The lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal (for which work function is $1.65 \ eV$) will be

The slope of the graph between the frequency of incident light and the stopping potential for a given surface 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.

Light described at a place by the equation $\varepsilon = (100 \text{ V/m}) [\sin(5 \times 10^{15} \text{ s}^{-1})t + \sin(8 \times 10^{15} \text{ s}^{-1})t]$ falls on a metal surface having a work function of $2 \text{ eV}$. Calculate the maximum kinetic energy of the photoelectrons in $\text{eV}$.

Two monochromatic light beams $A$ and $B$ of equal intensity are incident normally on a metallic surface per unit area. Their wavelengths are $\lambda_A$ and $\lambda_B$ respectively. The incident light is effective in ejecting photoelectrons. The ratio of the number of photoelectrons emitted by beam $A$ to that by beam $B$ is:

An electron and a proton are separated by a large distance. The electron starts approaching the proton with an initial kinetic energy of $3 \, eV$. The proton captures the electron and forms a hydrogen atom in the second excited state. The resulting photon is incident on a photosensitive metal with a threshold wavelength of $4000 \, Å$. What is the maximum kinetic energy of the emitted photoelectron? (In $eV$)