If resistivity of copper is 1.72 × 10^ -8 -m and number of free electrons in copper is 8.5 × 10^ 28 / m^3 . Find the mobility.

Correct answer: A. 4.25 × 10^ -3 ~m^2 / C Given, Resistivity of copper =1.72 × 10^ -8 -m Electrons density n= 8.5 × 10^ 26 / m^3 aligned Mobility (μ) & = 1 ne & = 1 1.72 × 10^ -8 × 8.5 × 10^ 28 × 1.6 × 10^ -19 & =4.25 × 10^ -3 m^2 / C aligned

JIPMER 2019

Exam: JIPMER
Year: 2019
Subject: Physics
Chapter: Current Electricity

JIPMER 2019 · Physics · Current Electricity

If resistivity of copper is 1.72 × 10^ -8 -m and number of free...

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.

Options

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 $3.4 \times 10^{-3} \mathrm{~m}^2 / \mathrm{C} \Omega$

Correct answer

A. $4.25 \times 10^{-3} \mathrm{~m}^2 / \mathrm{C} \Omega$

Step-by-step 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}

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