In photoelectric effect, the graph of stopping potential $(V_0)$ versus frequency $(\nu)$ is a straight line. The slope of this graph is . . . . . . .

If the frequency of incident light in a photoelectric experiment is doubled,then the stopping potential will

$A$ monochromatic light source of intensity $5 \, mW$ emits $8 \times 10^{15}$ photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this setup is $2.0 \, V$. The work function of the metal will be ............ $eV$. (in $.9$)

Light of wavelength $5000 \ \mathring A$ and intensity $4.68 \ mW/cm^2$ is incident on a photosensitive surface. If only $5\%$ of the incident photons produce photoelectrons,what is the number of photoelectrons emitted per unit area per unit time?

When light of frequency $v$ is incident on two metallic plates $A$ and $B$,photoelectrons are emitted. If the work function of $A$ is greater than that of $B$,which of the following curves correctly represents the relationship between the stopping potential $V$ and the incident frequency $v$?

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