Question

A conducting rod contains 8.5 × 1028 electrons/m3. Calculate its resistivity at room temperature and also the mobility of electrons if the collision time for electron scattering is 2 × 10–14 sec.   

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Answer 1

Given:

• Number of electrons $(n) = 8.5 \times 10^{28} \ electrons/m^3$

• Collision time $( \tau ) = 2 \times 10^{-14} sec.$

To Find:

• Electrical resistivity

• Mobility of electrons

Solution:

Electrical conductivity is given by

             $\sigma = \frac{n e^2 \tau}{m}$

             $\sigma = \frac{8.5 \times 10^{28} \times (1.6 \times 10^{-19})^2 \times 2 \times 10^{-14}}{9.11 \times 10^{-31}}$

             $\sigma = 4.77 \times 10^7 \Omega^{-1} m{-1}$

Electrical resistivity is given by

             $\rho = \frac{1}{\sigma}$

            $\rho = \frac{1}{4.77 \times 10^7 }$

           $\rho = 2.09 \times 10^{-8} \Omega m$

Mobility of electron is given by

            $\mu = \frac{\sigma}{ne}$

            $\mu = \frac{4.77 \times 10^7}{8.85 \time s10^{28} \times 1.6 \times 10^{-19}}$

            $\mu = 3.512 \times 10^{-3} m^2 V^{-1} s^{-1}$

Thus,

           Electrical resistivity $\rho = 2.09 \times 10^{-8} \Omega m$

          Mobility of electron $\mu = 3.512 \times 10^{-3} m^2 V^{-1} s^{-1}$