The maximum kinetic energies of photoelectrons emitted are $$\mathrm{K}_1$$ and $$\mathrm{K}_2$$ when lights of wavelengths $$\lambda_1$$ and $$\lambda_2$$ are incident on a metallic surface. If $$\lambda_1=3 \lambda_2$$ then

If the potential difference used to accelerate electrons is doubled, by what factor does the de-Broglie wavelength associated with electrons change?

The maximum kinetic energy of the photoelectrons varies

An electron accelerated through a potential difference '$$V_1$$' has a de-Broglie wavelength '$$\lambda$$'. When the potential is changed to '$$V_2$$' its de-Broglie wavelength increases by $$50 \%$$. The value of $$\left(\frac{\mathrm{V}_1}{\mathrm{~V}_2}\right)$$ is