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 find the electron mobility, we can follow these steps:

### Step 1: Convert the length of the conductor from centimeters to meters
The length of the conductor is given as 10 cm. We need to convert this to meters for consistency in units.

\[ \text{Length in meters} = \frac{10 \text{ cm}}{100} = 0.1 \text{ m} \]

### Step 2: Calculate the electric field (E)
The electric field (E) can be calculated using the formula:

\[ E = \frac{V}{d} \]

where \( V \) is the potential difference and \( d \) is the length of the conductor. Given \( V = 5 \text{ V} \) and \( d = 0.1 \text{ m} \):

\[ E = \frac{5 \text{ V}}{0.1 \text{ m}} = 50 \text{ V/m} \]

### Step 3: Use the formula for electron mobility (μ)
The electron mobility (μ) is defined as the ratio of the drift velocity (v_d) to the electric field (E):

\[ \mu = \frac{v_d}{E} \]

Given that the drift velocity \( v_d = 2.5 \times 10^{-4} \text{ m/s} \) and \( E = 50 \text{ V/m} \):

\[ \mu = \frac{2.5 \times 10^{-4} \text{ m/s}}{50 \text{ V/m}} \]

### Step 4: Calculate the mobility
Now, we can perform the division:

\[ \mu = \frac{2.5 \times 10^{-4}}{50} = 5 \times 10^{-6} \text{ m}^2 \text{ V}^{-1} \text{ s}^{-1} \]

### Conclusion
The electron mobility is:

\[ \mu = 5 \times 10^{-6} \text{ m}^2 \text{ V}^{-1} \text{ s}^{-1} \]