# Solution to Exercise 1.3-1

## 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!

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Exercise 1.3-1

© H. Föll