A potential difference of 5 V is applied across a conductor of length 10 cm . If drift velcoity of electrons is `2.5 xx 10^(-4) m//s` , then electron mobility will be

Options

A. `5xx10^(-4) m^(2) V^(-1) S^(-1)`
B. `5 xx 10^(-6) m^(2) V^(-1) s^(-1)`
C. `5 xx 10^(-2) m^(2) V^(-1) s^(-1)`
D. Zero

Answer

To solve the problem, we need to determine the electron mobility given the potential difference, length of the conductor, and the drift velocity of the electrons.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Potential Difference (V) = 5 V
- Length of Conductor (L) = 10 cm = 0.1 m (convert to meters for SI units)
- Drift Velocity of Electrons (Vd) = \(2.5 \times 10^{-4} \, \text{m/s}\)

2. **Calculate the Electric Field (E):**
The electric field (E) in a conductor can be calculated using the formula:
\[ E = \frac{V}{L} \]
Substituting the known values:
\[ E = \frac{5 \, \text{V}}{0.1 \, \text{m}} = 50 \, \text{V/m} \]

3. **Calculate Electron Mobility (μ):**
The mobility (μ) of electrons is defined as the ratio of the drift velocity (Vd) to the electric field (E):
\[ \mu = \frac{V_d}{E} \]
Substituting the values we have:
\[ \mu = \frac{2.5 \times 10^{-4} \, \text{m/s}}{50 \, \text{V/m}} \]

4. **Perform the Calculation:**
\[ \mu = \frac{2.5 \times 10^{-4}}{50} = 5 \times 10^{-6} \, \text{m}^2/\text{V s} \]

5. **Final Answer:**
The electron mobility is:
\[ \mu = 5 \times 10^{-6} \, \text{m}^2/\text{V s} \]