A $2.0\times {10}^{-3}\mathrm{V}/\mathrm{m}$electric field creates a $3.5\times {10}^{17}$electrons/s current in a $1.0mm$diameter aluminum wire. What are (a) the drift speed and (b) the mean time between collisions for electrons in this wire?

Electric field$=2.0\times {10}^{-3}\mathrm{V}/\mathrm{m}$

Time$=3.5\times {10}^{17}$electron/s

Aluminum wire diameter$=1.0mm$

The drift speed can be found as

$i_e=n_{e}A{v}_d\Rightarrow v_d=\frac{i_e}{n_{e}A}$

Since the cross-section is a circle, the area is $A=\frac{\pi D^2}{4}$,

$v_d=\frac{4i_e}{\pi n_e{D}^2}$

Substitute the values,

role="math" localid="1648880875567" $v_d=\left(\frac{4\times 3.5·{10}^{17}e/s}{\pi \times 6\times {10}^{28}e\times 0.001m{m}^2}\right)$

$=7.4\mu \mathrm{m}$

Hence, the drift speed is $7.4\mu m.$

Electric field$=2.0\times {10}^{-3}\mathrm{V}/\mathrm{m}$

Time $=3.5\times {10}^{17}$electron/s

Aluminum wire diameter$=1.0\mathrm{mm}$

We Know the drift speed, we can find the mean time as,

$v_d=\frac{e\tau }{m}E\Rightarrow \tau =\frac{m{v}_d}{eE}$

However, if we trust our previous solution, we could use the parametric solution in place of $v_d,$

role="math" localid="1648881487003" $\tau =\left(\frac{9.1\times {10}^{-31}kg\times 7.4·{10}^{-6}\mu m}{1.6\times {10}^{-19}c\times 0.002V/m}\right)$

$=2.1·{10}^{-14}\mathrm{s}.$

Hence, the mean time between collisions for electrons in this wire is$21fs.$