According to Einstein's photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is

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 $2$?

Consider the following statements regarding the photoelectric effect experiment:
$(I)$ Photoelectrons are emitted as soon as the metal is exposed to light.
$(II)$ There is a minimum frequency below which no photocurrent is observed.
$(III)$ The stopping potential is proportional to the frequency of light.
$(IV)$ The photocurrent varies linearly with the intensity of the light.
Which of the above statements indicate that light consists of quanta (photons) with energy proportional to frequency?

When a metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $3 \lambda$,the stopping potential is $V/4$. The threshold wavelength for the metallic surface is.

The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)

The maximum velocity of emitted photoelectrons is $4.8 \ m/s$. If the $e/m$ ratio of an electron is $1.76 \times 10^{11} \ C/kg$,then the stopping potential is ......