Question

Two particles move at right angle to each other. Their de-Broglie wavelengths are \(\lambda_1\) and \(\lambda_2\) respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength \(\lambda_2\) of the final particle, is given by :

• A. \(\lambda = \frac{\lambda_1 + \lambda_2}{2}\)
• B. \(\frac{1}{\lambda^2} = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}\)
• C. \(\lambda = \sqrt{\lambda_1\lambda_2}\)
• D. \(\frac{2}{\lambda} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}\)

Hint:

Momentum is conserved vectorially; the two initial momenta \(h/\lambda_1\) and \(h/\lambda_2\) are at right angles, so the final momentum is their Pythagorean sum.

Answer 1

The Correct Option is B

\(p_1 = \frac{h}{\lambda_1}\), \(p_2 = \frac{h}{\lambda_2}\), at right angles.
In a perfectly inelastic collision the particles stick together and momentum is conserved: \(\vec{p} = \vec{p}_1 + \vec{p}_2\), so \(p^2 = p_1^2 + p_2^2\).
\(\left(\frac{h}{\lambda}\right)^2 = \left(\frac{h}{\lambda_1}\right)^2 + \left(\frac{h}{\lambda_2}\right)^2 \Rightarrow \frac{1}{\lambda^2} = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}\).
Answer: (2) \(\frac{1}{\lambda^2} = \frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}\)