When light falls on a metal surface,the maximum kinetic energy of the emitted photo-electrons depends upon

Initially,a photon of wavelength $\lambda_1$ falls on a photocathode and emits an electron of maximum energy $E_1$. If the wavelength of the incident photon is changed to $\lambda_2$,the maximum energy of the electron emitted becomes $E_2$. Then the value of $hc$ ($h=$ Planck's constant,$c=$ velocity of light) is

The work function of caesium metal is $2.14 \; eV$. When light of frequency $6 \times 10^{14} \; Hz$ is incident on the metal surface,photoemission of electrons occurs. What is the
$(a)$ maximum kinetic energy of the emitted electrons,
$(b)$ stopping potential,and
$(c)$ maximum speed of the emitted photoelectrons?

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.

In a photoelectric effect experiment,the graph of stopping potential $V$ versus the reciprocal of wavelength $(1/\lambda)$ is shown in the figure. As the intensity of incident radiation is increased:

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