A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelength `'lambda_(x)`' and `'lambda_(y)` respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength 'P' is :

Options

A. `(lambda_(x)lambda_(y))/(lambda_(x)+lambda_(y))`
B. `(lambda_(x)lambda_(y))/(|lambda_(x)-lambda_(y)|)`
C. `lambda_(x)-lambda_(y)`
D. `lambda_(x)+lambda_(y)`

Answer

To solve the problem of finding the de-Broglie wavelength of particle P formed from a completely inelastic collision of particles x and y, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding de-Broglie Wavelength**:
The de-Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

2. **Momentum of Particles x and y**:
For particles x and y, their momenta can be expressed in terms of their respective wavelengths:
\[ p_x = \frac{h}{\lambda_x} \quad \text{and} \quad p_y = \frac{h}{\lambda_y} \]
Since they are moving in opposite directions, we assign the direction of particle x as positive and that of particle y as negative:
\[ p_x = \frac{h}{\lambda_x} \quad \text{and} \quad p_y = -\frac{h}{\lambda_y} \]

3. **Conservation of Momentum**:
In a completely inelastic collision, the total momentum before the collision equals the total momentum after the collision. Let \( p \) be the momentum of the combined particle P:
\[ p_x + p_y = p \]
Substituting the expressions for \( p_x \) and \( p_y \):
\[ \frac{h}{\lambda_x} - \frac{h}{\lambda_y} = p \]

4. **Expressing Momentum of Particle P**:
The momentum of particle P can also be expressed in terms of its wavelength:
\[ p = \frac{h}{\lambda} \]
Therefore, we can equate the two expressions for momentum:
\[ \frac{h}{\lambda_x} - \frac{h}{\lambda_y} = \frac{h}{\lambda} \]

5. **Canceling Planck's Constant**:
Since \( h \) is a common factor, we can cancel it from both sides:
\[ \frac{1}{\lambda_x} - \frac{1}{\lambda_y} = \frac{1}{\lambda} \]

6. **Finding the Wavelength of Particle P**:
Rearranging the equation gives:
\[ \frac{1}{\lambda} = \frac{1}{\lambda_x} - \frac{1}{\lambda_y} \]
To combine the fractions on the right side:
\[ \frac{1}{\lambda} = \frac{\lambda_y - \lambda_x}{\lambda_x \lambda_y} \]
Taking the reciprocal gives:
\[ \lambda = \frac{\lambda_x \lambda_y}{\lambda_y - \lambda_x} \]

### Final Result:
Thus, the de-Broglie wavelength of particle P is given by:
\[ \lambda = \frac{\lambda_x \lambda_y}{\lambda_y - \lambda_x} \]