$A$ photon incident on a metal of work function $2 \text{ eV}$ produces photoelectrons of maximum kinetic energy of $2 \text{ eV}$. The wavelength associated with the photon is: (in $\text{ Å}$)

$A$ certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for the photoelectric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

The stopping potential as a function of frequency of incident radiation is plotted for two different photoelectric surfaces $A$ and $B$. The graph shows that the work function of $A$ is

According to Einstein's photoelectric equation, the graph of the kinetic energy of the emitted photoelectrons versus the frequency of incident radiation gives a straight line whose slope

The surface of a metal is illuminated alternately with photons of energies $E_{1} = 4 \ eV$ and $E_{2} = 2.5 \ eV$ respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is $2$. The work function of the metal in $eV$ is:

When radiation of wavelength $\lambda$ is incident on a photocell,the maximum velocity of the photoelectrons is $v$. What will be the maximum velocity of the photoelectrons when radiation of wavelength $3\lambda/4$ is incident on the photocell?