$A$ metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value,then the maximum $KE$ of the emitted photoelectrons would be

The work function of a metal is $1.6 \times 10^{-19} \, J$. When the metal surface is illuminated by light of wavelength $6400 \, \mathring{A}$,the maximum kinetic energy of the emitted photo-electrons will be. (Planck's constant $h = 6.4 \times 10^{-34} \, Js$)

$A$ certain metallic surface is illuminated by monochromatic radiation of wavelength $\lambda$. The stopping potential for photoelectric current for this radiation is $3V_{0}$. If the same surface is illuminated with a radiation of wavelength $2\lambda$,the stopping potential is $V_{0}$. The threshold wavelength of this surface for the photoelectric effect is $n\lambda$. Find the value of $n$.

The maximum kinetic energy of emitted photoelectrons depends on

The threshold frequency of a metal with work function $6.63 \ eV$ is:

Light of frequency $4\nu_0$ is incident on a metal surface with threshold frequency $\nu_0$. The maximum kinetic energy of the emitted photoelectrons is: