MSE 200A
Survey of Materials Science
Fall, 2008
Problem Set No. 1
Problem 1:
In the diamond modification of solid carbon each atom has exactly four nearest
neighbors that are configured in a tetrahedral coordination around it. In the graphite
modification of carbon the carbon atoms form planar sheets that are relatively widely
separated from one another. Each carbon atom is bonded to three near neighbors in the
plane, and has a weak interaction with atoms in adjacent planes.
(a) Diamond is an electrical insulator and is one of the hardest substances known.
Interpret these properties in terms of its bonding.
Since each carbon atom has four carbon neighbors, its four sp3 hybrid bonds are
saturated. All electron states in the bonds are filled, so there can be no net electron motion
from bond to bond. The only mechanism available to conduct electricity is the excitation of
valence electrons into excited orbitals. However, this requires a significant activation energy and is, consequently, a rare event at all normal temperatures. The conductivity is very
low; diamond is an insulator.
The hardness of diamond is largely due to the rigidity of its bonding. In order for
the carbon to deform its atoms must move with respect to one another. The local distortion
of the atom configuration requires some reconfiguration of the valence electron distribution.
However, these electrons are in very strong, saturated bonds that are difficult to distort.
(b) Graphite is an anisotropic conductor. It has high conductivity in the plane of
the sheet, and very low conductivity perpendicular to the sheet plane. Explain this
behavior in terms of a simple chemical bonding model.
The structure of graphite is one in which carbon atoms form sheets in which the
atoms are hexagonally configured as shown in Fig. 1.1. The sheets are stacked in a configuration in which successive sheets are weakly bonded to one another. Assuming that the
valence bonding of carbon is in the plane of the sheet only, each carbon atom has three
neighbors, and hence its bonds are unsaturated. Graphite is a good conductor in the plane
of its sheet.
On the other hand, graphite is a very poor conductor in the direction perpendicular
to the sheet. Since the sheets are relatively far apart the bonds between them are nearly of
the Van der Waals (polarization) type, and do not permit electrical conduction.
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MSE200A: Fall, 2008
Problem Set No. 1
Fig. 1.1: Atom configuration in a plane of graphite.
(c) In its usual microstructure, small three-dimensional particles made of stacked
planar sheets, graphite deforms easily, and is used in pencil lead and lubricants. Given
that the carbon sheets themselves are very rigid, explain why graphite is so easily
deformed.
The flakes of graphite slide easily over one another, much like cards in a deck of
playing cards, so that small particles that are stacks of planar sheets are relatively easy to
deform.
(d) Graphite fiber is made by modifying the microstructure into long, thin ribbons
in which the carbon sheets parallel the plane of the ribbon. The ribbons are then rolled,
somewhat as one would roll a newspaper, to form the graphite fiber. While normal
graphite is weak, graphite fiber is extremely strong when pulled along the fiber axis.
Explain in terms of the microstructure and the bonding in graphite.
When graphite sheets are rolled into tubes to create fibers they are very strong when
pulled along the fiber axis. To deform a rolled sheet one must stretch the bonds in the sheet
or reconfigure the atoms. Since the carbon atoms in graphite are tightly bound in the sheet
plane, this is very difficult to do.
Problem 2:
Let two atoms bond together so that the potential energy of the pair is described
by equation 2.1 of the notes.
(a) Find the equilibrium separation of the atoms, r0, in terms of the constants A,
B, m and n.
From eq. 2.1 the potential energy of the bond is
A B
Ï(r) = m - n
r
r
2.1
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MSE200A: Fall, 2008
Problem Set No. 1
The equilibrium separation, r0, is at the minimum of the potential, Ï(r). The condition
dÏ
dr = 0
(at r = r0)
2.2
yields the result
1
mA
r0 =  nB m-n
2.3
(b) Show that the molecule behaves like a simple spring for small displacements,
∂r = r - r0, away from the equilibrium separation; that is, show that the force that acts to
restore the equilibrium separation is given approximately by the relation
F = - K∂r
2.4
Evaluate the spring constant, K.
