A charged particle having drift velocity of `7.5xx10^(-4)ms^(-1)` in electric field of `3xx10^(-10)Vm^(-1)` mobility is

Options

A. `6.5xx10^(6)m^(2)V^(-1)s^(-1)`
B. `2.5xx10^(6)m^(2)V^(-1)s^(-1)`
C. `2.5xx10^(4)m^(2)V^(-1)s^(-1)`
D. `6.5xx10^(4)m^(2)V^(-1)s^(-1)`

Explanation

To find the mobility of a charged particle given its drift velocity and the electric field, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Drift velocity (\( V_d \)) = \( 7.5 \times 10^{-4} \, \text{m/s} \)
- Electric field (\( E \)) = \( 3 \times 10^{-10} \, \text{V/m} \)

2. **Understand the Formula for Mobility**:
- The mobility (\( \mu \)) of a charged particle is defined by the equation:
\[ \mu = \frac{V_d}{E} \]

3. **Substitute the Given Values into the Formula**:
- Plug in the values of \( V_d \) and \( E \):
\[ \mu = \frac{7.5 \times 10^{-4} \, \text{m/s}}{3 \times 10^{-10} \, \text{V/m}} \]

4. **Perform the Calculation**:
- Calculate the mobility:
\[ \mu = \frac{7.5 \times 10^{-4}}{3 \times 10^{-10}} = 2.5 \times 10^{6} \, \text{m}^2 \, \text{V}^{-1} \, \text{s}^{-1} \]

5. **State the Final Answer**:
- The mobility of the charged particle is:
\[ \mu = 2.5 \times 10^{6} \, \text{m}^2 \, \text{V}^{-1} \, \text{s}^{-1} \]