A praticle of mass M at rest decays into two particle of masses `m_1` and `m_2`, having non-zero velocities. The ratio of the de Broglie wavelength of the particles `(lamda_1)/(lamda_2)` is

Options

A. `(m_1)/(m_2)`
B. `(m_2)/(m_1)`
C. 1
D. `(sqrtm_2)/(sqrtm_1)`

Explanation

To solve the problem, we need to analyze the decay of a particle of mass \( M \) into two particles of masses \( m_1 \) and \( m_2 \) and find the ratio of their de Broglie wavelengths \( \frac{\lambda_1}{\lambda_2} \).

### Step-by-Step Solution:

1. **Understanding the Conservation of Momentum**:
- Initially, the particle of mass \( M \) is at rest, so its initial momentum is \( 0 \).
- After decay, the two particles \( m_1 \) and \( m_2 \) will have velocities \( v_1 \) and \( v_2 \) respectively.
- By conservation of momentum:
\[ M \cdot 0 = m_1 v_1 + m_2 v_2 \]
This simplifies to:
\[ m_1 v_1 + m_2 v_2 = 0 \]
- Rearranging gives:
\[ m_1 v_1 = -m_2 v_2 \]

2. **Finding the de Broglie Wavelengths**:
- The de Broglie wavelength \( \lambda \) of a particle is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the particle.
- For particle 1 (mass \( m_1 \)):
\[ \lambda_1 = \frac{h}{m_1 v_1} \]
- For particle 2 (mass \( m_2 \)):
\[ \lambda_2 = \frac{h}{m_2 v_2} \]

3. **Finding the Ratio of the Wavelengths**:
- Now, we need to find the ratio \( \frac{\lambda_1}{\lambda_2} \):
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{m_1 v_1}}{\frac{h}{m_2 v_2}} = \frac{m_2 v_2}{m_1 v_1} \]

4. **Substituting the Momentum Relation**:
- From the momentum conservation equation \( m_1 v_1 = -m_2 v_2 \), we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = -\frac{m_1}{m_2} v_1 \]
- Substituting this into the ratio:
\[ \frac{\lambda_1}{\lambda_2} = \frac{m_2 \left(-\frac{m_1}{m_2} v_1\right)}{m_1 v_1} = -1 \]

5. **Final Result**:
- Since we are interested in the absolute value of the ratio, we conclude:
\[ \frac{\lambda_1}{\lambda_2} = 1 \]

### Conclusion:
The ratio of the de Broglie wavelengths of the two particles is \( \frac{\lambda_1}{\lambda_2} = 1 \).