Two particles `A_1 and B_2` of masses `m_1 and m_2 (m_1 gt m_2)` have the same de Broglie wavelength . Then

Options

A. their momenta are the same.
B. their energies are the same
C. the energy of `A_1` is less than the energy of `A_2`
D. the 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 masses of two particles and their de Broglie wavelengths. Let's break it down step by step:

### Step 1: Understanding de Broglie Wavelength
The de Broglie wavelength (\( \lambda \)) 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.

### Step 2: Momentum of the Particles
The momentum (\( p \)) of a particle can also be expressed in terms of its mass (\( m \)) and velocity (\( v \)):
\[ p = mv \]
Thus, for two particles \( A_1 \) and \( B_2 \) with masses \( m_1 \) and \( m_2 \) (where \( m_1 > m_2 \)), we can write:
\[ \lambda_1 = \frac{h}{m_1 v_1} \quad \text{and} \quad \lambda_2 = \frac{h}{m_2 v_2} \]

### Step 3: Setting the Wavelengths Equal
Since both particles have the same de Broglie wavelength, we can set the equations equal to each other:
\[ \frac{h}{m_1 v_1} = \frac{h}{m_2 v_2} \]
By canceling \( h \) from both sides, we get:
\[ \frac{1}{m_1 v_1} = \frac{1}{m_2 v_2} \]
Cross-multiplying gives us:
\[ m_2 v_2 = m_1 v_1 \]

### Step 4: Analyzing the Relationship
From the equation \( m_2 v_2 = m_1 v_1 \), we can express the velocities in terms of each other:
\[ v_2 = \frac{m_1}{m_2} v_1 \]
Since \( m_1 > m_2 \), it follows that \( v_2 > v_1 \). This implies that the particle with the smaller mass (particle \( B_2 \)) must have a greater velocity to maintain the same de Broglie wavelength.

### Step 5: Kinetic Energy Comparison
The kinetic energy (\( K \)) of a particle is given by:
\[ K = \frac{1}{2} mv^2 \]
Substituting for \( v \) from our earlier relationship:
\[ K_1 = \frac{1}{2} m_1 v_1^2 \quad \text{and} \quad K_2 = \frac{1}{2} m_2 v_2^2 \]
Substituting \( v_2 = \frac{m_1}{m_2} v_1 \) into the kinetic energy of particle \( B_2 \):
\[ K_2 = \frac{1}{2} m_2 \left(\frac{m_1}{m_2} v_1\right)^2 = \frac{1}{2} m_2 \frac{m_1^2}{m_2^2} v_1^2 = \frac{1}{2} \frac{m_1^2}{m_2} v_1^2 \]
Now, comparing \( K_1 \) and \( K_2 \):
\[ K_1 = \frac{1}{2} m_1 v_1^2 \quad \text{and} \quad K_2 = \frac{1}{2} \frac{m_1^2}{m_2} v_1^2 \]
Since \( m_1 > m_2 \), it follows that \( K_2 > K_1 \).

### Conclusion
Thus, we conclude that:
- The momenta of both particles are equal.
- The kinetic energy of particle \( B_2 \) is greater than that of particle \( A_1 \).

### Final Answer
The correct conclusion is that the kinetic energy of \( B_2 \) is greater than that of \( A_1 \).

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