Two radiations of photon energies $1 \; eV$ and $2.5 \; eV$ successively illuminate a photosensitive metallic surface of work function $0.5 \; eV$. The ratio of the maximum speeds of the emitted electrons is

$A$ metallic surface is illuminated with radiation of wavelength $\lambda$,the stopping potential is $V_0$. If the same surface is illuminated with radiation of wavelength $2 \lambda$,the stopping potential becomes $\frac{V_0}{4}$. The threshold wavelength for this metallic surface will be -

The work function of Cesium is $2.27 eV$. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of $600 nm$ wavelength is

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with energy $E$. If $E$ is to be increased to twice the original value,the wavelength changes to $\lambda_1$. Which of the following is correct?

When light is incident on a surface,photoelectrons are emitted. For these photoelectrons:

In a photocell circuit,the stopping potential $V_0$ is a measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency $\nu$ of incident light. The values of Planck's constant and the work function as determined from the graph are (taking the magnitude of electronic charge to be $e = 1.6 \times 10^{-19} \, C$):