MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2007
Midterm Examination - Solutions
Thursday, February 15, 8:00 – 9:15 AM
Place all answers in a 8.5" x 11" Bluebook
Allowed: 1, double-sided 8.5" x 11" page of notes – must return with exam.
Required: Picture ID when returning the completed exam.
Note: 5 questions (front and back of this page)
1. Miller Indices (20 points)
(a) Draw the (1 1 2) plane in a cubic unit cell. Clearly label the axes.
(b) Draw two directions in the (1 1 2) plane, which are separated by 90°. Properly label the
directions.
Solutions
a) plane (1-12) is outlined
b) Any two directions which satisfy:
a. The scalar product of the two directions is 0
b. The scalar product of each direction with the normal to the plane (1 -1 2) is 0
Z
[-1 1 1]
(1 -1 2)
Y
X
[1 1 0]
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2007
2. Atomic Bonding (15 points)
(a) Calculate the atomic packing factor for SC (simple cubic)
(b) Calculate the metallic radius of a polonium (Po) atom given that Po has a SC structure, a unit
cell dimension of 3.352 Å, and an atomic weight of 209 g/mol. Calculate the density of Po.
Solutions
a.) Atomic Packing Factor for Simple Cubic
Atoms contact along unit cell edges.
2R atoms along edge length a.
Volume of Atoms 4/3·π·R3
1 Atom per unit cell
Unit Cell volume a3=8R3
Packing Factor = (4/3·π·R3)/(8R3) = π/6 = 0.523%
b.) Metallic Radius of Polonium:
Unit Cell 3.352 Å` Æ radius = 1.676 Å
Density: 209 g/mol
209 (g/mol) / NV (mol/atom) = 3.471 × 10-22 g/atom
1 atom per unit cell Æ 3.471 × 10-22 g/unit cell
Unit Cell Volume = 3.3523=37.6627 Å3
(3.471 × 10-22 ) / (37.6627 x 10-24) = 9.216 g/cm3
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2007
3. Point Defects (25 points)
a. Calculate the equilibrium number of vacancies per cubic centimeter (cm3) in aluminum
(FCC packing) at 600 °C and at 298 °C. The vacancy formation energy in Al is 0.625
eV/atom.
b. Make a simple estimate of the average spacing between vacancies at 298 °C
c. Calculate the number of interstitials per cubic centimeter (cm3) in aluminum (FCC
packing) at 600 °C and at 298 °C. The interstitial formation energy in Al is 1.58 eV/atom.
d. Make a simple estimate of the average spacing between interstitials at 298 °C
The density and AW for aluminum are 2.70 g/cm3 and 26.98 g/mol, respectively.
Note: Boltzmann’s constant = 8.6174x10-5 eV K-1
Note: Avogrado’s # = 6.0221 x 1023/mol
Solutions
a.)
N sites
g ⎞⎛
atoms ⎞
⎛
⎜ 2.7 3 ⎟⎜ 6.022e23
⎟
ρN A ⎝ cm ⎠⎝
mole ⎠
=
=
g
Mw
26.98
mole
atoms
N sites = 6.026 e22
cm 3
⎛ −Q ⎞
⎟⎟
N defects = N sites exp⎜⎜
⎝ k BT ⎠
⎛
⎞
− 0.625
⎟⎟
N defects = 6.026e22 exp⎜⎜
⎝ 8.6174e − 5 * (298 + 273) ⎠
N defects (298°C ) = 1.84e17
vacancies
cm 3
⎛
⎞
− 0.625
⎟⎟
N defects = 6.026e22 exp⎜⎜
⎝ 8.6174e − 5 * (600 + 273) ⎠
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
N defects (600°C ) = 1.49e19
vacancies
cm 3
b.)
⎛ 1
D=⎜
⎜N
⎝ defects
⎞
⎟
⎟
⎠
1/ 3
⎛ 1 ⎞
D=⎜
⎟
⎝ 1.84e17 ⎠
D = 1.76e − 6cm
D = 176nm
1/ 3
WINTER 2007
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
c.)
N defects
⎛ −Q ⎞
⎟⎟
N defects = N sites exp⎜⎜
k
T
⎝ B ⎠
⎛
⎞
− 1.58
⎟⎟
= 6.026e22 exp⎜⎜
⎝ 8.6174e − 5 * (298 + 273) ⎠
N defects (298°C ) = 6.90e8
defects
cm 3
⎛
⎞
− 1.58
⎟⎟
N defects = 6.026e22 exp⎜⎜
⎝ 8.6174e − 5 * (600 + 273) ⎠
N defects (600°C ) = 4.59e13
d.)
⎞
⎟
⎟
⎠
1/ 3
⎛ 1 ⎞
D=⎜
⎟
⎝ 6.90e8 ⎠
1/ 3
⎛ 1
D=⎜
⎜N
⎝ defects
D = 1.13e − 3cm
D = 11.3µm
defects
cm 3
WINTER 2007
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2007
4. Polymers (20 points)
The weight average molecular weight of polystyrene (PS) is 600,000 g/mol.
(a) Compute the weight average degree of polymerization. See the mer unit for PS below:
Note AWs: H: 1 g/mol; C: 12.01 g/mol; U: 238.029 g/mol
(b) Do you expect PS to be a thermoplastic polymer or a thermosetting polymer?
Solutions
(a)
For each polystyrene unit there are 2 carbon atoms, 3 hydrogen atoms and 1 benzene Ring (or
equivalently, 5 more hydrogen atoms and 6 more carbon atoms). Therefore the mer molecular
weight is given as
+5
1
2
= 8 (12.01 g/mol ) + 8 (1.0 g/mol )
m = 8 ( AC ) + 8 ( AH )
= 104.08 g/mol
Now the degree of polymerization (DP) can be computed from Equation 14.6,
+5
DP =
M w 600, 000 g/mol
+5
=
= 5764.8 ≈ 5800
m
104.08 g/mol
Where the weight average molecular weight M w was specified in the question.
(b)
Based on the weight average molecular weight we would expect polystyrene to be a
thermoplastic this is because most often such materials are high molecular weight polymers.
Moreover, polystyrene is also a relatively soft material and forms relatively linear chains with
minimal branching, which make it quite flexible.
+5
MATERIALS 101
INTRODUCTION TO STRUCTURE AND PROPERTIES
WINTER 2007
5. Mechanical Properties (20 points)
A cylindrical specimen of a hypothetical metal alloy is stressed in compression (what is the
sign of the stress?). This material, Goletum, has a Young’s modulus of 100 GPa and a shear
modulus of 40 GPa. If the original and final diameters are 100.0 mm and 100.10 mm,
respectively, and its final length is 215.5 mm, compute the following:
Compressive stress
Poisson’s ratio
Bulk modulus
Original length (assuming purely elastic deformation)
Solution