Question 17.2: Drift Speed of Electrons Goal Calculate a drift speed and co...
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 \times 10^{-6} m² carries a current of 10.0 A. (a) Assuming that 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 20.0°C. The density of copper is 8.92 g/cm³, and its atomic mass is 63.5 u.
Strategy All the variables in Equation 17.2 are known except for n, the number of free charge carriers per unit volume. We can find n by recalling that one mole of copper contains an Avogadro’s number \left(6.02 \times 10^{23}\right) 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.
I=\frac{\Delta Q}{\Delta t}=n q v_{d} A
(a) Find the drift speed of the electrons.
Calculate the volume of one mole of copper from its density and its atomic mass:
V=\frac{m}{\rho}=\frac{63.5 g }{8.92 g / cm ^{3}}=7.12 cm ^{3}
Convert the volume from cm³ to m³:
7.12 cm ^{3}\left(\frac{1 m }{10^{2} cm }\right)^{3}=7.12 \times 10^{-6} m ^{3}
Divide Avogadro’s number (the number of electrons in one mole) by the volume per mole to obtain the number density:
\begin{aligned}n &=\frac{6.02 \times 10^{23} \text { electrons } / \text { mole }}{7.12 \times 10^{-6} m ^{3} / mole } \\&=8.46 \times 10^{28} \text { electrons } / m ^{3}\end{aligned}
Solve Equation 17.2 for the drift speed, and substitute:
\begin{aligned}&v_{d}=\frac{I}{n q A} \\&=\frac{10.0 C / s }{\left(8.46 \times 10^{28} \text { electrons } / m ^{3}\right)\left(1.60 \times 10^{-19} C \right)\left(3.00 \times 10^{-6} m ^{2}\right)} \\&v_{d}=2.46 \times 10^{-4} m / s\end{aligned}
(b) Find the rms speed of a gas of electrons at 20.0°C.
Apply Equation 10.18:
v_{ rms }=\sqrt{\overline{v^{2}}}=\sqrt{\frac{3 k_{B} T}{m}}=\sqrt{\frac{3 R T}{M}}
v_{ rms }=\sqrt{\frac{3 k_{B} T}{m_{e}}}
Convert the temperature to the Kelvin scale, and substitute values:
\begin{aligned}v_{ rms } &=\sqrt{\frac{3\left(1.38 \times 10^{-23} J / K \right)(293 K )}{9.11 \times 10^{-31} kg }} \\&=1.15 \times 10^{5} m / s\end{aligned}
Remarks The drift speed of an electron in a wire is very small—only about one-billionth of its random thermal speed.