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(C) Let the momenta of the two electrons be $\vec{p}_1 = \frac{h}{\lambda_1} \hat{i}$ and $\vec{p}_2 = \frac{h}{\lambda_2} \hat{j}$.Since both are ele

Vedclass · Accepted Answer

$\lambda_{CM} = \frac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}}$

Vedclass · Answer

(C) Let the momenta of the two electrons be $\vec{p}_1 = \frac{h}{\lambda_1} \hat{i}$ and $\vec{p}_2 = \frac{h}{\lambda_2} \hat{j}$.Since both are electrons,they have the same mass $m$.The velocity of the centre of mass is $\vec{V}_{CM} = \frac{\vec{p}_1 + \vec{p}_2}{2m} = \frac{h}{2m\lambda_1} \hat{i} + \frac{h}{2m\lambda_2} \hat{j}$.The velocity of the first electron relative to the centre of mass is $\vec{v}_{1,CM} = \vec{v}_1 - \vec{V}_{CM} = \frac{\vec{p}_1}{m} - \frac{\vec{p}_1 + \vec{p}_2}{2m} = \frac{\vec{p}_1 - \vec{p}_2}{2m} = \frac{h}{2m\lambda_1} \hat{i} - \frac{h}{2m\lambda_2} \hat{j}$.The momentum of the electron in the $CM$ frame is $\vec{p}_{CM} = m \vec{v}_{1,CM} = \frac{h}{2\lambda_1} \hat{i} - \frac{h}{2\lambda_2} \hat{j}$.The magnitude of this momentum is $p_{CM} = \sqrt{(\frac{h}{2\lambda_1})^2 + (-\frac{h}{2\lambda_2})^2} = \frac{h}{2} \sqrt{\frac{1}{\lambda_1^2} + \frac{1}{\lambda_2^2}} = \frac{h}{2} \frac{\sqrt{\lambda_1^2 + \lambda_2^2}}{\lambda_1 \lambda_2}$.The de Broglie wavelength in the $CM$ frame is $\lambda_{CM} = \frac{h}{p_{CM}} = \frac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}}$.

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are $\lambda_1$ and $\lambda_2$,their de Broglie wavelength in the frame of reference attached to their centre of mass is

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