Question

Two identical non-relativistic particles A and B move at right angles to each other, processing de Broglie wavelengths ${\lambda }_1$ and ${\lambda }_2$ respectively. The de Broglie wavelength of each particle in their centre 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. ${\lambda }_1{\lambda }_2/\left(\sqrt{|{\lambda }_1^2+{\lambda }_2^2|}\right)$
• D. $({\lambda }_1+{\lambda }_2)/2$

Answer 1

The correct option is B $2{\lambda }_1{\lambda }_2/\left(\sqrt{{\lambda }_1^2+{\lambda }_2^2}\right)$

Two identical non-relativistic particles A and B move at right angles to each other with say velocity $v_1,v_2$. along X and Y directions respectively (say).Therefore Velocity of center of mass (COM) ; $v_c=v_1\hat{i}+v_2\hat{j}$Velocity of 1st particle with respect to COM; $v_{1c}=v_1-v_c=\frac{v_1\hat{i}-v_2\hat{j}}{2}$
Velocity of 2nd particle with respect to COM; $v_{2c}=v_2-v_c=\frac{v_2\hat{j}-v_1\hat{i}}{2}$De Broglie wavelength of 1st particle w.r.t COM
${\lambda }_{1c}=\frac{h}{m{v}_{1c}}=\frac{2h}{m\sqrt{v_1^2+v_2^2}}$${\lambda }_{1c}=\frac{2h}{m\sqrt{(\frac{h}{m{\lambda }_1}{)}^2+(\frac{h}{m{\lambda }_2}{)}^2}}$${\lambda }_{1c}=\frac{2{\lambda }_1{\lambda }_2}{\sqrt{{\lambda }_1^2+{\lambda }_2^2}}$Similarly,De Broglie wavelength of 2nd particle w.r.t COM${\lambda }_{2c}=\frac{h}{m{v}_{2c}}=\frac{2h}{m\sqrt{v_1^2+v_2^2}}$${\lambda }_{2c}=\frac{2h}{m\sqrt{(\frac{h}{m{\lambda }_1}{)}^2+(\frac{h}{m{\lambda }_2}{)}^2}}$${\lambda }_{2c}=\frac{2{\lambda }_1{\lambda }_2}{\sqrt{{\lambda }_1^2+{\lambda }_2^2}}$