When ultraviolet rays are incident on a metal plate, the photoelectric effect does not occur. It occurs by the incidence of:

The figure showing the correct relationship between the stopping potential $V_0$ and the frequency $\nu$ of light for potassium and tungsten is

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum kinetic energies of the photoelectrons emitted in the two cases are in the ratio of $1:k$, then the threshold frequency of the metallic surface is:

The electric field of a light wave is given as $\vec E = 10^{-3} \cos \left( \frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14} t \right) \hat x \, N/C$. This light falls on a metal plate with a work function of $2 \, eV$. The stopping potential of the photoelectrons is ................ $V$.

Given below are two statements:
Statement $I$: Stopping potential in photoelectric effect does not depend on the power of the light source.
Statement $II$: For a given metal,the maximum kinetic energy of the photoelectron depends on the wavelength of the incident light.
In the light of the above statements,choose the most appropriate answer from the options given below.

The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)