Materials 100B Problem Set #1 Solutions
Winter Quarter 2001
Problem 1
The bonds between pairs of atoms in a solid can be
a)
b)
c)
d)
e)
ionic bonds
covalent bonds
metallic bonds
van der Waals (induced dipole-dipole interaction)
hydrogen bonds
Which type(s) of bond has, at long distance, and attractive force that varies as 1/r2
between pairs?
Ans: a
Attractive forces varying with 1/r2 imply a Coulombic interaction between
particles of opposite charge. Coulomb’s Law gives the force of this interaction:
F =k
q1q2
r2
Ionic bonding, where attractive electrostatic interactions are present, is exactly
this sort of interaction.
Which type(s) of bond cannot occur in pure solid elements?
Ans: a and e
Ionic bonding exists because one atom in the pair removes electronic charge from
the other to form an ionic pair. This requires that the two atoms be dissimilar.
In pure solid elements this is not the case. Hydrogen bonding requires a similar
condition, but for different reasons. Hydrogen bonding acts between an atom of
hydrogen and, typically, one of oxygen, nitrogen or sulfur. In pure elemental
solids this cannot be the case.
Which type of bonding occurs between Ne atoms in solid neon?
Ans: d
Elements in group 18 of the periodic table (He, Ne, Ar, etc.), are the noble gases.
These elements are highly inert and stable in elemental form. This rules out
covalent bonding. Because we are considering solid Ne, ionic bonding and
hydrogen bonding are eliminated (these require dissimilar elements). Metallic
bonding is unreasonable because Ne is not a metal. Finally we are left with van
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der Waals interaction, and this accounts for the low melting point of solid neon,
because van der Waals forces are so weak.
Which type(s) of bond is found is solid Si?
Ans: b
Silicon is similar to carbon in that it forms tetrahedral bonding arrangements.
Again, the other bonding types are eliminated. Silicon is not a metal. Pure solids
cannot exhibit hydrogen or ionic bonding. And, van der Waals interactions are
negligible compared to the covalent interactions of Si.
Which type of bond is the weakest (i.e has the lowest spring constant)?
Ans: d
Van der Waals bonding describes a dipolar attraction between uncharged atoms.
This interaction is the result of unsymmetrical electron density about the nucleus,
which causes a momentary dipole (charge separation) in the atom. This dipole
can then interact electrostatically with neighboring atoms of opposite dipole.
This interaction is very weak in comparison to the other bonding types.
Which type(s) of bonds are important in ice crystals? (Skiers take note)
Ans: e
Hydrogen bonding occurs between pairs of atoms consisting of 1) hydrogen, and
2) an electronegative element (e.g. oxygen, nitrogen, or sulfur). Water molecules
bound together in the solid are arranged such that the hydrogen of one molecule
is coordinated with the oxygen of neighboring molecules in a hydrogen bonding
interaction. To note, this accounts for the interesting property of water in which
the liquid-phase is more dense that the solid-phase.
In solids with which type of bonding are valence electrons free to move quickly over
large distances?
Ans: c
Large long-range electron mobility implies conductivity. This is characteristic of
metals and thus metallic bonding is the correct answer.
Which type(s) of bond is highly directional?
Ans: b (hydrogen bonding was also accepted as somewhat directional)
Covalent bonding arises from the overlap of spatially arranged electronic
orbitals, which surround the positive nuclei of the atoms. It is said that electrons
“orbit” the nucleus in these orbitals making covalent bonding highly dependent
upon their orientation. Covalent bonding is thus highly directional in nature.
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Hydrogen bonding can also be considered directional to a smaller extent,
whereas the other bonding interactions are non-directional.
Problem 2
The bonds between two atoms in a solid have the potential energy En versus distance r
curve given by the equation:
  a  6  a 12 
En ( r ) = C −   +   
  r   r  
where C = 1.0 * 10-20 J/bond and a = 0.38 nm
a) Plot the attractive and repulsive terms of this potential energy versus distance
between the nuclei of the atoms.
Total Energy
J
-21
3´ 10
2´ 10-21
1´ 10-21
r m
4.27́ 10-10
-1´ 10-21
-21
-2´ 10
Attractive Energy J
r m
4.27́ 10-10
-5´ 10-21
-1´ 10-20
-1.5´ 10-20
-2´ 10-20
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Repulsive Energy J
2.5́ 10-20
2´ 10-20
1.5́ 10-20
1´ 10-20
5´ 10-21
r m
4.27́ 10-10
b) Which of the terms in En(r) is dominant at large distances? Which of the terms is
dominant at short distances?
The attractive term,
a
Eattractive ( r ) = −C  
r
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is dominant at long distances, whereas the repulsive term,
a
E repulsive ( r ) = C  
r
12
is dominant at short distances.
c) Find the equilibrium distance ro between the atoms if no force is applied to the bond.
Ans: 4.27 *10-10 m
From observation of the En(r) versus r plot we observe a minimum in the energy
at a unique value of r. We need to determine the value of this distance such that
the bond energy is minimized. This value of r is the equilibrium bond distance.
This value can be directly obtained from the plot, but a more meaningful
mathematical solution is available. Notice, that the slope at the energy minimum
is zero. This allows us to compute the first derivative of the energy function, set it
to zero, and solve for r.
a
dE n C 
= − 12
dr
ro 
 ro

