Question

When a 5 $V$ potential difference is applied across a wire of length $0.1m$, the drift speed of electrons is $2.5\times {10}^{-4}m{s}^{-1}$. If the electron density in the wire is $8\times {10}^{28}{m}^{-3}$, the resistivity of the material is close to:

• A. $1.6\times {10}^{-8}\Omega m$
• B. $1.6\times {10}^{-7}\Omega m$
• C. $1.6\times {10}^{-6}\Omega m$
• D. $1.6\times {10}^{-5}\Omega m$

Answer 1

The correct option is D $1.6\times {10}^{-5}\Omega m$

We know,
$i=neA{v}_d$
where symbol have their usual meaning.
From Ohm's law,
$i=\frac{V}{R}=neA{v}_d$-------(1)
Where $R=\rho \frac{L}{A}$
Putting value of $R$ in equation (1) we get,
$\frac{V}{\rho \frac{L}{A}}=neA{v}_d$
$\rho =\frac{V}{ne{v}_{d}L}$
$\rho =\frac{5}{8\times {10}^{28}\times 1.6\times {10}^{-19}\times 2.5\times {10}^{-4}\times 0.1}$
$\rho =1.6\times {10}^{-5}\Omega m$