$A$ photocell stops emission if it is maintained at $2V$ negative potential. The energy of the most energetic photoelectron is:

Assertion : Photoelectric saturation current increases with the increase in frequency of incident light.
Reason : Energy of incident photons increases with increase in frequency and as a result photoelectric current increases.

The number of photo-electrons emitted per second from a metal surface increases when:

What is the phenomenon of the photoelectric effect?

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:

$A$ monochromatic beam of light of wavelength $400\,nm$ is incident normally upon a photosensitive surface ($25\%$ reflects and the rest is absorbed). This produces a radiation pressure of $5 \times 10^{-7}\,N/m^2$ on the surface. If $0.1\%$ of the incident photons produce photoelectrons,the corresponding saturation current will be $\dots \mu A$. (Consider the area of the photosensitive surface $= 5\,cm^2$)