Photoelectrons are emitted from a metal surface for frequencies $\nu_1$ and $\nu_2$. If the ratio of the maximum kinetic energy of the emitted photoelectrons is $1 : k$,then the threshold frequency of the metal surface is:

$A$ photosensitive metallic surface is illuminated alternately with lights of wavelength $3100 \mathring A$ and $6200 \mathring A$. It is observed that the maximum speeds of the photoelectrons in the two cases are in the ratio $2:1$. The work function of the metal is $(hc = 12400 \, eV \mathring A)$.

$A$ 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 $.49$)

The work functions for the following metals are given:
$Na: 2.75\;eV; K: 2.30\;eV; Mo: 4.17\;eV; Ni: 5.15\;eV$. Which of these metals will not give photoelectric emission for a radiation of wavelength $3300\,\mathring{A}$ from a $He-Cd$ laser placed $1\;m$ away from the photocell? What happens if the laser is brought nearer and placed $50\;cm$ away?

The stopping potential for photoelectrons from a metal surface is $V_{1}$ when monochromatic light of frequency $v_{1}$ is incident on it. The stopping potential becomes $V_{2}$ when monochromatic light of another frequency is incident on the same metal surface. If $h$ is the Planck's constant and $e$ is the charge of an electron,then the frequency of light in the second case is

The threshold wavelength of a metal surface is $5 \times 10^{-10} \, m$. When it is illuminated by light of wavelength $2 \times 10^{-10} \, m$,the stopping potential is $V_0$. What will be the stopping potential if the wavelength of the incident light is doubled?