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For obvious reasons some
of the symbols deviate a little from the symbols used in the text; e.g. we have
hFP instead of HFP. |
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We start with the system of equations that came
from the mass action
law |
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We start with the calculation of
cv(C): |
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Inserting the first and the second equation into the third
equation yields: |
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That was the
first equation for
cV(C). Next we calculate
ci(C). |
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Start with the third equation and eliminate cv(A)
using the second. We have the final result after a series of mathematical
manipulations: |
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That was the
third equation. Next we
calculate cV(A). |
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Start with the third equation and eliminate ci(C)
using the first, we obtain |
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That's it. Nothing
to it. ;-) |
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Well, not exactly. I myself certainly cannot solve problems like this
without making some dumb mistakes in breaking down the math. Almost everybody
does. |
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However, I usually notice that I made a stupid mistake because the result
just can't be true. And I can, if I really employ myself, get the right result
eventually - because I did some exercises like this before. And that is why you should do it, too!. |
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As a last comment we may note that
solving equations coming from the mass action law can become rather tedious
very quickly - compare the example in the link, which is about as
simple as it could be. |
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Now we look at the limiting cases of pure Schottky
or pure Frenkel disorder. |
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For pure Frenkel disorder we must have
hFP << hS, and
cV(A) = 0. |
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For pure Schottky disorder we must have
hFP >> hS, and
ci(C) = 0. |
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For the first case - pure Frenkel
disorder - just look at the expression |
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( |
1 + |
N
N' |
· exp |
hS hFP
kT |
) |
1/2 |
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For hS >>
hFP, the exponential in this case is positive which means
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N
N' |
· exp |
hS hFP
kT |
>> 1 |
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So you may neglect the 1 in the above expression and
replace the whole square root by |
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This gives for ci(C) |
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ci(C) |
= |
N
N' |
· |
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= |
N
N' |
· exp |
hFP
kT |
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This is the result as as it should be. |
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With this we immediately obtain |
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cV(C) |
= |
N
N' |
· exp |
hFP
2kT |
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cV(A) |
= |
0 |
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This is so because |
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N
N' |
· exp |
hS hFP
kT |
>> |
1 |
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Contrariwise, if
hS << hFP, 1 + N/N'
· exp[(hS hFP)/kT]
» 1 obtains. |
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Because hS
2hFP is a large negative number we get |
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ci(C) |
= |
N
N' |
· exp |
hS 2hFP
2kT |
» |
0 |
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The expressions for cV(C) and
cV(A) immediately reduce to the proper equation |
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cV(C) |
= cV(A) = exp |
hS
2kT |
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© H. Föll