Light of frequency $4 \times 10^{14} \,Hz$ is incident on a metal surface of work function $2.14 \,eV$, resulting in photoemission of electrons. The maximum kinetic energy of the emitted electrons is $\left[h=6.63 \times 10^{-34} \,J-s\right]$ (in $\,eV$)

When a photon of energy $4.25 \, eV$ strikes the surface of a metal $A$,the ejected photoelectrons have a maximum kinetic energy $T_A \, eV$ and de-Broglie wavelength ${\lambda _A}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by a photon of energy $4.70 \, eV$ is $T_B = (T_A - 1.50) \, eV$. If the de-Broglie wavelength of these photoelectrons is ${\lambda _B} = 2{\lambda _A}$,then:

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 the wavelength of an incident photon is decreased, then:

Photoelectric emission is observed from a metallic surface for frequencies ${\nu _1}$ and ${\nu _2}$ of the incident light rays $({\nu _1} > {\nu _2})$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $1:k$,then the threshold frequency of the metallic surface is

The work function of a metal is $2.3 \ eV$ and the wavelength of the incident photon is $4.84 \times 10^{-7} \ m$. What is the maximum kinetic energy of the photoelectrons in $eV$?