Solution to Exercise 2.1-2

Derive numbers for v0, vD, t, and l

First Task: Derive a number for v0 (at room temperature). We have:
v0   =  æ
ç
è 		3kT
m 		ö
÷
ø 		1/2  =  æ
ç
è 		3 · 8.6 · 10–5 · 300
9.1 · 10–31		 eV · K
K · kg		 ö
÷
ø 		1/2  =   2.92 · 1014 · æ
ç
è 		eV
kg		 ö
÷
ø 		1/2
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 · 10–19 J = 1.6 · 10–19 kg · m2 · s–2, which gives us
v0   =  2.92 · 1014 · æ
ç
è 		1.6 · 10–19 kg · m2
kg · s2		 ö
÷
ø 		1/2  =  1.17 · 105 m/s =  4.21 · 105 km/h
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.
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 1200 K, let's say, we would just double the average speed of the electrons.
 
Second Task: Derive a number for t. We have:
t  =   s · m
n · e2
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 (i.e, the atomic density) we have to divide the density by the atomic weight.
Atom Density
[kg · m–3]		 Atomic weight
[1.66 · 10–27 kg]		 Conductivity s
[107 W–1 · m–1]		 Atomic dens.
[1028 m–3]
Na 970 23 2.4 2.54
Cu 8,920 64 5.9 8.40
Au 19,300 197 4.5 5.90
So let's take 5 · 1028 m–3 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 · 107 W–1 m–1. We obtain
t  =   5 · 107 · 9.1 · 10–31
5 · 1028 · (1.6 · 10–19)2		 kg · m3
W · m · A2 · s2		  =  3.55 · 10–14 kg · m2
V · A · s2
Well, somehow the whole thing would look much better with the unit s. So let's see if we can remedy the situation.
Easy: volt times ampere equals watt, which is power, i.e. energy per time, with the unit J · s–1 = kg · m2 · s–3. Insertion yields
t  =  3.55 · 10–14 kg · m2 · s3
kg · m2 · s2		  =  3.55 · 10–14 s =  36 fs
The backbone thus is right again. The scattering time is in the order of femtoseconds, which is a short time indeed. Since all variables enter the equation linearly, looking at somewhat other carrier densities (e.g. more than 1 electron per atom) or conductivities does not really change the general picture very much.
 
Third Task: Derive a number for vD. We have (for a field strength E = 100 V/m = 1 V/cm):
|vD|   =   E · e · t
m		  =   100 · 1.6 · 10–19 · 3.55 · 10–14
9.1 · 10–31 		V · C · s
m · kg		  =   6.24 · 10–1 V · A · s2
m · kg
                 
 =   6.24 · 10–1 kg · m2 · s2
m · kg · s3		  =  6.24 · 10–1 m/s  =  624 mm/s
This is somewhat larger than the value given in the backbone text.
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.
It is actually j = s · E = 5 · 107 W–1 m–1 · 100 V/m = 5 · 109 A/m2 = 5 · 105 A/cm2!
For a more "reasonable" current density of 103 A/cm2 we have to reduce E hundredfold and then end up with |vD| = 6.24 mm/s – and that is slow indeed!
 
Fourth Task: Derive a number for l. We have:
l  = 2 · v0 · t  =  2 · 1.17 · 105 · 3.55 · 10–14 m   =  8.31 · 10–9 m   =  8.31 nm
Right again! If we add the comparatively miniscule vD, nothing would change.
Note, however, that the last equation does NOT mean that decreasing the temperature would lower l to eventually zero! Why? Because t isn't a constant but scales with the conductivity – look at the starting equation of the second task! And since the conductivity increases at lower temperatures, so does the mean free path.
 

Mit Frame Mit Frame as PDF

gehe zu Exercise 2.1-1: Derive and discuss numbers for µ and vD

gehe zu Exercise 2.1-2

© H. Föll (MaWi 2 Skript)