E344 2013 Summer
Solution Set 2
1. Silicon Carbide (SiC) is a ceramic material that is often referred to as a wide-bandgap semiconductor.
It is a polymorphic material that can be found in a large number of different crystallographic structures.
The bandgap of the so-called 3C form of SiC is 2.2 eV whereas the bandgap of the so-called 6H form of
SiC is 3.0 eV.
A) Draw a well-labeled electron energy-band diagram characteristic of solid 3C-SiC and another
characteristic of 6H-SiC.
B) Calculate the range of visible light wavelengths that can be transmitted by 6H-SiC but not by 3C-SiC.
Show your work.
Solution
A)
B)
2. 3.5 Show that the atomic packing factor for BCC is 0.68.
Solution
The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or
APF =
VS
VC
Since there are two spheres associated with each unit cell for BCC

4R3  8R3
VS = 2(sphere volume) = 2 
 3 
=
3


Also, the unit cell has cubic symmetry, that is VC = a3. But a depends on R according to Equation 3.3, and

3
4R 
64 R3
VC =   =
 3 
3 3
Thus,

APF =

VS
VC
=
8 R3 /3
64 R3 /3 3
= 0.68
3. Using information about the crystal structure, the lattice parameter, and the atomic mass,
determine the theoretical density of gold (Au). Show your work.
Solution
Gold has a FCC crystal structure. Therefore, in a unit cell there are 4 lattice points (Au atoms). The lattice
constant a = 0.40788 nm. Vunitcell = a3 = 0.067857 nm3.
Theoretical Density = 4×MAu/NA/Vunitcell = 4× 196.967g·mole-1/6.02×1023/0.067857nm3
= 1.9287×10-20g/nm3 =19.287g/cm3, which is very close to the 19.30g/cm3 density given in wiki.
4. 3.16 Iodine has an orthorhombic unit cell for which the a, b, and c lattice parameters are 0.479, 0.725, and
0.978 nm, respectively.
(a) If the atomic packing factor and atomic radius are 0.547 and 0.177 nm, respectively, determine the
number of atoms in each unit cell.
(b) The atomic weight of iodine is 126.91 g/mol; compute its theoretical density.
Solution
(a) For indium, and from the definition of the APF
4

n   R3 
V
3

APF = S =
VC
abc
we may solve for the number of atoms per unit cell, n, as

n =
(APF) abc
4
 R3
3
Incorporating values of the above parameters provided in the problem state leads to

=

(0.547)(4.79  10-8 cm)(7.25  10-8 cm) (9.78  10-8 cm)
4
 (1.77  10-8 cm) 3
3
= 8.0 atoms/unit cell
(b) In order to compute the density, we just employ Equation 3.5 as
 =
=

nAI
abc N A
(8 atoms/unit cell)(126.91 g/mol)
(4.79  10-8 cm)(7.25  10-8 cm) (9.78  10-8 cm)/ unit cell(6.022  1023 atoms/mol)
= 4.96 g/cm3
5. One way to describe a crystal structure is to use a simple unit cell with a multi-atom basis. The "basis"
represents the set of atoms that is translated to each point in a unit cell to create the crystal structure. For
example, silver (Ag) is FCC with a one-atom basis. That one atom would have coordinates of (0,0,0).
Similarly, silicon (Si) has the diamond-cubic crystal structure, which can be described as FCC with a twoatom basis. The basis would consists of a Si atom at (0,0,0) and another Si atom at (¼, ¼, ¼ ). If these
two atoms are translated to all of the corner and face positions of the FCC unit cell, the diamond cubic
lattice is created. The rock salt structure (see fig. 12.2 in Callister) is as an FCC unit cell with a two-atom
basis. For NaCl, define the two atoms and their positions in this basis. Sketch the (100) face of the NaCl
unit cell and show that the Na and Cl positions can be generated by moving the two-atom basis to each of
the FCC positions on this face.
Solution
Solution
Cl (0,0,0)
Na(1/2,1/2,1/2), or any one of the positions that have Na atoms in the unit cell.
The (100) face is labeled in white dotted line. Blue dots are Cl, green dots are Na. Movements of Cl /Na are labeled
with red/lightblue arrows, respectively.
6. 3.30 Within a cubic unit cell, sketch the following directions:
(c)
[01 2],
(e)
[1 1 1] ,
Solution

7. 3.31 Determine the indices for the directions shown in the following cubic unit cell:
Solution
Direction C is a [112] direction, which determination is summarized as follows. We first of all position the origin of
the coordinate system at the tail of the direction vector; then in terms of this new coordinate system
Projections
Projections in terms of a, b, and c
Reduction to integers

Enclosure

x
y
a
2
1
b
2
1
2
2
1

1

[112]
z
c
1
2
Direction D is a [112 ] direction, which determination is summarized as follows. We first of all position
the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system

