Two identical non relativistic particles move at right angles to each other, possessing de- Broglie wavelengths $λ_{1}$ and $λ_{2}$ . Find the de-Broglie wavelength of each particle in the frame of their centre of mass.

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$λ_{1} + λ_{2}$

$\frac{λ_{1} λ_{2}}{\sqrt{λ_{1}^{2} + λ_{2}^{2}}}$

$\frac{λ_{1} λ_{2}}{λ_{1} + λ_{2}}$

$\frac{2 λ_{1} λ_{2}}{\sqrt{λ_{1}^{2} + λ_{2}^{2}}}$

answer is D.

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Detailed Solution

$v_{cm} = \frac{m_{1} v_{1} {+mv}_{2}}{m+m} = \frac{v_{1} {+v}_{2}}{2}$
Question Image
$v_{1cm} = v_{1} - v_{cm} = v_{1} - \frac{v_{1} {+v}_{2}}{2} = \frac{v_{1} - v_{2}}{2}$
$| v_{1,cm} | = \frac{1}{2} \sqrt{v_{1}^{2} + v_{2}^{2} - {2v}_{1} v_{2} \cos 90^{\circ}} = \frac{1}{2} \sqrt{v_{1}^{2} + v_{2}^{2}}$
So, required wavelength is
$λ = \frac{h}{m | v_{1,cm} |} = \frac{2h}{\sqrt{v_{1}^{2} {+v}_{2}^{2}}}$
$= \frac{2h}{\sqrt{{(\frac{h}{λ_{1}})}^{2} + {(\frac{h}{λ_{2}})}^{2}}} = \frac{{2λ}_{1} λ_{2}}{\sqrt{λ_{1}^{2} {+λ}_{2}^{2}}}$