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First Task: Derive a number for
v0 (at room temperature). We have |
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v0 |
= |
æ
ç
è |
3kT
m |
ö
÷
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1/2 |
= |
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ç
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8,6 · 105 · 300
9,1 · 1031 |
eV · K
K · kg |
ö
÷
ø |
1/2 |
= |
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1,68 · 1014 · |
æ
ç
è |
eV
kg |
ö
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1/2 |
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The dimension "square root of
eV/kg" does not look so good - for a velocity we would like to have
m/s. In looking at the energies we equated kinetic energy with the
classical dimension [kg · m2/s2] = [J] with
thermal energy kT expressed in [eV]. So let's convert
eV to J (use the
link) and
see if that solves the problem. We have 1 eV = 1,6 ·
1019 J = 1,6 · 1019 kg ·
m2 · s2 which gives us |
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v0 |
= |
1,68 · 1014 · |
æ
ç
è |
1,6 · 1019 kg · m2
kg · s2 |
ö
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1/2 |
= 5,31 · 104 m/s = 1,91 ·
105 km/hr |
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Possibly a bit surprising - those electrons are no
sluggards but move around rather fast. Anyway, we have shown that a value of
» 104 m/s
as postulated in the
backbone is really OK. |
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Of course, for T ®
0, we would have v0® 0 -
which should worry us a bit ???? If instead of room temperature (T =
300 K) we would go to let's say 1200 K , we would just double the
average speed of the electrons. |
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Second Task: Derive a number for
t. We have |
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First we need some number for the concentration of
free electrons per m3. For that we complete the
table given, noting that for the
number of atoms per m3 we have to divide the density by the
atomic weight. |
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Atom |
Density
[kg · m3] |
Atomic weight
× 1,66 · 1027 kg |
Conductivity s
× 105 [W1 ·
m1] |
No. Atoms [m3]
× 10 28 |
Na |
970 |
23 |
2,4 |
2,54 |
Cu |
8.920 |
64 |
5,9 |
8,40 |
Au |
19.300 |
197 |
4,5 |
5,90 |
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So let's take 5 · 1028
m3 as a good order of magnitude guess for the number of
atoms in a m3, and for a first estimate some average value
s = 5 · 105 [W1 · m1]. We
obtain |
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t = |
5 · 105 · 9,1 · 1031
5 · 1028 · (1,6 ·
1019)2 |
kg · m3
W · m · A2 ·
s2 |
= 3,55 · 1016 |
kg · m2
V · A · s2 |
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Well, somehow the whole thing would look much
better with the unit [s]. So let's see if we can remedy the
situation. |
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Easy: Volts times Amperes equals Watts which is power, e.g. energy per time, with the
unit [J · s1] = kg · m2 ·
s3. Insertion yields |
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t |
= 1,42 · 1028 |
kg · m2 · s3
kg · m2 · s2 |
= 3,55 · 1016 s = 0.35 fs |
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The backbone thus is right again. The scattering
time is in the order of
femtosecond
which is a short time indeed. Since all variables enter the equation linearly,
looking at somewhat other carra ier densities (e.g. more than 1 electron
per atom) or conductivities does not really change the general picture very
much. |
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Third Task: Derive a number for
vD . We have (for a field strength E = 100 V/m = 1
V/cm) |
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|vD| |
= |
E · e · t
m |
= |
100 · 1,6 · 1019 · 3,55 ·
1016
9,1 · 1031 |
V · C · s
m · kg |
= |
6,24 · 103 |
V · A · s2
m · kg |
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= |
6,24 · 103 |
kg · m2 · s2
m · kg · s3 |
= 6,24 · 103 m/s = 6,24
mm/s |
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This is somewhat larger than the
value given in the backbone
text. |
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However - a field strength of 1 V/cm applied to a
metal is huge! Think about the current
density j you would get if you apply 1 V to a piece of
metal 1 cm thick. |
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It is actually j = s
· E = 5 · 107 [W1 · m1] · 100
V/m = 5 · 109 A/m2 = 5 · 105
A/cm2! |
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For a more "reasonable" current density of
103 A/cm2 we have to reduce E
hundredfold and then end up with |vD| = 0,0624 mm/s - and
that is slow indeed! |
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© H. Föll