Question

A particle 'A' of mass ' 2m' and having initial velocity \({}'u_{A}'\) collides with particle B of mass ' m / 2' at rest. The collision is head on and elastic. If \(\lambda _{A}\) and \(\lambda _{B}\) are the de-Broglie wavelength of the particles 'A' and 'B' after collision, such that \(\lambda _{A}=x\lambda _{B}\). The value of \(x\) is

• A. \(\frac{2}{3}\)
• B. \(\frac{1}{2}\)
• C. \(1\)
• D. \(\frac{1}{3}\)

Hint:

Find both velocities after the head-on elastic collision, then use that wavelength is inversely proportional to momentum.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
For a head-on elastic collision with B at rest, \(v_{A}=\frac{m_{A}-m_{B}}{m_{A}+m_{B}}u_{A}\) and \(v_{B}=\frac{2m_{A}}{m_{A}+m_{B}}u_{A}\).

Step 2: Calculation
With \(m_{A}=2m\) and \(m_{B}=\frac{m}{2}\): \(v_{A}=\frac{1.5m}{2.5m}u_{A}=\frac{3u_{A}}{5}\) and \(v_{B}=\frac{4m}{2.5m}u_{A}=\frac{8u_{A}}{5}\). Momenta: \(p_{A}=2m\times\frac{3u_{A}}{5}=\frac{6mu_{A}}{5}\) and \(p_{B}=\frac{m}{2}\times\frac{8u_{A}}{5}=\frac{4mu_{A}}{5}\).

Step 3: Wavelength ratio
\(\lambda=\frac{h}{p}\), so \(\frac{\lambda_{A}}{\lambda_{B}}=\frac{p_{B}}{p_{A}}=\frac{4}{6}=\frac{2}{3}\).

Final Answer:
The value of x is two-thirds. \[ \boxed{\text{(A) }\dfrac{2}{3}} \]