The stopping potential as a function of frequency of incident radiation is plotted for two different photoelectric surfaces $A$ and $B$. The graph shows that the work function of $A$ is

The work function of a metal surface is $6.2 \ eV$. If the stopping potential for radiation incident on this surface is $5 \ V$,then the incident radiation will be in the ...... region.

The work functions for the following metals are given:
$Na: 2.75\;eV; K: 2.30\;eV; Mo: 4.17\;eV; Ni: 5.15\;eV$. Which of these metals will not give photoelectric emission for a radiation of wavelength $3300\,\mathring{A}$ from a $He-Cd$ laser placed $1\;m$ away from the photocell? What happens if the laser is brought nearer and placed $50\;cm$ away?

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

$A$ photon released by the transition of an electron from the second excited state to the ground state of a Hydrogen atom is incident on the surface of a metal with a work function of $3.1 \ eV$. The de Broglie wavelength of the most energetic electron emitted from that metal surface is nearly:

$(i)$ 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?