# Solution to Exercise 2.1-1

## Derive and Discuss numbers for µ

| First Task: Derive numbers for the mobility µ.
|  | First we need typical conductivities and electron densities in metals, which we can take from the table in the link.
|  | At the same time we expand the table a bit
Material | r [W cm] | s [W–1 cm–1] | Density d × 103 [kg m–3] | Atomic weight w [× 1u = 1,66 · 10–27 kg] | n = d/w [m–3]
--- | --- | --- | --- | --- | ---
Silver Ag | 1,6·10–6 | 6.2·105 | 10,49 | 107,9 | 5,85 · 1028
Copper Cu | 1,7·10–6 | 5.9·105 | 8,92 | 63,5 | 8,46 · 1028
Lead Pb | 21·10–6 | 4.8·104 | 11,34 | 207,2 | 3,3 · 1028
| For the mobility µ we have the equation
µ | = | s q · n
--- | --- | ---
| With q = elementary charge = 1,60 10–19 C we obtain, for example for µAg
µ | Ag = | 6,2 · 105 1,6 · 10–19 · 5,85 · 1028 |  | m3 C · W · cm | = | 66,2 |  | cm2 C · W
--- | --- | --- | --- | --- | --- | --- | --- | ---
| The unit is a bit strange, but rembering that [C] = [A · s] and [W] = [V/A], we obtain
µAg | = | 66,2 |  | cm2 Vs
--- | --- | --- | --- | ---
µCu | = | 43.6 |  | cm2 Vs
µPb | = | 9,1 |  | cm2 Vs

| Second Task: Derive numbers for the drift velocity vD by considering a reasonable field strength.
|  | The mobility µ was defined as
µ | = | vD E
--- | --- | ---
| or |
vD | = | µ · E
| So what is a reasonable field strength in a metal?
|  | Easy. Consider a cube with side lengt l = 1 cm. Its resistance R is given by
R | = | r · l F | = | r W
--- | --- | --- | --- | ---
|  | A Cu or Ag cube thus would have a resistance of about 1,5 ·10–6 W. Applying a voltage of 1 V, or equivalently a field strength of 1 V/cm thus produces a current of I = U/R » 650 000 A or a current density j = 650 000 A/cm2
|  | That seems to be an awfully large current. Yes, but it is the kind of current density encountered in integrated circuits! Think about it!
|  | Nevertheless, the wires in your house carry at most about 30 A (above that the fuse blows) with a cross section of about 1 mm2; so a reasonable current density is 3000 A/cm2, which we will get for about U = 1,5 ·10–6 W · 3000 A = 4,5 mV.
|  | For a rough estimate we then take a field strength of 5 mV/cm and a mobility of 50 cm2/Vs and obtain
vD | = | 50 · 5 |  | mV · cm2 cm · V · s | = 0,25 | cm s | = 2,5 | mm s
--- | --- | --- | --- | --- | --- | --- | --- | ---
| That should come as some surprise! The electrons only have to move v e r y s l o w l y on average in the current direction (or rather, due to sign conventions, against it).
|  | Is that true, or did we make a mistake?
|  | It is true! However, it does not mean, that electrons will not run around like crazy inside the crystal, at very high speeds. It only means that their net movement in current anti-direction is very slow.
|  | Think of an single fly in a fly swarm. Even better read the module that discusses this analogy in detail. The flies are flying around at high speed like crazy - but the fly swarm is not going anywhere as long as it stays in place. There is then no drift velocity and no net fly current!

With frame

Averaging Vectors

Exercise 2.1-1

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