The ratio of de-Broglie wavelength of two par | vedclass

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(C) According to de-Broglie's wavelength,$\lambda = \frac{h}{mv}$.For particle $A$,$\lambda_A = \frac{h}{m_A v_A}$.For particle $B$,$\lambda_B = \frac

Vedclass · Accepted Answer

$1: 5$

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(C) According to de-Broglie's wavelength,$\lambda = \frac{h}{mv}$.For particle $A$,$\lambda_A = \frac{h}{m_A v_A}$.For particle $B$,$\lambda_B = \frac{h}{m_B v_B}$.The ratio of wavelengths is given as $\frac{\lambda_A}{\lambda_B} = \frac{2}{1}$.Substituting the expressions,we get $\frac{\lambda_A}{\lambda_B} = \frac{m_B v_B}{m_A v_A} = 2$.Given $v_A = 0.05 \ ms^{-1}$ and $v_B = 0.02 \ ms^{-1}$,we have $\frac{m_B 	imes 0.02}{m_A 	imes 0.05} = 2$.$\frac{m_B}{m_A} 	imes \frac{2}{5} = 2$.$\frac{m_B}{m_A} = 2 	imes \frac{5}{2} = 5$.Therefore,$\frac{m_A}{m_B} = \frac{1}{5}$,which is $1: 5$.

The ratio of de-Broglie wavelength of two particles $A$ and $B$ is $2: 1$. If the velocities of $A$ and $B$ are $0.05 \ ms^{-1}$ and $0.02 \ ms^{-1}$,respectively,then the ratio of their masses $m_A: m_B$ must be

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