In an experiment on the photoelectric effect,the frequency $\nu$ of the incident light is plotted against the stopping potential $V_0$. The work function of the photoelectric surface is given by ($e$ is the electronic charge):

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 maximum wavelength of light which causes photoelectric emission from a photosensitive metal surface is $\lambda_0$. Two light beams of wavelengths $\frac{\lambda_0}{3}$ and $\frac{\lambda_0}{9}$ are incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is

The threshold wavelength of a metal is $400 \ nm$. The maximum kinetic energy of the emitted photoelectrons is $1.5 \ eV$. Find the wavelength of the incident photon in $\mathring{A}$.

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

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: