Photoelectrons are emitted from a photosensitive surface for light of wavelengths $\lambda_{1} = 360 \ nm$ and $\lambda_{2} = 600 \ nm$. What is the ratio of the work functions for the lights of wavelengths $\lambda_{1}$ and $\lambda_{2}$?

$A$ photon of energy $6 \ eV$ is incident on a metal surface. The work function of the metal is $2 \ eV$. The minimum reverse potential required to stop the emission of electrons is ........ $V$.

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 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 $2 \, s$.

From a metallic surface,photoelectric emission is observed for frequencies $v_1$ and $v_2$ $(v_1 > v_2)$ of the incident light. The maximum kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1:x$. Hence,the threshold frequency of the metallic surface is

Monochromatic radiation of wavelength $640.2 \;nm$ $(1 \;nm = 10^{-9} \;m)$ from a neon lamp irradiates a photosensitive material made of caesium on tungsten. The stopping voltage is measured to be $0.54 \;V$. The source is replaced by an iron source and its $427.2 \;nm$ line irradiates the same photo-cell. Predict the new stopping voltage (in $V$). (in $;V$)