Solution to Exercise 2.1-2

Derive numbers for v₀, v_D, 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 8,6 · 10–5 · 300 9,1 · 10–31 eV · K K · kg ö ÷ ø 1/2  =   1,68 · 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   =  1,68 · 1014 · æ ç è 1,6 · 10–19 kg · m2 kg · s2 ö ÷ ø 1/2  =  5,31 · 104 m/s =  1,91 · 105 km/hr
	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 let's say 1200 K , 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 we have to divide the density by the atomic weight.
		Atom Density [kg · m–3] Atomic weight × 1,66 · 10–27 kg Conductivity s × 105 [W–1 · m–1] No. Atoms [m–3] × 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
		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 · 105 [W–1 · m–1]. We obtain
		t  =   5 · 105 · 9,1 · 10–31 5 · 1028 · (1,6 · 10–19)2 kg · m3 W · m · A2 · s2  =  3,55 · 10–16 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: Volts times Amperes equals Watts which is power, e.g. energy per time, with the unit [J · s–1] = kg · m2 · s–3. Insertion yields
		t  =  1,42 · 10–28 kg · m2 · s3 kg · m2 · s2  =  3,55 · 10–16 s =  0.35 fs
	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.

	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 · 10–16 9,1 · 10–31 V · C · s m · kg  =   6,24 · 10–3 V · A · s2 m · kg                    =   6,24 · 10–3 kg · m2 · s2 m · kg · s3  =  6,24 · 10–3 m/s  =  6,24 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| = 0,0624 mm/s - and that is slow indeed!

	Fourth Task: Derive a number for l. We have
		lmin  =  2 · v0 · t  =  2 · 5,31 · 104 · 3,55 · 10–16 m   =  3,77 · 107 m   =  0,0377 nm
		Right again! If we add the comparatively miniscule vD, nothing would change. Decreasing the temperature would lower l to eventually zero, or more precisely, to 2 · vD · t and thus to a value far smaller than an atom..

With frame

Exercise 2.1-2

© H. Föll (Advanced Materials B, part 1 - script)