In the graph given below,if the slope is $4.12 \times 10^{-15} \ V-sec$,then the value of $h$ should be:

The figure represents the graph of photocurrent $I$ versus applied voltage $(V)$. The maximum kinetic energy of the emitted photoelectrons is:

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 $eV$)

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$ : The photoelectric effect does not take place,if the energy of the incident radiation is less than the work function of a metal.
Reason $R$ : Kinetic energy of the photoelectrons is zero,if the energy of the incident radiation is equal to the work function of a metal.
In the light of the above statements,choose the most appropriate answer from the options given below.

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

The figure showing the correct relationship between the stopping potential $V_0$ and the frequency $\nu$ of light for potassium and tungsten is