The value of Plank's constant, if the slope of the graph of stopping potential vs frequency of incident light is $4\times {10}^{-15} \mathrm{V} \mathrm{s}$ is (given charge of an electron $=1.6\times {10}^{-19} \mathrm{C}$)

Electrons are emitted with kinetic energy $T$ from a metal plate by an irradiation of light of intensity $J$ and frequency $v.$ Then, which of the following will be true?

In a photoelectric effect experiment, the graph of stopping potential $V$ versus reciprocal of wavelength obtained is shown in the figure. As the intensity of incident radiation is increased : 

In photoelectric effect experiment, the slope of the graph of the stopping potential versus frequency gives the value of:

The stopping potential $V_o$ (in volt) as a function of frequency $\left(v\right)$ for a sodium emitter, is shown in the figure. The work function of sodium, form the data plotted in the figure, will be:
(Given: Planck’s constant $h=6.63\times {10}^{-34}\mathrm{ }\mathrm{J}\mathrm{ }\mathrm{s}$ , electron charge $e=1.6\times {10}^{-19}\mathrm{ }\mathrm{C})$

In an experimental observation of the photoelectric effect, the stopping potential was plotted against the incident light frequency as shown in the figure below:

If the work function of the metal is given by ${\varphi }_0$, the angle $\alpha$ is given by

(Here, $h$ and $e$ are Planck's constant and charge of electron respectively).

Guess the shape of curve which shows the variation of $v_{\circ }$ with v in case of photoelectric emission shown by the relation where symbols have their usual meaning.

$V_{\circ }=\left(\frac{h}{e}\right)v-\frac{{\varphi }_{\circ }}{e}$