Energy conversion in a photoelectric cell takes place from

The number of photoelectrons emitted varies with which physical quantity?

The figure represents the graph of photoelectric current $I$ versus applied voltage $(V)$. The maximum kinetic energy of the emitted photoelectrons is

Two metallic plates $A$ and $B$,each of area $5 \times 10^{-4} \, m^2$,are placed parallel to each other at a separation of $1 \, cm$. Plate $B$ carries a positive charge of $33.7 \, pC$. $A$ monochromatic beam of light,with photons of energy $5 \, eV$ each,starts falling on plate $A$ at $t = 0$,so that $10^{16}$ photons fall on it per square meter per second. Assume that one photoelectron is emitted for every $10^6$ incident photons. Also assume that all the emitted photoelectrons are collected by plate $B$ and the work function of plate $A$ remains constant at $2 \, eV$. The electric field between the plates at the end of $10 \, seconds$ is:

$A$ photocell is illuminated by a small bright light source placed at a distance of $1 \ m$. When the same light source is placed at a distance of $\frac{1}{2} \ m$,the number of electrons emitted by the photocathode will be:

$A$ photocell is illuminated by a small source of light placed at a distance of $1 \ m$. When the same source of light is placed at a distance of $1/2 \ m$,the number of electrons emitted by the photocathode will .......