Photons of energy $6\, eV$ are incident on a metal surface whose work function is $4\, eV$. The minimum kinetic energy of the emitted photo-electrons will be :-

The graph of stopping potential $(V_{s})$ against frequency $(\nu)$ of incident radiation is plotted for two different metals '$P$' and '$Q$' as shown in the graph. If $\phi_{P}$ and $\phi_{Q}$ are the work functions of metals '$P$' and '$Q$' respectively,then which of the following is correct?

$A$ beam of light has two wavelengths of $4972\,\mathring{A}$ and $6216\,\mathring{A}$ with a total intensity of $3.6 \times 10^{-3}\,\text{W/m}^2$ equally distributed among the two wavelengths. The beam falls normally on an area of $1\,\text{cm}^2$ of a clean metallic surface of work function $2.3\,\text{eV}$. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photoelectrons liberated in $2\,\text{s}$ is approximately:

Two photons of energy $2.5 eV$ and $3.5 eV$ fall on a metal surface of work function $1.5 eV$. The ratio of the maximum velocities of the photoelectrons emitted from the metal surface is

The graph shows the variation of stopping potential $V_o$ with the frequency $\nu$ of the incident radiation for three photosensitive metals $X_1, X_2$ and $X_3$. Which metal will emit photoelectrons with greater kinetic energy, for the same wavelength of incident radiation?

When light of frequency $v$ is incident on a metal,photoelectrons are emitted. This is because...