Question

The drift speed of electrons in a material is found to be \( 0.3 \, \text{ms}^{-1} \) when an electric field of \( 2 \, \text{Vm}^{-1} \) is applied across it. The electron mobility (in \( \text{m}^2 \text{V}^{-1} \text{s}^{-1} \)) in the material is

• A. \( 60 \times 10^{-2} \)
• B. \( 15 \times 10^{-2} \)
• C. \( 1350 \times 10^6 \)
• D. \( 5400 \times 10^6 \)

Hint:

Direct formula application. Ensure units are consistent (SI units are used here).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Electron mobility \( \mu \) is defined as the magnitude of drift velocity per unit electric field.
Step 2: Key Formula or Approach:
\[ \mu = \frac{v_d}{E} \] where \( v_d \) is drift velocity and \( E \) is the electric field.
Step 3: Detailed Explanation:
Given: \( v_d = 0.3 \, \text{m/s} \) \( E = 2 \, \text{V/m} \) Calculate \( \mu \): \[ \mu = \frac{0.3}{2} = 0.15 \, \text{m}^2 \text{V}^{-1} \text{s}^{-1} \] Expressing in scientific notation to match options: \[ 0.15 = 15 \times 10^{-2} \]
Step 4: Final Answer:
The electron mobility is \( 15 \times 10^{-2} \).