Two particles `A_(1) and A_(2)` of masses `m_(1), m_(2) (m_(1)gtm_(2))` have the same de-broglie wavelength. Then

Options

A. their momenta are the same.
B. their energies are the same.
C. momentum of `A_(1)` is less than the momentum of `A_(2)`.
D. energy of `A_(1)` is more than the energy of `A_(2)`.

Explanation

To solve the problem, we need to analyze the relationship between the de Broglie wavelength, momentum, and kinetic energy of the two particles A1 and A2.

### Step-by-Step Solution:

1. **Understanding de Broglie Wavelength**:
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the particle.

2. **Setting up the relationship for both particles**:
Since both particles A1 and A2 have the same de Broglie wavelength, we can write:
\[ \lambda_1 = \lambda_2 \]
This implies:
\[ \frac{h}{p_1} = \frac{h}{p_2} \]
Hence, we can conclude:
\[ p_1 = p_2 \]

3. **Relating momentum to kinetic energy**:
The momentum \(p\) of a particle is related to its kinetic energy \(K\) by the equation:
\[ p = \sqrt{2mK} \]
For particles A1 and A2, we can write:
\[ p_1 = \sqrt{2m_1K_1} \]
\[ p_2 = \sqrt{2m_2K_2} \]

4. **Equating the momenta**:
Since \(p_1 = p_2\), we can set the two equations equal to each other:
\[ \sqrt{2m_1K_1} = \sqrt{2m_2K_2} \]
Squaring both sides gives:
\[ 2m_1K_1 = 2m_2K_2 \]
Simplifying this, we have:
\[ m_1K_1 = m_2K_2 \]

5. **Analyzing the relationship**:
Given that \(m_1 > m_2\), we can rearrange the equation:
\[ \frac{K_1}{K_2} = \frac{m_2}{m_1} \]
Since \(m_2 < m_1\), it follows that:
\[ K_1 < K_2 \]
This means that the kinetic energy of particle A1 is less than that of particle A2.

### Conclusion:
1. The momenta of both particles are equal: \(p_1 = p_2\).
2. The kinetic energy of particle A1 is less than that of particle A2: \(K_1 < K_2\).

### Final Answer:
- The correct option is that the momenta are the same, and the energy of A1 is less than the energy of A2.