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 m
B. 0.7 mm
C. 7 cm
D. 0.7 cm

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the electric force acting on the electron
The electric force \( F \) acting on a charged particle in an electric field \( E \) is given by the formula:
\[ F = qE \]
where \( q \) is the charge of the electron and \( E \) is the electric field intensity.

Given:
- Charge of the electron \( q = 1.6 \times 10^{-19} \, \text{C} \)
- Electric field intensity \( E = 1 \times 10^{3} \, \text{N/C} \)

Substituting the values:
\[ F = (1.6 \times 10^{-19}) \times (1 \times 10^{3}) = 1.6 \times 10^{-16} \, \text{N} \]

### Step 2: Calculate the acceleration of the electron
Using Newton's second law, the acceleration \( a \) can be calculated using the formula:
\[ a = \frac{F}{m} \]
where \( m \) is the mass of the electron.

Given:
- Mass of the electron \( m = 9 \times 10^{-31} \, \text{kg} \)

Substituting the values:
\[ a = \frac{1.6 \times 10^{-16}}{9 \times 10^{-31}} \approx 1.78 \times 10^{14} \, \text{m/s}^2 \]

### Step 3: Use kinematic equations to find the distance traveled before coming to rest
We will use the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) is the final velocity (0 m/s, since the electron comes to rest),
- \( u \) is the initial velocity (\( 5 \times 10^{6} \, \text{m/s} \)),
- \( a \) is the acceleration (which will be negative since it is retardation),
- \( s \) is the distance traveled.

Rearranging the equation to solve for \( s \):
\[ 0 = (5 \times 10^{6})^2 + 2(-a)s \]
\[ (5 \times 10^{6})^2 = 2as \]
\[ s = \frac{(5 \times 10^{6})^2}{2a} \]

Substituting the values:
\[ s = \frac{(5 \times 10^{6})^2}{2 \times 1.78 \times 10^{14}} \]
Calculating \( (5 \times 10^{6})^2 \):
\[ (5 \times 10^{6})^2 = 25 \times 10^{12} \]
Now substituting:
\[ s = \frac{25 \times 10^{12}}{2 \times 1.78 \times 10^{14}} = \frac{25 \times 10^{12}}{3.56 \times 10^{14}} \approx 0.0704 \, \text{m} \]

### Final Answer
The distance traveled by the electron before coming to rest is approximately:
\[ s \approx 0.0704 \, \text{m} \approx 7.04 \, \text{cm} \]