The wavelengths of light incident on a photocell are $400 \ nm$ and $250 \ nm$. The velocities of the emitted photoelectrons are $v$ and $2v$ respectively. What is the work function of the metal?

In the photoelectric effect,when photons of energy $h \nu$ fall on a photosensitive surface (work function $h \nu_0$),electrons are emitted from the metallic surface. It is possible to say that:

When radiation of frequency $6 \times 10^{14} \ Hz$ is incident on a metal surface with a work function of $2 \ eV$,the maximum kinetic energy of the emitted photoelectrons is ........ $eV$ (Given: $h = 6.63 \times 10^{-34} \ J \cdot s$,$1 \ eV = 1.6 \times 10^{-19} \ J$). (in $.49$)

Radiation of wavelength $332 \ nm$ is incident on a metal surface having a work function of $1.07 \ eV$. The stopping potential required to stop the emission of photoelectrons from the metal surface is ............ $V$. $(h = 6.6 \times 10^{-34} \ J s, c = 3 \times 10^8 \ m/s, 1 \ eV = 1.6 \times 10^{-19} \ J)$

Light of wavelength $5000 \; \mathring A$ falls on a sensitive plate with photoelectric work function of $1.9 \; eV$. The kinetic energy of the photoelectron emitted will be ............ $eV$.

$A$ metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value,then the maximum $KE$ of the emitted photoelectrons would be