Question

The potential difference between the ends of a straight conductor of length 20 cm is 16 V. If the drift speed of the electrons is \( 2.4 \times 10^{-4} \) m/s, the electron mobility in \( m^2 V^{-1} s^{-1} \) is:

• A. \( 3.6 \times 10^{-6} \)
• B. \( 2.4 \times 10^{-6} \)
• C. \( 2 \times 10^{-6} \)
• D. \( 3 \times 10^{-6} \)

Hint:

Use the formula \( \mu_e = \frac{v_d}{E} \) to calculate electron mobility.
- Convert all units properly before substitution (e.g., cm to meters).

Answer 1

The Correct Option is D

The electron mobility \( \mu_e \) is given by the relation: \[ \mu_e = \frac{v_d}{E} \] where: - \( v_d \) is the drift velocity (\( 2.4 \times 10^{-4} \) m/s), - \( E \) is the electric field intensity. 1. Calculate Electric Field \( E \): The electric field is given by: \[ E = \frac{V}{L} \] where: - \( V = 16 \) V (potential difference), - \( L = 20 \) cm = 0.2 m. \[ E = \frac{16}{0.2} = 80 \text{ V/m} \] 2. Calculate Electron Mobility \( \mu_e \): \[ \mu_e = \frac{2.4 \times 10^{-4}}{80} \] \[ = 3 \times 10^{-6} \text{ m}^2 \text{V}^{-1} \text{s}^{-1} \] Thus, the correct answer is \(\boxed{3 \times 10^{-6}}\).