A particle of mass 4m at rest decays into two particles of masses m and 3m having nonzero velocities. The ratio of the de Broglie wavelengths of the particles 1 and 2 is (take h = `6.6 xx 10^(-34) m^2 kg s^-1` )

Options

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

Answer

To solve the problem of finding the ratio of the de Broglie wavelengths of the two particles resulting from the decay of a particle of mass \(4m\), we will follow these steps:

### Step 1: Understand the system
We have a particle of mass \(4m\) at rest that decays into two particles of masses \(m\) and \(3m\). Since the initial particle is at rest, the total initial momentum is zero.

### Step 2: Apply conservation of momentum
According to the law of conservation of momentum, the total momentum before the decay must equal the total momentum after the decay. Thus, we can write:

\[ P_{\text{initial}} = P_{\text{final}} \]

Since the initial momentum is zero, we have:

\[ 0 = P_1 + P_2 \]

Where \(P_1\) is the momentum of the particle with mass \(m\) and \(P_2\) is the momentum of the particle with mass \(3m\). This implies:

\[ P_1 = -P_2 \]

### Step 3: Express momentum in terms of mass and velocity
The momentum of each particle can be expressed as:

\[ P_1 = m v_1 \quad \text{and} \quad P_2 = 3m v_2 \]

Substituting these into the momentum conservation equation gives:

\[ m v_1 = -3m v_2 \]

Taking magnitudes, we have:

\[ v_1 = 3 v_2 \]

### Step 4: Calculate de Broglie wavelengths
The de Broglie wavelength \(\lambda\) of a particle is given by the formula:

\[ \lambda = \frac{h}{P} \]

Thus, we can write the de Broglie wavelengths for both particles:

\[ \lambda_m = \frac{h}{P_1} = \frac{h}{m v_1} \]
\[ \lambda_{3m} = \frac{h}{P_2} = \frac{h}{3m v_2} \]

### Step 5: Find the ratio of the de Broglie wavelengths
Now, we can find the ratio of the de Broglie wavelengths:

\[ \frac{\lambda_{3m}}{\lambda_m} = \frac{\frac{h}{3m v_2}}{\frac{h}{m v_1}} = \frac{m v_1}{3m v_2} = \frac{v_1}{3 v_2} \]

### Step 6: Substitute \(v_1\) in terms of \(v_2\)
From step 3, we know that \(v_1 = 3 v_2\). Substituting this into the ratio gives:

\[ \frac{\lambda_{3m}}{\lambda_m} = \frac{3 v_2}{3 v_2} = 1 \]

### Conclusion
Thus, the ratio of the de Broglie wavelengths of the particles of masses \(m\) and \(3m\) is:

\[ \frac{\lambda_{3m}}{\lambda_m} = 1 \]

### Final Answer
The ratio of the de Broglie wavelengths of the two particles is \(1\).

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