Two particles move at right angle to each other. Their de Broglie wavelenghts are `lambda_(1)` and `lambda_(2)` respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelenght `lambda`, of the final particle, is given by:

Options

A. `(1)/(lambda^(2))=(1)/(lambda_(1)^(2))+(1)/(lambda_(2)^(2))`
B. `lambda=sqrt(lambda_(1)lambda_(2))`
C. `lambda=(lambda_(1)+lambda_(2))/(2)`
D. `(2)/(lambda)=(1)/(lambda_(1))+(1)/(lambda_(2))`

Answer

To solve the problem, we need to find the de Broglie wavelength \( \lambda \) of the final particle after two particles collide perfectly inelastically. The two particles are moving at right angles to each other with de Broglie wavelengths \( \lambda_1 \) and \( \lambda_2 \).

### Step-by-Step Solution:

1. **Understand the Momentum Relation**:
The de Broglie wavelength \( \lambda \) of a particle is related to its momentum \( p \) by the formula:
\[ p = \frac{h}{\lambda} \]
where \( h \) is Planck's constant.

2. **Define the Momenta of the Particles**:
For the two particles, we can express their momenta as:
\[ p_1 = \frac{h}{\lambda_1} \quad \text{(for particle 1)} \]
\[ p_2 = \frac{h}{\lambda_2} \quad \text{(for particle 2)} \]

3. **Consider the Collision**:
Since the particles collide perfectly inelastically, they stick together after the collision. The momentum of the system before the collision is conserved.

4. **Calculate the Resultant Momentum**:
The two momenta \( p_1 \) and \( p_2 \) are perpendicular to each other. Therefore, the magnitude of the resultant momentum \( p \) can be calculated using the Pythagorean theorem:
\[ p = \sqrt{p_1^2 + p_2^2} = \sqrt{\left(\frac{h}{\lambda_1}\right)^2 + \left(\frac{h}{\lambda_2}\right)^2} \]

5. **Substituting the Momentum Values**:
Substitute the expressions for \( p_1 \) and \( p_2 \):
\[ p = \sqrt{\left(\frac{h}{\lambda_1}\right)^2 + \left(\frac{h}{\lambda_2}\right)^2} = \sqrt{\frac{h^2}{\lambda_1^2} + \frac{h^2}{\lambda_2^2}} \]

6. **Factor Out \( h^2 \)**:
Factor out \( h^2 \) from the square root:
\[ p = h \sqrt{\frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}} \]

7. **Relate to the Final Wavelength**:
The momentum of the resultant particle after the collision can also be expressed in terms of its de Broglie wavelength \( \lambda \):
\[ p = \frac{h}{\lambda} \]

8. **Equate the Two Expressions for Momentum**:
Set the two expressions for momentum equal to each other:
\[ \frac{h}{\lambda} = h \sqrt{\frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}} \]

9. **Cancel \( h \)**:
Since \( h \) is common in both sides, we can cancel it out:
\[ \frac{1}{\lambda} = \sqrt{\frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}} \]

10. **Square Both Sides**:
Square both sides to eliminate the square root:
\[ \frac{1}{\lambda^2} = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2} \]

11. **Final Result**:
Thus, we have derived the relationship:
\[ \frac{1}{\lambda^2} = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2} \]

### Final Answer:
The de Broglie wavelength \( \lambda \) of the final particle after the perfectly inelastic collision is given by:
\[ \frac{1}{\lambda^2} = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2} \]