Photoelectrons are emitted from a photosensitive surface for light of wavelengths $\lambda_{1} = 360 \ nm$ and $\lambda_{2} = 600 \ nm$. What is the ratio of the work functions for the lights of wavelengths $\lambda_{1}$ and $\lambda_{2}$?

Radiations of wavelength $400 \,nm$ are incident on a photosensitive material with a work function of $2.2 \,eV$. The stopping potential is nearly: (in $\,V$)

Let $v_1$ and $v_2$ be the maximum velocities of the emitted electrons when the surface of a metal is illuminated with light waves of energy $E_1 = 4 \text{ eV}$ and $E_2 = 2.5 \text{ eV}$,respectively. If the work function of the metal is $2 \text{ eV}$,then the ratio $\frac{v_1}{v_2}$ is

Two identical photocathodes receive light of frequencies $f_{1}$ and $f_{2}$ respectively. If the velocities of the photo-electrons emitted are $v_{1}$ and $v_{2}$ respectively,then:

The maximum velocity of the photoelectron emitted by the metal surface is $V$. The charge and mass of the photoelectron are denoted by $e$ and $m$ respectively. The stopping potential in volts is:

$A$ metal surface is illuminated by radiation of wavelength $4500 \; \mathring A$. The ejected photoelectron enters a constant magnetic field of $2 \; mT$ making an angle of $90^{\circ}$ with the magnetic field. If it starts revolving in a circular path of radius $2 \; mm$,the work function of the metal is approximately ............. $eV$.