Kinetic energy with which the electrons are emitted from the metal surface due to photoelectric effect is

Average force exerted on a non-reflecting surface at normal incidence is $2.4 \times 10^{-4} \mathrm{~N}$. If $360 \mathrm{~W} / \mathrm{cm}^2$ is the light energy flux during span of $1$ hour $30$ minutes. Then the area of the surface is:

What is energy and momentum of photon having frequency $v$ ? 

Light of wavelength $5000\,\,\mathop A\limits^o $ falling on a sensitive surface. If the surface has received $10^{-7}\,J$ of energy, then the number of photons falling on the surface will be

The energy of a photon is $E = hv$ and the momentum of photon $p = \frac{h}{\lambda }$, then the velocity of photon 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 ?