When a metal is irradiated by monochromatic light,the maximum kinetic energy of the photoelectrons is $1.2 \ eV.$ If the frequency of the light is increased by $50 \%$,then the maximum kinetic energy of the photoelectrons is $3.6 \ eV.$ Evaluate the work function of the metal.

When the wavelength of an incident photon is decreased, then:

$A$ 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 $.49$)

Energy of the incident photons on the metal surface is initially $4W$ and then $6W$,where $W$ is the work function of that metal. The ratio of the maximum velocities of the emitted photoelectrons is:

Photoelectrons are emitted when photons of energy $4.2 \text{ eV}$ are incident on a photosensitive metallic sphere of radius $10 \text{ cm}$ and work function $2.4 \text{ eV}$. The number of photoelectrons emitted before the emission is stopped is
$\left[\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text{ SI unit; } e=1.6 \times 10^{-19} \text{ C}\right]$

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: