A potential difference of 10V is applied across a conductor of length 0.1 m. If the drift velocity of electrons is 2 x1`0^(-4)` m/s, the electron mobility is _______ `m^2V^(-1)s^(-1)`.

Options

A. `1 xx 10^(-6)`
B. `2 xx 10^(-6)`
C. `3 xx 10^(-6)`
D. `4xx 10^(-6)`

Answer

To find the electron mobility, we can use the formula:

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

where:
- \( \mu \) is the electron mobility (in \( m^2 V^{-1} s^{-1} \)),
- \( v_d \) is the drift velocity (in \( m/s \)),
- \( E \) is the electric field (in \( V/m \)).

### Step 1: Calculate the Electric Field (E)

The electric field \( E \) can be calculated using the formula:

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

where:
- \( V \) is the potential difference (in volts),
- \( L \) is the length of the conductor (in meters).

Given:
- \( V = 10 \, V \)
- \( L = 0.1 \, m \)

Substituting the values:

\[ E = \frac{10 \, V}{0.1 \, m} = 100 \, V/m \]

### Step 2: Substitute the Values into the Mobility Formula

Now that we have \( E \), we can substitute \( v_d \) and \( E \) into the mobility formula. Given:
- \( v_d = 2 \times 10^{-4} \, m/s \)

Substituting the values:

\[ \mu = \frac{2 \times 10^{-4} \, m/s}{100 \, V/m} \]

### Step 3: Calculate the Mobility (μ)

Now, calculating \( \mu \):

\[ \mu = \frac{2 \times 10^{-4}}{100} = 2 \times 10^{-6} \, m^2 V^{-1} s^{-1} \]

### Final Answer

The electron mobility is:

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

---