The graph shows the variation of stopping potential $V_o$ with the frequency $\nu$ of the incident radiation for three photosensitive metals $X_1, X_2$ and $X_3$. Which metal will emit photoelectrons with greater kinetic energy, for the same wavelength of incident radiation?

Two photons having energies twice and thrice the work function of a metal are incident one after another on the metal surface. Then the ratio of maximum velocities of the photoelectrons emitted in the two cases is respectively:

For the photoelectric effect,the maximum kinetic energy $E_k$ of the emitted photoelectrons is plotted against the frequency $\nu$ of the incident photons as shown in the figure. The slope of the curve gives

$A$ light whose frequency is equal to $6 \times 10^{14} \, Hz$ is incident on a metal whose work function is $2 \, eV$. $[h = 6.63 \times 10^{-34} \, Js, 1 \, eV = 1.6 \times 10^{-19} \, J]$. The maximum kinetic energy of the emitted electrons will be ............ $eV$. (in $.49$)

$A$ 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:

$A$ metal surface is illuminated by light of two different wavelengths $248 \ nm$ and $310 \ nm$. The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$,respectively. If the ratio $u_1: u_2 = 2: 1$ and $hc = 1240 \ eV \ nm$,the work function of the metal is nearly: (in $eV$)