How does the maximum kinetic energy of a photoelectron depend on the frequency of incident radiation? On which factors does it not depend?

Light rays consisting of photons with energy $1.8 \ eV$ are incident on a metal surface with a work function of $1.2 \ eV$. What is the stopping potential required to stop the emitted electrons in $eV$?

The stopping potential $V_0$ for photoelectric emission from a metal surface is plotted along the $Y-$ axis and the frequency $\nu$ of incident light along the $X-$ axis. $A$ straight line is obtained as shown. Planck's constant $h$ is given by:

In a photo-emissive cell,with exciting wavelength $\lambda$,the maximum kinetic energy of the electron is $K$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the kinetic energy of the fastest emitted electron will be:

With reference to the observations in photoelectric effect,identify the correct statements from below:
$A.$ The square of maximum velocity of photoelectrons varies linearly with frequency of incident light.
$B.$ The value of saturation current increases on moving the source of light away from the metal surface.
$C.$ The maximum kinetic energy of photoelectrons decreases on decreasing the power of $LED$ (light emitting diode) source of light.
$D.$ The immediate emission of photoelectrons out of metal surface can not be explained by particle nature of light/electromagnetic waves.
$E.$ Existence of threshold wavelength can not be explained by wave nature of light/electromagnetic waves.
Choose the correct answer from the options given below:

When a piece of metal is illuminated by a monochromatic light of wavelength $\lambda$,the stopping potential is $3 V_{s}$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential becomes $V_{s}$. The value of the threshold wavelength for photoelectric emission is: