Two identical photocathodes receive light of frequencies $v_1$ and $v_2$. If the velocities of the photoelectrons (of mass $m$) coming out are $v_1$ and $v_2$ respectively,then:

When light of frequency $v$ is incident on two metallic plates $A$ and $B$,photoelectrons are emitted. If the work function of $A$ is greater than that of $B$,which of the following curves correctly represents the relationship between the stopping potential $V$ and the incident frequency $v$?

The threshold wavelengths for photoelectric emission from two metals $A$ and $B$ are $400 \ nm$ and $800 \ nm$ respectively. The ratio of their work functions,$\phi_{A} : \phi_{B}$ is:

Photoelectrons are emitted from a photosensitive surface for light of wavelengths $\lambda_{1} = 360 \ nm$ and $\lambda_{2} = 600 \ nm$. What is the ratio of the work functions for the lights of wavelengths $\lambda_{1}$ and $\lambda_{2}$?

$A$ photon of energy $8\,eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15}\,Hz$. The maximum kinetic energy of photoelectrons emitted is .......... $eV$.
(Take $h = 6.6 \times 10^{-34}\,J\cdot s$; $1\,eV = 1.6 \times 10^{-19}\,J$)

$A$ photoelectric cell is connected to a source of variable potential difference and the resulting photoelectric current $(\mu A)$ is plotted against the applied potential difference $(V)$. The graph with the broken line represents one situation for a given frequency and intensity of the incident radiation. If the frequency is increased and intensity is reduced,which of the following graphs of unbroken line represents the new situation?