Radiation of wavelength $300 \ nm$ and intensity $100 \ W \ m^{-2}$ falls on the surface of a photosensitive material. If $2 \%$ of the incident photons produce photoelectrons,the number of photoelectrons emitted from an area of $2 \ cm^2$ of the surface per second is nearly:

In the photoelectric effect,when photons of energy $h \nu$ fall on a photosensitive surface (work function $h \nu_0$),electrons are emitted from the metallic surface. It is possible to say that:

The work function of a photosensitive metal surface is $hv_0$. If a photon of energy $2hv_0$ is incident on this surface,electrons are emitted with a maximum velocity of $4 \times 10^6 \, m/s$. What will be the maximum velocity of the photoelectrons when the energy of the incident photon is increased to $5hv_0$?

$A$ 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 the emitted photoelectrons is ............... $eV$ (Given $h = 6.6 \times 10^{-34} Js$).

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 $2 \, s$.

Two identical photo-cathodes receive light of frequencies $f_1$ and $f_2$. If the velocities of the photoelectrons (of mass $m$) coming out are respectively $v_1$ and $v_2$,then: