The frequency of incident light falling on a photosensitive material is doubled. The kinetic energy $(K.E.)$ of the emitted photoelectrons will be:

When photons of energies twice and thrice the work function of a metal are incident on the metal surface one after other,the maximum velocities of the photoelectrons emitted in the two cases are $V_1$ and $V_2$ respectively. The ratio $V_1: V_2$ is

Write the slope of the $eV_0 - \nu$ graph.

If light of wavelength $\lambda_1$ is allowed to fall on a metal,then the kinetic energy of the photoelectrons emitted is $E_1$. If the wavelength of light changes to $\lambda_2$,then the kinetic energy of the electrons changes to $E_2$. Then the work function of the metal is:

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:

$A$ photosensitive surface made of cesium on tungsten is irradiated by monochromatic radiation of wavelength $640.2 \ nm$ $(1 \ nm = 10^{-9} \ m)$ from a neon bulb. The measured stopping potential is $0.54 \ V$. If the source is replaced by another source and the same photocell is irradiated by a line of $427.2 \ nm$,what will be the new stopping potential in $V$?