question_answer

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 :-             [JEE Main 9-4-2019 Afternoon]

A) \[{{\lambda }_{x}}+{{\lambda }_{y}}\]                

B) \[\frac{{{\lambda }_{x}}{{\lambda }_{y}}}{{{\lambda }_{x}}+{{\lambda }_{y}}}\]

C) \[\frac{{{\lambda }_{x}}{{\lambda }_{y}}}{|{{\lambda }_{x}}-{{\lambda }_{y}}|}\]

D) \[{{\lambda }_{x}}-{{\lambda }_{y}}\]

Correct Answer: C

Solution :

By momentum conservation \[{{P}_{x}}-{{P}_{y}}={{P}_{P}}\] \[\frac{h}{{{\lambda }_{x}}}-\frac{h}{{{\lambda }_{y}}}=\frac{h}{{{\lambda }_{p}}}\] \[{{\lambda }_{p}}=\frac{{{\lambda }_{x}}{{\lambda }_{y}}}{\left| {{\lambda }_{y}}-{{\lambda }_{x}} \right|}.\]