If the light from the Balmer series of hydrogen is used to eject photoelectrons from a metal, then the maximum work function of the metal can be: (in $\text{ eV}$)

In a photoelectric emission experiment,the stopping potential for a given metal is $V$ volt,when radiation of wavelength $\lambda$ is used. If radiation of wavelength $2 \lambda$ is used with the same metal,then the stopping potential (in volt) will be. [Given: $c = \text{velocity of light}$,$e = \text{charge on electron}$,$h = \text{Planck's constant}$]

In case of photoelectric emission from a certain metal, the cutoff frequency is $\nu$. If radiation of frequency $3\nu$ is incident on the metal plate, the maximum possible velocity of the emitted electrons will be ($m$ = mass of electron, $h$ = Planck's constant).

When a point source of monochromatic light is at a distance of $0.2 \ m$ from a photoelectric cell,the cut-off voltage and the saturation current are $0.6 \ V$ and $18 \ mA$ respectively. If the same source is placed $0.6 \ m$ away from the photoelectric cell,then:

Two photons with energies three times and nine times the work function of a metal surface are incident on it. What is the ratio of the maximum velocities of the photoelectrons emitted in the two cases?

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 $e = 1.6 \times 10^{-19} \, C$)