Explanation:
: Given,
For, particle 'A'-
$\lambda_{\mathrm{A}}=1 \AA$
Let, the mass of particle $\mathrm{A}\left(\mathrm{m}_{\mathrm{A}}\right)=\mathrm{x}$
Velocity of particle $A\left(V_A\right)=y$
For, particle 'B'-
Mass of particle $(B)=\frac{x \times 25}{100}=\frac{x}{4}$
Velocity of particle $(B)=\frac{y \times 75}{100}=\frac{3 y}{4}$
de-Broglie equation for the particle 'A', we get-
$\begin{aligned}
& \lambda_{\mathrm{A}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}} \\
& 1 \times 10^{-10}=\frac{\mathrm{h}}{\mathrm{x} \times \mathrm{y}}
\end{aligned}$
de-Broglie equation for the particle ' $\mathrm{B}$ ', we get-
$\begin{aligned}
& \lambda_{\mathrm{B}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}} \\
& \lambda_{\mathrm{B}}=\frac{\mathrm{h}}{\left(\frac{\mathrm{x}}{4}\right)\left(\frac{3 \mathrm{y}}{4}\right)}
\end{aligned}$
or $\quad \lambda_B=\frac{16}{3} \frac{\mathrm{h}}{\mathrm{xy}}$
From equation (i) and (ii), we get-
$\begin{aligned}
\lambda_B & =\frac{16}{3} \times 1 \times 10^{-10} \\
& \lambda_{\mathrm{B}}=5.33 \times 10^{-10} \\
\text { or } \quad \lambda_{\mathrm{B}} & =5.33 \AA
\end{aligned}$
