A nucleus A, With a finite de-broglie wavelength `lambda_(A)`, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, While C flies in the opposite direction with a velocity equal to half of that of B . The de-Broglie wavelength `lambda_(B)` and `lambda_(B)` and C are respectively:

Options

A. `lambda_(A),2lambda_(A)`
B. `2lambda_(A),lambda_(A)`
C. `lambda_(A),lambda_(A)/2`
D. `lambda_(A)/2,lambda_(A)`

Answer

To solve the problem, we need to analyze the situation step by step, applying the principles of momentum conservation and the de Broglie wavelength concept.

### Step 1: Understand the system
A nucleus A with a de Broglie wavelength \( \lambda_A \) undergoes spontaneous fission into two nuclei B and C of equal mass. Nucleus B moves in the same direction as A, while nucleus C moves in the opposite direction with a velocity equal to half of that of B.

### Step 2: Define the initial conditions
Let:
- The mass of nucleus A be \( m \).
- The initial velocity of nucleus A be \( v_0 \).
- The mass of nuclei B and C be \( \frac{m}{2} \) each (since they are of equal mass).
- The velocity of nucleus B be \( v \).
- The velocity of nucleus C be \( \frac{v}{2} \) (since it moves in the opposite direction).

### Step 3: Apply conservation of momentum
According to the law of conservation of momentum:
\[ \text{Initial momentum} = \text{Final momentum} \]
\[ m v_0 = \frac{m}{2} v - \frac{m}{2} \left(\frac{v}{2}\right) \]
Cancelling \( m \) from both sides gives:
\[ v_0 = \frac{1}{2} v - \frac{1}{4} v \]
This simplifies to:
\[ v_0 = \frac{1}{4} v \]
From this, we can solve for \( v \):
\[ v = 4 v_0 \]

### Step 4: Calculate the de Broglie wavelengths
The de Broglie wavelength \( \lambda \) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum.

#### For nucleus B:
The momentum of nucleus B is:
\[ p_B = \frac{m}{2} v = \frac{m}{2} (4 v_0) = 2 m v_0 \]
Thus, the de Broglie wavelength of B is:
\[ \lambda_B = \frac{h}{p_B} = \frac{h}{2 m v_0} \]

#### For nucleus C:
The momentum of nucleus C is:
\[ p_C = \frac{m}{2} \left(\frac{v}{2}\right) = \frac{m}{2} \left(2 v_0\right) = m v_0 \]
Thus, the de Broglie wavelength of C is:
\[ \lambda_C = \frac{h}{p_C} = \frac{h}{m v_0} \]

### Step 5: Relate the wavelengths to \( \lambda_A \)
We know that:
\[ \lambda_A = \frac{h}{m v_0} \]
From this, we can express \( \lambda_B \) and \( \lambda_C \) in terms of \( \lambda_A \):
- For nucleus B:
\[ \lambda_B = \frac{h}{2 m v_0} = \frac{1}{2} \lambda_A \]
- For nucleus C:
\[ \lambda_C = \frac{h}{m v_0} = \lambda_A \]

### Final Answer
Thus, the de Broglie wavelengths are:
- \( \lambda_B = \frac{1}{2} \lambda_A \)
- \( \lambda_C = \lambda_A \)