Atoms & Nuclei

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Question 141 of 239

Two electrons are moving with non-relativistic speeds perpendicular to each other. If corresponding de Broglie wavelengths are λ1\lambda_1 and λ2\lambda_2, their de Broglie wavelength in the frame of reference attached to their centre of mass is:

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Answer: (C) λCM=2λ1λ2λ12+λ22\lambda_{C M}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}

Momentum (p) of each electron hλ1i^\frac{\mathrm{h}}{\lambda_1} \hat{\mathrm{i}} and hλ2j^\frac{\mathrm{h}}{\lambda_2} \hat{\mathrm{j}}
Velocity of centre of mass
Vcm=h2 mλ1i^+h2 mλ2j^(∵p=mv)\mathrm{V}_{\mathrm{cm}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_1} \hat{\mathrm{i}}+\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_2} \hat{\mathrm{j}} \quad(\because \mathrm{p}=\mathrm{mv})
Velocity of 1 st particle about centre of mass
V1 cm=h2 mλ1i^−h2 mλ2j^V_{1 \mathrm{~cm}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_1} \hat{\mathrm{i}}-\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_2} \hat{\mathrm{j}}
λcm=hh24λ12+h24λ22=2λ1λ2λ12+λ22(∵λ=hp)\lambda_{\mathrm{cm}}=\frac{\mathrm{h}}{\sqrt{\frac{\mathrm{h}^2}{4 \lambda_1^2}+\frac{\mathrm{h}^2}{4 \lambda_2^2}}}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}} \quad\left(\because \lambda=\frac{\mathrm{h}}{\mathrm{p}}\right)

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