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(A) Given: Velocity of particle $A$ $(v_A)$ = $0.05 \ ms^{-1}$ and velocity of particle $B$ $(v_B)$ = $0.02 \ ms^{-1}$.Let the mass of particle $A$ $(

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$2: 1$

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(A) Given: Velocity of particle $A$ $(v_A)$ = $0.05 \ ms^{-1}$ and velocity of particle $B$ $(v_B)$ = $0.02 \ ms^{-1}$.Let the mass of particle $A$ $(m_A)$ = $m$.Then,the mass of particle $B$ $(m_B)$ = $5m$.According to the de-Broglie equation,$\lambda = \frac{h}{mv}$.For particle $A$: $\lambda_A = \frac{h}{m_A v_A} = \frac{h}{m 	imes 0.05}$.For particle $B$: $\lambda_B = \frac{h}{m_B v_B} = \frac{h}{5m 	imes 0.02}$.Taking the ratio $\frac{\lambda_A}{\lambda_B} = \frac{h}{m 	imes 0.05} 	imes \frac{5m 	imes 0.02}{h}$.$\frac{\lambda_A}{\lambda_B} = \frac{5 	imes 0.02}{0.05} = \frac{0.1}{0.05} = \frac{2}{1}$.Thus,the ratio is $2: 1$.

The velocities of two particles $A$ and $B$ are $0.05 \ ms^{-1}$ and $0.02 \ ms^{-1}$ respectively. The mass of $B$ is five times the mass of $A$. The ratio of their de-Broglie's wavelength is

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