The ratio of de-Broglie wavelength of two particles $A$ and $B$ is $2: 1$. If the velocities of $A$ and $B$…
The ratio of de-Broglie wavelength of two particles $A$ and $B$ is $2: 1$. If the velocities of $A$ and $B$ are $0.05 \mathrm{~ms}^{-1}$ and $0.02 \mathrm{~ms}^{-1}$, respectively, then the ratio of their masses $m_A: m_B$ must be
$5: 1$
$10: 1$
$1: 5$
$1: 8$
Solution
According to de-Broglie's wavelength, $\lambda=\frac{h}{m v}$ Now, de-Broglie's wavelength of particle $A$ is
$
\lambda_A=\frac{h}{m_A v_A}
$
Similarly, de-Broglie's wavelength of particle $B$ is
$
\lambda_B=\frac{h}{m_B v_B}
$
$\therefore$ Ratio of de-Broglie wavelength of $A$ and $B$ is $2: 1$ i.e.,
$
\begin{aligned}
\frac{\lambda_A}{\lambda_B} & =\frac{h}{m_A v_A} \times \frac{m_B v_B}{h} \\
\frac{2}{1} & =\frac{h}{m_A \times 0.05} \times \frac{m_B \times 0.02}{h} \\
\frac{2}{1} & =\frac{m_B \times 0.02}{m_A \times 0.05} \\
\frac{2}{0.02} & =\frac{m_B}{m_A \times 0.05} \\
\frac{1}{0.01} & =\frac{m_B}{m_A \times 0.05} \Rightarrow \frac{m_A}{m_B}=\frac{1}{5}
\end{aligned}
$
Hence, the ratio of their masses $m_A: m_B$ is $1: 5$