Photons of frequencies equal to the frequencies of $H_\beta$ and $H_{\infty}$ lines of hydrogen are incident on a photosensitive plate,whose threshold frequency is equal to the frequency of the $H_\alpha$ line of hydrogen. The ratio of the maximum kinetic energies of the emitted electrons is

If the energy of a photon is $25\, eV$ and the work function of the material is $7\, eV$,then the value of the stopping potential is :-................. $V$

The given figure shows a few data points in a photoelectric effect experiment for a certain metal. The minimum energy for the ejection of an electron from its surface is $....... eV$. (Planck's constant $h = 6.62 \times 10^{-34} \, J \cdot s$)

The stopping potential $V_0$ (in $volt$) as a function of frequency $(\nu)$ for a sodium emitter is shown in the figure. The work function of sodium,from the data plotted in the figure,will be: ................. $eV$
(Given: Planck's constant $(h) = 6.63 \times 10^{-34} \, Js$,electron charge $e = 1.6 \times 10^{-19} \, C$)

In the following diagram,if $V_2 > V_1$,then:

$(i)$ In the explanation of the photoelectric effect,we assume one photon of frequency $f$ collides with an electron and transfers its energy. This leads to the equation for the maximum kinetic energy $E_{max}$ of the emitted electron as $E_{max} = hf - \phi_0$ (where $\phi_0$ is the work function of the metal). If an electron absorbs $2$ photons (each of frequency $f$),what will be the maximum energy for the emitted electron?
$(ii)$ Why is this fact (two-photon absorption) not taken into consideration in our discussion of the stopping potential?