Photoelectric effect is observed on a surface for frequencies $5.5 \times 10^8 \text{ MHz}$ and $4.5 \times 10^8 \text{ MHz}$. If the ratio of the maximum kinetic energy of the emitted photoelectrons is $1:5$,then the threshold frequency for the metal surface is:

If the frequency of incident light falling on a metallic surface is doubled, what happens to the maximum kinetic energy of the emitted photoelectrons?

Light source having wavelength $331 \text{ nm}$ is used to generate photo-electrons whose stopping potential is $0.2 \text{ V}$. The work function of the used metal in the experiment is $\alpha \times 10^{-19} \text{ J}$. The value of $\alpha$ is . . . . . . . ($h = 6.62 \times 10^{-34} \text{ J s}$, $e = 1.6 \times 10^{-19} \text{ C}$ and $c = 3 \times 10^8 \text{ m/s}$) (in $.68$)

In a photoelectric experiment,the collector plate is at $2.0\,V$ with respect to the emitter plate made of copper $(\phi = 4.5\,eV)$. The emitter is illuminated by a source of monochromatic light of wavelength $200\,nm$.

If a metal sheet is irradiated with radiations of frequencies $v_1$ and $v_2$ and the kinetic energies of the emitted photoelectrons are in the ratio $1 : x$, then the threshold frequency of the metal is:

When a metal plate is illuminated with light of wavelengths $400 \ nm$ and $250 \ nm$,the maximum velocities of the emitted photoelectrons are $v$ and $2v$,respectively. The work function of the metal is ($h$ = Planck's constant; $c$ = speed of light in vacuum):