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

Yellow light of wavelength $557 \ nm$ is incident on a cesium surface. When the cathode-anode voltage drops below $0.25 \ V$,no photoelectrons flow in the circuit. What is the threshold wavelength of the cesium surface in $nm$?

Light from a hydrogen discharge tube is incident on the cathode of a photoelectric cell. The work function of the cathode surface is $4.2 \ eV$. In order to reduce the photo-current to zero,the voltage of the anode relative to the cathode must be made ........... $V$.

According to Einstein's photoelectric equation,the graph of the kinetic energy of photoelectrons emitted from a metal versus the frequency of incident radiation is a straight line whose slope:

Light of frequency $4\nu_0$ is incident on a metal surface with a threshold frequency $\nu_0$. The maximum kinetic energy of the emitted photoelectrons is:

$A$ light source is placed at a distance of $20 \ cm$ from a photocell,and the measured stopping potential is $0.6 \ V$. If the distance of the source is changed to $40 \ cm$,what will be the stopping potential in $V$?