The stopping potential as a function of frequency of incident radiation is plotted for two different photoelectric surfaces $A$ and $B$. The graph shows that the work function of $A$ is

$A$ beam of light of wavelength $\lambda$ and with illumination $L$ falls on a clean surface of sodium. If $N$ photoelectrons are emitted each with kinetic energy $E$,then

$A$ metal plate of area $1 \times 10^{-4} \, m^2$ is illuminated by radiation of intensity $16 \, mW/m^2$. The work function of the metal is $5 \, eV$. The energy of the incident photons is $10 \, eV$ and only $10\%$ of the incident photons produce photoelectrons. The number of emitted photoelectrons per second and their maximum kinetic energy, respectively, will be: $[1 \, eV = 1.6 \times 10^{-19} \, J]$

When ultraviolet rays are incident on a metal plate, the photoelectric effect does not occur. It occurs by the incidence of:

Which one of the following four graphs showing lines $P, Q, R$ and $S$ between maximum kinetic energy $(E)$ and intensity of incident light $(I)$ is correct?

$A$ photoelectric surface is illuminated successively by monochromatic light of wavelengths $\lambda$ and $\lambda/2$. If the maximum kinetic energy of the emitted photoelectrons in the second case is three times that in the first case,the work function of the surface is