MSE280: Introduction to Engineering Materials
Homework
4: Solutions
1. Problem 5.2 from Callister
(8 points)
This problem calls for the computation of the activation energy for vacancy formation in silver.
Upon examination of Equation 5.1, all parameters besides
Q
v
are given except
N
, the total
number of atomic sites.
However,
N
is related to the density, (
),
Avogadro's number (
N
A
), and
the atomic weight (
A
) according to Equation 5.2 as
N
=
N
A
ρ
Pb
A
Pb
=
(
6.02
×
10
23
atoms /mol
)(
9.5 g /cm
3
)
107.9 g /mol
= 5.30
×
10
22
atoms/cm
3
= 5.30
×
10
28
atoms/m
3
Now, taking natural logarithms of both sides of Equation 5.1,
ln
N
v
= ln
N
−
Q
v
kT
and, after some algebraic manipulation
Q
v
=
−
kT
ln
N
v
N
Now, inserting values for the parameters given in the problem statement leads to
Q
v
=
−
(
8.62
×
10
5
eV/atom K
)
(800
°
C + 273 K) ln
3.60
×
10
23
m
−
3
5.30
×
10
28
m
−
3
= 1.10 eV/atom
Problem 5.5 from Callister
In this problem we are asked to cite which of the elements listed form with Ni the three
possible solid solution types.
For complete substitutional solubility the following criteria must
be met:
1) the difference in atomic radii between Ni and the other element (
R
%
) must be less
(8 points)
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than ±15%, 2) the crystal structures must be the same, 3) the electronegativities must be
similar, and 4) the valences should be the same, or nearly the same.
Below are tabulated, for
the various elements, these criteria.
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 Fall '08
 Johnson
 Materials Science, Atom, Chemical element, Burgers vector

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