The photoelectric effect was successfully explained first by

For a photosensitive material, the work function is $W_0$ and the stopping potential is $V$. What is the wavelength of the incident radiation? ($h$ = Planck's constant, $c$ = velocity of light, $e$ = electronic charge)

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
$Assertion$ $A$ : Number of photons increases with increase in frequency of light.
$Reason$ $R$ : Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.
In the light of the above statements,choose the most appropriate answer from the options given below :

If the maximum kinetic energy of emitted electrons in the photoelectric effect is $3.2 \times 10^{-19} \text{ J}$ and the work function for the metal is $6.63 \times 10^{-19} \text{ J}$,then the stopping potential and threshold wavelength respectively are:
[Planck's constant $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$]
[Velocity of light $c = 3 \times 10^{8} \text{ m/s}$]
[Charge on electron $e = 1.6 \times 10^{-19} \text{ C}$]

$A$ light wave described by $E = 60 \sin(3 \times 10^{15} t) + \sin(12 \times 10^{15} t)$ (in $SI$ units) falls on a metal surface of work function $2.8 \text{ eV}$. The maximum kinetic energy of the ejected photoelectron is (approximately) . . . . . . $\text{eV}$.

In the photoelectric emission process from a metal with a work function of $1.8\, eV$,the kinetic energy of the most energetic electrons is $0.5\, eV$. The corresponding stopping potential is ......... $V$.