Photoelectric emission takes place from a certain metal at threshold frequency $v$. If the radiation of frequency $4v$ is incident on the metal plate,the maximum velocity of the emitted photoelectrons will be ($m=$ mass of photoelectron,$h=$ Planck's constant).

According to Einstein's photoelectric equation,the graph of the kinetic energy of photoelectrons emitted from a metal versus the frequency of incident radiation is a straight line whose slope:

Given below are two statements:
Statement-$I$: The figure shows the variation of stopping potential $(V_0)$ with frequency $(v)$ for two photosensitive materials $M_1$ and $M_2$. The slope gives the value of $\frac{h}{e}$,where $h$ is Planck's constant and $e$ is the charge of an electron.
Statement-$II$: $M_2$ will emit photoelectrons of greater kinetic energy for incident radiation having the same frequency.
In the light of the above statements,choose the most appropriate answer from the options given below.

The work function of metals is in the range of $2 eV$ to $5 eV$. Find which of the following wavelengths of light cannot be used for the photoelectric effect (in $nm$)? (Consider, Planck constant $= 4 \times 10^{-15} eVs$, velocity of light $= 3 \times 10^{8} m/s$)

In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):

According to the photoelectric effect,the plot of kinetic energy of the emitted photo-electrons from a metal versus the frequency of the incident radiation gives a straight line whose slope