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

• Correct answer
  A

  $1.56\times {10}^{-5} \mathrm{\Omega } \mathrm{m}$
• B

  $1.6\times {10}^{-8} \mathrm{\Omega } \mathrm{m}$
• C

  $1.6\times {10}^{-7} \mathrm{\Omega } \mathrm{m}$
• D

  $1.6\times {10}^{-6} \mathrm{\Omega } \mathrm{m}$

Solution

Potential difference$= 5 \mathrm{V}$, length $=\text{0.1 m}=l$ and the electron speed $=$ drift velocity ${\mathit{\text{v}}}_{\text{d}}=\text{2.5}\times 10^{-4} {\text{m s}}^{-1}$, electron density $\left(\text{n}\right)=8\times 10^{28}{\text{m}}^{-3}$ and the charge on each electron(e) $=\text{1.6}\times 10^{-19} \text{C}$. We know $\mathit{\text{i}}=\mathit{\text{nAe}}{\mathit{v}}_{\text{d}}\mathit{ }\mathit{ }\mathit{.}\mathit{.}\mathit{.}\left(\mathrm{i}\right)$. And $V=iR ...\left(\mathrm{ii}\right)$



Resistance $R$ is also equal to, $\frac{\rho l}{\text{A}}$$\text{⇒}\mathit{\text{R}}=\frac{\rho l}{\mathit{\text{A}}}$

$\rho =\frac{\mathit{\text{AR}}}{l}$       $\left[\rho =\text{Resistivity}\right]$

$=\frac{\text{A}}{l}\times \frac{V}{\mathit{\text{i}}}$     [from (ii)]$=\frac{\mathit{\text{Av}}}{l\times \mathit{\text{nA}}\text{ e}{\text{v}}_{\text{d}}}$ & [from (i)]$=\frac{\mathit{\text{V}}}{l\times \mathit{\text{n}}\times \mathit{\text{e}}\times {\mathit{\text{v}}}_{\text{d}}}$

$=\frac{5}{\text{0.1}\times 8\times 10^{28}\times \text{1.6}\times 10^{-19}\times \text{2.5}\times 10^{-4}}$

$=\text{0.156}\times 10^{-4} \mathrm{\Omega } \text{m}=\text{1.56}\times 10^{-5} \mathrm{\Omega } \text{m}$

Hence, the correct answer is option: 'A'.