a
2
 ro
6

 = 1

ro = a ( 2)
4
12
6

a 
 + 6   = 0
 r  

1
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d) Find the energy that must be supplied to pull the two atoms from ro to a very large
distance apart. This energy is called the binding energy, Eb.
Ans: 2.5*10-21 J/bond
From part (c) we can compute the bond energy at ro. We then compute the energy
of the bond at a very large distance, say infinity. The difference between the
energy at infinity and the energy at equilibrium provides us the binding energy.
Eb = En (∞) − E n ( ro ) = 0 + 2.50 * 10 −21
e) Compare the binding energy with thermal energy kBT at room temperature, where kB
is Boltzmann’s constant. If kBT >>Eb the material will not be a solid or even a liquid
at room temperature (thermal agitation will break the bond). At what temperature
will the thermal energy be equal to Eb?
Ans: kBT = 4.12*10-21 J (room temp, 298 K)
kBT= Eb when T = 181 K
Thermal energy is always given by Boltzmann’s constant multiplied by the
temperature. Comparison of the binding energy, Eb, from part (e) with the
thermal energy at room temperature (298 K), shows that the energy from thermal
agitation is not significantly greater than the binding energy of the solid. This
implies that the solid will remain in a condensed state at 298 K (solid or liquid).
f) Determine the force f versus distance curve for the bond and plot it. What is the
“spring constant” of the bond?
Ans: S = 0.97 N/m
From eq. 4.6 in Ashby and Jones, we recognize that the force between bonded
atoms is given by the first derivative of the energy with respect to r.
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6
dE n C 
a
a 
= − 12  + 6  
F=
dr
r 
r
 r  
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Force N
2´ 10-11
1´ 10-11
r m
4.27́ 10-10
-1´ 10-11
-2´ 10-11
-3´ 10-11
Then according to eq. 4.8 in Ashby and Jones, we recognize that the spring
constant, S, is a constant when the bond stretching is small, and we can evaluate
the equation at the equilibrium distance, ro.
d 2 En
S=
=C
dr 2
156a 12 42a 6 
 14 − 8 
r 
 r
g) If no thermal energy is available (T = 0 K), what is the maximum force the bond can
support?
Ans: 1.58 *10-11 N
In order to solve this problem, we must first calculate the maximum bonding
distance, rmax. This will give us the distance at which the restoring force is the
greatest; anything beyond this results in bond rupture. This is carried out by
setting the derivative of the force equal to zero and solving for r.
156a12 42a 6 
d 2 En
=
C
 14 − 8  = 0
dr 2
r 
 r
rmax = 4.73 *10-10 m
This value is then substituted into our expression for force, and a final answer is
obtained.
h) The strain in simple tension is usually defined as (L-Lo)/Lo where Lo is the initial
length and L is the final length along the axis of tension. Apply this definition to the
bond under tension and determine the tensile strain to break the bond.
Ans: 0.11 (unitless)
This problem simply requires substitution of ro for Lo and rmax (computed in part
(g)) for L.
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Problem 3
Nickel metal has a face centered cubic crystal structure and an atomic radius of 0.1246
nm. Compute the volume of its fcc unit cell.
Ans: 43.8 angstroms3= 4.38*10-23 cm3
For the fcc crystal structure we find that the atoms make contact along the cube
face diagonal allowing us to relate atomic radius to edge length.
Then, using the rule of right triangles:
a 2 + a 2 = ( 4r ) 2
a = 2r 2
V = a 3 = ( 2r 2 ) 3
Substitution or r (0.1246 nm) yields the final result.
Problem 4
The unit cell of an elemental solid has the following structure (all angles of the corners
are 90o):
a) To what crystal system does a crystal with this unit cell belong?
Ans: body centered tetragonal (bct)
See viewgraph 2 from Dr. Kramer’s 3rd lecture (Crystal Structure Atomic
Packing, Fri. Jan 12th). This is a table of the 14 Bravais point lattices.
b) If the atomic weight of the material is 114 g/mol, compute the density of the solid
with this unit cell.
Ans: 10.52 g/cm3
This problem requires conversion of the atomic weight to material density. This
is done when we recognize Avogadros number (6.022*1023 atoms/mol), and the
fact that the unit cell has a total of 2 atoms (1 in the middle, and 1/8 * 8 for the
corners). We also know the cell dimensions such that we can calculate the unit
cell volume (0.3*0.3 *0.4 = 0.036 nm3). The conversion is then carried out as
follows.
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  2 atoms   unit cell   (107 ) 3 nm 3 
mol
 114 g  

