The energy of the incident photon on a metal surface is $3W$ and then $5W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

Photons of energy $10 eV$ are incident on a photosensitive surface of threshold frequency $2 \times 10^{15} Hz$. The kinetic energy in $eV$ of the photoelectrons emitted is [Planck's constant $h = 6.63 \times 10^{-34} Js$]. (in $eV$)

The electric field associated with a light wave is given by $E = E_0 \sin [1.57 \times 10^7 (x - ct)]$,where $x$ is in meters and $t$ is in seconds. If this light is used to produce photoelectric emission from a metal surface with a work function of $1.9 \ eV$,what will be the stopping potential in $V$?

Energy of a photoelectron depends on which factors?

In the experiment of $P.E.E.$ (Photoelectric Effect), the saturation current is $5\,mA$ and the stopping potential is $10\,V$. If the intensity and frequency of the incident light are both doubled, then what will be the new saturation current $(i_s)$ and stopping potential $(V_s)$?

Electrons are emitted with kinetic energy $T$ from a metal plate by an irradiation of light of intensity $J$ and frequency $v$. Then, which of the following will be true?