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

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential is $4.8 \ V$. If the same surface is illuminated with radiation of double the wavelength,then the stopping potential becomes $1.6 \ V$. Then the threshold wavelength for the surface is

$A$ certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$. The stopping potential for the photoelectric current for this light is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. The threshold wavelength for this surface for the photoelectric effect is:

Radiation $p$ with wavelength $6561 \; \mathring{A}$ falls on a metal surface to produce photoelectrons. The electrons are made to enter a uniform magnetic field of $3 \times 10^{-4} \; T$. If the radius of the largest circular path followed by the electrons is $10 \; mm$,the work function of the metal is close to ............... $eV$.

$A$ photon of energy $4 eV$ imparts all its energy to an electron that leaves a metal surface with $1.1 eV$ of kinetic energy. The work function of the metal is (in $eV$)

When light of wavelength $\lambda$ is incident on a photosensitive surface,the stopping potential is $V$. When light of wavelength $3 \lambda$ is incident on the same surface,the stopping potential is $\frac{V}{6}$. Then the threshold wavelength for the surface is: