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, focusing on the concepts of momentum conservation and de Broglie wavelength.

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

### Step 2: Define the velocities
- Let the velocity of nucleus B be \( V_B \).
- Therefore, the velocity of nucleus C will be \( V_C = \frac{1}{2} V_B \).

### Step 3: Apply conservation of momentum
Before the fission, the momentum of nucleus A is:
\[ P_A = m_A V_A \]
After the fission, the momentum of nuclei B and C can be expressed as:
\[ P_B + P_C = m_B V_B - m_C V_C \]
Since the masses of B and C are equal and denoted as \( m \), we can write:
\[ P_B + P_C = m V_B - m \left(-\frac{1}{2} V_B\right) = m V_B + \frac{1}{2} m V_B = \frac{3}{2} m V_B \]
Setting the initial momentum equal to the final momentum gives:
\[ m_A V_A = \frac{3}{2} m V_B \]

### Step 4: Solve for \( V_B \)
Assuming \( m_A = 2m \) (since A splits into two equal masses):
\[ 2m V_A = \frac{3}{2} m V_B \]
Dividing both sides by \( m \):
\[ 2 V_A = \frac{3}{2} V_B \]
Thus,
\[ V_B = \frac{4}{3} V_A \]

### Step 5: Calculate the de Broglie wavelengths
The de Broglie wavelength is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum.

#### For nucleus B:
The momentum of nucleus B is:
\[ p_B = m V_B = m \left(\frac{4}{3} V_A\right) = \frac{4m}{3} V_A \]
Thus, the de Broglie wavelength of B is:
\[ \lambda_B = \frac{h}{p_B} = \frac{h}{\frac{4m}{3} V_A} = \frac{3h}{4m V_A} \]

#### For nucleus C:
The momentum of nucleus C is:
\[ p_C = m V_C = m \left(-\frac{1}{2} V_B\right) = -\frac{1}{2} m \left(\frac{4}{3} V_A\right) = -\frac{2m}{3} V_A \]
Thus, the de Broglie wavelength of C is:
\[ \lambda_C = \frac{h}{|p_C|} = \frac{h}{\frac{2m}{3} V_A} = \frac{3h}{2m V_A} \]

### Step 6: Relate the wavelengths to \( \lambda_A \)
Since \( \lambda_A = \frac{h}{m V_A} \):
- For nucleus B:
\[ \lambda_B = \frac{3h}{4m V_A} = \frac{3}{4} \lambda_A \]

- For nucleus C:
\[ \lambda_C = \frac{3h}{2m V_A} = \frac{3}{2} \lambda_A \]

### Final Results
Thus, the de Broglie wavelengths are:
\[ \lambda_B = \frac{3}{4} \lambda_A, \quad \lambda_C = \frac{3}{2} \lambda_A \]