91Ó°ÊÓ

Gold wire diameter$=1.0mm$

Electron drift speed$=5.0×{10}^{-5}\mathrm{m}/\mathrm{s}$

Number of electrons$=1mol$

A wire's cross-sectional area, divided by its electron current per unit of time, is equal to its electron flux per unit of time.$I_e=n_e{v}_{\mathrm{D}}A$

Where $n_e=5.9·{10}^{28}{\mathrm{m}}^{-3}$is the electron density in gold (see Table 27.1), $v_{\mathrm{D}}=5.0·{10}^{-5}\mathrm{m}/\mathrm{s}$is the drift speed and $A=\frac{1}{4}\mathrm{π}{d}^2$is the area of the cross-section of the wire ( $d=1.0\mathrm{mm}$is the diameter).

It gives,

$I_e=\frac{1}{4}{n}_e{v}_{\mathrm{D}}\mathrm{Ï€}{d}^2$

The total number of electrons that flow through the cross-section of the wire in time $t$is equal to the electron current multiplied by $t$:

$N=I_{e}t$

Set $N=6.0·{10}^{23}$which is equal to Avogadro's number and determines when one mole has passed.

We get,

localid="1648800315312" $t=\frac{N}{I_e}=\frac{4N}{n_e{v}_{\mathrm{D}}\mathrm{Ï€}{d}^2}$

$t=\left(\frac{4(6.0×{10}^{23})}{5.9×{10}^{28}e×5×{10}^{-5}{\mathrm{s}×\mathrm{π}×1.0}^{2}mm}\right)$

$=260000\mathrm{s}≈72\mathrm{h}13\min$

Hence, the time required is$260000\mathrm{s}$or approximately $72\mathrm{h}$and $13\min$for 1 mole of electrons to flow through a cross-section of the wire.