Light described at a place by the equation $\varepsilon = (100 \text{ V/m}) [\sin(5 \times 10^{15} \text{ s}^{-1})t + \sin(8 \times 10^{15} \text{ s}^{-1})t]$ falls on a metal surface having a work function of $2 \text{ eV}$. Calculate the maximum kinetic energy of the photoelectrons in $\text{eV}$.

Two photons of energies twice and thrice the work function of a metal are incident on the metal surface. Then,the ratio of maximum velocities of the photoelectrons emitted in the two cases respectively,is

When photons of energy $h \nu$ fall on a photosensitive surface of work function $E_0$,photoelectrons of maximum kinetic energy $k$ are emitted. If the frequency of radiation is doubled,the maximum kinetic energy will be equal to ($h=$ Planck's constant).

In a photoemission experiment,the maximum kinetic energies of photoelectrons from metals $P, Q$ and $R$ are $E_P, E_Q$ and $E_R$,respectively,and they are related by $E_P = 2E_Q = 2E_R$. In this experiment,the same source of monochromatic light is used for metals $P$ and $Q$,while a different source of monochromatic light is used for metal $R$. The work functions for metals $P, Q$ and $R$ are $4.0 \ eV$,$4.5 \ eV$ and $5.5 \ eV$,respectively. The energy of the incident photon used for metal $R$,in $eV$,is:

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 $.68$)

Assuming photoemission takes place,the factor by which the maximum velocity of the emitted photoelectrons changes when the wavelength of the incident radiation is increased four times,is