Q127 / 180 · JIPMER 2019 · Physics

If resistivity of copper is $1.72 \times 10^{-8} \Omega-\mathrm{m}$ and number of free electrons in copper is $8.5 \times 10^{28} / \mathrm{m}^3$. Find the mobility.

  • Correct answer
    A
    $4.25 \times 10^{-3} \mathrm{~m}^2 / \mathrm{C} \Omega$
  • B
    $6.8 \times 10^{-3} \mathrm{~m}^2 / \mathrm{C} \Omega$
  • C
    $8.5 \times 10^{-3} \mathrm{~m}^2 / \mathrm{C} \Omega$
  • D
    $3.4 \times 10^{-3} \mathrm{~m}^2 / \mathrm{C} \Omega$

Solution

Given, Resistivity of copper $\rho=1.72 \times 10^{-8} \Omega-\mathrm{m}$ Electrons density $\mathrm{n}=8.5 \times 10^{26} / \mathrm{m}^3$ $\begin{aligned} \therefore \quad \text { Mobility }(\mu) & =\frac{1}{\rho \text { ne }} \\ & =\frac{1}{1.72 \times 10^{-8}} \times 8.5 \times 10^{28} \times 1.6 \times 10^{-19} \\ & =4.25 \times 10^{-3} \mathrm{~m}^2 / \mathrm{C} \Omega \end{aligned}$

Hence, the correct answer is option: 'A'.

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