The following graphs show the variation of stopping potential $(V_s)$ corresponding to the frequency of incident radiation $(f)$ for a given metal. The correct variation is shown in graph ($f_0 =$ Threshold frequency):

The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3v}{2}$. Then the work function of the metal must be

Photons of energy $6 eV$ are incident on a metal surface whose work function is $4 eV$. The minimum kinetic energy of the emitted photo-electrons will be ........... $eV$.

The photoelectric threshold wavelength of Tungsten is $2300\; \mathring{A}$. The kinetic energy of the electrons ejected from the surface by ultraviolet light of wavelength $1800\; \mathring{A}$ is $.............\,eV$.

If the electron in a hydrogen atom jumps from the third Bohr orbit to the ground state directly and the difference between the energies of the two states is radiated in the form of photons. If the work function of the material is $4.1 \text{ eV}$, then the stopping potential is nearly:
$\left[\text{Energy of electron in } n^{\text{th}} \text{ orbit} = \frac{-13.6}{n^2} \text{ eV}\right]$ (in $\text{ V}$)

For the photoelectric effect,which of the following statements are true?
$I$ The kinetic energies of the photoelectrons do not depend on the frequency of light.
$II$ The photoelectric effect will always occur for highly intense light.
$III$ The maximum kinetic energy of a photoelectron does not depend upon the intensity of the light.
$IV$ The escaping electron's kinetic energy is larger for a larger frequency.