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 $4100 \mathring{A}$?

Radiation of two photon energies,twice and five times the work function of a metal,are incident successively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases is:

When a photon of energy $4.25 \, eV$ strikes the surface of a metal $A$,the ejected photoelectrons have a maximum kinetic energy $T_A \, eV$ and de-Broglie wavelength ${\lambda _A}$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by a photon of energy $4.70 \, eV$ is $T_B = (T_A - 1.50) \, eV$. If the de-Broglie wavelength of these photoelectrons is ${\lambda _B} = 2{\lambda _A}$,then:

The maximum kinetic energy of the photo-electrons depends only on

In the photoelectric effect,the kinetic energy $(K.E.)$ of electrons emitted from the metal surface depends upon:

When photons of energy $1 \ eV$ and $2.5 \ eV$ are incident on a metal surface with a work function of $0.5 \ eV$,what is the ratio of the maximum kinetic energies of the emitted photoelectrons?