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ENSC 3313, Spring 2003
TEST 1  KEY
Name_________________________
Answer all 10 questions, show all work, 2points each
1.
Radio carbon dating is based on the principle that the naturally occurring C
14
isotopes in materials such as
wood will decay exponentially after the organism dies.
Calculate the age of a piece of wood that has 10% of its
C
14
remaining.
The halflife of C
14
is 5730 years.
First need to figure out the decay constant.
Use the halflife for this.
τ
/
t
o
e
m
m
−
=
t
m
m
o
−
=
ln
()
years
8267
5730
5
.
0
ln
=
⇒
−
=
Now solve for the age:
years
t
t
035
,
19
10
.
0
ln
8267
=
−
=
2.
Calculate the surface energy (i.e, number of unbonded atoms per unit area) on the (001) face for single
crystal, FCC aluminum.
Give your answer in J/m
2
.
The radius of an aluminum atom is R
Al
= 0.143 nm and the
AlAl bond energy is 2.21x10
19
J/bond.
Hint: First, calculate the planar density of atoms on the (001) plane.
The side length of an FCC unit cell is
Al
R
a
2
2
=
The area of the (001) face of the unit cell is
2
19
2
9
2
2
10
64
.
1
)
10
143
.
0
(
8
8
m
x
m
x
R
a
Al
−
−
=
=
=
The number of broken bond on this (001) face is 2 per unit cell, therefore
2
2
19
19
/
7
.
2
10
64
.
1
)
10
21
.
2
(
2
m
J
m
x
J
x
E
surface
=
=
−
−
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3.
Identify (with integers) the lattice plane and direction drawn in the sketch.
Choose the origin as shown.
The direction is [
10
1]
.
The intersection of the plane with the xaxis is 1 by
extrapolation, with the yaxis is 1, and with the zaxis is –1/2.
Therefore, the plane is (
2
11 ).
4.
Calculate the atomic packing factor and density (g/cm
3
) of FCC gold.
R
Au
=0.144 nm, Atomic wt. = 196.97 amu, and Avagadro’s number N
A
= 6.023x10
23
The atomic packing factor (APF) is given by:
cell
unit
atoms
V
V
APF
=
.
()
74
.
0
2
3
,
2
2
,
3
4
4
3
3
=
=
∴
=
=
π
APF
R
V
R
V
Au
cell
unit
Au
atoms
Density:
=
⇒
⋅
=
=
ρ
3
7
23
)
10
1
1
144
.
0
2
2
(
/
10
023
.
6
)
/
97
.
196
(
4
nm
x
cm
nm
mol
atoms
x
mol
g
atoms
V
m
cell
unit
atoms
19.36 g/cm
3
5.
What is the weight % carbon in an ironcarbon alloy with 1 atom% carbon?
The atomic weight of carbon is
12 g/mol and of iron is 56 g/mol.
Assume 100 moles of atoms total
Total moles of carbon is 1 mole,
Total weight of carbon is 12 g
Total moles of iron is 99 moles,
Total weight of iron is 99*56 = 5346 g
Weight % carbon = 12/(12+5346) x100% = 0.22%
x
y
z
1/2
1/2
origin
3
6.
From the data below, predict the relative solubility of Ni, Re, and Cr in Co.
(i.e., which are more soluble and
why?)
Radius (nm)
Crys. Type
Elec. Neg.
Val.
Co
0.125
HCP
1.8
2+
Ni
0.1246
FCC
1.8
2+
Cr
0.125
BCC
1.6
3+
Re
0.137
HCP
1.9
2+
The radius and crystal type favor neither Ni nor Cr over the other.
However, the electronegativity and valence
state both favor Ni over Cr.
Therefore, Ni should be more soluble than Cr in Co.
Rhenium is within 10% in size of cobalt and all other aspects match well, so (probably) Re has the highest
solubility in Co.
7.
List 4 types of crystal defects based on their dimension (i.e., 0D, 1D, 2D, 3D) giving an example of each.
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This note was uploaded on 01/10/2012 for the course ENSC 3313 taught by Professor Staff during the Fall '11 term at Oklahoma State.
 Fall '11
 staff

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