Projections
Projections in terms of a, b, and c
Reduction to integers

Enclosure

x
y
z
a
2
1
b
2
1
–c
2
2
1



1
[112 ]
–1
–2
8. 3.39 What are the indices for the two planes drawn in the sketch below?
Solution
Plane 1 is a (020) plane. The determination of its indices is summarized below.
x
y
z
Intercepts
a
b/2
c
Intercepts in terms of a, b, and c

1/2

Reciprocals of intercepts
0
2
0
Enclosure
(020)
Plane 2 is a (221) plane, as summarized below.
x
y
z
Intercepts
a/2
-b/2
c
Intercepts in terms of a, b, and c
1/2
-1/2
1
2
-2
1

Reciprocals of intercepts
(221)
Enclosure

9. 3.40 Sketch within a cubic unit cell the following planes:
(a) (01 1 ) ,
(e) (1 11 ) ,
Solution
 are plotted in the cubic unit cells shown below.
Theplanes called for
10. 4.10 What is the composition, in atom percent, of an alloy that contains 98 g tin and 65 g of lead?
Solution
The concentration of an element in an alloy, in atom percent, may be computed using Equation 4.5.
However, it first becomes necessary to compute the number of moles of both Sn and Pb, using Equation 4.4. Thus,
the number of moles of Sn is just
nmSn =
Likewise, for Pb
'
mSn
=
ASn
98 g
= 0.826 mol
118.71 g / mol

nmPb =
65 g
= 0.314 mol
207.2 g / mol
Now, use of Equation 4.5 yields

' =
CSn
nmSn
nmSn  nmPb
 100
0.826 mol
=
 100 = 72.5 at%
 0.826 mol  0.314 mol
Also,

' = =
CPb

0.314 mol
 100 = 27.5 at%
0.826 mol  0.314 mol
11. 4.29 (a) For a given material, would you expect the surface energy to be greater than, the same as, or
less than the grain boundary energy? Why?
Solution
(a) The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms
on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore,
there will be fewer unsatisfied bonds along a grain boundary.
12. 12.7 Compute the atomic packing factor for the cesium chloride crystal structure in which rC/rA = 0.732.
Solution
This problem asks that we compute the atomic packing factor for the cesium chloride crystal structure
when rC/rA = 0.732. From Equation 3.2
APF =
VS
VC
With regard to the sphere volume, VS, there are 1 cation and 1 anion spheres per unit cell. Thus,

4

4

VS = (1) rA3  + (1) rC3 
3

3

But, since rC/rA = 0.732


4
VS = rA3 1  (0.732) 3 = (5.832) rA3
3
Now, for rC/rA = 0.732, the corner anions in Table 12.2 just touch one another along the diagonal of the cubic unit
cell such that
VC = a 3 = [
(2rA  2rc ) 3
] 
3
 8


(rA  0.732rA )3  8rA3
3 3

Thus
APF =
VS (5.832) rA3
=
= 0.729
VC
8rA3
13. 13.10 What is the distinction between glass transition temperature and melting temperature?
Solution
The glass transition temperature is, for a noncrystalline ceramic, that temperature at which there is a change
of slope for the specific volume versus temperature curve (Figure 13.6).
The melting temperature is, for a crystalline material and upon cooling, that temperature at which there is a
sudden and discontinuous decrease in the specific-volume-versus-temperature curve.
14. 13.13 The viscosity η of a glass varies with temperature according to the relationship
Q 
  A exp vis 
 RT 
where Qvis is the energy of activation for viscous flow, A is a temperature-independent constant, and R and T are,

respectively, the gas constant and the absolute temperature. A plot of ln η versus l/T should be nearly linear, and
with a slope of Qvis/R. Using the data in Figure 13.7, (a) make such a plot for the borosilicate glass, and (b)
determine the activation energy between temperatures of 500 and 900°C.
Solution
(a) Below is shown the logarithm viscosity versus reciprocal of temperature plot for the borosilicate glass,
using the data in Figure 13.7.
The dashed line has been drawn through the data points corresponding to
temperatures between 500 and 900ºC (as stipulated in the problem statement).
(b) The activation energy, Qvis, may be computed according to
Qvis
where R is the gas constant, and









 ln 
ln   ln 2

= R
 R  1
1 
 1  1 
 



 T1 T2 
 T 

 ln 
is the slope of the dashed line that has been constructed. Taking 1/T1 and
1 
  
T 
1/T2 as 0.8  10-3 and 1.3  10-3 K-1, respectively, then the corresponding values of ln 1 and ln 2 are 10.59 and
32.50. Therefore,

Qvis







ln   ln 2
10.59  32.50
 (8.31 J/mol K) 
 R  1

 1  1 
0.8  103 K1  1.3  103 K1 


 T1 T2 
= 364,000 J/mol
Solution
A)
B)
C) The viscosities of Du2O3 and Du2O3-10% Sw2O3 at a given temperature will be very similar, because
Sw will not create non-bridging oxygens and it will simply substitute for the Du.
16. E
17. C
18. D
19. C
20. E
21. B