If a ball A of mass `m_A = m/2` moving along x-axis collides elastically with ball B of mass `m_B = m/3` initially at rest and they move along x-axis. If initially velocity of ball A was `v_0`. And initially de-broglie wavelength of ball A is `lambda_0`. Find change in debroglie wavelength `delta lambda` of ball A in terms of `lambda_0`

Options

A. 3/2 `lambda_0`
B. 5/3 `lambda_0`
C. 4 `lambda_0`
D. 3 `lambda_0`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
- Ball A has mass \( m_A = \frac{m}{2} \) and an initial velocity \( v_0 \).
- Ball B has mass \( m_B = \frac{m}{3} \) and is initially at rest.

### Step 2: Write the initial momentum
The initial momentum of the system is given by:
\[ p_{\text{initial}} = m_A \cdot v_0 + m_B \cdot 0 = \frac{m}{2} \cdot v_0 \]

### Step 3: Write the final momentum
After the elastic collision, let the final velocities of balls A and B be \( v_A \) and \( v_B \) respectively. The final momentum is:
\[ p_{\text{final}} = m_A \cdot v_A + m_B \cdot v_B = \frac{m}{2} v_A + \frac{m}{3} v_B \]

### Step 4: Apply conservation of momentum
According to the conservation of momentum:
\[ \frac{m}{2} v_0 = \frac{m}{2} v_A + \frac{m}{3} v_B \]
Dividing through by \( m \) (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v_0 = \frac{1}{2} v_A + \frac{1}{3} v_B \]

### Step 5: Apply conservation of kinetic energy
Since the collision is elastic, kinetic energy is also conserved:
\[ \frac{1}{2} m_A v_0^2 = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 \]
This simplifies to:
\[ \frac{1}{2} \cdot \frac{m}{2} v_0^2 = \frac{1}{2} \cdot \frac{m}{2} v_A^2 + \frac{1}{2} \cdot \frac{m}{3} v_B^2 \]
Dividing through by \( \frac{m}{2} \):
\[ v_0^2 = v_A^2 + \frac{2}{3} v_B^2 \]

### Step 6: Solve the equations
Now we have two equations:
1. \( \frac{1}{2} v_0 = \frac{1}{2} v_A + \frac{1}{3} v_B \) (momentum conservation)
2. \( v_0^2 = v_A^2 + \frac{2}{3} v_B^2 \) (energy conservation)

From the first equation, we can express \( v_B \) in terms of \( v_A \):
\[ v_B = 3v_0 - 3v_A \]

Substituting \( v_B \) into the second equation and solving for \( v_A \) will yield the final velocities.

### Step 7: Calculate the de Broglie wavelength
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{mv} \]
Initially, for ball A:
\[ \lambda_0 = \frac{h}{m_A v_0} = \frac{h}{\frac{m}{2} v_0} = \frac{2h}{mv_0} \]

After the collision, the new de Broglie wavelength \( \lambda' \) is:
\[ \lambda' = \frac{h}{m_A v_A} = \frac{h}{\frac{m}{2} v_A} = \frac{2h}{mv_A} \]

### Step 8: Find the change in de Broglie wavelength
The change in de Broglie wavelength \( \Delta \lambda \) is given by:
\[ \Delta \lambda = \lambda' - \lambda_0 = \frac{2h}{mv_A} - \frac{2h}{mv_0} \]
Factoring out \( \frac{2h}{m} \):
\[ \Delta \lambda = \frac{2h}{m} \left( \frac{1}{v_A} - \frac{1}{v_0} \right) \]

### Step 9: Express in terms of \( \lambda_0 \)
Using the relationship between \( v_A \) and \( v_0 \) derived from the previous steps, we can express \( \Delta \lambda \) in terms of \( \lambda_0 \).

### Final Result
After solving, we find that:
\[ \Delta \lambda = 4 \lambda_0 \]