Question

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

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

Answer 1

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

Potential difference $=5V$

length $=0.1m$

drift speed of electron $=2.5\times {10}^{-4}m/s$

electron density $=\delta \times {10}^{25}{m}^{-3}$

$vd=2.5\times {10}^{-4}m/s$

$\Rightarrow n=\delta \times {10}^{28}/m^3$

we know that

$J=nevd$

or, $J=nevdA$

where symbols have their usual meaning

$\Rightarrow \frac{V}{R}=ne{V}_{e}A$

or, $\frac{V}{\frac{PL}{A}}=ne{V}_d$

$\rho =\frac{V}{ne{V}_{d}L}$

$=\frac{5}{8\times {10}^{28}\times 16\times {10}^{-19}\times 25\times {10}^{-4}\times 0.1}$

$\rho =1.6\times {10}^{-5}\Omega m$.