When a point source of light is at a distance of $50 \, cm$ from a photoelectric cell,the cut-off voltage is found to be $V_0$. If the same source is placed at a distance of $1 \, m$ from the cell,then the cut-off voltage will be

The work function of a metal is $2 \ eV$. If a radiation of wavelength $3000 \ \text{Å}$ is incident on it,the maximum kinetic energy of the emitted photoelectrons is (Planck's constant $h=6.6 \times 10^{-34} \ \text{Js}$; velocity of light $c=3 \times 10^8 \ \text{m/s}$; $1 \ \text{eV}=1.6 \times 10^{-19} \ \text{J}$).

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

$A$ certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for the photoelectric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

The electric field associated with a light wave is given by $E = E_0 \sin [1.57 \times 10^7 (x - ct)]$,where $x$ is in meters and $t$ is in seconds. If this light is used to produce photoelectric emission from a metal surface with a work function of $1.9 \ eV$,what will be the stopping potential in $V$?

When light of wavelength '$\lambda$' is incident on a photosensitive surface, the stopping potential is '$V$'. When a light of wavelength $1.5\lambda$ is incident on the same surface, the stopping potential is '$\frac{V}{4}$'. Threshold wavelength for the surface is