The force on the molecule is related to the potential by the equation
dÏ
mA
nB
F = - dr = m+1 - n+1
r
r
2.5
It follows that F vanishes when r = r0, is positive when r < r0, and is negative when r > r0.
Hence the molecule behaves as a spring; a displacement away from r0 induces a force that
acts to restore the equilibrium separation.
To find the "spring constant" of the molecule we compute the restoring force for a
small displacement, ∂r = r - r0. A Taylor expansion of the force, F(r), about the point, r0,
gives
dF 
F(∂r) = F(r0) +  dr  (∂r) + ...
r0
2.6
where the subscript on the derivative indicates that it is to be evaluated at r0, and we neglect
higher order terms in the expansion since they depend on (∂r)2 or higher powers of ∂r,
which is taken to be small. Since F(r0) = 0 we have
d2Ï
dF 
F(∂r) =  dr  (∂r) = -  2  (∂r)
dr r0
r0
2.7
The force on a spring that is displaced from equilibrium by the amount, ∂r, is
F = - K ∂r
2.8
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MSE200A: Fall, 2008
Problem Set No. 1
where K is the spring constant. It follows that, for small displacements, the molecule behaves like a spring with a spring constant
d2Ï
K =  2 
dr r0
2.9
Since
d2Ï m(m+1)A n(n+1)B
=
- n+2
dr2
rm+2
r
2.10
the spring constant is found to be, after some algebra,
d2Ï
 1 n+2
K =  2  = r 
[(m-n)B]
dr r0
0
2.11
(This is the simplest expression I can find.) Note that K is positive, as it should be for a
stable molecule.
(c) Find the force that would have to be applied to the molecule to break it in two.
This is called the ultimate tensile strength of the molecule.
If the potential and force are plotted against (r) in the same coordinate frame the plot
appears as shown below. The force is the negative gradient of the potential. When r < r0
the force is positive, when r > r0 it is negative, so the force that develops between the particles always acts to restore their equilibrium configuration. However, as shown in the
diagram, the force has a minimum value that is reached at the separation, rm. For r > rm the
force increases again (decreases in magnitude) and asymptotes to zero at large interatomic
distances.
Let an external force, Fa, be applied to the molecule, and let the force be positive
(tensile) so that it tends to separate the atoms. The equilibrium separation in the presence of
the force, Fa, is the value of r at which
F(r) = - Fa
2.12
so that the total force is zero. Now suppose that Fa is gradually increased. The atoms separate further until Fa exceeds - F(rm). Since this is the minimum value of F(r), it is not
possible to find a solution to equation 2.12 when Fa > - F(rm), and the atom flies apart.
Hence the tensile strength of the atom is - F(rm).
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MSE200A: Fall, 2008
Fig. 1.2:
Problem Set No. 1
Plot of the potential, Ï(R), and the force, F(R), required to
separate a binary molecule to the distance, R. The maximum of
this force, F(max), is the ultimate tensile strength of the
molecule.
We find rm from the condition that the derivative of F(r) vanishes there. Then
dF
d2Ï
=
=0
dr
dr2
2.13
m(m+1)A 1
m+1 1


m-n
rm =  n(n+1)B 
= r0  n+1 m-n
2.14
which gives the result
which shows that rm > r0. The tensile strength is
1n+1 mA

F = - F(rm) =  r   m-n - nB
r
1n+1 
n+1 
=  r  nB1 - m+1
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2.15
MSE200A: Fall, 2008
Problem Set No. 1
Problem 3:
(a) Consider a two-dimensional crystal made up of spherical atoms that are
packed together as tightly as possible in the plane. Draw the Bravais lattice for this
crystal, and the primitive cell defined by the unit vectors of the Bravais lattice.
Fig. 1.3 shows the planar distribution of close-packed atoms, and shows the
primitive unit cell that is a parallelogram with atoms at each of its corners. Since each of the
corner atoms is shared by four cells, there is one atom/cell. A unit cell that contains only
one atom is called a primitive cell.
Fig. 1.3: Close packing of atoms with a unit cell in the form of a parallelogram
The Bravais lattice can also be chosen as the hexagonal cell shown in Fig. 1.4, wich reveals
its hexagonal symmetry. As you can show by geometric construction, or can simply
reason from the number of atoms/cell, that there are three cells of the parallelogram type
in each hexagonal unit cell.
