Question

Two identical particles move at right angles to each other, possessing de-Broglie wavelengths \( \lambda_1 \) and \( \lambda_2 \). The de-Broglie wavelength of each of the particles in their centre of mass frame will be

• A. \(\sqrt{\dfrac{\lambda_1^2+\lambda_2^2}{2}}\)
• B. \(\dfrac{\lambda_1+\lambda_2}{2}\)
• C. \(\dfrac{2\lambda_1\lambda_2}{\lambda_1+\lambda_2}\)
• D. \(\dfrac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}\)

Hint:

Use momentum–wavelength relation and treat perpendicular velocities vectorially in COM frame.

Answer 1

The Correct Option is D

Step 1: De-Broglie wavelength is related to momentum by \[ \lambda=\frac{h}{p}. \]
Step 2: For identical masses with perpendicular momenta \(p_1\) and \(p_2\), in the centre of mass frame each particle effectively has momentum equal to the vector average magnitude: \[ p=\frac{\sqrt{p_1^2+p_2^2}}{2}. \]
Step 3: Express \(p_1=\frac{h}{\lambda_1}\) and \(p_2=\frac{h}{\lambda_2}\).
Step 4: Resultant momentum: \[ p=\frac{h}{2}\sqrt{\frac1{\lambda_1^2}+\frac1{\lambda_2^2}} =\frac{h}{2}\frac{\sqrt{\lambda_1^2+\lambda_2^2}}{\lambda_1\lambda_2}. \]
Step 5: Corresponding wavelength: \[ \lambda=\frac{h}{p} =\frac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}. \] This matches option (D).