• # 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 $n=0,\,\,1,\,\,2,...$) $10-\frac{\lambda }{2}=(2n)\frac{\lambda }{2},\,\,n=0,\,\,1,\,2,...$           $10=(2n+1)\frac{\lambda }{2},\,n=0,\,1,\,\,2,...$

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