Fig. 1.4: Close-packed planar configuration of atoms with hexagonal unit
cell.
(b) Show that the unit cell can be drawn as a hexagon with an atom in the center
and atoms at each of the six corners. How many atoms are there per cell?
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MSE200A: Fall, 2008
Problem Set No. 1
The hexagonl cell is shown in Fig. 1.4. Recalling that the structure is twodimensional, the hexagonal cell has 1 atom in its interior plus 6 corner atoms that are
shared between three cells. Hence the total number of atoms/cell is 3
(c) Consider a two-dimensional crystal that has a hexagonal unit cell with an
empty center and atoms located at the corners of the hexagon. Draw the atom positions
generated by several neighboring cells. Show that the unit cell used in part (b) is a
satisfactory cell for this structure, but the cell used in part (a) is not.
The hexagonal configuration is drawn in Fig. 1.5. It is not an unphysical configuration. The graphite modification of carbon, which is, in fact, the stable form of carbon
at normal temperature and pressure, is a stacking of atom planes that have this configuration. If the cell used in Fig. 1.3 is drawn in the structure shown in Fig. 1.5, it has an
empty corner. It is not an acceptable unit cell since the lattice points it defines do not
identify identical groups of atoms. However, the hexagonal cell drawn in Fig. 1.4 is a
satisfactory choice since all atoms can be located by repeating identical hexagons through
space.
Fig. 1.5: A two-dimensional hexagonal configuration of atoms
(d) Draw a proper set of lattice vectors for the structure in (c). Indicate the basis
vectors that are needed to complete the description of the structure. Given your choice of
lattice vectors, what atom group is repeated at each lattice point?
One correct choice for the unit cell is shown in Fig. 3.10 of the notes, which is
repeated in Fig. 1.6 below. With this choice the repeated atom group is a two-atom arrangement in which the atoms are displaced from the lattice point by the vectors τ 1 and τ 2 .
The lattice point is vacant. [Note that this problem does not have a unique answer. You
can consistently choose a different set or lattice vectors and/or a different origin for the
lattice vectors. In this case the basis vectors will change as well. The critical thing is that
the lattice vectors must generate a periodic lattice of points, each of which has exactly the
same atom group associated with it.]
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MSE200A: Fall, 2008
Problem Set No. 1
a2
a2
a1
a1
Fig. 1.6:
†2
†1
A simple hexagonal arrangement of atoms showing the hexagonal unit cell and location and basis vectors of the two atoms
in the cell.
Problem 4:
(a) Prove that there is a [–101] direction in the (111) plane.
The vector [111] is perpendicular to the (111) plane. It is also perpendicular to the
vector [–101] since
[–101] • [111] = 1 - 1 = 0
4.1
It follows that the [–101] direction is in the (111) plane.
(b) Prove that there is no [101] direction in the (111) plane.
Since
[101] • [111] = 1 + 1 ≠ 0
4.2
the [101] direction does not lie in the (111) plane.
(c) Find the angle at which the (111) and (–111) planes intersect. [Hint: this is a
very difficult problems in geometry unless you remember that the [hkl] direction is a
vector perpendicular to the (hkl) plane and recall that the scalar product of two vectors, a
and b, obeys the relation
a^b = |a||b| cos(œ)
With this relation the problem becomes simple.]
Since
[–111] • [111] = 1 = 3 cos(œ) 4.3
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MSE200A: Fall, 2008
Problem Set No. 1
we have
œ = cos-1 (1/3) = 1.23 radians = 71º
4.4
Problem 5:
(a) Let the lattice sites of the FCC bravais lattice be filled with spherical atoms
that are close-packed in the sense that they just touch one another. What is the radius of
the largest sphere that will fit inside an octahedral void without distorting the structure?
What is the radius of the largest sphere that will fit in a tetrahedral void without
distortion? [Hint: the geometry of the tetrahedral void is much easier to treat if one first
shows that the center of the void lies at the center of a cube that is 1/8 the size of the FCC
unit cell.]
