A particle X of massm and initial velocity u collide with another particle Y of mass `(3m)/(4)` which is at rest, The collision is head on and perfectly elastic. The ratio of de-Broglie wavelengths `lambda_(Y)` and `lambda_(X)` after the collision is

Answer

To solve the problem step by step, we will analyze the elastic collision between the two particles and then calculate the de-Broglie wavelengths to find the desired ratio.

### Step 1: Understand the initial conditions
- Particle X has mass \( m \) and initial velocity \( u \).
- Particle Y has mass \( \frac{3m}{4} \) and is initially at rest (velocity = 0).

### Step 2: Apply conservation of momentum
The total momentum before the collision must equal the total momentum after the collision.

Initial momentum:
\[ p_{\text{initial}} = m \cdot u + \frac{3m}{4} \cdot 0 = mu \]

Let \( v_x \) be the final velocity of particle X and \( v_y \) be the final velocity of particle Y after the collision. The final momentum is:
\[ p_{\text{final}} = m \cdot v_x + \frac{3m}{4} \cdot v_y \]

Setting initial momentum equal to final momentum:
\[ mu = mv_x + \frac{3m}{4}v_y \]

Dividing through by \( m \):
\[ u = v_x + \frac{3}{4}v_y \quad \text{(Equation 1)} \]

### Step 3: Apply conservation of kinetic energy
In a perfectly elastic collision, the total kinetic energy before the collision equals the total kinetic energy after the collision.

Initial kinetic energy:
\[ KE_{\text{initial}} = \frac{1}{2} m u^2 + 0 = \frac{1}{2} m u^2 \]

Final kinetic energy:
\[ KE_{\text{final}} = \frac{1}{2} m v_x^2 + \frac{1}{2} \cdot \frac{3m}{4} v_y^2 \]

Setting initial kinetic energy equal to final kinetic energy:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m v_x^2 + \frac{3m}{8} v_y^2 \]

Dividing through by \( \frac{1}{2} m \):
\[ u^2 = v_x^2 + \frac{3}{4} v_y^2 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously
From Equation 1:
\[ v_x = u - \frac{3}{4}v_y \]

Substituting \( v_x \) into Equation 2:
\[ u^2 = \left(u - \frac{3}{4}v_y\right)^2 + \frac{3}{4}v_y^2 \]

Expanding the left side:
\[ u^2 = u^2 - \frac{3}{2}uv_y + \frac{9}{16}v_y^2 + \frac{3}{4}v_y^2 \]

Combining the terms:
\[ u^2 = u^2 - \frac{3}{2}uv_y + \frac{9}{16}v_y^2 + \frac{12}{16}v_y^2 \]
\[ u^2 = u^2 - \frac{3}{2}uv_y + \frac{21}{16}v_y^2 \]

Cancelling \( u^2 \) from both sides:
\[ 0 = -\frac{3}{2}uv_y + \frac{21}{16}v_y^2 \]

Factoring out \( v_y \):
\[ v_y \left(-\frac{3}{2}u + \frac{21}{16}v_y\right) = 0 \]

Thus, either \( v_y = 0 \) (not possible since Y is moving after collision) or:
\[ -\frac{3}{2}u + \frac{21}{16}v_y = 0 \]
\[ v_y = \frac{3}{2}u \cdot \frac{16}{21} = \frac{24u}{21} = \frac{8u}{7} \]

Substituting \( v_y \) back to find \( v_x \):
\[ v_x = u - \frac{3}{4} \cdot \frac{8u}{7} = u - \frac{6u}{7} = \frac{1u}{7} = \frac{u}{7} \]

### Step 5: Calculate de-Broglie wavelengths
The de-Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{mv} \]

For particle Y:
\[ \lambda_Y = \frac{h}{\frac{3m}{4} \cdot \frac{8u}{7}} = \frac{h \cdot 4 \cdot 7}{3m \cdot 8u} = \frac{28h}{24mu} = \frac{7h}{6mu} \]

For particle X:
\[ \lambda_X = \frac{h}{m \cdot \frac{u}{7}} = \frac{7h}{mu} \]

### Step 6: Find the ratio \( \frac{\lambda_Y}{\lambda_X} \)
\[ \frac{\lambda_Y}{\lambda_X} = \frac{\frac{7h}{6mu}}{\frac{7h}{mu}} = \frac{1}{6} \]

Thus, the ratio of the de-Broglie wavelengths \( \lambda_Y : \lambda_X \) is:
\[ \boxed{1:6} \]