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  • Two identical non-relativistic particles move at right angles to each other, possessing DeBroglie wavelengths <img alt="\lambda_1" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clambda_1"

Two identical non-relativistic particles move at right angles to each other, possessing DeBroglie wavelengths \lambda_1 and \lambda_2. The De-Broglie wavelength of each particle is 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 \mathrm{\vec{P}_2} are the momentum corresponding to wave lengths \mathrm{\lambda _1} and \mathrm{\lambda _2}. The velocity of centre of mass


 \begin{aligned} \mathrm{\left(\vec{v}_{\mathrm{cm}}\right)} & =\mathrm{\frac{\vec{p}_1+\vec{p}_2}{m+m} }\\ \mathrm{2 \vec{l}_{\mathrm{Cm}} }& =\mathrm{\sqrt{p_1^2+p_2{ }^2}}\\ \mathrm{or ~2 P_{c m}}&=\mathrm{\sqrt{P_1^2+P_2^2}}\\ \mathrm{\frac{h}{\lambda} }& =\mathrm{\sqrt{\left(\frac{h}{\lambda_1}\right)^2+\left(\frac{h}{\partial_2}\right)^2}} \\ \lambda & =\mathrm{\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}} \end{aligned}

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Nehul

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