$(i)$ 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?

(A) $(i)$ Suppose one electron absorbs two photons each having frequency $f$. Out of this total energy $2hf$,it spends $W$ amount of energy to overcome the work function for its emission,and the remaining amount $(2hf - W)$ is possessed by it in the form of kinetic energy after being emitted.
Hence,$K = 2hf - W$.
Therefore,$K_{max} = 2hf - W_{min}$.
Since $W_{min} = \phi_0$,we get $K_{max} = 2hf - \phi_0$.
$(ii)$ If the above assumption were true,then according to the work-energy theorem,taking $K_{max} = V_0 e$ (where $V_0$ is the stopping potential):
$2hf - \phi_0 = V_0 e$
$V_0 = (\frac{2h}{e})f - (\frac{\phi_0}{e})$
This equation represents a straight line $y = mx + c$. If we plot a graph of $V_0$ versus $f$ experimentally,the slope should be $(\frac{2h}{e})$. However,experimental results consistently show the slope to be $(\frac{h}{e})$. Thus,the assumption of two-photon absorption is incorrect,which is why it is not considered in the discussion of stopping potential.