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 need to determine the change in the de Broglie wavelength of ball A after it collides elastically with ball B.

### Step-by-Step Solution:

1. **Identify initial conditions**:
- Mass of ball A: \( m_A = \frac{m}{2} \)
- Mass of ball B: \( m_B = \frac{m}{3} \)
- Initial velocity of ball A: \( v_0 \)
- Initial velocity of ball B: \( u_B = 0 \) (at rest)
- Initial de Broglie wavelength of ball A: \( \lambda_0 \)

2. **Calculate initial momentum of ball A**:
\[ p_{A_i} = m_A \cdot v_0 = \frac{m}{2} \cdot v_0 \]

3. **Calculate initial de Broglie wavelength of ball A**:
Using the formula for de Broglie wavelength:
\[ \lambda_0 = \frac{h}{p_{A_i}} = \frac{h}{\frac{m}{2} \cdot v_0} = \frac{2h}{m v_0} \]

4. **Apply conservation of momentum**:
The total momentum before the collision equals the total momentum after the collision:
\[ m_A v_0 + m_B \cdot 0 = m_A v_A + m_B v_B \]
Substituting the values:
\[ \frac{m}{2} v_0 = \frac{m}{2} v_A + \frac{m}{3} v_B \]
Dividing through by \( m \):
\[ \frac{1}{2} v_0 = \frac{1}{2} v_A + \frac{1}{3} v_B \]
Multiplying through by 6 to eliminate fractions:
\[ 3v_0 = 3v_A + 2v_B \quad \text{(Equation 1)} \]

5. **Apply conservation of kinetic energy** (since the collision is elastic):
\[ \frac{1}{2} m_A v_0^2 = \frac{1}{2} m_A v_A^2 + \frac{1}{2} m_B v_B^2 \]
Dividing through by \( \frac{1}{2} m \):
\[ \frac{1}{2} v_0^2 = \frac{1}{2} v_A^2 + \frac{1}{3} v_B^2 \]
Multiplying through by 6:
\[ 3v_0^2 = 3v_A^2 + 2v_B^2 \quad \text{(Equation 2)} \]

6. **Solve the equations simultaneously**:
From Equation 1:
\[ v_B = \frac{3v_0 - 3v_A}{2} \]
Substitute \( v_B \) into Equation 2:
\[ 3v_0^2 = 3v_A^2 + 2\left(\frac{3v_0 - 3v_A}{2}\right)^2 \]
Expanding and simplifying:
\[ 3v_0^2 = 3v_A^2 + \frac{2(9v_0^2 - 18v_0v_A + 9v_A^2)}{4} \]
\[ 3v_0^2 = 3v_A^2 + \frac{9v_0^2 - 18v_0v_A + 9v_A^2}{2} \]
Multiplying through by 2 to eliminate the fraction:
\[ 6v_0^2 = 6v_A^2 + 9v_0^2 - 18v_0v_A + 9v_A^2 \]
Rearranging gives:
\[ 0 = 3v_0^2 - 18v_0v_A + 15v_A^2 \]

7. **Solve the quadratic equation** for \( v_A \):
Using the quadratic formula:
\[ v_A = \frac{18v_0 \pm \sqrt{(18v_0)^2 - 4 \cdot 15 \cdot 3v_0^2}}{2 \cdot 15} \]
Simplifying gives the values of \( v_A \) and \( v_B \).

8. **Calculate final momentum of ball A**:
\[ p_{A_f} = m_A v_A \]

9. **Calculate final de Broglie wavelength of ball A**:
\[ \lambda_f = \frac{h}{p_{A_f}} = \frac{h}{m_A v_A} \]

10. **Determine change in de Broglie wavelength**:
\[ \Delta \lambda = \lambda_f - \lambda_0 \]

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