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

The threshold wavelength for the photoelectric effect of a metal is $6500 \mathring{A}$. The work function of the metal is approximately .......... $eV$.

In a photoelectric experiment,the potential difference $V$ that must be maintained between the illuminated surface and the collector so as just to prevent any electron from reaching the collector is determined for different frequencies $f$ of the incident illumination. The graph obtained is shown. The maximum kinetic energy of the electrons emitted at frequency $f_1$ is

When light of wavelength $300 \ nm$ falls on a photoelectric emitter,photoelectrons are just liberated. For another emitter,light of wavelength $600 \ nm$ is just sufficient for liberating photoelectrons. The ratio of the work function of the two emitters is

When green light is incident on the surface of a metal, it emits photo-electrons, but there is no such emission with yellow light. Which one of the following colours can produce emission of photo-electrons?

If the maximum kinetic energy of emitted electrons in the photoelectric effect is $3.2 \times 10^{-19} \text{ J}$ and the work function for the metal is $6.63 \times 10^{-19} \text{ J}$,then the stopping potential and threshold wavelength respectively are:
[Planck's constant $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$]
[Velocity of light $c = 3 \times 10^{8} \text{ m/s}$]
[Charge on electron $e = 1.6 \times 10^{-19} \text{ C}$]