Let the radius of a spherical atom be R and let the edge of the unit cell have length,
a. If the lattice atoms just touch, then they touch along the face diagonal. Hence 2 a =
4R, and a = 2.83R.
The diameter of the octahedral void is equal to the gap between lattice atoms along
the cube edge. Hence d = 0.83R, or r, the radius of the largest atom that will fit in the octahedral hole, is
roct = 0.414R
5.1
To obtain the radius of a tetrahedral void note that it lies at the center of an octet of
the cube, that is, a small cube of edge (a/2) centered on the corner, as shown in Fig. 3.29 of
the notes. An atom in the tetrahedral hole would first touch a lattice atom along the
diagonal of the octet. Hence, letting d' be the length of the diagonal of the octet, d'/2 =
R+rtet. Since
3
d' = 2 a = .866a = 2.45R
5.2
rtet = (1.225 - 1)R = 0.225R
5.3
(b) Let a crystal have the hexagonal close-packed structure, and consist of
spherical atoms that just touch one another. Find the number of octahedral and
tetrahedral voids per atom, and compute the sizes of the largest spheres that will fit in
each type of site without distortion in terms of the radius, R, of the atoms on the HCP
lattice sites. [Hint: assuming that you have solved parts (a) and (b), if you have to
compute anything at all to answerthis question you are approaching it wrong.]
As one can see by considering the stacking of close-packed planes, the coordination
of voids in fcc and hcp are the same. Hence the numbers and sizes of the voids are
identical.
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MSE200A: Fall, 2008
Problem Set No. 1
Problem 6:
(a) Let the lattice sites of the BCC bravais lattice be filled with spherical atoms
that are close-packed in the sense that they just touch one another. What is the radius of
the largest sphere that will fit inside an “octahedral” void without distorting the
structure? What is the radius of the largest sphere that will fit in a “tetrahedral” void
without distortion?
Let a BCC unit cell be made of spherical atoms that are packed as densely as
possible. In that case the atoms touch along the cube diagonal. If the atom radius is R and
the cube edge is a,
a =(4/√3)R
6.1
There is an octahedral void in BCC at the center of each edge. If there is a spherical atom
or radius r in that void, the largest that will fit has
Roct = (a/2) – R
= (2/√3 - 1)R
= 0.158R
6.2
There is a tetrahedral void at positions like (0. 1/2, 1/4) that has neighboring atoms at
(0,0,0) and (0,1,0). The largest sphere that can fit in this void has radius, r, such that (r+R)
is the hypotenuse of a right triangle with sides a/2 and a/4. It follows that
(r+R)2 = (a/2)2 + (a/4)2
= (5/16)a2 = (5/3)R2
from which
rtet = .291R
6.3
(b) The common interstitial solutes in BCC metals, in particular, C and N in Fe,
are found in the “octahedral” interstices rather than the tetrahedral ones. Given your
answer to part (a), how do you explain this?
While the “radius” of the octahedral site is smaller than that of the tetrahedral site, its
volume is larger. In fact, since there are 6 octahedral voids and 12 tetrahedral voids per
unit cell, the free volume available per octahedral void is twice as large (roughly 0.16V
vs. 0.08V, where V is the volume of a lattice atom.)
Since the whole electron cloud of the interstitial atom must be squeezed into the
interstitial hole, the volume of the hole is more important than its shape, and interstitials
prefer the octahedral site in BCC.
Problem 7:
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MSE200A: Fall, 2008
Problem Set No. 1
(a) The perovskite crystal structure, which is a common structure of ferroelectric
oxides like BaTiO3, is usually drawn in a cubic cell with Ti atoms at the corners, Ba at
the center, and O at the centers of each of the edges. Draw the structure and show that it
has the stoichiometric formula BaTiO3.
Ti
Ba
O
Fig. 1.7: The perovskite structure of BaTiO3
The unit cell is drawn in Fig. 1.7. The unit cell contains 1 Ba atom at the cell center,
a net of 1 Ti atom (8 corner atoms, each 1/8 in the cell), and a net of 3 O atoms (12 edge
atoms, each 1/4 in the unit cell). Hence the formula is BaTiO3, and is an electrically neutral
ionic compound if Ti has valence +4, Ba has +2, and O has -2. These are common
valences for these elements.
