The photoelectric threshold wavelength for a metal surface is $6600 \, \mathring{A}$. The work function for this is:

Photons of energy $6\, eV$ are incident on a metal surface whose work function is $4\, eV$. The minimum kinetic energy of the emitted photo-electrons will be :-

In a photoemission experiment,the maximum kinetic energies of photoelectrons from metals $P, Q$ and $R$ are $E_P, E_Q$ and $E_R$,respectively,and they are related by $E_P = 2E_Q = 2E_R$. In this experiment,the same source of monochromatic light is used for metals $P$ and $Q$,while a different source of monochromatic light is used for metal $R$. The work functions for metals $P, Q$ and $R$ are $4.0 \ eV$,$4.5 \ eV$ and $5.5 \ eV$,respectively. The energy of the incident photon used for metal $R$,in $eV$,is:

The maximum kinetic energies of photoelectrons emitted are $K_1$ and $K_2$ when lights of wavelengths $\lambda_1$ and $\lambda_2$ are incident on a metallic surface. If $\lambda_1 = 3 \lambda_2$,then:

$V$ (stopping potential) is plotted against $\frac{1}{\lambda}$,where $\lambda$ is the wavelength of incident radiation,for two metals.

The metallic surface is illuminated with monochromatic light of wavelength $\lambda$ and the stopping potential for the photoelectric current is $5V_0$. When the same metallic surface is illuminated with light of wavelength $2\lambda$,the stopping potential is $V_0$. What is the threshold wavelength for the surface?