DRIFT SPEED OF ELECTRONS GOAL Calculate a drift speed and compare it with the rms speed of an electron gas. PROBLEM A copper wire of cross-sectional area 3.00×10^−6 m² carries a current of 10.0 A. (a) Assuming each copper atom contributes one free electron to the metal, find the drift speed of the

Question

DRIFT SPEED OF ELECTRONS

GOAL Calculate a drift speed and compare it with the rms speed of an electron gas.

PROBLEM A copper wire of cross-sectional area [latex]3.00 imes 10^{-6} \mathrm{~m}^{2}[/latex] carries a current of [latex]10.0 \mathrm{~A}[/latex]. (a) Assuming each copper atom contributes one free electron to the metal, find the drift speed of the electrons in this wire. (b) Use the ideal gas model to compare the drift speed with the random rms speed an electron would have at [latex]20.0^{\circ} \mathrm{C}[/latex]. The density of copper is [latex]8.92 \mathrm{~g} / \mathrm{cm}^{3}[/latex], and its atomic mass is [latex]63.5 \mathrm{u}[/latex].

STRATEGY All the variables in Equation [latex]17.2[/latex]

[latex]I = \lim_{\Delta t o 0} \frac{\Delta Q}{\Delta t} = n q v_d A[/latex] [17.2]

are known except for [latex]n[/latex], the number of free charge carriers per unit volume. We can find [latex]n[/latex] by recalling that one mote-of copper contains an Avogadro's number [latex]\left(6.02 imes 10^{23} ight)[/latex] of atoms, and each atom contributes one charge carrier to the metal. The volume of one mole can be found from copper's known density and atomic mass. The atomic mass is the same, numerically, as the number of grams in a mole of the substance.

Accepted Answer

(a) Find the drift speed of the electrons.

Calculate the volume of one mole of copper from its density and its atomic mass:

[latex]V=\frac{m}{ ho}=\frac{63.5 \mathrm{~g} / \mathrm{mol}}{8.92 \mathrm{~g} / \mathrm{cm}^{3}}=7.12 \mathrm{~cm}^{3} / \mathrm{mol}[/latex]

Convert the volume from [latex]\mathrm{cm}^{3}[/latex] to [latex]\mathrm{m}^{3}[/latex] :

[latex]7.12 \mathrm{~cm}^{3} / \operatorname{mol}\left(\frac{1 \mathrm{~m}}{10^{2} \mathrm{~cm}} ight)^{3}=7.12 imes 10{ }^{-6} \mathrm{~m}^{3} / \mathrm{mole}[/latex]

Divide Avogadro's number (the number of electrons in one mole) by the volume per mole to obtain the number density:

[latex]n=\frac{6.02 imes 10^{23} ext { electrons } / ext { mole }}{7.12 imes 10^{-6} \mathrm{~m}^{3} / ext { mole }}[/latex]

[latex]\begin{aligned}& =8.46 imes 10^{28} ext { electrons } / \mathrm{m}^{3}\end{aligned}[/latex]

Solve Equation [latex]17.2[/latex] for the drift speed and substitute:

[latex]\begin{aligned}v_{d} &=\frac{I}{n q A} \&=\frac{10.0 \mathrm{C} / \mathrm{s}}{\left(8.46 imes 10^{28} ext { electrons } / \mathrm{m}^{3} ight) \left(1.60 imes 10{ }^{-19} \mathrm{C} ight)\left(3.00 imes 10{ }^{-6} \mathrm{~m}^{2} ight)} \y_{d} &=2.46 imes 10^{-4} \mathrm{~m} / \mathrm{s}\end{aligned}[/latex]

(b) Find the rms speed of a gas of electrons at [latex]20.0^{\circ} \mathrm{C}[/latex].

Apply Equation 10.18:

[latex]\begin{aligned}& v_{\mathrm{rms}}=\sqrt{\frac{3 k_{\mathrm{B}} T}{m_{e}}} \end{aligned}[/latex]

Convert the temperature to the Kelvin scale and substitute values:

[latex]\begin{aligned}& v_{\mathrm{rms}}=\sqrt{\frac{3\left(1.38 imes 10^{-23} \mathrm{~J} / \mathrm{K} ight)(293 \mathrm{~K})}{9.11 imes 10^{-31} \mathrm{~kg}}} \& =1.15 imes 10^{5} \mathrm{~m} / \mathrm{s}\end{aligned}[/latex]

REMARKS The drift speed of an electron in a wire is very small, only about one-billionth of its random thermal speed.

Question 17.2: DRIFT SPEED OF ELECTRONS GOAL Calculate a drift speed and co...

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