When a metallic surface is illuminated with light of wavelength $\lambda$,the stopping potential is $x \ V$. When the same surface is illuminated by light of wavelength $2 \lambda$,the stopping potential is $\frac{x}{3} \ V$. The threshold wavelength for the metallic surface is ..........

$(i)$ In the explanation of the photoelectric effect,we assume one photon of frequency $f$ collides with an electron and transfers its energy. This leads to the equation for the maximum kinetic energy $E_{max}$ of the emitted electron as $E_{max} = hf - \phi_0$ (where $\phi_0$ is the work function of the metal). If an electron absorbs $2$ photons (each of frequency $f$),what will be the maximum energy for the emitted electron?
$(ii)$ Why is this fact (two-photon absorption) not taken into consideration in our discussion of the stopping potential?

When light is incident on a surface,photoelectrons are emitted. For these photoelectrons,which of the following is true?

When radiations of wavelength $\lambda$ are incident on a metallic surface,the stopping potential required is $4.8 \ V$. If the same surface is illuminated with radiations of double the wavelength,the required stopping potential becomes $1.6 \ V$. The value of the threshold wavelength for the surface is:

The frequency of monochromatic light incident on an emitting surface is $f$. If the threshold frequency for the surface is $f_0$,then the maximum kinetic energy of the emitted photoelectrons is .......

$A$ and $B$ are two metals with threshold frequencies $1.8 \times 10^{14} \ Hz$ and $2.2 \times 10^{14} \ Hz$. Two identical photons of energy $0.825 \ eV$ each are incident on them. Then photoelectrons are emitted by (Take $h = 6.6 \times 10^{-34} \ J \cdot s$)