The ratio of work functions of two metals is $1:2$. If light of frequencies $f$ and $2f$ are incident on them respectively,what is the ratio of the maximum kinetic energy of the emitted photoelectrons?

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 ............ $V$. $(h = 6.6 \times 10^{-34} \ J s, c = 3 \times 10^8 \ m/s, 1 \ eV = 1.6 \times 10^{-19} \ J)$

Photons of wavelength $\lambda$ emitted by a source of power $P$ are incident on a photocell. If the current produced in the cell is $I$,then the percentage of incident photons which produce current in the photocell is: (where $h$ is Planck's constant and $c$ is the speed of light in vacuum)

The work functions for metals $A, B$ and $C$ are $1.92 eV, 2.0 eV$ and $5 eV$ respectively. According to Einstein's photoelectric equation,which of these metals will emit photoelectrons for an incident radiation of wavelength $4100 \mathring{A}$?

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

Electrons are emitted with kinetic energy $T$ from a metal plate by an irradiation of light of intensity $J$ and frequency $v$. Then, which of the following will be true?