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 $E_1 = 4 \text{ eV}$ and $E_2 = 2.5 \text{ eV}$,respectively. If the work function of the metal is $2 \text{ eV}$,then the ratio $\frac{v_1}{v_2}$ is

In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):

The threshold frequency of a photosensitive material is equal to the frequency of the $H_{\alpha}$ line of hydrogen. If a photon whose frequency is equal to the frequency of the $H_{\beta}$ line of hydrogen is incident on this photosensitive material,the maximum kinetic energy of the emitted photoelectrons is ($R$ = Rydberg's constant,$h$ = Planck's constant,and $c$ = speed of light in vacuum).

Light of wavelength $5000 \; \mathring A$ falls on a sensitive plate with photoelectric work function of $1.9 \; eV$. The kinetic energy of the photoelectron emitted will be ............ $eV$.

When radiation of wavelength $400 \ nm$ is incident on a metal surface,the kinetic energy of the emitted photoelectrons is $1.68 \ eV$. The work function of the metal is .......... $eV$.

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelengths $\lambda$ and $\lambda/2$. If the maximum kinetic energy of the emitted photoelectrons in the second case is three times that in the first case,the work function of the surface is