The work function of caesium is $2.14 \ eV$. Find the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of $0.60 \ V$.

The correct graph between the maximum kinetic energy $(K_{\max})$ of a photoelectron and the inverse of the wavelength $(1/\lambda)$ of the incident radiation is given by the curve:

If the light from the Balmer series of hydrogen is used to eject photoelectrons from a metal, then the maximum work function of the metal can be: (in $\text{ eV}$)

$A$ beam of light has two wavelengths of $4972\,\mathring{A}$ and $6216\,\mathring{A}$ with a total intensity of $3.6 \times 10^{-3}\,\text{W/m}^2$ equally distributed among the two wavelengths. The beam falls normally on an area of $1\,\text{cm}^2$ of a clean metallic surface of work function $2.3\,\text{eV}$. Assume that there is no loss of light by reflection and that each capable photon ejects one electron. The number of photoelectrons liberated in $2\,\text{s}$ is approximately:

The ratio of work functions of two metals $A$ and $B$ is $1:2$. If radiations of frequencies $f$ and $2f$ are incident on the surfaces of $A$ and $B$ respectively,the ratio of the maximum kinetic energies of the emitted photoelectrons will be: (Given $f >$ threshold frequency of $A$ and $2f >$ threshold frequency of $B$)

Mercury violet light $(\lambda = 4558 \mathring{A})$ is falling on a photosensitive material $(\phi = 2.5 \text{ eV})$. The speed of the ejected electrons is,in $\text{m/s}$,about: