The stopping potential $(V_0)$ versus frequency $(\nu)$ plot of a substance is shown in the figure. The threshold wavelength is:

The threshold frequency of a metal is $F_0$. When light of frequency $2F_0$ is incident on the metal plate,the maximum velocity of the photoelectrons is $V_1$. When the frequency of incident radiation is increased to $5F_0$,the maximum velocity of the photoelectrons emitted is $V_2$. The ratio of $V_1$ to $V_2$ is:

In the photoelectric effect,if the intensity of light is doubled,then the maximum kinetic energy of the photoelectrons will become:

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 $nm$?

Which one of the following graphs correctly represents the variation of maximum kinetic energy $(E_{k})$ of the emitted electrons with frequency $(\nu)$ of incident light in the photoelectric effect?

If the wavelengths of incident radiation are $2500 \ \mathring A$ and $5000 \ \mathring A$ respectively,and the work function of the metal surface is $2 \ eV$,find the approximate ratio of the stopping potentials for the emitted photoelectrons. (in $:1$)