Name: Meter Bridge
Uploaded: 2019-07-19
Duration: 7 min 2 s
Description: Measurement, by comparison, is always accurate. The video talks about a metre bridge device that helps us find the unknown resistance by comparing it with a...

Question

Compare the average drift speed of conduction electrons in a copper wire of cross-sectional area

3.0 \times 10^{- 7} m^{2}

carrying a current of

5 A

with the thermal speed of copper atoms at ordinary temperatures. Assume that each copper atom contributes roughly one conduction electron. The density of copper is

9.0 \times 10^{3} kg m^{- 3}

and its atomic mass is

63.5 u

.

Accepted Answer

Given that,

The cross-sectional area of the copper wire is $A = 3.0 \times 10^{- 7} m^{2}$ .

The current through the wire is $I = 5 A$ .

Charge on an electron is $e = 1.6 \times 10^{- 19} C$

The density of copper is $ρ = 9.0 \times 10^{3} kg m^{- 3} = 9.0 g {cm}^{- 3}$ .

Since $6.0 \times 10^{23}$ copper atoms have a mass of $63.5 g$ and there is one conduction electron per atom of copper, the number density of electrons is $n = \frac{6 \times 10^{23}}{63.5} \times 9.0 = 8.5 \times 10^{22} {cm}^{- 3} = 8.5 \times 10^{28} m^{- 3}$ .

The average drift speed of conduction electrons is $v_{d} = \frac{I}{A n e} = \frac{5 C s^{- 1}}{3.0 \times 10^{- 7} m^{2} \times 8.5 \times 10^{28} m^{- 3} \times 1.6 \times 10^{- 19} C}$

$\Rightarrow v_{d} = 0.0012 m s^{- 1} = 1.2 \times 10^{- 3} m s^{- 1}$ .

The Boltzmann constant, $k = 1.38 \times 10^{- 23} J {mole}^{- 1} K^{- 1}$ .

The thermal speed of electron of mass $m = 9.1 \times 10^{- 31} kg$ at temperature $T = 25 ℃ = 298 K$ is $v = \sqrt{\frac{3 k T}{m}} = \sqrt{\frac{3 \times 1.38 \times 10^{- 23} \times 298}{9.1 \times 10^{- 31}}} = 1.16 \times 10^{5} m s^{- 1}$

Thus, $\frac{v_{d}}{v} = \frac{1.2 \times 10^{- 3}}{1.16 \times 10^{5}} = 1.03 \times 10^{- 8} \approx 10^{- 8}$ .

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