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

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$2: 1$

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