A particle of mass `4m` at rest decays into two particles of masses `m` and `3m` having non-zero velocities. The ratio of the de-Broglie wavelengths of the particles 1 and 2 is

Options

A. `(1)/(2)`
B. `(1)/(4)`
C. `2`
D. `1`

Explanation

To solve the problem, we need to find the ratio of the de-Broglie wavelengths of two particles resulting from the decay of a particle of mass `4m` into two particles of masses `m` and `3m`.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- We have a particle of mass `4m` at rest. Therefore, its initial momentum is zero.

2. **Apply Conservation of Momentum**:
- After the decay, the two particles have masses `m` and `3m`, and their velocities will be denoted as `v1` and `v2`, respectively.
- According to the conservation of momentum:
\[ 0 = mv_1 + 3mv_2 \]
- From this equation, we can simplify to:
\[ mv_1 = -3mv_2 \]
- Dividing through by `m` (assuming `m` is not zero):
\[ v_1 = -3v_2 \]

3. **Calculate the de-Broglie Wavelengths**:
- The de-Broglie wavelength \(\lambda\) is given by:
\[ \lambda = \frac{h}{p} \]
- Where \(p\) is the momentum of the particle, which can be expressed as \(p = mv\).
- Therefore, the wavelengths for the two particles can be expressed as:
\[ \lambda_1 = \frac{h}{mv_1} \]
\[ \lambda_2 = \frac{h}{3mv_2} \]

4. **Find the Ratio of the Wavelengths**:
- We want to find the ratio \(\frac{\lambda_1}{\lambda_2}\):
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{mv_1}}{\frac{h}{3mv_2}} = \frac{3mv_2}{mv_1} \]
- The \(h\) and \(m\) cancel out:
\[ \frac{\lambda_1}{\lambda_2} = \frac{3v_2}{v_1} \]
- Substituting \(v_1 = -3v_2\):
\[ \frac{\lambda_1}{\lambda_2} = \frac{3v_2}{-3v_2} = -1 \]

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

### Conclusion:
The ratio of the de-Broglie wavelengths of the particles is \(1\).