Two particles of masses m and 2m have equal k | vedclass

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(D) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $E = \frac{p^2}

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$\sqrt{2}: 1$

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(D) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p}$,where $p$ is the momentum.Since kinetic energy $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.Substituting this into the wavelength formula,we get $\lambda = \frac{h}{\sqrt{2mE}}$.Given that $h$ and $E$ are constants for both particles,$\lambda \propto \frac{1}{\sqrt{m}}$.For masses $m_1 = m$ and $m_2 = 2m$,the ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{\sqrt{m_2}}{\sqrt{m_1}} = \frac{\sqrt{2m}}{\sqrt{m}} = \frac{\sqrt{2}}{1}$.Thus,the ratio is $\sqrt{2}: 1$.

Two particles of masses $m$ and $2m$ have equal kinetic energies. The de-Broglie wavelengths are in the ratio of

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