When a metallic surface is illuminated by light of wavelength $\lambda$,the stopping potential for the photoelectric current is $3 \text{ V}$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential is $1 \text{ V}$. The threshold wavelength for this surface is:

For a photoelectric cell,the graph showing the variation of stopping potential $(V_o)$ with frequency $(\nu)$ of incident light is best represented by:

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 $e = 1.6 \times 10^{-19} \text{ C}$]

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

When light of frequencies ${\nu _1}$ and ${\nu _2}$ $({\nu _1} > {\nu _2})$ is incident on a metal surface,the ratio of the maximum kinetic energy of the emitted photoelectrons is $1:k$. What is the threshold frequency of the metal?

Two light beams of different frequencies,having energies $1 \ eV$ and $2.5 \ eV$ respectively,are incident on a metal surface with a work function of $0.5 \ eV$. The ratio of the maximum kinetic energies of the emitted photoelectrons is: