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 of an electron moving in an electric field, we can follow these steps:

### Step 1: Calculate the Electric Force on the Electron
The electric force \( F \) acting on the electron can be calculated using the formula:
\[ F = qE \]
where:
- \( q \) is the charge of the electron \( (1.6 \times 10^{-19} \, \text{C}) \)
- \( E \) is the electric field intensity \( (1 \times 10^{3} \, \text{N/C}) \)

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 2: Calculate the Initial Kinetic Energy of the Electron
The initial kinetic energy \( KE \) of the electron can be calculated using the formula:
\[ KE = \frac{1}{2} mv^2 \]
where:
- \( m \) is the mass of the electron \( (9 \times 10^{-31} \, \text{kg}) \)
- \( v \) is the initial speed of the electron \( (5 \times 10^{6} \, \text{m/s}) \)

Substituting the values:
\[ KE = \frac{1}{2} (9 \times 10^{-31} \, \text{kg}) (5 \times 10^{6} \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} (9 \times 10^{-31}) (25 \times 10^{12}) = \frac{1}{2} (2.25 \times 10^{-18}) = 1.125 \times 10^{-18} \, \text{J} \]

### Step 3: Relate Work Done to Change in Kinetic Energy
The work done \( W \) by the electric force is equal to the change in kinetic energy. Since the electron comes to rest, the final kinetic energy is 0:
\[ W = KE_{final} - KE_{initial} = 0 - KE_{initial} = -KE_{initial} \]
The work done can also be expressed as:
\[ W = -F \cdot d \]
where \( d \) is the distance traveled by the electron before coming to rest.

### Step 4: Set Up the Equation
Setting the two expressions for work equal gives:
\[ -F \cdot d = -KE_{initial} \]
This simplifies to:
\[ F \cdot d = KE_{initial} \]
Substituting for \( F \):
\[ (1.6 \times 10^{-16} \, \text{N}) \cdot d = 1.125 \times 10^{-18} \, \text{J} \]

### Step 5: Solve for Distance \( d \)
Rearranging the equation to solve for \( d \):
\[ d = \frac{1.125 \times 10^{-18}}{1.6 \times 10^{-16}} \]
Calculating this gives:
\[ d = 7.03125 \times 10^{-3} \, \text{m} = 0.00703125 \, \text{m} = 7.03125 \, \text{cm} \]

### Final Result
Thus, the distance traveled by the electron before coming to rest is approximately:
\[ \boxed{7 \, \text{cm}} \]