A particle A of mass \(m\) and initial velocity \(v\) collides with a particle B of mass \(m/2\) which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths \(\lambda_A\) to \(\lambda_B\) after the collision is:
Show Hint
- \(\frac{\lambda_A}{\lambda_B}=\frac{1}{3}\)
- \(\frac{\lambda_A}{\lambda_B}=2\)
- \(\frac{\lambda_A}{\lambda_B}=\frac{2}{3}\)
- \(\frac{\lambda_A}{\lambda_B}=\frac{1}{2}\)
The Correct Option is B
Solution and Explanation
Momentum conservation: \(mv = mv_A + \frac{m}{2}v_B\), i.e. \(2v = 2v_A + v_B\).
Elastic head-on collision (relative speed of separation = relative speed of approach): \(v_B - v_A = v\).
Substituting \(v_B = v + v_A\): \(2v = 2v_A + v + v_A\), so \(v_A = \frac{v}{3}\) and \(v_B = \frac{4v}{3}\).
Momenta: \(p_A = m\cdot\frac{v}{3} = \frac{mv}{3}\), \(p_B = \frac{m}{2}\cdot\frac{4v}{3} = \frac{2mv}{3}\).
de Broglie wavelength \(\lambda = \frac{h}{p}\), so \(\frac{\lambda_A}{\lambda_B} = \frac{p_B}{p_A} = \frac{2mv/3}{mv/3} = 2\).
Answer: (2) \(\frac{\lambda_A}{\lambda_B}=2\)
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