3.31 Determine the indices for the directions shown in the following cubic unit cell:
+z
A
1
1 1
,
2 2
2
1
B
2
+y
+x
[ ]
Direction A is the 110 direction.
the direction vector.
To identify it, we position the origin at the tail of
Then, in terms of this new coordinate system:
x
y
Projections!
Projections in terms of a,b, and c
Reduction to integers
Enclosure
-a
-1
z
b
1
not necessary
110
0c
0
[ ]
Direction B is the [121] direction. This vector passes through the origin of the
coordinate system, and thus no translation is necessary. Therefore:
!
x
y
z
Projections
Projections in terms of a, b, and c
a/2
1/2
b
1
c/2
1/2
Reduction to integers
Enclosure
1
2
[121]
1
x
y
z
0a
0
0
−b/2
-1/2
-1
–c
-1
-2
Direction C is the [ 012 ] direction:
Projections
Projections in terms of a, b, and c
Reduction to integers
Enclosure
[0 12 ]
Direction D is the [1 2 1] direction:
Projections
Projections in terms of a, b, and c
Reduction to integers
Enclosure
x
a/2
1/2
1
y
–b
-1
-2
[1 2 1]
z
c/2
1/2
1
3.39 Sketch within a cubic unit cell the following planes:
(a)
(b)
(c)
(d)
(10 1 ),
(2 1 1),
(012),
(3 1 3),
(e) (1 1 1 ),
(f) ( 2 1 2 ),
(g) (31 2),
(h) (301).
3.40
Determine the Miller indices for the planes shown in the following unit cell:
1
+z
3
1
2
B
A
+y
+x
3.53 (a) Derive planar density expressions for FCC(100) and (111) planes in terms of the
atomic radius R.
(b) Compute and compare planar density values for these same two planes for molybdenum.
a)
Z
uv
B
(110)
uv
C
X
Y
uv
A
uv
Suppose vector A = [100]
Choose two vectors within plane (100):
uv
A
uv
A
uv
B = [1 0 0 ] [ 0 01 ]= 0 + 0 + 0 = 0
uv
B = [1 0 0 ] [ 01 0 ]= 0 + 0 + 0 = 0
!
uv
uv
B = [001], C =[010]
[100] ! (100)
b)
Z
uv
A
[111]
(111)
uv
C
uv
B
Y
X
uv
Suppose vector A = [111]
Choose two vectors within plane (111) :
uv
A
uv
A
uv
B = [111 ]
uv
B = [111]
! 0 11 " = 0 +1+ (-1) = 0
#
$
! 11 0 " = (%1) + 1 + 0 = 0
#
$
!
uv
uv
B = !#0 11"$ , C = !# 11 0 "$
[ 111 ]! ( 111 )
6 InSb (Indium Antimonide) is a so-called narrow bandgap semiconductor. Its gap
energy is 0.17eV. It is often used as part of a detector for infrared light in night-vision
systems, for example.
A) sketch a well-labeled energy-band diagram characteristic of InSb.
B) Calculate the maximum wavelength of light that could be absorbed by a crystal of
pure InSb by exciting an electron from its valence band to its conduction band
A)
E=0
Empty Band
Energy Gap
Eg = 0.17 eV
Full Valence Band
B)
hc
E=
!
!
hc
!=
=
E
1eV
(6.63 #10"34 J " s )(
)(3 #108 m / s )
1.602 #10"19
= 7.3 µ m
0.17eV
7. Crystallographic properties of cubic SiC
(a) Percent ionicity =1 ! exp[!(0.25)( x A ! xB ) 2 ] "100
xSi = 1.8
xC = 2.5
! % Ionicity = 1 ! exp[!(0.25)(1.8 ! 2.5) 2 ] " 100 = 11.6%
A weak dipole would form because the carbon side of the bond would on average have a net
negative charge while the Si side of the bond would have a net positive charge.
(b) In an FCC unit cell, there are four atoms. Since SiC is FCC with a 2-atom basis, i.e. a
pair of Si-C is placed with the Si atoms at each corner and at each face. There will be 4Si
and 4C in each SiC unit cell.
ρ=
mass
volume
(4 atoms)(12.01g/mol)(
=
=
1mol
1mol
) + (4atoms)(28.09g/mol)(
)
23
6.022 "10 atoms
6.022 "1023 atoms
(0.43596 "10!7 cm)3
2.664 "10!22 g
= 3.215g/cm3
!23
3
8.286 "10 cm
A handbook value for ρ ofβSiC is 3.208 g/cm3.
8. For a ceramic compound, what are the two characteristics of the component ions that
determine the crystal structure.
The two characteristics of component ions that determine the crystal structure of a ceramic
compound are: 1) the magnitude of the electrical charge on each ion, and 2) the relative
sizes of the cations and anions.
9. Iron oxide (FeO) has the rock salt crystal structure and a density of 5.70g/cm3.
(a) Determine the unit cell edge length.
(b) How does this result compare with the edge length as determined from the radii in Table 12.3,
assuming that the Fe+ and O2- ions just touch each other along the edges?
The rock salt structure can be constructed from the FCC unit cell with a 2-atom basis. One atom is
1
at (0,0,0), and the other is at (0,0, ). In the case of FeO, the unit cell contains 4 Fe atoms and 40
2
atoms.
M
mass of the 8 atoms
(a) ρ= =
V
volume of the unit cell
5.70g/cm3 =
(4)(55.85 g / mol + 16.00 g / mol )(
a3 =
a3
1mol
atoms )
6.022 !1023
4.773 "10!22 g
5.70 g / cm3
a = 4.375 "10!8 m
a = 0.4375 nm
(b)
2+
r Fe ionic = 0.077 nm (The cation )
rO
2!
ionic
= 0.140 nm (The anion)
Assuming the cell edge is close-packed (The [100] direction)
a
FeO
o
!2
Fe +2
o
!2
aFeO = (2)(0.140nm) + (2)(0.077nm) = 0.434 nm
In good agreement.