In the photoelectric effect,the kinetic energy $(K.E.)$ of electrons emitted from the metal surface depends upon:

Light described at a place by the equation $\varepsilon = (100 \text{ V/m}) [\sin(5 \times 10^{15} \text{ s}^{-1})t + \sin(8 \times 10^{15} \text{ s}^{-1})t]$ falls on a metal surface having a work function of $2 \text{ eV}$. Calculate the maximum kinetic energy of the photoelectrons in $\text{eV}$.

When a metallic surface is illuminated with monochromatic light of wavelength $\lambda$,the stopping potential is $5V_0$. When the same surface is illuminated with light of wavelength $3\lambda$,the stopping potential is $V_0$. Then the work function of the metallic surface is:

$A$ photon of energy $6 \ eV$ is incident on a metal surface. The work function of the metal is $2 \ eV$. The minimum reverse potential required to stop the emission of electrons is ........ $V$.

Which of the following graphs represents the variation of stopping potential $(V_0)$ with the frequency $(\nu)$ of incident light for a photoelectric cell?

$A$ 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-fourth that in the second case,the work function of the surface of the material is ($c=$ speed of light,$h=$ Planck's constant).