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Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are λ1 and λ2, their de Broglie wavelength in the frame of reference attached to their centre of mass is :
Option 1)
$\lambda_{CM}=\lambda _{1}=\lambda _{2}$

Option 2)
$\lambda_{CM}=\frac{2\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}$
Option 3)
$\frac{1}{\lambda_{CM}}= \frac{1}{\lambda _{1}}+\frac{1}{\lambda_{2}}$
Option 4)
$\lambda _{CM}=\left ( \frac{\lambda_{1}+\lambda_{2}}{2} \right )$

Answers (2)

As we learned

De - Broglie wavelength -

$\lambda = \frac{h}{p}= \frac{h}{mv}= \frac{h}{\sqrt{2mE}}$

- wherein

$h= plank's\: constant$

$m= mass \: of\: particle$

$v= speed \: of\: the \: particle$

$E= Kinetic \: energy \: of \: particle$

$p_{1} = \frac{h}{\lambda _{1}}\hat{i}, p_{2}=\frac{h}{\lambda _{2}}\hat{j}$

$\vec{V_{cm}}=\frac{\vec{p_{1}}+\vec{p_{2}}}{M}=\frac{h}{2m\lambda _{1}}\hat{i}+\frac{h}{2m\lambda }\hat{j}$

velocity of first electron about centre of mass

$\vec{V}_{cm}=$ $\vec{V}_{1}-\vec{V}_{cm}$ $=\frac{\lambda }{m\lambda _{1}}\hat{i}-\left [ \frac{h}{2m\lambda _{1}}\hat{i} +\frac{h}{2m\lambda _{2}} \hat{j}\right ]$

$= \frac{h}{2m\lambda _{1}}\hat{i} -\frac{h}{2m\lambda _{2}} \hat{j}$

$\Rightarrow\lambda _{cm}= \frac{h}{\sqrt{\frac{h^{2}}{4\lambda {_{1}}^{2}}+\frac{h^{2}}{4\lambda {_{2}}^{2}}}}$ $\Rightarrow\lambda _{cm}= \frac{2\lambda _{1}\lambda _{2}}{\sqrt{\lambda _{1}^{2}+\lambda _{2}^{2}}}$

Option 1)

$\lambda_{CM}=\lambda _{1}=\lambda _{2}$

Option 2)
$\lambda_{CM}=\frac{2\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}$
Option 3)

$\frac{1}{\lambda_{CM}}= \frac{1}{\lambda _{1}}+\frac{1}{\lambda_{2}}$

Option 4)

$\lambda _{CM}=\left ( \frac{\lambda_{1}+\lambda_{2}}{2} \right )$

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