The lowest frequency of light that will cause the emission of photoelectrons from the surface of a metal (for which work function is $1.65 \ eV$) will be

If the work function of a metal is $\phi$ and the frequency of the incident light is $v$,there is no emission of photoelectrons for:

Photoelectrons are ejected from a metal when light of frequency $v$ falls on it. Pick out the wrong statement from the following:

Two photons having energies twice and thrice the work function of a metal are incident one after another on the metal surface. Then the ratio of maximum velocities of the photoelectrons emitted in the two cases is respectively:

Two identical metal plates show the photoelectric effect. Light of wavelength ${\lambda _A}$ falls on plate $A$ and light of wavelength ${\lambda _B}$ falls on plate $B$,where ${\lambda _A} = 2{\lambda _B}$. The maximum kinetic energy is:

The magnetic field associated with a light wave is given,at the origin,by $B = B_0 [\sin(3.14 \times 10^7 ct) + \sin(6.28 \times 10^7 ct)]$. If this light falls on a silver plate having a work function of $4.7 \ eV$,what will be the maximum kinetic energy of the photoelectrons? (in $eV$)