Energy of a quanta of frequency ${10^{15}}Hz$ and $h = 6.6 \times {10^{ - 34}}J{\rm{ - }}\sec $ will be

$(i)$ 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 ? 

Both the nucleus and the atom of some element are in their respective first excited states. They get de-excited by emitting photons of wavelengths $\lambda _N,\,\lambda _A$  respectively. The ratio $\frac{{{\lambda _N}}}{{{\lambda _A}}}$  is closest to

A photon, an electron and a uranium nucleus all have the same wavelength. The one with the most energy

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

Where $m$ is the rest mass of the photon

A source $S_1$ is producing, $10^{15}$ photons per second of wavelength $5000 \;\mathring A.$ Another source $S_2$ is producing $1.02 \times 10^{15}$ photons per second of wavelength $5100\;\mathring A$. Then, $($ power of $S_2)/$  $($ power of $S_1)$ is equal to