The stopping potential in the context of the photoelectric effect depends on the following property of incident electromagnetic radiation:

The threshold wavelength of a metal surface is $5 \times 10^{-10} \, m$. When it is illuminated by light of wavelength $2 \times 10^{-10} \, m$,the stopping potential is $V_0$. What will be the stopping potential if the wavelength of the incident light is doubled?

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?

The longest wavelength of light that can initiate the photoelectric effect in a metal with a work function of $9 \ eV$ is:

The threshold frequency of a photoelectric metal is $\nu_0$. If light of frequency $4\nu_0$ is incident on this metal,then the maximum kinetic energy of emitted electrons will be $.......h\nu_0$.

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $\frac{\lambda}{2}$. If the maximum kinetic energy of the emitted photoelectrons in the first case is one-third that in the second case,the work function of the surface of the material is ($c=$ speed of light,$h=$ Planck's constant).