$A$ photon of $5.5 \ eV$ energy falls on the surface of a metal,emitting photoelectrons with a maximum kinetic energy of $4.0 \ eV$. The stopping voltage required for these electrons is .......... $V$.

In a photoelectric experiment,the stopping potential was measured to be $V_1$ and $V_2$ volts with incident light of wavelength $\lambda$ and $\frac{\lambda}{2}$ respectively. The value of $V_2$ is: [where $\phi=$ work function,$e=$ electronic charge]

Two monochromatic light beams $A$ and $B$ of equal intensity are incident normally on a metallic surface per unit area. Their wavelengths are $\lambda_A$ and $\lambda_B$ respectively. The incident light is effective in ejecting photoelectrons. The ratio of the number of photoelectrons emitted by beam $A$ to that by beam $B$ is:

In a photoelectric effect experiment, the frequency of incident radiation is increased while keeping other factors constant $(f > f_0)$. The stopping potential

Two identical photocathodes receive light of frequencies $n_1$ and $n_2$. If the velocities of the emitted photoelectrons of mass $m$ are $V_1$ and $V_2$ respectively,then ($h=$ Planck's constant):

$(i)$ 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?