When a photon of energy $6 \ eV$ is incident on a metal surface with a work function of $4 \ eV$,the minimum kinetic energy of the emitted photoelectrons is ......... $eV$.

Energy of the incident photons on the metal surface is initially $4W$ and then $6W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

The frequency and work function of an incident photon are $\nu$ and $\phi_0$. If $\nu_0$ is the threshold frequency,then the necessary condition for the emission of photoelectrons is:

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:

The work function of nickel is $5 \text{ eV}$. When light of wavelength $2000 \text{ Å}$ falls on it,it emits photoelectrons. The potential difference necessary to stop the fastest emitted electrons is (given $h = 6.67 \times 10^{-34} \text{ J-s}$): (in $\text{ V}$)

$A$ 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: