Two identical non-re,aticivistic particles move at right angle to each other. Possesing de Broglie wavelength `lambda_(1)` and `lambda_(2)` Find the Broglie wavelength of each particle in the frame of their centre of inertia.

Options

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

Answer

To find the de Broglie wavelength of each particle in the frame of their center of inertia, we can follow these steps:

### Step-by-Step Solution:

1. **Define the velocities of the particles**:
Let the velocities of the two identical particles be:
- Particle 1: \( \vec{v_1} = v_1 \hat{i} \)
- Particle 2: \( \vec{v_2} = v_2 \hat{j} \)

2. **Calculate the center of mass velocity**:
The velocity of the center of mass (\( \vec{v_{cm}} \)) of the two particles is given by:
\[ \vec{v_{cm}} = \frac{\vec{v_1} + \vec{v_2}}{2} = \frac{v_1 \hat{i} + v_2 \hat{j}}{2} \]

3. **Determine the velocities of the particles with respect to the center of mass**:
The velocity of each particle with respect to the center of mass can be calculated as:
- For Particle 1:
\[ \vec{v_{1,cm}} = \vec{v_1} - \vec{v_{cm}} = v_1 \hat{i} - \frac{v_1 \hat{i} + v_2 \hat{j}}{2} = \left(v_1 - \frac{v_1}{2}\right) \hat{i} - \frac{v_2}{2} \hat{j} = \frac{v_1}{2} \hat{i} - \frac{v_2}{2} \hat{j} \]
- For Particle 2:
\[ \vec{v_{2,cm}} = \vec{v_2} - \vec{v_{cm}} = v_2 \hat{j} - \frac{v_1 \hat{i} + v_2 \hat{j}}{2} = -\frac{v_1}{2} \hat{i} + \left(v_2 - \frac{v_2}{2}\right) \hat{j} = -\frac{v_1}{2} \hat{i} + \frac{v_2}{2} \hat{j} \]

4. **Calculate the magnitudes of the velocities with respect to the center of mass**:
- For Particle 1:
\[ v_{1,cm} = \sqrt{\left(\frac{v_1}{2}\right)^2 + \left(-\frac{v_2}{2}\right)^2} = \frac{1}{2} \sqrt{v_1^2 + v_2^2} \]
- For Particle 2:
\[ v_{2,cm} = \sqrt{\left(-\frac{v_1}{2}\right)^2 + \left(\frac{v_2}{2}\right)^2} = \frac{1}{2} \sqrt{v_1^2 + v_2^2} \]

5. **Apply the de Broglie wavelength formula**:
The de Broglie wavelength (\( \lambda \)) is given by:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant and \( m \) is the mass of the particle. For both particles, we can write:
- For Particle 1:
\[ \lambda_{1,cm} = \frac{h}{m \cdot v_{1,cm}} = \frac{h}{m \cdot \frac{1}{2} \sqrt{v_1^2 + v_2^2}} = \frac{2h}{m \sqrt{v_1^2 + v_2^2}} \]
- For Particle 2:
\[ \lambda_{2,cm} = \frac{h}{m \cdot v_{2,cm}} = \frac{h}{m \cdot \frac{1}{2} \sqrt{v_1^2 + v_2^2}} = \frac{2h}{m \sqrt{v_1^2 + v_2^2}} \]

6. **Express the result in terms of the given wavelengths**:
If we let \( \lambda_1 = \frac{h}{mv_1} \) and \( \lambda_2 = \frac{h}{mv_2} \), we can relate the new wavelengths to the original wavelengths:
\[ \lambda_{1,cm} = \frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}} \]
\[ \lambda_{2,cm} = \frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}} \]

### Final Result:
Thus, the de Broglie wavelengths of both particles in the center of mass frame are:
\[ \lambda_{1,cm} = \lambda_{2,cm} = \frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}} \]