Current Electricity

JEE Main level, single correct, medium

31 of 380

Question 31 of 380

In a conductor, if the number of conduction electrons per unit volume is 8.5×1028m−38.5\times {10}^{28} {m}^{-3} and mean free time is 25fs25 fs (femto second), it’s approximate resistivity is: (me=9.1×10−31kg)\left({m}_{e}=9.1\times {10}^{-31} kg\right)

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Answer: (D) 10−8Ωm{10}^{-8} \Omega m

We know i=neAvdi=ne{Av}_{d}
⇒i=VR=VAρl=EAρ\Rightarrow i=\dfrac{V}{R}=\dfrac{VA}{\rho l}=\dfrac{EA}{\rho }
ρ=EAi=EAneAvd\rho =\dfrac{EA}{i}=\dfrac{EA}{ne{Av}_{d}}
Drift velocity can be written as vd=eEmeτ{v}_{d}=\dfrac{eE}{{m}_{e}}\tau
ρ=EAmeneAeEτ=mene2τ\rho =\dfrac{EA{m}_{e}}{neAeE\tau }=\dfrac{{m}_{e}}{n{e}^{2}\tau }
=9.1×10−318.5×1028×(1.6×10−19)2×25×10−15=\dfrac{9.1\times 1{0}^{-31}}{8.5\times 1{0}^{28}\times {\left(1.6\times {10}^{-19}\right)}^{2}\times 25\times 1{0}^{-15}}
≈1.65×10−8Ωm\approx 1.65\times 1{0}^{-8}\Omega m

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