question_answer

Two particles move at right angle to each other. Their de-Broglie wavelengths are \[{{\lambda }_{1}}\]and \[{{\lambda }_{2}}\] respectively. The particles suffer perfectly inelastic collision. The de-Broglie wavelength \[\lambda \], of the final particle, is given by : [JEE Main 8-4-2019 Morning]

A) \[\lambda =\frac{{{\lambda }_{1}}+{{\lambda }_{2}}}{2}\]                  

B) \[\frac{2}{\lambda }=\frac{1}{{{\lambda }_{1}}}+\frac{1}{{{\lambda }_{2}}}\]

C) \[\lambda =\sqrt{{{\lambda }_{1}}{{\lambda }_{2}}}\]               

D) \[\frac{1}{{{\lambda }^{2}}}=\frac{1}{\lambda _{1}^{2}}+\frac{1}{\lambda _{2}^{2}}\]

Correct Answer: D

Solution :

\[{{\vec{P}}_{1}}=\frac{h}{{{\lambda }_{1}}}\hat{i}\]&\[{{\vec{P}}_{2}}=\frac{h}{{{\lambda }_{2}}}\hat{j}\] Using momentum conservation \[\vec{P}={{\vec{P}}_{1}}+{{\vec{P}}_{2}}\] \[=\frac{h}{{{\lambda }_{1}}}\hat{i}+\frac{h}{{{\lambda }_{2}}}\hat{j}\]             \[\left| {\vec{P}} \right|=\sqrt{{{\left( \frac{h}{{{\lambda }_{1}}} \right)}^{2}}+{{\left( \frac{h}{{{\lambda }_{2}}} \right)}^{2}}}\]             \[\frac{h}{\lambda }=\sqrt{{{\left( \frac{h}{{{\lambda }_{1}}} \right)}^{2}}+{{\left( \frac{h}{{{\lambda }_{2}}} \right)}^{2}}}\]             \[\frac{1}{{{\lambda }^{2}}}=\frac{1}{\lambda _{1}^{2}}+\frac{1}{\lambda _{2}^{2}}\]