If the work function of a metal is $\phi$ and the frequency of the incident light is $v$,there is no emission of photoelectrons for:

$A$ 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:

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 coming from a discharge tube filled with hydrogen falls on the cathode of the photoelectric cell. The work function of the surface of the cathode is $4 \ eV$. Which one of the following values of the anode voltage (in Volts) with respect to the cathode will likely make the photocurrent zero?

The stopping potential $V_0$ for photoelectric emission from a metal surface is plotted along the $Y-$ axis and the frequency $\nu$ of incident light along the $X-$ axis. $A$ straight line is obtained as shown. Planck's constant $h$ is given by:

In the photoelectric effect, the stopping potential depends on: