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Two identical non-relativistic particles more right angles to each other, possessing DeBroglie wavelengths \lambda_1 and \lambda_2 The De-Broglie wavelength of each particle in the frame of their centre of mass.

Option: 1

\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1{ }^2+\lambda_2{ }^2}}


Option: 2

\frac{\lambda_1 \lambda_2}{\sqrt{\lambda_1{ }^2+\lambda_2{ }^2}}


Option: 3

\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1{ }^2-\lambda_2{ }^2}}


Option: 4

\frac{2 \lambda_1 \lambda_2}{\lambda_1{ }^2+\lambda_2{ }^2}


Answers (1)

best_answer

If \mathrm{ \vec{P}_1 and \vec{P}_2 } are the momentum corresponding to wave lengths \mathrm{ \lambda_1 } and \mathrm{ \lambda_2}. The velocity of centre of mass \mathrm{ \left(\vec{v}_{c m}\right)=\frac{\vec{p}_1+\vec{p}_2}{m+m} }\mathrm{ \left(\vec{v}_{c m}\right)=\frac{\vec{p}_1+\vec{p}_2}{m+m} }
                        \mathrm{ \begin{aligned} 2 \vec{p}_{c m} & =\sqrt{P_1^2+P_2^2} \\ 2 P_{c m} & =\sqrt{P_1^2+P_2^2} \\ 2 \cdot \frac{h}{\lambda} & =\sqrt{\left(\frac{h}{\lambda_1}\right)^2+\left(\frac{h}{\lambda_2}\right)^2} \\ \lambda & =\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}} \end{aligned} Ans }

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Gaurav

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