question_answer

A particle A of mass m and initial velocity v collides with a particle B of mass\[\frac{m}{2}\]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:    JEE Main Solved Paper-2017

A) \[\frac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\frac{2}{3}\]                             

B)  \[\frac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\frac{1}{3}\]

C)  \[\frac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\frac{1}{3}\]                            

D)  \[\frac{{{\lambda }_{A}}}{{{\lambda }_{_{B}}}}=2\]

Correct Answer: D

Solution :

By conservation of linear momentum \[mv=m{{v}_{1}}+\frac{m}{2}{{v}_{2}}\]                \[2v=2{{v}_{1}}+{{v}_{2}}\]                                          ?. by law of collision \[e=\frac{{{v}_{2}}-{{v}_{1}}}{{{u}_{1}}-{{u}_{2}}}\] \[u={{v}_{2}}-{{v}_{1}}\]                               ?. By equation [a] and [b] \[{{v}_{1}}=\frac{v}{3};\]                              \[{{v}_{2}}=\frac{4v}{3}\] \[{{\lambda }_{1}}=\frac{h}{{{p}_{1}}};\]               \[{{\lambda }_{2}}=\frac{h}{{{o}_{2}}}\] \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{2}{1}\] Option [d]