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

Answer

To find the ratio of the de Broglie wavelengths of particles A and B after an elastic collision, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (\( \lambda \)) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is the Planck's constant and \( p \) is the momentum of the particle. The momentum \( p \) can be expressed as:
\[ p = mv \]
where \( m \) is the mass and \( v \) is the velocity of the particle.

### Step 2: Write the expressions for the de Broglie wavelengths of A and B
For particle A (mass \( m \), velocity \( v_A \)):
\[ \lambda_A = \frac{h}{m v_A} \]
For particle B (mass \( \frac{m}{2} \), velocity \( v_B \)):
\[ \lambda_B = \frac{h}{\frac{m}{2} v_B} = \frac{2h}{m v_B} \]

### Step 3: Find the ratio of the de Broglie wavelengths
The ratio of the de Broglie wavelengths \( \frac{\lambda_A}{\lambda_B} \) is:
\[ \frac{\lambda_A}{\lambda_B} = \frac{\frac{h}{m v_A}}{\frac{2h}{m v_B}} = \frac{v_B}{2 v_A} \]

### Step 4: Use conservation of momentum
In an elastic collision, the total momentum before and after the collision is conserved. Initially, particle A has momentum \( mv \) and particle B is at rest. Therefore, the conservation of momentum gives us:
\[ mv = m v_A + \frac{m}{2} v_B \]
This simplifies to:
\[ v = v_A + \frac{1}{2} v_B \quad \text{(Equation 1)} \]

### Step 5: Use the coefficient of restitution
For elastic collisions, the coefficient of restitution \( e \) is defined as:
\[ e = \frac{\text{relative velocity after collision}}{\text{relative velocity before collision}} = 1 \]
This gives us:
\[ v_B - v_A = v \quad \text{(Equation 2)} \]

### Step 6: Solve the equations
From Equation 1:
\[ v_B = 2v - 2v_A \quad \text{(substituting into Equation 2)} \]
Substituting \( v_B \) into Equation 2:
\[ (2v - 2v_A) - v_A = v \]
This simplifies to:
\[ 2v - 3v_A = v \]
Thus, we find:
\[ 3v_A = v \implies v_A = \frac{v}{3} \]

Now substituting \( v_A \) back into Equation 1:
\[ v = \frac{v}{3} + \frac{1}{2} v_B \]
This leads to:
\[ v_B = \frac{4v}{3} \]

### Step 7: Substitute \( v_A \) and \( v_B \) back into the ratio
Now we can substitute \( v_A \) and \( v_B \) into the ratio:
\[ \frac{\lambda_A}{\lambda_B} = \frac{v_B}{2 v_A} = \frac{\frac{4v}{3}}{2 \cdot \frac{v}{3}} = \frac{4v/3}{2v/3} = \frac{4}{2} = 2 \]

### Final Answer
Thus, the ratio of the de Broglie wavelengths is:
\[ \frac{\lambda_A}{\lambda_B} = 2 \]