The work function of a metal is $2.51 eV$. Its threshold frequency is:

Energy of the incident photons on the metal surface is initially $4W$ and then $6W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

In a photocell circuit,the stopping potential $V_0$ is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency $\nu$ of incident light. The values of Planck's constant and the work function as determined from the graph are (taking the magnitude of electronic charge to be $e = 1.6 \times 10^{-19} \, C$):

In the photoelectric effect experiment, if the frequency of incident radiation $(v)$ is increased, keeping all other factors constant, what happens to the stopping potential $(V_s)$, given $(v > v_0)$ where $(v_0)$ is the threshold frequency?

From a metallic surface,photoelectric emission is observed for frequencies $v_1$ and $v_2$ $(v_1 > v_2)$ of the incident light. The maximum kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1:x$. Hence,the threshold frequency of the metallic surface is

The work function of caesium is $2.14 \ eV$. Find
$(a)$ the threshold frequency for caesium,and
$(b)$ the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of $0.60 \ V$.