91Ó°ÊÓ

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

Time$=3.5×{10}^{17}$electron/s

Aluminum wire diameter$=1.0mm$

The drift speed can be found as

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

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

$v_d=\frac{4i_e}{\mathrm{Ï€}{n}_e{D}^2}$

Substitute the values,

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

$=7.4\mathrm{μ}\mathrm{m}$

Hence, the drift speed is $7.4\mathrm{μ}m.$

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

Time $=3.5×{10}^{17}$electron/s

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

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

$v_d=\frac{e\mathrm{τ}}{m}E⇒\mathrm{τ}=\frac{m{v}_d}{eE}$

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

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

$=2.1·{10}^{-14}\mathrm{s}.$

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