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 $\,eV$)

$A$ sodium surface is illuminated by light of wavelength $3000\ \mathring A$. The work function of sodium is $2.6\ eV$. The maximum $K.E.$ of the emitted electrons is ........ $eV$.

When radiation of wavelength $\lambda$ is incident on a photoelectric cell,the maximum velocity of the emitted photoelectrons is $\upsilon$. If the wavelength of the incident radiation is changed to $3\lambda / 4$,the maximum velocity of the emitted photoelectrons will be ..........

Energy of the incident photons on the metal surface is initially $4W$ and then $6W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

For the photoelectric effect,the maximum kinetic energy $(E_{k})$ of the photoelectrons is plotted against the frequency $(\nu)$ of the incident photons as shown in the figure. The slope of the graph gives:

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.