(i) In the explanations of photoelectric effect, we assume one photon of frequency `nu` collides with an electron and transfer its energy. This leads to the equation for the maximum energy `E_(max)` of the emitted electron as `E_(max)=hnu-phi_(0)` Where `phi_(0)` 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?

Explanation

Here it is given that, an electron absorbs 2 photons each of frequency v then v = 2v where, v is the frequency of emitted electron.
Given, `" "E_(max)=hv-phi_(0)`
Now, maximum energy for emitted electrons is
`" "E_(max)=h(2v)-phi_(0)-2hv-phi_(0)`
(ii) The probability of absorbing 2 photons by the same electron is very low. Hence, such emission will be negligible.