$UV$ light of $4.13 eV$ is incident on a photosensitive metal surface having work function $3.13 eV$. The maximum kinetic energy of ejected photoelectrons will be : (in $eV$)

For photoelectric emission from a certain metal,the cutoff frequency is $\nu$. If radiation of frequency $2\nu$ impinges on the metal plate,the maximum possible velocity of the emitted electron will be ($m$ is the electron mass).

Light of two different frequencies with photon energies of $1 \ eV$ and $2.5 \ eV$ are incident one after another on a metal surface having a work function of $0.5 \ eV$. The ratio of the maximum kinetic energy of the emitted photoelectrons will be:

Einstein's photoelectric equation states that ${E_k} = h\nu - \phi$. In this equation,${E_k}$ refers to:

If the frequency of incident light falling on a metallic surface is doubled, what happens to the maximum kinetic energy of the emitted photoelectrons?

The stopping potential of the photoelectrons from a photocell is: