When a photon of energy $8 \ eV$ is incident on a metal surface with a threshold frequency of $1.6 \times 10^{15} \ Hz$,the maximum kinetic energy of the emitted photoelectrons is .......... $eV$. (Given: $h = 6.6 \times 10^{-34} \ Js, 1 \ eV = 1.6 \times 10^{-19} \ J$)

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.

Light of wavelength $4000 \ \mathring A$ is incident on a photosensitive surface. If a potential of $-2 \ V$ is required to stop the emitted electrons,the work function of the material is: $(h = 6.6 \times 10^{-34} \ J \cdot s, e = 1.6 \times 10^{-19} \ C, c = 3 \times 10^8 \ m/s)$ (in $eV$)

The maximum velocity of electrons emitted from a metal surface is $V$,when the frequency of light falling on it is $f$. What is the maximum velocity when the frequency becomes $4f$?

If the frequency of incident light in a photoelectric experiment is doubled,then the stopping potential will

The threshold wavelengths for photoelectric emission from two metals $A$ and $B$ are $400 \ nm$ and $800 \ nm$ respectively. The ratio of their work functions,$\phi_{A} : \phi_{B}$ is: