A particle 'P' is formed due to a completely inelastic collision of particles 'x' and 'y' having de-Broglie wavelengths '\(\lambda_x\)' and '\(\lambda_y\)' respectively. If x and y were moving in opposite directions, then the de-Broglie wavelength of 'P' is :-
Show Hint
- \(\lambda_x-\lambda_y\)
- \(\dfrac{\lambda_x\lambda_y}{|\lambda_x-\lambda_y|}\)
- \(\lambda_x+\lambda_y\)
- \(\dfrac{\lambda_x\lambda_y}{\lambda_x+\lambda_y}\)
The Correct Option is B
Solution and Explanation
The particles move in opposite directions, and momentum is conserved in the perfectly inelastic collision: \(p_P=\left|\dfrac{h}{\lambda_x}-\dfrac{h}{\lambda_y}\right|=h\,\dfrac{|\lambda_y-\lambda_x|}{\lambda_x\lambda_y}\).
\(\lambda_P=\dfrac{h}{p_P}=\dfrac{\lambda_x\lambda_y}{|\lambda_x-\lambda_y|}\).
Answer: (2) \(\dfrac{\lambda_x\lambda_y}{|\lambda_x-\lambda_y|}\)
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