$A$ sodium surface is illuminated by light of wavelength $3000\ \mathring A$. The work function of sodium is $2.6\ eV$. The maximum $K.E.$ of the emitted electrons is ........ $eV$.

$(i)$ 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?

Radiation of two photon energies,twice and five times the work function of a metal,are incident successively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted in the two cases is:

$A$ monochromatic point source of light is placed at a distance $d$ from a metal surface. Photoelectrons are ejected at a rate $n$ per second,and with maximum kinetic energy $E$. If the source is brought nearer to a distance $d / 2$,the rate and the maximum kinetic energy per photoelectron become nearly

$A$ beam of photons with an energy of $10.5 \ eV$ strikes a metal plate. The photoelectrons are emitted with a maximum velocity of $1.6 \times 10^6 \ m \ s^{-1}$. The work function of the metal is (Assume mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$). (in $eV$)

The work function of a metal is $6.825 \ eV$. Its threshold wavelength is approximately: (Given $c = 3 \times 10^8 \ m/s$)