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

The variation of stopping potential for metals $A$,$B$,$C$ and $D$ with the frequency of incident radiation is shown in the figure. For which metal is the stopping potential higher for a given frequency of incident radiation $(v)$ if the threshold frequency is lower $(\nu_o)$?

The maximum kinetic energy of a photoelectron is $E$ when the wavelength of incident radiation is $\lambda$. If the wavelength of the incident radiation is reduced to $\frac{\lambda}{3}$,the maximum kinetic energy becomes $4E$. The work function of the metal is:

In an experiment to study the photoelectric effect,the observed variation of stopping potential $(V_0)$ with the frequency $(\nu)$ of incident radiation is as shown in the figure. The slope and $y$-intercept are respectively

When photons of wavelength $\lambda_1$ are incident on an isolated sphere,the corresponding stopping potential is found to be $V$. When photons of wavelength $\lambda_2$ are used,the corresponding stopping potential is thrice that of the above value. If light of wavelength $\lambda_3$ is used,then find the stopping potential for this case.

The threshold frequency of cesium is $5.16 \times 10^{14} \ Hz$. Then its work function is . . . . . . $eV$.