 * 
 * 
*

 * 
23
3  
cm 3
 mol   6.022 * 10 atoms   unit cell   0.036 nm  

Problem 5
Atomic radii for elements can be determined by assuming that the atoms touch along the
closest packed directions in the unit cell of their crystal structures, e.g. along the face
diagonal in the face centered cubic (fcc) unit cell and along the body (cube) diagonal in
the body centered (bcc) unit cell. Rhodium has an atomic radius of 0.1345 nm and a
density of 12.41 g/cm3. Determine whether rhodium has a fcc or bcc crystal structure.
Ans: the crystal structure is fcc.
We first need to recognize the relationship between the atomic radii and the unit
cell edge length.
We then need to know the atomic weight of rhodium (102.91 g/mol). Also needed
is the number of atoms per unit cell. For the fcc, this is 4 (8*1/8 for the corners,
and 6*1/2 for the faces) and this is 2 for the bcc. From the diagrams we observe
the following:
(fcc)
(bcc)
a = 2r 2
a=
4r
3
Volumes of the unit cells are then simply given by a3, where r = 0.1345 nm. The
conversion is similar to that of problem 4b, and a comparison of the results with
that of the known rhodium unit cell density (12.41 g/cm3) yields the crystal lattice
type.
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Problem 6
Draw the unit cell of the fcc structure. How many close packed planes with different
orientations are there in the unit cell? (Don’t count planes that are parallel to one
another.) Sketch them on the unit cell surface.
Ans: 4 (the <111> family)
It may not seem immediately intuitive, but slicing
the structure along the (111) direction gives the
close packed planes. These can be drawn by
connecting a line from the upper corner point to
each of the two lower corners which makeup the
face diagonal. Essentially you are slicing off a
corner of the cube. Because we are not counting
parallel planes, it becomes apparent that there are
4 unique close packed planes in the structure.
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