# 3.31 Determine the indices for the directions shown in the following ```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  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

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
(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 = 
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 = , C =
 ! (100)
b)
Z
uv
A

(111)
uv
C
uv
B
Y
X
uv
Suppose vector A = 
Choose two vectors within plane (111) :
uv
A
uv
A
uv
B = [111 ]
uv
B = 
! 0 11 &quot; = 0 +1+ (-1) = 0
#
\$
! 11 0 &quot; = (%1) + 1 + 0 = 0
#
\$
!
uv
uv
B = !#0 11&quot;\$ , C = !# 11 0 &quot;\$
[ 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&quot;34 J &quot; s )(
)(3 #108 m / s )
1.602 #10&quot;19
= 7.3 &micro; m
0.17eV
7. Crystallographic properties of cubic SiC
(a) Percent ionicity =1 ! exp[!(0.25)( x A ! xB ) 2 ] &quot;100
xSi = 1.8
xC = 2.5
! % Ionicity = 1 ! exp[!(0.25)(1.8 ! 2.5) 2 ] &quot; 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 &quot;10 atoms
6.022 &quot;1023 atoms
(0.43596 &quot;10!7 cm)3
2.664 &quot;10!22 g
= 3.215g/cm3
!23
3
8.286 &quot;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 &quot;10!22 g
5.70 g / cm3
a = 4.375 &quot;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  direction)
a
FeO
o
!2
Fe +2
o
!2
aFeO = (2)(0.140nm) + (2)(0.077nm) = 0.434 nm
In good agreement.
```