A particle A of mass m and initial velocity v collides with a particle of mass `m//2` which is at rest. The collision is head on, and elastic. The ratio of the de-broglie wavelength `lambda_(A) and lambda_(B)` after the collision is

Options

A. `(lambda_(A))/(lambda_(B)) =2/3`
B. `(lambda_(A))/(lambda_(B)) =1/2`
C. `(lambda_(A))/(lambda_(B)) =1/3`
D. `(lambda_(A))/(lambda_(B)) =2`

Explanation

To solve the problem, we need to find the ratio of the de Broglie wavelengths of two particles A and B after a head-on elastic collision.

### Step-by-Step Solution:

1. **Identify the Masses and Initial Velocities:**
- Let the mass of particle A be \( m \) and its initial velocity be \( v \).
- The mass of particle B is \( \frac{m}{2} \), and it is initially at rest (velocity = 0).

2. **Conservation of Momentum:**
- According to the law of conservation of momentum:
\[ m v + \frac{m}{2} \cdot 0 = m v_A' + \frac{m}{2} v_B' \]
- This simplifies to:
\[ mv = m v_A' + \frac{m}{2} v_B' \]
- Dividing through by \( m \):
\[ v = v_A' + \frac{1}{2} v_B' \quad \text{(Equation 1)} \]

3. **Conservation of Kinetic Energy:**
- For an elastic collision, the total kinetic energy is conserved:
\[ \frac{1}{2} m v^2 + \frac{1}{2} \cdot \frac{m}{2} \cdot 0^2 = \frac{1}{2} m v_A'^2 + \frac{1}{2} \cdot \frac{m}{2} v_B'^2 \]
- This simplifies to:
\[ \frac{1}{2} mv^2 = \frac{1}{2} m v_A'^2 + \frac{1}{4} mv_B'^2 \]
- Dividing through by \( \frac{1}{2} m \):
\[ v^2 = v_A'^2 + \frac{1}{2} v_B'^2 \quad \text{(Equation 2)} \]

4. **Solving the Equations:**
- From Equation 1, we can express \( v_A' \) in terms of \( v_B' \):
\[ v_A' = v - \frac{1}{2} v_B' \]
- Substitute \( v_A' \) into Equation 2:
\[ v^2 = \left(v - \frac{1}{2} v_B'\right)^2 + \frac{1}{2} v_B'^2 \]
- Expanding the left-hand side:
\[ v^2 = v^2 - v v_B' + \frac{1}{4} v_B'^2 + \frac{1}{2} v_B'^2 \]
- Combining terms:
\[ v^2 = v^2 - v v_B' + \frac{3}{4} v_B'^2 \]
- Rearranging gives:
\[ 0 = -v v_B' + \frac{3}{4} v_B'^2 \]
- Factoring out \( v_B' \):
\[ v_B' \left(\frac{3}{4} v_B' - v\right) = 0 \]
- Thus, \( v_B' = \frac{4}{3} v \).

5. **Finding \( v_A' \):**
- Substitute \( v_B' \) back into Equation 1:
\[ v_A' = v - \frac{1}{2} \cdot \frac{4}{3} v = v - \frac{2}{3} v = \frac{1}{3} v. \]

6. **Calculating the De Broglie Wavelength:**
- The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{mv} \]
- For particle A:
\[ \lambda_A = \frac{h}{m \cdot \frac{1}{3}v} = \frac{3h}{mv} \]
- For particle B:
\[ \lambda_B = \frac{h}{\frac{m}{2} \cdot \frac{4}{3}v} = \frac{3h}{2mv} \]

7. **Finding the Ratio:**
- The ratio of the de Broglie wavelengths is:
\[ \frac{\lambda_A}{\lambda_B} = \frac{\frac{3h}{mv}}{\frac{3h}{2mv}} = \frac{3h}{mv} \cdot \frac{2mv}{3h} = 2. \]

### Final Answer:
The ratio of the de Broglie wavelengths after the collision is:
\[ \frac{\lambda_A}{\lambda_B} = 2. \]