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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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The correct answer is:
$\lambda_{C M}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}$
$\lambda_{C M}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}}$
Momentum (p) of each electron $\frac{\mathrm{h}}{\lambda_1} \hat{\mathrm{i}}$ and $\frac{\mathrm{h}}{\lambda_2} \hat{\mathrm{j}}$
Velocity of centre of mass
$$
\mathrm{V}_{\mathrm{cm}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_1} \hat{\mathrm{i}}+\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_2} \hat{\mathrm{j}} \quad(\because \mathrm{p}=\mathrm{mv})
$$
Velocity of 1 st particle about centre of mass
$$
V_{1 \mathrm{~cm}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_1} \hat{\mathrm{i}}-\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_2} \hat{\mathrm{j}}
$$
$\lambda_{\mathrm{cm}}=\frac{\mathrm{h}}{\sqrt{\frac{\mathrm{h}^2}{4 \lambda_1^2}+\frac{\mathrm{h}^2}{4 \lambda_2^2}}}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}} \quad\left(\because \lambda=\frac{\mathrm{h}}{\mathrm{p}}\right)$
Velocity of centre of mass
$$
\mathrm{V}_{\mathrm{cm}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_1} \hat{\mathrm{i}}+\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_2} \hat{\mathrm{j}} \quad(\because \mathrm{p}=\mathrm{mv})
$$
Velocity of 1 st particle about centre of mass
$$
V_{1 \mathrm{~cm}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_1} \hat{\mathrm{i}}-\frac{\mathrm{h}}{2 \mathrm{~m} \lambda_2} \hat{\mathrm{j}}
$$
$\lambda_{\mathrm{cm}}=\frac{\mathrm{h}}{\sqrt{\frac{\mathrm{h}^2}{4 \lambda_1^2}+\frac{\mathrm{h}^2}{4 \lambda_2^2}}}=\frac{2 \lambda_1 \lambda_2}{\sqrt{\lambda_1^2+\lambda_2^2}} \quad\left(\because \lambda=\frac{\mathrm{h}}{\mathrm{p}}\right)$
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