Current Electricity 1 Question 5
5. In a conductor, if the number of conduction electrons per unit volume is $8.5 \times 10^{28} m^{-3}$ and mean free time is $25 fs$ (femto second), it’s approximate resistivity is (Take, $m _e=9.1 \times 10^{-31} kg$ )
(a) $10^{-7} \Omega-m$
(b) $10^{-5} \Omega-m$
(c) $10^{-6} \Omega-m$
(d) $10^{-8} \Omega-m$
(2019 Main, 9 April II)
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Solution:
- Resistivity of a conductor is
$$ \rho=\frac{m _e}{n e^{2} \tau} $$
where, $m _e=$ mass of electron $=9.1 \times 10^{-31} kg$,
$n=$ free charge density $=8.5 \times 10^{28} m^{-3}$,
$\tau=$ mean free time $=25 fs=25 \times 10^{-15} s$
and
$$ e=\text { charge of electron }=1.6 \times 10^{-19} C $$
Substituting values in Eq. (i), we get
$$ \begin{aligned} \rho & =\frac{9.1 \times 10^{-31}}{8.5 \times 10^{28} \times\left(1.6 \times 10^{-19}\right)^{2} \times 25 \times 10^{-15}} \\ & =\frac{9.1 \times 10^{-6}}{8.5 \times 2.56 \times 25}=0.016 \times 10^{-6} \\ & =1.6 \times 10^{-8} \Omega-m \simeq 10^{-8} \Omega-m \end{aligned} $$
