A physicist was performing experiments to study the velociyt and wavelength of the electron. In one case, the electron was accelerated through a potential differences of 1 KV and in 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.

Answer

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

### Step-by-Step Solution

1. **Understand the relationship between kinetic energy and potential difference**:
The kinetic energy (KE) acquired by an electron when it is accelerated through a potential difference (V) is given by:
\[ KE = eV \]
where \( e \) is the charge of the electron.

2. **Express kinetic energy in terms of velocity**:
The kinetic energy can also be expressed in terms of the mass (m) and velocity (v) of the electron:
\[ KE = \frac{1}{2} mv^2 \]

3. **Set the two expressions for kinetic energy equal**:
Since both expressions represent the kinetic energy of the electron, we can set them equal to each other:
\[ eV = \frac{1}{2} mv^2 \]

4. **Rearrange the equation to solve for velocity**:
Rearranging the equation to solve for velocity \( v \):
\[ v^2 = \frac{2eV}{m} \]
Taking the square root of both sides gives:
\[ v = \sqrt{\frac{2eV}{m}} \]

5. **Determine the relationship between velocity and potential difference**:
From the equation \( v = \sqrt{\frac{2eV}{m}} \), we can see that the velocity \( v \) is directly proportional to the square root of the potential difference \( V \):
\[ v \propto \sqrt{V} \]

6. **Calculate the velocities for the given potential differences**:
Let \( V_1 = 1 \, \text{kV} \) and \( V_2 = 2 \, \text{kV} \). The velocities \( v_1 \) and \( v_2 \) can be expressed as:
\[ v_1 = k\sqrt{V_1} = k\sqrt{1} = k \]
\[ v_2 = k\sqrt{V_2} = k\sqrt{2} \]
where \( k = \sqrt{\frac{2e}{m}} \).

7. **Find the ratio of the velocities**:
The ratio of the velocities \( v_1 \) and \( v_2 \) can be calculated as follows:
\[ \frac{v_1}{v_2} = \frac{k}{k\sqrt{2}} = \frac{1}{\sqrt{2}} \]
Therefore, we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = v_1 \sqrt{2} \approx 1.414 v_1 \]

### Final Answer
The velocity acquired by the electron when accelerated through 2 kV is approximately \( 1.414 \) times the velocity acquired when accelerated through 1 kV.