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 wavelength associated with the electron will be

Options

A. double in the second case than in the first case
B. double in the first case than in the second case
C. 1.4 times in the second case than in the first case
D. 1.4 times in the first case than in the second case

Explanation

To solve the problem of finding the wavelength associated with an electron 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) gained by an electron when accelerated through a potential difference (V) is given by:
\[ KE = eV \]
where \( e \) is the charge of the electron.

2. **Calculate kinetic energy for both potential differences**:
- For the first case (1 kV):
\[ KE_1 = e \times 1000 \, \text{V} \]
- For the second case (2 kV):
\[ KE_2 = e \times 2000 \, \text{V} \]

3. **Relate wavelength to momentum**:
The de Broglie wavelength (\( \lambda \)) of an electron is given by:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the electron.

4. **Express momentum in terms of kinetic energy**:
The momentum \( p \) can be expressed as:
\[ p = \sqrt{2m \times KE} \]
where \( m \) is the mass of the electron.

5. **Substitute kinetic energy into the wavelength formula**:
For the first case (1 kV):
\[ \lambda_1 = \frac{h}{\sqrt{2m \times KE_1}} = \frac{h}{\sqrt{2m \times e \times 1000}} \]
For the second case (2 kV):
\[ \lambda_2 = \frac{h}{\sqrt{2m \times KE_2}} = \frac{h}{\sqrt{2m \times e \times 2000}} \]

6. **Take the ratio of the wavelengths**:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2m \times e \times 1000}}}{\frac{h}{\sqrt{2m \times e \times 2000}}} \]
This simplifies to:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\sqrt{2000}}{\sqrt{1000}} = \sqrt{2} \]

7. **Calculate the relationship between the wavelengths**:
Therefore, we can express this as:
\[ \lambda_1 = \sqrt{2} \times \lambda_2 \]
Since \( \sqrt{2} \approx 1.414 \), we can say:
\[ \lambda_1 \approx 1.4 \times \lambda_2 \]

8. **Conclusion**:
The wavelength associated with the electron accelerated through 1 kV is approximately 1.4 times the wavelength associated with the electron accelerated through 2 kV.

### Final Answer:
The wavelength associated with the electron accelerated through 1 kV is approximately 1.4 times that of the wavelength associated with the electron accelerated through 2 kV.