According to question, $\frac{X_A}{X_B}=\frac{K}{1}$ ...(i)

$\frac{U_A}{U_B}=\frac{1}{K^2}$

So, $U_A=y$ and $U_B=K^{2}y$

According to de-Broglie wavelength,

$\lambda =\frac{h}{\sqrt{2m{K}_1}}$ .....(ii)

Here, $K_1$ is kinetic energy.

So, from Equations (i) and (ii),

$\Rightarrow \frac{\left(\frac{h}{\sqrt{2m{K}_{1A}}}\right)}{\left(\frac{h}{\sqrt{2m{K}_{1B}}}\right)}=\frac{K}{1}\Rightarrow \sqrt{\frac{K_{1B}}{K_{1A}}}=\frac{K}{1}$

$\Rightarrow \frac{K_{1B}}{K_{1A}}=K^2$

So, $K_{1B}=K^{2}x$ and $K_{1A}=x$

$\because$ Total energy, $E=K_1+U$

So, $E_A=K_{1A}+U_A$

and $E_B=K_{1B}+U_B$

or $\frac{E_A}{E_B}=\frac{K_{1A}+U_A}{K_{1B}+U_B}=\frac{x+y}{K^{2}x+K^{2}y}$

$\Rightarrow$ So, $\frac{E_A}{E_B}=\frac{1}{K^2} \left(\mathrm{o}\mathrm{r}\right) E_A :E_B=1 :K^2$