$A$ photon of energy $8 eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} Hz$. The maximum kinetic energy of the emitted photoelectrons is ............... $eV$ (Given $h = 6.6 \times 10^{-34} Js$).

For two different photosensitive materials having work functions $\phi$ and $2 \phi$ respectively,which are illuminated with light of sufficient energy to emit electrons,if the graph of stopping potential $(V_s)$ versus frequency $(\nu)$ is drawn for these two materials,what is the ratio of the slopes of the graphs for these two materials?

In the photoelectric effect,when photons of energy $h \nu$ fall on a photosensitive surface (work function $h \nu_0$),electrons are emitted from the metallic surface. It is possible to say that:

The surface of a metal is first illuminated with a light of wavelength $300 \ nm$ and later illuminated by another light of wavelength $500 \ nm$. It is observed that the ratio of maximum velocities of photoelectrons in the two cases is $3$. The work function of the metal is close to: (in $eV$)

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

The graph between the maximum speed $v_{\text{max}}$ of a photoelectron and frequency $(\nu)$ of the incident radiation,in the photoelectric effect,is correctly represented by: