An electron is accelerated under a potential difference of 182 V. the maximum velocity of electron will be (Given, charge of an electron is `1.6xx10^(-19)C` `ms^(-1)` and its mass is `9.1xx10^(-31)kg`)

Options

A. `5.65xx10^(6)ms^(-1)`
B. `4xx10^(6)ms^(-1)`
C. `8xx10^(6)ms^(-1)`
D. `16xx10^(-6)ms^(-1)`

Answer

To find the maximum velocity of an electron accelerated through a potential difference of 182 V, we can use the work-energy theorem. The work done by the electric field on the electron is equal to the change in its kinetic energy.

### Step-by-Step Solution:

1. **Understanding the Work-Energy Theorem**:
The work done (W) on the electron by the electric field is given by:
\[ W = Q \times V \]
where \( Q \) is the charge of the electron and \( V \) is the potential difference.

2. **Kinetic Energy Relation**:
According to the work-energy theorem, the work done is equal to the change in kinetic energy (KE):
\[ W = \Delta KE = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the electron and \( v \) is its final velocity.

3. **Setting Up the Equation**:
From the above two equations, we can equate the work done to the kinetic energy:
\[ Q \times V = \frac{1}{2} m v^2 \]

4. **Substituting Known Values**:
Given:
- Charge of the electron, \( Q = 1.6 \times 10^{-19} \, C \)
- Mass of the electron, \( m = 9.1 \times 10^{-31} \, kg \)
- Potential difference, \( V = 182 \, V \)

Substituting these values into the equation:
\[ 1.6 \times 10^{-19} \times 182 = \frac{1}{2} \times 9.1 \times 10^{-31} \times v^2 \]

5. **Calculating the Left Side**:
Calculate \( 1.6 \times 10^{-19} \times 182 \):
\[ 1.6 \times 182 = 291.2 \times 10^{-19} = 2.912 \times 10^{-17} \, J \]

6. **Rearranging for \( v^2 \)**:
Now, rearranging the equation to solve for \( v^2 \):
\[ v^2 = \frac{2 \times 2.912 \times 10^{-17}}{9.1 \times 10^{-31}} \]

7. **Calculating \( v^2 \)**:
Calculate the right side:
\[ v^2 = \frac{5.824 \times 10^{-17}}{9.1 \times 10^{-31}} \approx 6.3967 \times 10^{13} \]

8. **Finding \( v \)**:
Taking the square root to find \( v \):
\[ v = \sqrt{6.3967 \times 10^{13}} \approx 8.0 \times 10^{6} \, m/s \]

### Final Answer:
The maximum velocity of the electron will be approximately:
\[ v \approx 8.0 \times 10^{6} \, m/s \]