The work function of a metallic surface is $5.01\ eV$. Photoelectrons are emitted when light of wavelength $2000\ \mathring{A}$ falls on it. The minimum potential difference required to stop the fastest photoelectrons is ................. $V$.

$A$ 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 $.49$)

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays $(v_1 > v_2)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $1:k$,then the threshold frequency of the metallic surface is

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:

$A$ monochromatic point source of light is placed at a distance $d$ from a metal surface. Photoelectrons are ejected at a rate $n$ per second,and with maximum kinetic energy $E$. If the source is brought nearer to a distance $d / 2$,the rate and the maximum kinetic energy per photoelectron become nearly

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: