A potential difference of 5V is applied across a conductor of length 10 cm. IF drift of electron is `2.5 xx 10^(-4) m//s`, then electron mobility in SI unit is

Options

A. `5 xx 10^(-4)`
B. ` 5 xx 10^(-6) `
C. `5 xx 10^(-2)`
D. Zero

Answer

To find the electron mobility in SI units, we can follow these steps:

### Step 1: Identify the given values
- Potential difference (V) = 5 V
- Length of the conductor (L) = 10 cm = 0.1 m (convert cm to m)
- Drift velocity (Vd) = 2.5 x 10^(-4) m/s

### Step 2: Calculate the electric field (E)
The electric field (E) can be calculated using the formula:
\[ E = \frac{V}{L} \]

Substituting the known values:
\[ E = \frac{5 \, \text{V}}{0.1 \, \text{m}} = 50 \, \text{V/m} \]

### Step 3: Calculate the mobility (μ)
Mobility (μ) is defined as the ratio of drift velocity (Vd) to the electric field (E):
\[ \mu = \frac{V_d}{E} \]

Substituting the values we have:
\[ \mu = \frac{2.5 \times 10^{-4} \, \text{m/s}}{50 \, \text{V/m}} \]

### Step 4: Simplify the expression
Calculating the above expression:
\[ \mu = \frac{2.5 \times 10^{-4}}{50} = 5 \times 10^{-6} \, \text{m}^2/\text{s/V} \]

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

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