Question

If the number density of free electrons in a copper wire is $8.5\times10^{22}cm^{-3}$ and the relaxation time of free electrons in the wire is $2.25\times10^{-14}s$, then the electrical conductivity of copper is (Mass of the electron $=9\times10^{-31}$ kg and charge of the electron $=1.6\times10^{-19}C$)}

• A. $6.80\times10^{7}Sm^{-1}$
• B. $3.40\times10^{7}Sm^{-1}$
• C. $5.44\times10^{7}Sm^{-1}$
• D. $2.72\times10^{7}Sm^{-1}$ \bigskip

Hint:

Always convert $cm^{-3}$ to $m^{-3}$ before substituting in electrical conductivity problems.

Answer 1

The Correct Option is A

Concept: Electrical conductivity in Drude model is given by: \[ \sigma = ne^{2}\tau / m \]

Step 1: Convert number density into SI units.
\[ n = 8.5\times10^{22}cm^{-3} = 8.5\times10^{28}m^{-3} \]

Step 2: Substitute values in conductivity formula.
\[ \sigma = \frac{(8.5\times10^{28})(1.6\times10^{-19})^{2}(2.25\times10^{-14})}{9\times10^{-31}} \]

Step 3: Simplify powers.
\[ (1.6\times10^{-19})^{2} = 2.56\times10^{-38} \]

Step 4: Multiply numerator.
\[ 8.5\times2.56\times2.25 \approx 49.0 \]

Step 5: Final computation.
\[ \sigma \approx \frac{49\times10^{-24}}{9\times10^{-31}} \approx 6.8\times10^{7}Sm^{-1} \]