If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photo-electrons would be (given $h = 6.63 \times 10^{-34} \ J \cdot s$):

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.

$A$ photo-emissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_c$. The longest wavelength of radiation that can emit photoelectrons is $\lambda_0$. The expression for the de-Broglie wavelength is given by ($m$: mass of the electron,$h$: Planck's constant,and $c$: speed of light).

The threshold wavelength for photoelectric emission from a material is $5200 \, \mathring{A}$. Photo-electrons will be emitted when this material is illuminated with monochromatic radiation from a

The wave equation of an electric field at a point is $E = 100 \frac{V}{m} [\sin(7 \omega t) + \cos(10 \omega t) + \cos(15 \omega t)]$ at instant $t$. If the work function of the photocell is $\phi$,then the stopping potential is:

According to Einstein's photoelectric equation,the graph of the maximum kinetic energy of emitted photoelectrons versus the frequency of incident radiation is a straight line. Its slope . . . . . .