Let $v_1$ and $v_2$ be the maximum velocities of the emitted electrons when the surface of a metal is illuminated with light waves of energy $E_1=4 \mathrm{eV}$ and $E_2=2.5 \mathrm{eV}$,respectively. If the work function of the metal is 2 eV , then the ratio $\frac{v_1}{v_2}$ is

Photons of energy 2.4 eV and wavelength $\lambda$ fall on a metal plate and release photoelectrons with a maximum velocity $v$. By decreasing $\lambda$ by $50 \%$, the maximum velocity of photoelectrons becomes $3 v$. The work function of the material of the metal plate is