The threshold frequency for a certain metal is $3.3 \times 10^{14} \;Hz$. If light of frequency $8.2 \times 10^{14} \;Hz$ is incident on the metal,predict the cutoff voltage (in $eV$) for the photoelectric emission.

$A$ photon of energy $8\,eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15}\,Hz$. The maximum kinetic energy of photoelectrons emitted is .......... $eV$.
(Take $h = 6.6 \times 10^{-34}\,J\cdot s$; $1\,eV = 1.6 \times 10^{-19}\,J$)

For a photoelectric cell,the graph showing the variation of stopping potential $(V_o)$ with frequency $(\nu)$ of incident light is best represented by:

The work function of Cesium is $2.27 eV$. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of $600 nm$ wavelength is

Which one of the following graphs correctly represents the variation of maximum kinetic energy $(E_{k})$ of the emitted electrons with frequency $(\nu)$ of incident light in the photoelectric effect?

The work functions of metal $A$ and $B$ are in the ratio $1: 2$. If light of frequency $f$ and $2f$ is incident on surface $A$ and $B$ respectively,then the ratio of the maximum kinetic energies of the emitted photoelectrons is: