A physicist was performing experiments to study the effect of varying voltage on the velocity and wavelength of the electrons. In one case, the electron was accelerated through a potential difference of 1kV and in the second case, it was accelerated through a potential difference of 2kV
The velocity acquired by the electron will be

Options

A. double in the second case than in the first case
B. four times in the second case than in the first case
C. same in both cases
D. 1.4 times in the second case than in the first case

Explanation

To solve the problem of finding the velocity acquired by an electron when accelerated through different potential differences, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Kinetic Energy**:
The kinetic energy (KE) acquired by an electron when it is accelerated through a potential difference (V) is given by the formula:
\[ KE = e \cdot V \]
where \( e \) is the charge of the electron and \( V \) is the potential difference.

2. **Relating Kinetic Energy to Velocity**:
The kinetic energy can also be expressed in terms of velocity (v) as:
\[ KE = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron.

3. **Setting Up Equations for Each Case**:
For the first case (1 kV):
\[ \frac{1}{2} mv_1^2 = e \cdot 1000 \quad \text{(since 1 kV = 1000 V)} \]
For the second case (2 kV):
\[ \frac{1}{2} mv_2^2 = e \cdot 2000 \quad \text{(since 2 kV = 2000 V)} \]

4. **Taking the Ratio of the Two Cases**:
We can take the ratio of the two kinetic energy equations:
\[ \frac{\frac{1}{2} mv_1^2}{\frac{1}{2} mv_2^2} = \frac{e \cdot 1000}{e \cdot 2000} \]
This simplifies to:
\[ \frac{v_1^2}{v_2^2} = \frac{1000}{2000} = \frac{1}{2} \]

5. **Finding the Relationship Between Velocities**:
Taking the square root of both sides gives:
\[ \frac{v_1}{v_2} = \frac{1}{\sqrt{2}} \]
Rearranging this gives:
\[ v_2 = \sqrt{2} \cdot v_1 \]

6. **Calculating the Approximate Value**:
The value of \( \sqrt{2} \) is approximately 1.414. Thus:
\[ v_2 \approx 1.414 \cdot v_1 \]

### Final Answer:
The velocity acquired by the electron in the second case (2 kV) is approximately 1.414 times the velocity acquired in the first case (1 kV).