An electron moving with the speed `5xx10^(6)` per sec is shot parallel to the electric field of intensity `1xx10^(3)N//C`. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant (mass of `e= 9xx10^(-31)Kg` charge `= 1.6xx10^(-19)C)`

Options

A. `7 cm`
B. `0.7 mm`
C. `7 cm`
D. `0.7 cm`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given values
- Initial speed of the electron, \( u = 5 \times 10^6 \, \text{m/s} \)
- Electric field intensity, \( E = 1 \times 10^3 \, \text{N/C} \)
- Mass of the electron, \( m = 9 \times 10^{-31} \, \text{kg} \)
- Charge of the electron, \( q = 1.6 \times 10^{-19} \, \text{C} \)

### Step 2: Calculate the force acting on the electron
The force \( F \) acting on the electron due to the electric field can be calculated using the formula:
\[ F = qE \]
Substituting the values:
\[ F = (1.6 \times 10^{-19} \, \text{C}) \times (1 \times 10^3 \, \text{N/C}) = 1.6 \times 10^{-16} \, \text{N} \]

### Step 3: Calculate the acceleration of the electron
Using Newton's second law, the acceleration \( a \) can be calculated as:
\[ a = \frac{F}{m} \]
Substituting the values:
\[ a = \frac{1.6 \times 10^{-16} \, \text{N}}{9 \times 10^{-31} \, \text{kg}} \approx 1.78 \times 10^{14} \, \text{m/s}^2 \]

### Step 4: Use kinematic equations to find the distance traveled before coming to rest
We need to find the distance \( S \) traveled by the electron before it comes to rest. The kinematic equation we will use is:
\[ v^2 = u^2 - 2aS \]
Where:
- \( v = 0 \) (final velocity when the electron comes to rest)
- \( u = 5 \times 10^6 \, \text{m/s} \) (initial velocity)
- \( a = 1.78 \times 10^{14} \, \text{m/s}^2 \) (acceleration)

Rearranging the equation to solve for \( S \):
\[ 0 = (5 \times 10^6)^2 - 2(1.78 \times 10^{14})S \]
\[ (5 \times 10^6)^2 = 2(1.78 \times 10^{14})S \]
\[ S = \frac{(5 \times 10^6)^2}{2 \times 1.78 \times 10^{14}} \]

### Step 5: Calculate the distance \( S \)
Calculating \( S \):
\[ S = \frac{25 \times 10^{12}}{3.56 \times 10^{14}} = \frac{25}{3.56} \times 10^{-2} \approx 0.07 \, \text{m} \]
Converting to centimeters:
\[ S \approx 7 \, \text{cm} \]

### Final Answer
The distance traveled by the electron before coming to rest is approximately **7 cm**.

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