Question

Calculate the electrical conductivity of a copper wire if the number density of free electrons is 8.5x10\(^{28}\) electrons/m\(^3\) and the electron mobility is 0.0035 m\(^2\)/Vs.

• A. \( (4)76 \times 10^7 \) S/m
• B. \( 5.95 \times 10^7 \) S/m
• C. \( 6.80 \times 10^7 \) S/m
• D. \( (3)80 \times 10^7 \) S/m

Hint:

Electrical Conductivity Formula. \(\sigma = n e \mu\), where n = carrier density, e = elementary charge, \(\mu\) = carrier mobility. Unit is Siemens per meter (S/m).

Answer 1

The Correct Option is A

Electrical conductivity (\(\sigma\)) is related to the charge carrier density (\(n\)), the elementary charge (\(e\)), and the charge carrier mobility (\(\mu\)) by the formula: $$ \sigma = n e \mu $$ Given: Electron density \(n = 8.
5 \times 10^{28}\) m\(^{-3}\) Electron mobility \(\mu = 0.
0035\) m\(^2\)/Vs Elementary charge \(e = (1)602 \times 10^{-19}\) C Substitute the values: $$ \sigma = (8.
5 \times 10^{28} \, \text{m}^{-3}) \times ((1)602 \times 10^{-19} \, \text{C}) \times (0.
0035 \, \text{m}^2/\text{Vs}) $$ The unit C/(Vs) simplifies to S (Siemens).
The unit m\(^{-3} \cdot m^2\) = m\(^{-1}\).
So the final unit is S/m.
$$ \sigma = (8.
5 \times (1)602 \times 0.
0035) \times 10^{(28 - 19)} \, \text{S/m} $$ $$ \sigma = (1(3)617 \times 0.
0035) \times 10^9 \, \text{S/m} $$ $$ \sigma \approx 0.
04766 \times 10^9 \, \text{S/m} $$ $$ \sigma \approx (4)766 \times 10^7 \, \text{S/m} $$ This matches option (1).