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. **Recall the formula for mobility (\( \mu \))**:
- Mobility is defined as the ratio of drift velocity to the electric field:
\[ \mu = \frac{V_d}{E} \]

3. **Substitute the given values into the formula**:
\[ \mu = \frac{7.5 \times 10^{-4} \, \text{m/s}}{3 \times 10^{-10} \, \text{V/m}} \]

4. **Perform the division**:
- First, divide the coefficients:
\[ \frac{7.5}{3} = 2.5 \]
- Next, handle the powers of ten:
\[ \frac{10^{-4}}{10^{-10}} = 10^{-4 - (-10)} = 10^{-4 + 10} = 10^{6} \]
- Therefore, combining these results:
\[ \mu = 2.5 \times 10^{6} \, \text{m}^2/\text{V s} \]

5. **Write the final answer with the appropriate units**:
- The mobility of the charged particle is:
\[ \mu = 2.5 \times 10^{6} \, \text{m}^2/\text{V s} \]