An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If the photoelectron emitted from the surface has mass $m$ and de-Broglie wavelength $\lambda_{d}$,then

Photons of energy $2.4 \text{ eV}$ and wavelength $\lambda$ fall on a metal plate and release photoelectrons with a maximum velocity $v$. By decreasing $\lambda$ by $50 \%$, the maximum velocity of photoelectrons becomes $3 v$. The work function of the material of the metal plate is (in $\text{ eV}$)

The ratio of work functions of two metals $A$ and $B$ is $1:2$. If radiations of frequencies $f$ and $2f$ are incident on the surfaces of $A$ and $B$ respectively,the ratio of the maximum kinetic energies of the emitted photoelectrons will be: (Given $f >$ threshold frequency of $A$ and $2f >$ threshold frequency of $B$)

The retarding potential necessary to stop the emission of photoelectrons,when a target material of work function $1.24 eV$ is irradiated with light of wavelength $4.36 \times 10^{-7} m$ is (in $eV$)

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 ........... $eV$.

The beam of light has three wavelengths $4144 \,\mathring A$,$4972 \,\mathring A$,and $6216 \,\mathring A$ with a total intensity of $3.6 \times 10^{-5} \, W/m^2$ equally distributed amongst the three wavelengths. The beam falls normally on an area of $1 \, cm^2$ of a clean metallic surface with a work function of $2.3 \, eV$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $2 \, s$.