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 of ‘P’ is:

Options

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

Explanation

To solve the problem, we need to find the de-Broglie wavelength of the particle 'P' formed from the completely inelastic collision of particles 'x' and 'y'. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have two particles, 'x' and 'y', with de-Broglie wavelengths \( \lambda_x \) and \( \lambda_y \), respectively. They are moving towards each other and collide inelastically to form a new particle 'P'.

### Step 2: Apply Conservation of Momentum
In a completely inelastic collision, momentum is conserved. The initial momentum of the system (particles 'x' and 'y') must equal the final momentum of the new particle 'P'.

Let:
- \( p_x \) = momentum of particle 'x'
- \( p_y \) = momentum of particle 'y'
- \( p_P \) = momentum of particle 'P'

Since they are moving in opposite directions, we can write:
\[ p_{\text{initial}} = p_x - p_y = p_P \]

### Step 3: Relate Momentum to Wavelength
Using the de-Broglie relation, we know that momentum \( p \) is related to wavelength \( \lambda \) by:
\[ p = \frac{h}{\lambda} \]
where \( h \) is Planck's constant.

Thus, we can express the momenta of particles 'x' and 'y' as:
\[ p_x = \frac{h}{\lambda_x} \quad \text{and} \quad p_y = \frac{h}{\lambda_y} \]

### Step 4: Substitute into the Momentum Equation
Substituting the expressions for \( p_x \) and \( p_y \) into the momentum conservation equation gives:
\[ \frac{h}{\lambda_x} - \frac{h}{\lambda_y} = p_P \]

### Step 5: Express Momentum of Particle 'P'
The momentum of particle 'P' can also be expressed in terms of its wavelength \( \lambda' \):
\[ p_P = \frac{h}{\lambda'} \]

### Step 6: Set Up the Equation
Now we can set up the equation:
\[ \frac{h}{\lambda_x} - \frac{h}{\lambda_y} = \frac{h}{\lambda'} \]

### Step 7: Simplify the Equation
Dividing through by \( h \) (assuming \( h \neq 0 \)):
\[ \frac{1}{\lambda_x} - \frac{1}{\lambda_y} = \frac{1}{\lambda'} \]

### Step 8: Find \( \lambda' \)
To find \( \lambda' \), we can rearrange the equation:
\[ \frac{1}{\lambda'} = \frac{1}{\lambda_x} - \frac{1}{\lambda_y} \]
Taking the reciprocal gives:
\[ \lambda' = \frac{\lambda_x \lambda_y}{\lambda_y - \lambda_x} \]

### Step 9: Consider Absolute Values
Since we do not know which wavelength is larger, we take the absolute value:
\[ \lambda' = \frac{\lambda_x \lambda_y}{|\lambda_y - \lambda_x|} \]

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