Note that the structure is such that all the nearest neighbors of an ion of given
charge are ions of opposite charge. Referring to Fig. 3.1, the Ti+4 cations are located at the
corners of the unit cell. Their closest neighbors are the O-2 ions at the centers of the edges
of the cell. The Ba+2 ion is located in the center of the cell. It is closer to the O-2 ions on
the cell edges than to the Ti+4 ions at the cell corners. The closest neighbors of the O-2 ions
are the Ti+4 ions on the corners. The Ba+2 ions in the cell centers are only slightly further
away. O-2 ions are not neighbors of one another. It follows that this is a suitable structure
for an ionic crystal.
(b) To relate this structure to FCC most simply we consider lattice vacancies as a
component. By placing a vacancy at the center of each face of the cube show that
BaTiO3 can be regarded as the compound (Ti 3)(BaO3) where
is a vacancy. The
FCC lattice is filled by Ti and vacancies while the octahedral interstitial voids are filled
by Ba and O. Show that this structure is the NaCl structure if we ignore the difference
between Ti and , and between Ba and O. Show that Ti and
are distributed over the
FCC sites in the Cu3Au pattern, while Ba and O are distributed over the FCC lattice of
octahedral voids in a similar pattern.
The perovskite structure is re-drawn to show the positions of the vacancies in Fig.
1.8. If we assume a structure in which the Ti atoms and vacancies are joined into the
pseudo-atom Ti 3 and the Ba and O are joined into the pseudo-atom BaO3, the structure
is just the NaCl structure, as shown in Fig. 1.9.
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MSE200A: Fall, 2008
Problem Set No. 1
Fig. 1.8: The perovskite structure re-drawn to show the vacant FCC sites.
= Ti
3
= BaO 3
Fig. 1.9: The perovskite cell redrawn as an NaCl lattice of pseudo-atoms.
Fig. 1.10 shows the distribution of Ti atoms and vacancies in a perovskite unit cell
from which the Ba and O atoms have been removed. The distribution is precisely a
Cu3Au ordering over the FCC lattice sites.
= Ti
= vacancy
Fig. 1.10: The Cu3Au distribution of Ti atoms and vacancies in BaTiO3.
(c) Show that the perovskite structure of part (a) can also be understood as a BCC
lattice filled by Ba and Ti with O atoms in one-half of the octahedral voids. Show that
the Ba and Ti atoms are substitutionally ordered in a CsCl pattern.
The possibility of this description should be obvious on inspection of Fig. 1.8. The
octahedral voids on the edges of the BCC cell are filled. Those in the center of the faces are
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MSE200A: Fall, 2008
Problem Set No. 1
vacant. If we simply remove the O atoms from Fig. 1.8 the result is a unit cell of BaTi in
the CsCl ordered structure.
(d) While this representation may seem to be a simpler description of the
structure, in at least one important respect it is not. Show that the O atoms do not have a
simple BCC pattern, even when vacancies are included. [To treat the structure as a set of
BCC sublattices with CsCl order on all of them one must assume that the oxygen is
distributed over three separate sublattices, one associated with each type of octahedral
void (Ox, Oy, and Oz). The mixture of oxygen and vacancies on each of the oxygen
sublattices is ordered in a CsCl pattern. However, this leads to a much more complicated
picture of the structure.]
If we place vacancies in the centers of each face in the perovskite structure shown in
Fig. 1.8, which is equivalent to placing them in the face-centered octahedral voids of the
BCC-like CsCl lattice of Ba and Ti, move the origin to the center of one of the oxygen
atoms, and erase the Ba and Ti atoms, the resulting structure shows the distribution of
oxygen and vacancies with respect to the BCC reference frame, Fig. 1.11.
=O
= vacancy
Fig. 1.11: The distribution of oxygen atoms and vacancies in the perovskite
structure referred to a BCC cell.
This can be seen to be a set of three interpenetrating BCC lattices, each of which
contains oxygen atoms and vacancies in the CsCl pattern. One lattice includes the Ox interstitials in BCC, one the Oy, and one the Oz.
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