question_answer

Two identical non-relative particles A and B move right angles to each other, processing de Broglie wavelength \[{{\lambda }_{1}}\] and \[\,{{\lambda }_{2}}\], respectively. The de Broglie wavelength of each particle in their center of mass frame of reference is

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

B) \[2{{\lambda }_{1}}{{\lambda }_{2}}/\left( \sqrt{\lambda _{1}^{2}+\lambda _{2}^{2}} \right)\]

C) \[2{{\lambda }_{1}}{{\lambda }_{2}}/\left( \sqrt{\left| \lambda _{1}^{2}+\lambda _{2}^{2} \right|} \right)\]

D) \[\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)/2\]

Correct Answer: B

Solution :

[b] Let m be the mass of each particle, then \[{{\lambda }_{1}}=\left( h/m{{v}_{1}} \right)\] and \[{{\lambda }_{2}}=\left( h/m{{v}_{2}} \right)\] where \[{{v}_{1}}\] and \[{{v}_{2}}\]are the velocities of two particles as shown in the figure. \[{{\vec{v}}_{CM}}=\frac{m{{{\vec{v}}}_{1}}+m{{{\vec{v}}}_{2}}}{2m}=\frac{{{{\vec{v}}}_{1}}-{{{\vec{v}}}_{2}}}{2}\] \[{{\vec{v}}_{1c}}={{\vec{v}}_{1}}-{{\vec{v}}_{CM}}=\frac{{{{\vec{v}}}_{1}}-{{{\vec{v}}}_{2}}}{2}\] Velocity of A w.r.t. C frame is \[{{\vec{v}}_{1c}}={{\vec{v}}_{1}}-{{\vec{v}}_{CM}}=\frac{{{{\vec{v}}}_{1}}-{{{\vec{v}}}_{2}}}{2}\] \[\left| {{{\vec{v}}}_{1c}} \right|=\frac{\sqrt{{{{\vec{v}}}_{1}}-{{{\vec{v}}}_{2}}}}{2}=\,\left| {{{\vec{v}}}_{2c}} \right|\] So, required wavelength is \[\lambda =\frac{h}{3|{{{\vec{v}}}_{1c}}|}=\frac{h}{m}\times \frac{2}{\frac{h}{m}\sqrt{\frac{1}{\lambda _{1}^{2}}+\frac{1}{\lambda _{2}^{2}}}}\] \[=\frac{2{{\lambda }_{1}}{{\lambda }_{2}}}{\sqrt{\lambda _{1}^{2}+\lambda _{2}^{2}}}\]