating
sc lattice located a/2 behind the plane of
the front shaded points.
Prob. 1.9
(a) Find number of Si atoms/cm on (100) surface.
fee lattice with a = 5.43A
I number atoms per (100) surface = 4 - | + 1 = 2 atoms
a=5.43A
2
I atoms per (100) surface area ,- = 6 . 7 8 - 1 0 1 4 ^
2
(5.43A)
(b) Find the nearest neighbor distance in InP.
fee lattice with a = 5.87A
nearest neighbor distance = *.^i!ZA.V2-4.15A
,Prob. 1.10
Find Na CI density.
Na : atomic weight 23g/mol, radius
CI': atomic weight 35.5g/mol, radius 1.
unit cell with a = 2.8A by hard sphere approximation
1 atoms , Qg g _|_ 1 atoms . g g g g
Vi Na and lA CI atoms per unit cell = - 2 ^ ^ ^ '-^ = 4.86 -10"23 -\
23
6.02-10 ^gf ^
density = 4.86-10 2 3 ^ =22^_
(2.8-10-^)3^
The hard sphere approximation is comparable with the measured 2.17-^ density.
Prob. 1.11
Find packing fraction, B atoms per unit volume, and A atoms per unit area.
AhJ V rt A D
A rt
) Note: The atoms are the same size and touch each
other by the hard sphere approximation.
radii of A and B atoms are then lA
number of A atoms per unit cell = 8 • \ = 1
4A number of B atoms per unit cell = 1
volume of atoms per unit cell = l - ^ - ( l A ) 3 + l-^-(lA) 3 = f^A3
volume of unit cell = (4A)3 = 64A3
M 3
3
A 71
packing fraction = . , = — = 0.13 = 13%
64A3 24
B atoms volume density = ^- = 1.56-1022 ~
64A3
number of A atoms on (100) plane = 4-1 = 1
1 a
A atoms (100) aerial density = ^ ° m = 6.25-1014 -½
(4A)2
,Prob. 1.12
Find atoms/cell and nearest neighbor distance for sc, bcc, andfee lattices.
sc: atoms/cell = 8-j = 1
nearest neighbor distance = a
bcc: atoms/cell = 8 - 1 + 1 = 2
L->/3
nearest neighbor distance =
fee: atoms/cell = 8 - | + 6 - f = 4
i-V2
nearest neighbor distance =
Prob. 1.13
Draw cubes showing four {111} planes andfour {110} planes.
{111} planes
{110} planes
,Prob. 1.14
Findfraction occupiedfor sc, bcc, and diamond lattices.
sc: atoms/cell = 8-1 = 1
nearest neighbour = a -> radius = —
2
4TE Tt-a
atom sphere volume = (-)3
\2)
unit cell volume = a3
1 7c-a
ft 71
fraction occupied = ^— = — = 0.52
3
a 6
bcc: atoms/cell = 8 • { + 1 = 2
•V3
nearest neighbour = -> radius = -
4% (- >/3K\I TI-S-
atom sphere volume
v 4
y 16
unit cell volume = a3
fraction occupied = j ^ = = 0.68
a 8
diamond: atoms/cell = 4 (fee) + 4 (offset fee) = 8
nearest neighbour = — : — -> atom radius =
^a/2
4 8
471 <*•&* 7C-V3 •;
atom sphere volume =
128 -
v. " y
3
unit cell volume —= ar ,
128 Tt-A/3
fraction occupied = 0.34
16
,Prob. 1.15
Calculate densities ofGe andlnP.
The atomic weights of Ge, In, and P are 72.6,114.8, and 31, respectively.
Ge: a = 5.66-10"8 cm, 8 atoms/cell
8 atoms
8
= 4.41-1022 ^ r
a (5.66-l(r cm)
4.41-10 2 2 ^-72.6-!
mol
density = 23 = 5.32-^
6.02-10 _J_
mol
GaAs: a = 5.87 -10~8 cm, 4 each In, P atoms/cell
8
-=1.98-1022 ^ r
(5.87-10" cm)
1.98-10^^-(114.8+31)-^
J
density = , "* \ „, -^- = 4 . 7 9 ^
6-02-10 2 3 ^
Prob. 1.16
Sketch diamond lattice showing four atoms of interpenetratingfee in unit cell..
Full Interpenetrating Lattice Four Interpenetrating Atoms in Unit Cell
, Prob. 1.17
FindAlSbxAsi-x to lattice match InP and give band gap.
Lattice constants of AlSb, AlAs, and InP are 6.14k, 5.66k, and 5.87A, respectively from
Appendix III. Using Vegard's Law,
6.14A-x + 5.66A-(l-x) = 5.87A -> x = 0.44
AlSbo.44Aso.56 lattice matches InP and has Eg=1.9eV from Figure 1-13.
Find InxGai-x P to lattice match GaAs and give band gap.
Lattice constant of InP, GaP, and GaAs are 5.87A, 5.45A, and 5.65A, respectively
from Appendix IE. Using Vegard's Law,
5.87A-x + 5.45A-(l-x) = 5.65A -> x = 0.48
Ino.48Gao.52P lattice matches GaAs and has Eg=2.0eV from Figure 1-13.
Prob. 1.18
Find weight of As (kd=0.3) added to lkg Si in Czochralski growth for 1015 cm'3 doping.
atomic weight of As = 7 4 . 9 ^
15
1015 -J-
C =k d - ( ^ = 1 0 ^ ->CL = ^- = 3.33-1015-½
assume As may be neglected for overall melt weight and volume
i ° ° 2 « ® - 429.2cm' Si
2.33A
cm
3.33-10 15 -^ • 429.2cm3 = 1.43-1018As atoms
cm
1.43-1018atoms • 74.9-^
mol _ 1 0 1 A - 4 „ A „ - 1 0 1 A-7
= 1.8-10^g As = 1.8-10"'kg As
6.02-1023^
