Two identical capacitors are arranged as shown. The work function of plate $1$ is $\phi \ eV$. The $emf$ of the battery is $\frac{\phi}{e}$. If the energy of the incident photon is $hv$,then what is the maximum kinetic energy of the $e^-$ reaching plate $2$?

Two photons of energies twice and thrice the work function of a metal are incident on the metal surface. Then,the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively,is

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\frac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the first case is one-fourth that in the second case,the work function of the surface of the material is ($c=$ speed of light,$h=$ Planck's constant).

When radiation of frequency $6 \times 10^{14} \ Hz$ is incident on a metal surface with a work function of $2 \ eV$,the maximum kinetic energy of the emitted photoelectrons is ........ $eV$ (Given: $h = 6.63 \times 10^{-34} \ J \cdot s$,$1 \ eV = 1.6 \times 10^{-19} \ J$). (in $.49$)

If the wavelengths of incident radiation are $2500 \ \mathring A$ and $5000 \ \mathring A$ respectively,and the work function of the metal surface is $2 \ eV$,find the approximate ratio of the stopping potentials for the emitted photoelectrons. (in $:1$)

The maximum velocity of the photoelectron emitted by the metal surface is $V$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volts is: