The stopping potential for photoelectrons:

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 $\text{ eV}$)

Ultraviolet light of wavelength $300 \ nm$ and intensity $1.0 \ W/m^2$ is incident on the surface of a photosensitive material. If $1\%$ of the incident photons produce photoelectrons,the number of photoelectrons emitted from an area of $1.0 \ cm^2$ of the surface is:

The electric field at the point associated with a light wave is given by $E = 200 [\sin(6 \times 10^{15} t) + \sin(9 \times 10^{15} t)] \, Vm^{-1}$. Given $h = 4.14 \times 10^{-15} \, eVs$. If this light falls on a metal surface having a work function of $2.50 \, eV$,the maximum kinetic energy of the photoelectrons will be ........... $eV$.

The figure shows different graphs between stopping potential $(V_0)$ and frequency $(\nu)$ for photosensitive surfaces of cesium,potassium,sodium,and lithium. The plots are parallel. The correct ranking of the targets according to their work function,with the greatest first,is:

What is the slope of the $V_0 - \nu$ graph? Write its equation. Also,mention whether it depends on the type of material or not.