In a conductor, if the number of conduction electrons per unit volume is $8.5 \times 10^{28} \mathrm{~m}^{-3}$ and mean free time is 25fs (femto second), its approximate resistivity is: $(\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg})$

(D) $10^{-8} \Omega \mathrm{~m}$

(A) $10^{-5} \Omega \mathrm{~m}$

(B) $10^{-6} \Omega \mathrm{~m}$

(C) $10^{-7} \Omega \mathrm{~m}$

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Current Electricity

In a conductor, if the number of conduction electrons per unit volume is

8.5 \times 1 0^{28} m^{- 3}

and mean free time is 25fs (femto second), its approximate resistivity is:

(m_{e} = 9.1 \times 1 0^{- 31} kg)

Answer: (D)

Resistivity from the free-electron picture is

ρ = \frac{m _{e}}{n e ^{2} τ}

. Putting in

m_{e} = 9.1 \times 1 0^{- 31}

kg,

n = 8.5 \times 1 0^{28} m^{- 3}

e = 1.6 \times 1 0^{- 19}

C and

τ = 25

= 25 \times 1 0^{- 15}

s, the denominator is

8.5 \times 1 0^{28} \times (1.6 \times 1 0^{- 19})^{2} \times 25 \times 1 0^{- 15} \approx 5.4 \times 1 0^{- 23}

, so

ρ \approx \frac{9.1 \times 1 0 ^{- 31}}{5.4 \times 1 0 ^{- 23}} \approx 1.7 \times 1 0^{- 8} Ω m

, which is of the order of

1 0^{- 8} Ω m

JEE Main · 2019 · 09 Apr 2019 · Shift 2 · Single correct option