Question 36 of 380

When $5V$ 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$1.56\times {10}^{-5}\Omega m$B$1.6\times {10}^{-8}\Omega m$C$1.6\times {10}^{-7}\Omega m$D$1.6\times {10}^{-6}\Omega m$

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Answer: (A) $1.56\times {10}^{-5}\Omega m$

Potential difference$=5V$, length $=\text{0.1 m}=l$ and the electron speed $=$ drift velocity ${\text{v}}_{\text{d}}=\text{2.5}\times 1{0}^{-4}{\text{m s}}^{-1}$, electron density $\left(\text{n}\right)=8\times 1{0}^{28}{\text{m}}^{-3}$ and the charge on each electron(e) $=\text{1.6}\times 1{0}^{-19}\text{C}$. We know $\text{i}=\text{nAe}{v}_{\text{d}}...(i)$. And $V=iR...(ii)$



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

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

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

$=\dfrac{5}{\text{0.1}\times 8\times 1{0}^{28}\times \text{1.6}\times 1{0}^{-19}\times \text{2.5}\times 1{0}^{-4}}$

$=\text{0.156}\times 1{0}^{-4}\Omega \text{m}=\text{1.56}\times 1{0}^{-5}\Omega \text{m}$