An electron with an initial speed of `4.0 xx 10^6 ms^(-1)` is brought to rest by an electric field. The mass and charge of an electron are `9 xx 10^(-31)` kg and `1.6xx 10^(-19)C`, respectively. Identify the correct statement

Options

A. The electron moves from a region of lower potential to higher potential through a potential difference of 11.4 `muV`.
B. The electron moves from a region of higher potential to lower potential through a potential difference of 11.4 `muV`.
C. The electron moves from a region of lower potential to higher potential through a potential difference of 45 V.
D. The electron moves from a region of higher potential to lower potential through a potential difference of 45 V

Explanation

To solve the problem, we need to analyze the motion of the electron as it is brought to rest by an electric field. We will use the concepts of kinetic energy and electric potential energy.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- The initial speed of the electron, \( v_i = 4.0 \times 10^6 \, \text{m/s} \).
- The mass of the electron, \( m = 9 \times 10^{-31} \, \text{kg} \).
- The charge of the electron, \( q = 1.6 \times 10^{-19} \, \text{C} \).

2. **Calculate the initial kinetic energy (KE) of the electron:**
\[ KE = \frac{1}{2} mv^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \times (9 \times 10^{-31} \, \text{kg}) \times (4.0 \times 10^6 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times (9 \times 10^{-31}) \times (16 \times 10^{12}) \]
\[ KE = \frac{1}{2} \times (1.44 \times 10^{-18}) = 7.2 \times 10^{-19} \, \text{J} \]

3. **Relate the kinetic energy to electric potential energy (PE):**
When the electron comes to rest, all its kinetic energy is converted into electric potential energy:
\[ PE = qV \]
Where \( V \) is the potential difference. Setting \( KE = PE \):
\[ 7.2 \times 10^{-19} = (1.6 \times 10^{-19}) V \]

4. **Solve for the potential difference \( V \):**
\[ V = \frac{7.2 \times 10^{-19}}{1.6 \times 10^{-19}} = 4.5 \, \text{V} \]

5. **Determine the direction of the electron's motion:**
The electron, being negatively charged, moves from a region of higher potential to a region of lower potential when it is brought to rest by the electric field.

### Conclusion:
The correct statement is that the electron moves from a region of higher potential to a region of lower potential, and the potential difference required to stop the electron is \( 4.5 \, \text{V} \).