The photoelectric cut-off voltage in a certain experiment is $1.5 \; V$. What is the maximum kinetic energy of photoelectrons emitted?

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

When a metal surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $V/3$. The threshold wavelength for the surface is:

When a metallic surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential is $\frac{V}{3}$. The threshold wavelength for the metallic surface is:

The photoelectric threshold wavelength of silver is $3250 \times 10^{-10} \, m$. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536 \times 10^{-10} \, m$ is (Given $h = 4.14 \times 10^{-15} \, eV \cdot s$ and $c = 3 \times 10^8 \, m/s$):

The surface of a metal is illuminated alternately with photons of energies $E_{1} = 4 \ eV$ and $E_{2} = 2.5 \ eV$ respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is $2$. The work function of the metal in $eV$ is: