If `lambda_(1)` and `lambda_(2)` denote the de-Broglie wavelength of two particles with same masses but charges in the ratio of `1:2` after they are accelerated from rest through the same potential difference, then

Options

A. `lambda_(1)=lambda_(2)`
B. `lambda_(1)lt lambda_(2)`
C. `lambda_(1)gtlambda_(2)`
D. unpredictable

Answer

To solve the problem, we need to analyze the relationship between the de Broglie wavelengths of two particles that have the same mass but different charges. The charges are in the ratio of 1:2, and both particles are accelerated through the same potential difference.

### Step-by-Step Solution:

1. **Understanding de Broglie Wavelength**:
The de Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{\sqrt{2mK}} \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( K \) is the kinetic energy of the particle.

2. **Kinetic Energy Relation**:
The kinetic energy (\( K \)) of a charged particle accelerated through a potential difference (\( V \)) is given by:
\[ K = qV \]
where \( q \) is the charge of the particle.

3. **Substituting Kinetic Energy**:
Substituting the expression for kinetic energy into the de Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{\sqrt{2mqV}} \]

4. **Calculating Wavelengths for Two Particles**:
Let’s denote the two particles as particle 1 and particle 2. Their charges are \( q_1 \) and \( q_2 \) respectively, with the ratio:
\[ \frac{q_1}{q_2} = \frac{1}{2} \]
This implies \( q_2 = 2q_1 \).

5. **Expressing Wavelengths**:
For particle 1:
\[ \lambda_1 = \frac{h}{\sqrt{2m q_1 V}} \]
For particle 2:
\[ \lambda_2 = \frac{h}{\sqrt{2m q_2 V}} = \frac{h}{\sqrt{2m (2q_1) V}} = \frac{h}{\sqrt{4m q_1 V}} = \frac{h}{2\sqrt{m q_1 V}} \]

6. **Finding the Ratio of Wavelengths**:
Now, we can find the ratio of the two wavelengths:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2m q_1 V}}}{\frac{h}{2\sqrt{m q_1 V}}} = \frac{2\sqrt{m q_1 V}}{\sqrt{2m q_1 V}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

7. **Conclusion**:
Since \( \frac{\lambda_1}{\lambda_2} = \sqrt{2} \), we conclude that:
\[ \lambda_1 > \lambda_2 \]
Therefore, the correct option is that \( \lambda_1 \) is greater than \( \lambda_2 \).

### Final Answer:
\[ \lambda_1 > \lambda_2 \]