Given,

$\frac{E_A}{E_B}=\frac{1}{2},\frac{U_A}{U_B}=\frac{1}{2}$

So $E_A=x, E_B=2x$

And $U_A=y, U_B=2y$

$\because E_A=U_A+K_A$

And $E_B=U_B+K_B$

here $K_A$ and $K_B$ are kinetic energy of particles $A$ and $B$

So $K_A=E_A-U_A=\left(x-y\right)$ .....(i)

$K_B=E_B-U_B=2\left(x-y\right)$ ....(ii)

$\because$ de-Broglie wavelength,

$\lambda =\frac{h}{\sqrt{2mk}}$

So ${\lambda }_A=\frac{h}{\sqrt{2m{K}_A}}, {\lambda }_B=\frac{h}{\sqrt{2m{K}_B}}$

$\therefore \frac{{\lambda }_A}{{\lambda }_B}=\sqrt{\frac{K_B}{K_A}}$ .....(iii)

From Equation (i), (ii) and (iii),

$\frac{{\lambda }_A}{{\lambda }_B}=\sqrt{\frac{2\left(x-y\right)}{\left(x-y\right)}}=\frac{\sqrt{2}}{1}$