According to Einstein's photoelectric equation, the graph between the kinetic energy of photoelectrons ejected and the frequency of incident radiation is

$(i)$ In the explanation of photoelectric effect, we assume one photon of frequency v collides with an electron and transfers its energy. This leads to the equation for the maximum energy Emax of the emitted electron as $E_{max} = hf – \phi _0$ (where $\phi _0$ where do is the work function of the metal. If an electron absorbs $2$ photons (each of frequency $v$) 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 ? 

A $10\, kW$ transmitter emits radio waves of wavelength $500\, m$. The number of photons emitted per second by the transmitter is of the order of

If the momentum of a photon is $p$, then its frequency is

Where $m$ is the rest mass of the photon

Monochromatic light of frequency $6.0 \times 10^{14} \;Hz$ is produced by a laser. The power emitted is $2.0 \times 10^{-3} \;W$.

$(a)$ What is the energy of a photon in the light beam?

$(b)$ How many photons per second, on an average, are emitted by the source?