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Defects:
1.
Chung 2.5
As the temperature is raised from 800 to 1000K, the unit cell size of a given metal crystal (as
measured by xray diffraction) increases 0.05% due to thermal expansion (i.e., increased
equilibrium distance between atoms).
In the same temperature range, the density decreases by
0.16%, due to the combination of thermal expansion and vacancy production.
(a)
What is the expected percentage decrease in density due to normal thermal expansion
alone?
Hint: When the unit cell size increases by
x
% (<<100%), the volume per unit cell increases
by 3
x
%.
(b)
Hence evaluate the net density decrease due to vacancy production alone.
(c)
If the vacancy concentration at 800K is 10
6
, estimate the activation energy for vacancy
production.
Note:
Depending on how a vacancy is produced, equation 2.18 should be written as
exp
vv
NQ
A
N
kT
where A is a constant.
Therefore, to determine the value of
Q
, one needs to use data from
two different temperatures (see Appendix of Chapter 2).
Solution
(a) An increase of 0.05% in the lattice constant decreases the density by 0.15% (as given by the
hint above).
(b) The net density decrease due to vacancy production = 0.16%  0.15% = 0.01%
(c) The vacancy concentration at 800 K = 10
6
, while the vacancy concentration at 1000 K is 10
4
(= 0.01%).
We can write:
10
4
10
6
exp
Q
k
B
(1000)
Q
k
B
(800)
,
where
Q
is the activation energy for vacancy production, and
k
B
Boltzmann constant.
Taking natural log on both sides, we have:
4.605
Q
k
B
1
800
1
1000
Solving,
Q
= 1.59 eV
Alternate solution:
if Nv/N(800)/Nv/N(1000) = 0.0001, then Nv/N(1000) = 0.01, solving for
Q yields 3.17eV
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View Full Document2.
Chung 2.6
A TiO
2
crystal is grown with some Cr
2
O
3
impurities.
The valence of Ti is +4, and the valence of
Cr is +3.
Chromium ions occupy titanium sites substitutionally.
Explain what type of vacany will
be introduced.
When a Cr
3+
sits on a Ti
+4
that the site becomes relatively negative to what it would have been if
a Ti
+4
were occupying it.
Therefore, you need a vacancy which is relatively positively charged to
maintain neutrality.
In this case you would create oxygen vacancies because if you remove
something with a valence of 2 then the vacancy left behind has a relative charge of +2.
For every one Cr
2
O
3
added to the system, two Cr go onto two Ti sites, three oxygen go
onto oxygen sites, and one oxygen site remains empty.
(since in TiO
2
you have two oxygen sites
for every Ti site
–
if you fill 2 Ti sites you have 4 associated oxygen sites.).
Thus you have two Cr
on Ti sites with a relative charge of 1 each and one oxygen vacancy with a relative charge of +2,
so you maintain charge neutrality.
If you were to create Ti vacancies, they would have a
relative charge of 4 and would not help in your quest for neutrality.
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 Spring '08
 Chung

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