Two particles A and B of de-broglie wavelength `lambda_(1) and lambda_(2)` combine to from a particle C. The process conserves momentum. Find the de-Broglie wavelength of the particle C. (The motion is one dimensional).

Options

A. `lamda_(A)`
B. `lamda_(A)lamda_(B)//(lamda_(A)+lamda_(B))`
C. `lamda_(A)lamda_(B)//|lamda_(A)-lamda_(B)|`
D. both (b) and ( c)

Explanation

To find the de-Broglie wavelength of particle C formed by the combination of two particles A and B with de-Broglie wavelengths \( \lambda_1 \) and \( \lambda_2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding De-Broglie Wavelength**:
The de-Broglie wavelength \( \lambda \) of a particle is related to its momentum \( p \) by the equation:
\[ p = \frac{h}{\lambda} \]
where \( h \) is Planck's constant.

2. **Momentum of Particles A and B**:
For particles A and B, their momenta can be expressed as:
\[ p_A = \frac{h}{\lambda_1} \quad \text{and} \quad p_B = \frac{h}{\lambda_2} \]

3. **Conservation of Momentum**:
Since momentum is conserved in the process of combining particles A and B to form particle C, we can write:
\[ p_C = p_A + p_B \]

4. **Substituting for Momenta**:
Substituting the expressions for \( p_A \) and \( p_B \) into the conservation equation gives:
\[ p_C = \frac{h}{\lambda_1} + \frac{h}{\lambda_2} \]

5. **Factoring Out Planck's Constant**:
Factoring \( h \) out of the equation:
\[ p_C = h \left( \frac{1}{\lambda_1} + \frac{1}{\lambda_2} \right) \]

6. **Finding the Wavelength of Particle C**:
Now, using the relation for the de-Broglie wavelength of particle C:
\[ p_C = \frac{h}{\lambda_C} \]
Setting the two expressions for \( p_C \) equal to each other:
\[ \frac{h}{\lambda_C} = h \left( \frac{1}{\lambda_1} + \frac{1}{\lambda_2} \right) \]

7. **Canceling Planck's Constant**:
Canceling \( h \) from both sides:
\[ \frac{1}{\lambda_C} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} \]

8. **Finding \( \lambda_C \)**:
Taking the reciprocal gives us the expression for the de-Broglie wavelength of particle C:
\[ \lambda_C = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2} \]

### Final Result:
The de-Broglie wavelength of particle C is:
\[ \lambda_C = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2} \]