Photoelectric effect can be explained by

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 $\lambda_{2}$?

The work function of a metal is $1 \ eV$. If light of wavelength $3000 \ \mathring{A}$ is incident on the surface of this metal, the maximum velocity of the emitted photoelectrons is ........

$X$-rays of a particular wavelength are used to irradiate sodium and copper surfaces in two separate experiments and stopping potentials are determined. The stopping potentials are

In a photocell circuit,the stopping potential $V_0$ is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency $\nu$ of incident light. The values of Planck's constant and the work function as determined from the graph are (taking the magnitude of electronic charge to be $e = 1.6 \times 10^{-19} \, C$):

The work functions of cesium $(Cs)$ and lithium $(Li)$ metals are $1.9 \ eV$ and $2.5 \ eV$,respectively. If we incident light of wavelength $550 \ nm$ on these two metal surfaces,then the photoelectric effect is possible for which case?