question_answer

Two identical non-relativistic particles A and 6 move at right angles to each other, possessing de-Broglie wavelengths \[{{\lambda }_{1}}\] and \[{{\lambda }_{2}}\] respectively. The de-Broglie wavelength of each particle in their C frame of reference is

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

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

C)  \[\frac{{{\lambda }_{1}}{{\lambda }_{2}}}{\sqrt{|\lambda _{1}^{2}+\lambda _{2}^{2}|}}\]                  

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

Correct Answer: B

Solution :

 Let, m is the mass of each particle, then \[SD=\sqrt{{{60}^{2}}+{{25}^{2}}}\] at \[=\sqrt{4225}=65=DP\], where \[\Delta x=(SA+AP)-SP\] and \[\Rightarrow \] are the velocities of two particles as shown in figure. \[\Delta x=(65+65)-120\] Velocity of A w.r.t. C frame is \[\Rightarrow \] \[\Delta x=10\,m\] (as angle between \[\frac{\lambda }{2}\] and \[\lambda \] is \[=\left( 10-\frac{\lambda }{2} \right)\]) So, required wavelength is \[=(2n)\frac{\lambda }{2}\] (as \[n=0,\,\,1,\,\,2,...\]) \[10-\frac{\lambda }{2}=(2n)\frac{\lambda }{2},\,\,n=0,\,\,1,\,2,...\]           \[10=(2n+1)\frac{\lambda }{2},\,n=0,\,1,\,\,2,...\]