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

Maximum kinetic energy of photon is '$$E$$' when wavelength of incident radiation is '$$\lambda$$'. If wavelength of incident radiations is reduced to $$\frac{\lambda}{3}$$ then energy of photon becomes four times. Then work function of the metal is